CS计算机代考程序代写 Ex:11.1 (c) A=100V/VandAf =10 V/V

Ex:11.1 (c) A=100V/VandAf =10 V/V
Since A is not much greater than Af , we shall use the exact expression to determine β and hence R2/R1,
A
Vi =Vo = 10 =0.001V A 104
(f) A ⇒ 0.8×104 V/V
Af = 0.8 × 104 1+0.8×104×9×10−4
= 975.6 V/V
which is a change of 975.6 − 1000 × 100
1000
= −2.44% or a reduction of 2.44%.
Ex. 11.3 To constrain the corresponding change in Af to 0.1%, we need an amount-of-feedback of at least
1+Aβ= 10% =100 0.1%
Thus the largest obtainable closed-loop gain will be
Af = A =1000=10V/V 1 + Aβ 100
Each amplifier in the cascade will have a nominal gainof10V/Vandamaximumvariabilityof 0.1%; thus the overall voltage gain will be
(10)3 = 1000 V/V and the maximum variability willbe0.3%.
= 0.1
Exercise 11–1
Af =
10 = 100
Now, R1
1 + Aβ
1+100β ⇒ β = 0.09 V/V
= 0.09
R2 = 1 −1=10.11
R1 0.09
(d) Aβ=100×0.09=9
1 + Aβ = 10
⇒ 20 dB
(e) Vo =AfVs =10×1=10V Vf =βVo =0.09×10=0.9V
Vi = Vo = 10 =0.1V A 100
R + R 1 2
(f) A→80V/V
Af = 80 =9.756
R1 = 1 R1 + R2 1 + 9
1+80×0.09
achangeof 9.756−10 ×100=−2.44%ora
Ex. 11.4 β =
Aβ=104 ×0.1=1000
10 reduction of 2.44%.
1 + Aβ = 1001 Af = A
Ex.11.2 (c) A=104 V/VandAf =103 V/V Af = A
1 + Aβ 104
1+Aβ 103 = 104
Af = 1+104 ×0.1 =9.99V/V fHf =fH(1+Aβ)
1+104β ⇒β=9×10−4 V/V
= 100×1001 = 100.1 kHz
Ex.11.5 Signalatoutput=Vs 1+A A β
R
R1 + R2
R2 = 1
R1 9 × 10−4
A1A2 12
1 =9×10−4
1 × 100 1+1×100×1
= 1 ×
= 1 × 100 ≃ 1 V 101
A1
−1=1110.1
(d) Aβ=104 ×9×10−4 =9 1 + Aβ = 10
⇒ 20 dB
(e) Vs = 0.01 V
Vo =AfVs =103 ×0.01=10V
Vf =βVo =9×10−4 ×10=0.009V
Interferenceatoutput=Vn 1+A1A2β
= 1 × 1 ≃ 0.01 V
1+1×100×1
Thus S/I at the output becomes 1/0.01
= 100 or 40 dB
Since S/I at the input is 1/1=1 or 0 dB, the improvementis40dB.

Ex. 11.6 (a) Refer to Fig. 11.8(c). β= R1
Af = A = 1 + Aβ
36.36 =4.4V/V 1 + 36.36 × 0.2
Exercise 11–2
R1 + R2 (b)
RD
Q
If Aβ were ≫ 1, then
Af ≃ 1 = 1 = 5 V/V
Ex. 11.7 From the solution of Example 11.4, Aβ=6
1 + Aβ = 7
Thus,
fHf = (1 + Aβ)fH =7×1
= 7 kHz
Ex. 11.8 Refer to Fig. E11.8. The 1-mA bias current will split equally between the emitters of Q1 and Q2, thus
IE1 =IE2 =0.5mA
Transistor Q3 will be operating at an emitter
current
IE3 =5mA
determined by the 5-mA current source. Since the dc component of Vs = 0, the negative feedback will force the dc voltage at the output to be approximately zero. See Fig. 1 on next page.
The β circuit is shown in Fig. 1 together with the determination of β and of the loading effects of the β circuit on the A circuit,
β 0.2
Vd R2
􏰀􏰀
􏰒Vt Vr R1
􏰒
Figure 1
Figure 1 shows the circuit prepared for determining the loop gain Aβ. Observe that we have eliminated the input signal Vs, and opened the loop at the gate of Q where the input impedance is infinite obviating the need for a termination resistance at the right-hand side of the break. Now we need to analyze the circuit to determine
Aβ ≡ −Vr Vt
First, we write for the gain of the CS amplifier Q,
Vd = −gm[RD ∥ (R1 + R2)] Vt
(1)
R1 1
β=R +R =1+9=0.1V/V
12
R11 =R1 ∥R2 =1∥9=0.9k􏱹
R22 =R1 +R2 =1+9=10k􏱹
The A circuit is shown in Fig. 2. See figure on
then we use the voltage-divider rule to find Vr ,
Vr = R1 (2)
Vd R1 + R2
Combining Eqs. (1) and (2) gives
Aβ≡−Vr =gm[RD∥(R1+R2)] R1
Vt
which can be simplified to Aβ = g RDR1
next page.
re1 =re2 = VT
= 25mV =50􏱹 0.5 mA
R1 + R2
IE1,2
re3=VT =25mV=5􏱹
mR +R +R D12
IE3 5mA Vi
re1 +re2 + Rs +R11 β+1
Aβ (c) A = β
=g RD(R1+R2) m RD + R1 + R2
20
(d) β= R +R = 20+80 =0.2V/V
ie =
ie =
Vi
0.05 + 0.05 +
10+0.9 101
R1 12
⇒ ie = 4.81Vi
Rb3 =(β+1)[re3 +(R22 ∥RL)] = 101[0.005 + (10 ∥ 2)]
= 168.84 k􏱹
(1)
Aβ = 4 10 × 20 = 7.27 10+20+80
A = 7.27 = 36.36 V/V 0.2

These figures belong to Exercise 11.8.
R2
R2
Exercise 11–3
􏰀 1R2VR2􏰀V
1f11􏰒o 􏰒
b Vf 􏰔 R1
Vo R1 􏰀 R2
R2 0 R2
1 R1 2 1 R1 2
R11 􏰔R1􏰐􏰐R2
R22 􏰔R1 􏰀R2
RC
Figure 1
20 k􏰵
Q3 ib3 ie3
Rs 􏰔 10 k􏰵 aie
Q1 Q2 R Vo
b3
Vi 􏰀􏰒
i=αi RC
b3 e RC + Rb3
= 0.99 i 20
e 20 + 168.84
⇒ ib3 = 0.105ie
Vo = ie3(R22 ∥ RL)
= (β + 1)ib3(R22 ∥ RL)
= ib3 × 101(10 ∥ 2)
⇒ Vo = 168.33ib3 Combining (1)–(3), we obtain
ie
R11
R22 RL 􏰔 2 k􏰵
Ri
Ro
(2)
(3)
A≡Vo =85V/V Vi
β = 0.1 V/V Aβ = 8.5
1 + Aβ = 9.5
Af = 85 =8.95V/V 9.5
From the A circuit, we have
Ri = Rs +R11 +(β +1)(re1 +re2) = 10+0.9+101×0.1
=21k􏱹
Figure 2

Rif =Ri(1+Aβ)
= 21×9.5 = 199.5 k􏱹
Rin = Rif −Rs = 199.5−10 = 189.5 k􏱹 From the A circuit, we have
RD
Exercise 11–4
􏱾R􏱿 V Ro = RL ∥ R22 ∥ re5 + C o
β+1 􏱾 20 􏱿
R22
=2∥10∥ 0.005+ 101 = 181 􏱹
Q
R11
Ro
Rof = Ro 1+Aβ
= 181 =19.1􏱹 9.5
Rof = RL ∥ Rout 19.1 = 2 k􏱹 ∥ Rout ⇒Rout =19.2􏱹
Ex. 11.9 Figure 1 shows the β circuit together with the determination of β, R11 and R22.
Ri = 1 β=R+R gm
Vi 􏰀􏰒
Figure 2
Ri
R1 12
Ro =RD ∥R22
Rin =Rif =Ri(1+Aβ)
Rin= 1(1+Aβ) gm
Rout =Rof = Ro
1 + Aβ
R1
R1 +R2
β =
Af= A
1 + Aβ
From A circuit, we have
R11 =R1 ∥R2
R22 = R1 + R2
Figure 2 shows the A circuit. We can write Vo =gm(RD ∥R22)Vi
Thus,
Rout = RD ∥ (R1 + R2) VmD12 1+Aβ
A ≡ Vo = g [R ∥ (R + R )] i
This figure belongs to Exercise 11.9.
R2
R2
􏰀 1R2VR2􏰀V
1f11􏰒o 􏰒
b Vf 􏰔 R1
Vo R1 􏰀 R2
R2 0 R2 1R1 21R1 2
R11 􏰔R1􏰐􏰐R2
R22 􏰔R1 􏰀R2
Figure 1

Comparison with the results of Exercise 11.6 shows that the expressions for A and β are identical. However, Rin and Rout cannot be determined using the method of Exercise 11.6.
Ex. 11.10 From the solution to Example 11.6, we have
Aβ = 653.6
1 + Aβ = 654.6
A decrease in the op amp gain by 10% results in a decrease in A by 10% and a corresponding decrease in Af by
follows:
The open-loop gain A becomes
A = 0.9×653.6 = 588.24 mA/V
β=RF =1k􏱹 Af = 588.24
1+588.24×1 = 0.9983 mA/V
Change in Af = 0.9983 − 0.9985 = −0.0002
Percentage change in Af = −0.0002 × 100 =−0.02% 0.9983
Ex. 11.11 For a nominal closed-loop transconductance of 2 mA/V, we have
RF =β= 1 =0.5k􏱹 2 mA/V
From the solution to Example 12.6, we obtain A = μ gm(RF ∥ Rid ∥ ro2)
RF 1 + gm(RF ∥ Rid ∥ ro2) A = 1000 2(0.5 ∥ 100 ∥ 20)
0.5 1 + 2(0.5 ∥ 100 ∥ 20) A = 985.2 mA/V
Using Eq. (11.36), we obtain A1gm2
10% = 10% 1 + Aβ 654.6
Ri =Rs +Rid +RF
≃Rid +RF
= 100+0.2 = 100.2 k􏱹
From Eq. (11.35), we get
Aβ ≃ A1gm2RF = 200×2×0.2 = 80
1 + Aβ = 81
Rif = (1 + Aβ)Ri
= 81 × 100.2 ≃ 8.1 M􏱹
From Eq. (11.33), we have
Ro = ro2 +RL +RF
≃ro2+RF
= 20+0.2 = 20.2 k􏱹
Rof =Ro(1+Aβ)=20.2×81=1.64M􏱹 If gm2 drops by 50%, A drops by 50% to A=A1gm2 =200×1=200mA/V
and Af becomes
200
Af = 1+200×0.2 =4.878mA/V
Thus,
􏱺Af = 4.94 − 4.878 = −0.062
= 0.015%
A more exact solution (not using differential) is as
Exercise 11–5
Af ≃ 1 + A1 gm2 RF
= 200×2 = 4.94 mA/V
1+200×2×0.2 From Eq. (11.32), we have
􏱺Af Af
× 100 = −0.062 × 100 = −1.25% 4.94
Af ≡ Io = 985.2
Vs 1 + 985.2 × 0.5
Ex. 11.12 Af ≃ 5 mA/V β≃ 1 =0.2k􏱹=200􏱹
Af
RF =200􏱹
= 1.996 mA/V
Ex. 11.13 See figure on next page. Figure 1 shows the circuit for determining the loop gain. The figure also shows the analysis. We start by finding the current in the drain of Q2 as gm2Vg2 (this excludes the current in ro2). Since
ro2 ≫ RL + RF , most of gm2Vg2 will flow through RL and RF ∥ (Rid +Rs). Since RF ≪ Rid +Rs, the voltage across RF will be approximately −gm2Vg2RF . This voltage is amplified by A1 which provides at its output
Vr = −A1gm2RF Vg2
Thus, we find Aβ as
Vr
Aβ≡−V =A1gm2RF
g2

This figure belongs to Exercise 11.13.
Exercise 11–6
Rs
􏰀 A 1
Rid 􏰒
Rid 􏰀 Rs 􏰏􏰏 RF Figure 1
􏰒 G2
Q 2
RL
RF
r o 2 􏰏􏰏 R L 􏰀 R F
􏰀
V Vg2􏰀gm2Vg2 r􏰒
0
gm2Vg2 􏰀
􏰒gm2Vg2 RF 􏰒
Ex. 11.14 From Eq. (11.34), we obtain
Aβ= 􏱾R􏱿
􏱾 R id
􏱿􏱾 r 􏱿 o2
Rid +Rs +RF ForRF ≪ro2 andRid ≫Rs +RF,wehave
ro2
(A1gm2RF)
From Eq. (11.32), we obtain
ro2 +RL +RF
Rout = Rof − RL
=ro2 +RF +(A1gm2RF) id
Rid +Rs +RF
Ri =Rs +Rid +RF
Rif =Ri(1+Aβ)
=Ri+AβRi 􏱾
Rout ≃ro2(1+A1gm2RF) Ex. 11.16 To obtain
Af ≡Io ≃100mA/V Vs
Q.E.D.
ro2
s id F 1 m2 F id ro2 +RL +RF
􏱿
= R +R +R +(A g R )R
whentheloopgainislarge,weuse β≃ 1 =10􏱹
Rin =Rif −Rs
Rin ≃Rid +Agm2RFRid =Rid(1+Agm2RF) Q.E.D.
Ex. 11.15 From Eq. (11.34), we obtain
Aβ= 􏱾 R 􏱿􏱾 r 􏱿
􏱾r􏱿 Af
R =R +R +(Ag R )R o2
in id F m2 F id ro2 + RL + RF ForRF ≪Rid andro2 ≫RL +RF,wehave
But,
β=RE1 × RE2
RE2 +RF +RE1 ForRE1 =RE2 =100􏱹,
(A1gm2RF) id
Rid +Rs +RF
From Eq. (11.33), we get Ro =ro2 +RL +RF
Rof =Ro(1+Aβ)
= Ro + AβRo
= ro2+RL+RF +(A1gm2RF)
o2
ro2 +RL +RF
10= 100×100 200+RF
⇒RF =800􏱹 Vo = −IoRC1
Vs Vs
=−AfRC1 =−100×0.6=−60V/V
􏱾R􏱿 V id ro2 Aβ ≡ − r
Vt
Ex. 11.17 See figure on next page. Figure 1 shows the circuit prepared for the determination of the loop gain,
Rid + Rs + RF

This figure belongs to Exercise 11.17.
Exercise 11–7
RC1
RC2
RC3 Q3
Ic2 􏰀 Q2
Ib3
Q1 􏰒Ri3Ie3
Ic1
rp2 Vr Vt 􏰀
􏰒
Figure 1
Ie1 re1
RE1
We shall trace the signal around the loop as follows:
􏲜
(7) (8)
R F
If
RE2
Ic2 = gm2Vt I=I RC2
(1) (2)
Ie3 = 101Ib3
If = 0.13Ie3
Ie1 = 0.706If
Ic1 = 0.99Ie1
Vr = −1.957Ic1
Combining (9)–(15), we obtain
Aβ = −Vr = 249.3 Vt
Ex. 11.18
RF
􏰒
RR VVt􏰀 Rid􏰀􏰒mVtRL
b3 c2 RC2 + Ri3 where
(11) (12) (13) (14) (15)
Ri3 = 􏲛
(β+1) re3 +[RE2 ∥(RF +(RE1 ∥re1))]
(3) (4)
(5) (6)
Ie3 =(β+1)Ib3
I =I RE2
f e3 RE2 + RF + (RE1 ∥ re1) I=I RE1
e1 f RE1 +re1
Ic1 = αIe1
Vr = −Ic1(RC1 ∥ rπ2)
ro
Combining (1)–(7) gives Vr in terms of Vt and hence Aβ ≡ −Vr /Vt . We shall do this numerically using the values in Example 11.8:
􏰀
s id r 􏰒 􏰒
􏰒
􏰀
gm2 =40mA/V, RC2 =5k􏱹, β=100,
re3 =6.25􏱹, RE1 =RE2 =100􏱹, RF =640􏱹,
re1 =41.7􏱹, α1 =0.99, RC1 = 9k􏱹, andrπ2 =2.5k􏱹
Figure 1
R = 101􏲛0.00625 + [0.1 ∥ (0.64
Figure 1 shows the circuit prepared for determining the loop gain
Aβ≡−Vr Vt
Using the voltage-divider rule, we can write by inspection
i3
+ (0.1 ∥ 0.0417))]
􏲜
= 9.42 k􏱹
Ic2 = 40Vt (9) Ib3 = 0.347Ic2 (10)
Vr =
−μVt ro +􏲛RL ∥ [RF +(Rs ∥ Rid)]􏲜 RF +(Rs ∥ Rid)
RL ∥ [RF +(Rs ∥ Rid)] (Rs ∥ Rid)

Vr =
−μV RL(Rs ∥Rid)
tro[RL+RF+(Rs∥Rid)]+RL[RF+(Rs∥Rid)] Thus,
Aβ=−Vr = Vt
μRL(Rid ∥ Rs)
ro[RL +RF +(Rid ∥Rs)]+RL[RF +(Rid ∥Rs)]
Q.E.D.
Using the numerical values in Example 11.9, we get
Aβ = 104 × 1 × 1 0.1(1+10+1)+1(10+1)
=819.7
Ex. 11.19 See figure on next page. Figure 1(a) shows the feedback amplifier circuit. The β circuit is shown in Fig. 1(b), and the determination of β is shown in Fig. 1(c),
β=−1 RF
thus,
1 1
R =R [1+gm(ro∥RF)]
Exercise 11–8
in F ⇒Rin=
RF 1+gm(ro ∥RF)
Q.E.D.
Ro (d) Rout =Rof = 1+Aβ
ro ∥ RF
= 1 + gm (Rs ∥ RF )(ro ∥ RF )/RF
1 =1+1+gm(Rs∥RF) Rout ro RF RF
⇒ Rout = ro ∥ RF 1+gm(Rs ∥RF)
(e) A = −5(1 ∥ 10)(20 ∥ 10) A = −30.3 k􏱹
β=−1 =−1 =−0.1mA/V RF 10
Aβ = 3.03
1 + Aβ = 4.03
Af = A =−30.3=−7.52k􏱹 1 + Aβ 4.03
Q.E.D.
(a) For large loop gain, we have A ≃1=−R
f β F
(b) The determination of R11 and R22 is illustrated
(Compare to the ideal value of −10 k􏱹).
Ri =Rs ∥RF =1∥10=909􏱹
in Figs. 1(d) and (e), respectively: R11 =R22 =RF
Finally, the A circuit is shown in Fig. 1(f). We can writebyinspection
Ro =ro ∥RF =20∥10=6.67k􏱹 Ri 909
Ri = Rs ∥ R11 = Rs ∥ RF R=r∥R =r∥R
Rif = 1+Aβ = 4.03 =226􏱹 􏲈 􏱼 1 1 􏱽
o o 22 o F Vgs = IiRi
Vo = −gmVgs(ro ∥ R22) Thus,
Rin=1 R −R =291􏱹 if s
Rof = Ro = 6.67 = 1.66 k􏱹 1 + Aβ 4.03
Rout = Rof = 1.66 k􏱹
Ex. 11.20 From Eq. (11.54), we obtain A=−μ Ri R1∥R2∥ro2
R1 ∥R2 1/gm +(R1 ∥R2 ∥ro2) Forμ=100,R1 =10k􏱹,R2 =90k􏱹,
gm =5mA/V, ro2 =20k􏱹,wehave Ri = Rs ∥ Rid ∥ (R1 + R2)
= ∞ ∥ ∞ ∥ 100 = 100 k􏱹 A=−100 100 10∥90∥20
10∥90 0.2+(10∥90∥20) = −1076.4 A/A
A≡Vo =−gm(Rs∥RF)(ro∥RF) Ii
A = Vo =
A 1+Aβ
f
Af =
Is
gm(Rs ∥RF)(ro ∥RF)

(c) Rif = Ri
Q.E.D.
1+gm(Rs ∥RF)(ro ∥RF)/RF 1 + Aβ
Rif =
Rs ∥ RF
1+gm(Rs ∥RF)(ro ∥RF)/RF
1 = 1 + 1 +gm(ro ∥RF) RifRsRF RF
But,
1 = 1 + 1
Rif Rs Rin
β = − R1 = − R1 + R2
10 = −0.1 A/A 10 + 90
Af =− 1076.4 =−9.91A/A 1 + 107.64

This figure belongs to Exercise 11.19.
RF
􏰀
Is Rs Vgs
􏰒
gmVgs ro
Vo
Rout 􏰔 Rof
Exercise 11–9
Rif Rin
RF IfRF
(a)
1 2 1 2􏰀Vo 􏰒
(b)
RF
b􏰒 If 1 􏰔V 􏰔􏰒R
oF (c)
RF
1212
R11 􏰔RF
Ii
(d)
Ri
Rs
􏰀
R11 Vgs 􏰒
(f)
Figure 1
g V m gs
(e)
R22
ro
R22 􏰔RF Vo
Ro
Rif =
Ri
= 100 k􏱹 = 920 􏱹 108.64
SubstitutingR2 =0andRs =Rid =∞inEq. (11.50), we obtain
Ri = R1
and in Eq. (11.55), we obtain
Ro = ro2
and in Eq. (11.53), we obtain
1 + Aβ
Rin =Rif =920􏱹
Rout=Rof =Ro(1+Aβ)
Ro = ro2 +(R1 ∥ R2)+gmro2(R1 ∥ R2) = 929 k􏱹
Rout = 929 × 108.64 = 101 M􏱹
Ex. 11.21 With R2 = 0, Eq. (11.48) gives β = −1
Af = −1 A/A
A=−μ R1 1/gm
=−μgmR1
Now,
Af= A
1 + Aβ

Af =− μgmR1
1 + μgmR1
Rin=Rif =Ri/(1+Aβ) = R1
1+μgmR1
For μgmR1 ≫ 1, we have Rin ≃ 1/μgm
Rout=Rof =(1+Aβ)Ro = (1 + μgmR1)ro2
≃ μ(gmro2)R1
Ex. 11.22 Total phase shift will be 180◦ at the frequency ω180 at which the phase shift of each amplifier stage is 60◦. Thus,
tan−1 ω180 = 60◦ 104
ω180 = tan 60◦ × 104 √
= 3×104 rad/s At ω180, we have
1+A0β=1+105 ×0.01=1001
The pole will be shifted to a frequency
fPf =fP(1+A0β)
= 100×1001 = 100.1 kHz
If β is changed to a value that results in a nominal closed-loop gain of 1, then we obtain
β≃1
and
1 + A0β = 1 + 105 × 1 ≃ 105
then the pole will be shifted to a frequency fPf = 105 × 100 = 10 MHz
Ex. 11.24 From Eq. (11.68), we see that the poles coincide when
(ωP1 + ωP2)2 = 4(1 + A0β)ωP1ωP2
(104 +106)2 = 4(1+100β)×104 ×106
⇒ 1 + 100β = 25.5
⇒ β = 0.245
The corresponding value of Q = 0.5. This can also be verified by substituting in Eq. (11.70).
A maximally flat response is obtained when
| A| = = 125

Exercise 11–10
􏱾 10 􏱿3 √
1+3
Thus, the loop gain magnitude will be |Aβ|=125β
For stable operation, we require 125βcr < 1 ⇒βcr = 1 =0.008 125 β ≥ βcr will result in oscillations. Correspondingly, the minimum closed-loop gain for stable operation will be Q = 1/ 1 2. Substituting in Eq. (11.70), we obtain 􏲇 (1+100β)×104 ×106 104 + 106 ⇒ β = 0.5 In this case, the low-frequency closed-loop gain is Af (0) = A0 1 + A0β 100 = 1 + 100 × 0.5 = 1.96 V/V Ex. 11.25 The closed-loop poles are the roots of the characteristic equation √ = 2 103 Af = 1 + 103 β 103 1000 = 1+1000×0.008 = 9 cr Ex. 11.23 The feedback shifts the pole by a factor equal to the amount of feedback: 1 + s 104 = 111.1 1 + A(s)β = 0 ⎛⎞ 3 1+⎜⎝ 10 ⎟⎠β=0 To simplify matters, we normalize s by the factor 104, thus obtaining the normalized complex-frequency variable S = s/104, and the characteristic equation becomes (S+1)3+103β=0 (1) This equation has three roots, a real one and a pair that can be complex conjugate. The real pole can be found from From Fig. 1, we can easily obtain the loop gain as Aβ = A(s) × 0.01 = s ×0.01 Exercise 11–11 1+ = 1000 105 2π × 10 1+s (S+1)3 =−103β ⇒S =−1−10β1/3 =−􏲑1+10β1/3􏲅 (2) From this single-pole response (low-pass STC response) we can find the unity-gain frequency by inspection as f1 =fP ×1000 = 104 Hz The phase angle at f1 will be −90◦ and thus the phase margin is 90◦. Ex. 11.27 From Eq. (11.82), we obtain | Af (jω1)| = 1/|1 + e−jθ | 1/β = 1/|1+cos θ −j sin θ| (a) ForPM=30◦,θ =180−30=150◦,thus |Af (jω1)| ◦ ◦ 1/β = 1/|1+cos 150 −j sin 150 | = 1.93 (b) ForPM=60◦,θ =180−60=120◦,thus | Af (jω1)| = 1/|1 + cos 120◦ − j sin 120◦| 1/β =1 (c) ForPM=90◦,θ =180−90=90◦,thus | Af (jω1)| = 1/|1 + cos 90◦ − j sin 90◦| 1/β √ =1/ 2=0.707 Ex. 11.28 See figure on next page. To obtain guaranteed stable performance, the maximum rate of closure must not exceed 20 dB/decade. Thus we utilize the graphical construction in Fig. 1 to obtain the Dividing the characteristic polynomial in (1) by 􏲑S + 1 + 10β1/3􏲅 gives a quadratic whose two roots are the remaining poles of the feedback amplifier. After some straightforward but somewhat tedious algebra, we obtain S2 +􏲑10β1/3 −2􏲅S+􏲑1+100β2/3 −10β1/3􏲅 =0 (3) The pair of poles can now be obtained as S = 􏲑−1 + 5β1/3􏲅 ± j5√3 β1/3 Equations (1) and (3) describe the three poles shown in Fig. E11.25. From Eq. (2) we see that the pair of complex poles lie on the jω axis for the value of β that makes the coefficient of S equal to zero, thus 􏱾 2 􏱿3 βcr= 10 =0.008 Note that this is the same value found in the solution of Exercise 11.22. (4) 2π × 10 Ex. 11.26 R 99R 􏰒 A (s) 􏰀 Figure 1 This figure belongs to Exercise 11.28. dB 100 80 60 40 20 10􏰒1 100 101 vdifferentiator 􏰔 1τ |A| “rate of closure” 􏰔 20 dB/decade 􏰒20 dB/decade Exercise 11–12 􏰒10􏰒2 102 103 104 105 􏰒40 dB/decade 106 f (Hz) maximum value of the differentiator frequency as 1 Hz. Thus, 1 ≤ 2π × 1 Hz τ τ ≥ 1 s = 159 ms 2π Ex. 11.29 To obtain stable performance for closed-loop gains as low as 20 dB (which is 80 dB below A0, or equivalently 104 below A0), we must place the new dominant pole at 1 MHz/104 = 100 Hz. Ex. 11.30 The frequency of the first pole must be lowered from 1 MHz to a new frequency fD′ = 10MHz =1000Hz 104 that is, by a factor of 1000. Thus, the capacitance at the controlling node must be increased by a factor of 1000. Figure 1 􏰀20 dB/decade 20 log b1 for differentiator 11.1Af=1+Aβ 104 R2=100.033−1 =290k􏱹 (c) (i) A=1000(1−0.2)=800V/V 200 = 1 + 104β Chapter 11–1 A 􏱾1􏱿 ⇒β=4.9×10−3 If A changes to 103, then we get 800 1 + 800 × 0.099 Af = 1000 1 + 103 × 4.9 × 10−3 = 1000 = 169.5 5.9 Percentage change in Af = = −15.3% Af = = 9.975 V/V 11.2 (a) Because of the infinite input resistance of the op amp, the fraction of the output voltage Vo that is fed back and subtracted from Vs is determined by the voltage divider (R1 , R2 ), thus β= R1 R1 +R2 Thus, Af changes by 9.877 − 10 169.5−200 200 × 100 Thus, Af changes by = 9.975 − 10 × 100 = −0.25% 10 (ii) A=200(1−0.2)=160V/V 160 Af = 1+160×0.095 = 9.877 V/V = 10 × 100 = −1.23% (b) (i) A = 1000 V/V A (iii) A=15(1−0.2)=12V/V Af = 12 =8.574 1+12×0.033 Thus, Af changes by =8.575−10×100=−14.3% 10 We conclude that as A becomes smaller and hence the amount of feedback (1 + Aβ) is lower, the desensitivity of the feedback amplifier to changes in A decreases. In other words, the negative feedback becomes less effective as (1 + Aβ) decreases. 11.3 The direct connection of the output terminal to the inverting input terminal results in Vf = Vo andthus β=1 If A = 1000, then the closed-loop gain will be Af= A 1 + Aβ = 1000 = 0.999 V/V 1 + 1000 × 1 Amountoffeedback=1+Aβ = 1+1000×1 = 1001 or 60 dB ForVs =1V,weobtain Vo =AfVs =0.999×1=0.999V Vi = Vs −Vo = 1−0.999 = 0.001 V Af =1+Aβ 10 = 1000 1 + 1000β ⇒ β = 0.099 V/V R1 = 0.099 R1 + R2 1+R2= 1 R1 0.099 􏱾1􏱿 R2 =R1 0.099−1 􏱾1􏱿 −1 =91k􏱹 (ii) A = 200 V/V =10 0.099 10 = 200 1+200β ⇒ β = 0.095 V/V 􏱾 1 􏱿 R2 = R1 0.095 − 1 􏱾1􏱿 −1 =95.3k􏱹 (iii) A = 15 V/V =10 0.095 10 = 15 1+15β ⇒β=0.033V/V If A becomes 1000(1 − 0.1) = 900 V/V, then we get Af = 900 = 0.99889 1+900×1 Thus, Af changes by = 0.99889 − 0.999 × 100 = −0.011% 0.999 11.4A=Vo= 5V =500V/V Vi 10 mV Vf =Vs −Vi =1−0.01=0.99V β= Vf = 0.99 =0.198V/V Vo 5 A 11.5 (a) Af = 1 + Aβ The results obtained are as follows. Chapter 11–2 Case A (V/V) Af (V/V) for β = 0.00 Af (V/V) for β = 0.50 Af (V/V) for β = 1.00 (a) 1 1 0.667 0.500 (b) 10 10 1.667 0.909 (c) 100 100 1.961 0.990 (d) 1000 1000 1.996 0.999 (e) 10,000 10,000 1.9996 0.9999 Ideally, Af = 1 11.7 A= 5V =2500V/V 2 mV Af = 5V =50V/V 100 mV Amountoffeedback≡1+Aβ A 2500 =A = 50 =50 f or 34 dB Aβ = 49 β = 49 = 0.0196 V/V 2500 11.8 Anominal = 1000 Alow = 500 Ahigh = 1500 If we apply negative feedback with a feedback factor β, then Af , nominal = 1000 1 + 1000β β Af ideal−Af =β−1+Aβ = 1 + Aβ − Aβ = 1 (1 + Aβ)β (1 + Aβ)β Expressed as a percentage of the ideal gain 1/β, we have 􏲀􏲀 1 A 1 1 + Aβ Difference = Ideal For Aβ ≫ 1, Difference (b) For Af to be within: (i) 0.1% of ideal value, then 100 ≤ 0.1 Aβ ⇒Aβ≥1000 (ii) 1% of ideal value, then 100 ≤1 Aβ ⇒ Aβ ≥ 100 (iii) 5% of ideal value, then 100 ≤ 5 Aβ ⇒ Aβ ≥ 20 11.6 For each value of A given, we have three different values of β: 0.00, 0.50, and 1.00. To obtain Af , we use Af= A 1 + Aβ × 100% 100 Ideal ≃ Aβ % Af,low = Af,high = 500 1 + 500β 1500 1 + 1500β It is required that Af , low ≥ 0.99Af , nominal (1) and Af , high ≤ 1.01Af , nominal (2) If we satisfy condition (1) with equality, we can determine the required value of β. We must then clock that condition (2) is satisfied. Thus, 500 = 0.99 × 1000 1+500β 1+1000β ⇒ β = 0.098 For this value of β, we obtain 1000 Af,nominal = 1+1000×0.098 and break the loop at the input terminals of the op amp. To keep the circuit unchanged, we must place a resistance equal to Rid at the left-hand side of the break. This is shown in Fig. 2, where a test signal Vt is applied at the right-hand side of the break. To determine the returned voltage Vr , we use the voltage-divider rule as follows: V=−μV R1∥Rid r 1(R1 ∥Rid)+R2 Substituting V1 = Vt and rearranging, we obtain Vr R1 ∥ Rid R1 + R2 R1 1+500×0.098 A = 1500 f , high 1 + 1500 × 0.098 Chapter 11–3 = 10.101 Af,low = 500 =10 = 10.135 Thus, the low value of the closed-loop gain is 0.101 below nominal or −1%, and the high value is 0.034 above nominal or 0.34%. Thus, our amplifier meets specification and the nominal value of closed-loop gain is 10.1. This is the highest possible closed-loop gain that can be obtained while meeting specification. Now, if three closed-loop amplifiers are placed in cascade, the overall gain obtained will be Nominal Gain = (10.1)3 = 1030 Lowest Gain = 103 = 1000 Highest Gain = (10.135)3 = 1041 Thus, the lowest gain will be approximately 3% below nominal, and the highest gain will be 1% above nominal. Aβ≡−V =μ(R ∥R )+R t 1 id 2 Since β= R1 R1 + R2 we get R1 ∥ Rid A=μ(R1 ∥Rid)+R2 =μ =μ Rid/(R1 +Rid) R1Rid/(R1 +Rid)+R2 Rid(R1 +R2) R1Rid +R2Rid +R1R2 (R1 +R2) Q.E.D. 11.9 Vs 􏰀 􏰒 Thus, A = μ Rid Rid +(R1 ∥R2) X 􏰀 􏰀 Rid V1 mV1 􏰀 V 􏰒􏰒 dA/A =1+Aβ 11.10 From Eq. (11.10), we have 􏰒 o dAf/Af 1 X' R2 R1 Since −40 dB is 0.01, we have 0.01 = 1 1 + Aβ ⇒ Aβ = 99 Figure 1 For dAf /Af = 1 dA/A 5 􏰀􏰀 we have 1 + Aβ = 5 ⇒ Aβ = 4 R1 mV1 􏰒􏰒 R2 Figure 2 11.11 For A = 1000 V/V, we have Af =10= 1000 RidVrVt􏰀􏰒RidV1 􏰀􏰒 Figure 1 shows the given circuit with the op amp replaced with its equivalent circuit model. To determine the loop gain Aβ, we short circuit Vs Aβ = 99 β= 99 =0.099V/V 1000 1 + Aβ ⇒ Densensitivity factor ≡ 1 + Aβ = 100 For A = 500 V/V, we have Af =10= 500 1 + Aβ ⇒ Densensitivity factor ≡ 1 + Aβ = 50 Chapter 11–4 Substituting in (1) yields A = 24.75 × 45.9 = 1136 and β = 44.9 = 0.0395 1136 11.14 Let the gain of the ideal (nonvarying) driver amplifier be denoted μ. Then, the open-loop gain A will vary from 2μ to 12μ. Correspondingly, the closed-loop gain will vary from 95 V/V to 105 V/V. Substituting these quantities into the closed-loop gain expression, we obtain 95 = 2μ (1) 1 + 2μβ 105 = 12μ (2) 1+12μβ Dividing Eq. (2) by Eq. (1) yields 1.105= 6(1+2μβ) 11.12 A =10V/V f 1+12μβ β= 49 =0.098V/V 500 If the A = 1000 amplifiers have a gain uncertainty of ±10%, the gain uncertainty of the closed-loop amplifiers will be = ±10% = ±0.1% 100 If we require a gain uncertainty of ±0.1% using the A = 500 amplifiers, then Gain uncertainty of A = 500 amplifiers 50 ±0.1% = ⇒ Gain uncertainty = ±5% 1+Aβ= ±10% =100 ±0.1% 10= A 100 1.105+1.105×12μβ=6+12μβ ⇒ μβ = 3.885 Substituting in Eq. (1) yields μ= 95(1+2×3.885) =416.6V/V 100−1 2 ⇒ A = 1000 V/V β = 1000 = 0.099 V/V 11.13 The open-loop gain varies from A to 10A with temperature and time. Correspondingly, Af varies from (25 − 1%) i.e. 24.75 V/V to (25 + 1%) or 25.25 V/V. Substituting these quantities into the formula for the closed-loop gain Af = A 1+Aβ we obtain 3.885 −3 β= =9.33×10 V/V 416.6 If Af is to be held to within ±0.5%, Eqs. (1) and (2) are modified to 2μ 99.5 = 1 + 2μβ (3) 100.5 = 12μ (4) 1 + 12μβ Dividing (4) by (3) yields 6(1+2μβ) 24.75 = 25.25 = Substituting into (3) provides μ = 99.5(1 + 2 × 49.92) 2 = 5016.8 V/V which is more than a factor of 10 higher than the gain required in the less constrained case. The value of β required is β = 49.92 = 9.95 × 10−3 V/V 5016.8 A (1) 1 + Aβ 10A (2) 1 + 10Aβ 1.01= 1+12μβ ⇒ μβ = 49.92 Dividing Eq. (2) by Eq. (1), we obtain 1.02=10 1+Aβ 1 + 10Aβ 1.02 + 10.2Aβ = 10 + 10Aβ ⇒ Aβ = 44.9 Repeating for Af = 10 V/V (a factor of 10 lower than the original case): (a) For ±5% maximum variability, Eqs. (1) and (2) become and the cascade of two stages will thus show a variability of ±0.6%, well within the required ±1%. Thus two stages will suffice. We next investigate the design in more detail. Each stage will have a nominal gain of 10 and thus = 100 ⇒ β = 0.099 Since A ranges from 700 V/V to 1300 V/V, the 2μ 9.5= 1+2μβ 10.5 = 12μ 1+12μβ Dividing (6) by (5) yields 1.105 = 6(1 + 2μβ) 1+12μβ ⇒ μβ = 3.885 (5) (6) 1000 10 Chapter 11–5 1 + Aβ = ⇒ Aβ = 99 which is identical to the first case considered, and μ= 9.5(1+2×3.885) =41.66V/V 1+12μβ which is a factor of 10 lower than the value required when the gain required was 100. The feedback factor β is β = 3.885 = 9.33 × 10−2 V/V 41.66 which is a factor of 10 higher than the case with Af =10. (b) Finally, for the case Af = 10 ± 0.5% we can write by analogy μβ = 49.92 μ = 501.68 V/V β = 9.95 × 10−2 V/V 11.15 If we use one stage, the amount of feedback required is 1+Aβ= A =1000=10 Af 100 Thus the closed-loop amplifier will have a variability of Variability of Af = ±30% = ±3% 10 which does not meet specifications. Next, we try using two stages. For a nominal gain of 100, each stage will be required to have a nominal gain of 10. Thus, for each stage the amount of feedback required will be 1 + Aβ = 1000 = 100 10 Thus, the closed-loop gain of each stage will have a variability of ±30% = 100 = ±0.3% Af,low = 1+700×0.099 =9.957V/V andahighvalueof gain of each stage will range from 700 1300 Af,high = 1+1300×0.099 =10.023V/V Thus, the cascade of two stages will have a range of Lowest gain = 9.9572 = 99.14 V/V Highest gain = 10.0232 = 100.46 V/V which is −0.86% to +0.46% of the nominal 100 V/V gain, well within the required ±1%. 11.16 If the nominal open-loop gain is A, then we require that as A drops to (A/2) the closed-loop gain drops from 10 to a minimum of 9.8. Substituting these values in the expression for the closed-loop gain, we obtain 10 = 9.8 = A (1) 1 + Aβ A/2 1 (2) 1 + 2 Aβ Dividing Eq. (1) by Eq. (2) yields 􏱾1􏱿 2 1+2Aβ 1.02 = 1.02 = 2 + Aβ 1 + Aβ 1 + Aβ =1+ 1 1 + Aβ ⇒1+Aβ= 1 =50 0.02 Substituting in Eq. (1) gives A = 10 × 50 = 500 V/V and 50 − 1 β= 500 =0.098V/V If β is accurate to within ±1%, to ensure that the minimum closed-loop gain realized is 9.8 V/V, we have AM s/(s + ωL) 1 + AM βs/(s + ωL) Chapter 11–6 = =s+ω +sA β AMs L M 9.8=A/2 1+ 1A×0.098×1.01 2 ⇒ A = 653.4 V/V = = 1+AMβ s+ωL/(1+AMβ) AM s s(1 + AM β) + ωL AM s 11.17 Af = 100 = A A 1+Aβ Thus, AMf= AM 1+AMβ ωLf= ωL 1 + AM β Thus, both the midband gain and the 3-dB frequency are lowered by the amount of feedback, (1 + AM β). 11.19 1+AMβ= 1000 =100 10 Thus, fHf =(1+AMβ)fH = 100 × 10 = 1000 kHz = 1 MHz fLf= fL 1+AMβ = 100 =1Hz 100 11.20 To capacitively couple the output signal to an 8-􏱹 loudspeaker and obtain fL = 100 Hz, we need a coupling capacitor C, C= 1 2πfL ×8 = 1 = 198.9 μF ≃ 200 μF 2π ×100×8 If closed-loop gain AM f of 10 V/V is obtained from an amplifier whose open-loop gain AM = 1000 V/V, then 1 + AM β = 1000 = 100 10 99 = 1 + Aβ 0.1A 1 + 0.1Aβ (1) (2) Dividing Eq. (1) by Eq. (2) gives 1.01= 10(1+0.1Aβ) 1 + Aβ = 10 + Aβ 1+Aβ = 1 + 9 1 + Aβ 9 ⇒ 1+Aβ =0.01 1 + Aβ = 900 Aβ = 899 Substituting (1 + Aβ) = 900 into Eq. (1) yields A = 100 × 900 = 90, 000 V/V The value of β is β= 899 =9.989×10−3 V/V 90,000 If A were increased tenfold, i.e, A = 900, 000, we obtain A = 900, 000 = 100.1 V/V f 1 + 8990 If A becomes infinite, we get A Af =1+Aβ =1=1 and fLf =100=100=1Hz 1+β β A fL 100 If the required fLf is 50 Hz, then 1 = 9.989×10−3 =100.11V/V 11.18 A=A s M s + ωL Af=A 1+Aβ fL =50×(1+AMβ) =50×100=5000Hz, and the coupling capacitor C will have a value of C= 1 ≃4μF 2π ×5000×8 11.21 Let’s first try N = 2. The closed-loop gain of each stage must be √ Af = 1000 = 31.6 V/V Thus, the amount-of-feedback in each stage must be 1+Aβ= A =1000=31.6 Af 31.6 The 3-dB frequency of each stage is f3dB|stage = (1 + Aβ)fH = 31.6×20 = 632 kHz Thus, the 3-dB frequency of the cascade amplifier is f3dB |cascade = 632 21/2 − 1 = 406.8 kHz which is less than the required 1 MHz. Next, we try N = 3. The closed-loop gain of each stage is Af =(1000)1/3 =10V/V and thus each stage will have an amount-of-feedback 1 + Aβ = 1000 = 100 10 which results in a stage 3-dB frequency of f3dB|stage = (1 + Aβ)fH = 100×20 = 2000 kHz = 2 MHz The 3-dB frequency of the cascade amplifier Af = A1A2 1 + A1A2β 10 = 0.9A2 9 Chapter 11–7 ⇒A2 =100V/V β = 8 = 0.089 V/V 0.9 × 100 To reduce Vo ripple to 10 mV, 0.01 = 1× ⇒ 1 + A1A2β = 90 0.9 1 + A1A2β 􏲇 Af = A1A2 1 + A1A2β 10 = 0.9A2 90 A2 = 1000 V/V 89 β= 0.9×1000 =0.099V/V will be 􏲇 To reduce Vo ripple to 1 mV, 0.001 = 1 × 0.9 1 + A1A2β ⇒ 1 + A1A2β = 900 10 = 0.9A2 900 ⇒ A2 = 10,000 V/V β = 899 = 0.0999 V/V 0.9 × 10, 000 f3dB|cascade = 2 21/3 − 1 =1.02MHz which exceeds the required value of 1 MHz. Thus, we need three identical stages, each with a closed-loop gain of 10 V/V, an amount-of-feedback of 100, and a loop gain Aβ = 99 Thus, β = 0.099 V/V A1A2 11.23 Af = 1+A1A2β 100 = 10A2 (1) 1 + A1A2β (1 + A1A2β) × 8 = 40 kHz ⇒ 1 + A1A2β = 5 Substituting in (1) gives A2 = 100×5 =50V/V 10 1 + 10 × 50 × β = 5 ⇒ β = 0.008 V/V 80 fLf =1+A1A2β 80 = 5 =16Hz 11.22 V o ripple = V A1 n 1 + A1A2β To reduce Vo ripple to 100 mV, 0.1=1× 0.9 1 + A1A2β ⇒1+A1A2β=9 11.24 vS 􏰀 􏰀 ve m 􏰔 100 v 􏰒􏰒I 􏰀V 􏰒V and that for the second segment is Af2 = 100 (2) Chapter 11–8 vO Af1 =1.1 Af 2 1+100β We require Thus, dividing Eq. (1) by Eq. (2) yields 1.1 = 10 1+100β Figure 1 vO ⇒ β = 0.089 Af1 = 1000 =11.1V/V 1 + 1000β 1.1+1100β=10+1000β 1+1000×0.089 100 =10.1V/V The first segment ends at |vO| = 10 mV × 1000 = 10 V. This corresponds to vS = 10 V = 10 = 0.9 V Af 1 11.1 The second segment ends at |vO| = 10 + 0.05 × 100 = 15 V. This corresponds to vS =0.9+15−10 Af 2 =0.9+ 5 =1.4V 10.1 Thus, the transfer characteristic of the feedback amplifier can be described as follows: For |vS| ≤ 0.9 V, vO/vS = 11.1 V/V For0.9V≤|vS|≤1.4V, vO/vS =10.1V/V For|vS|≥1.4V, vO =±15V The transfer characteristic is shown in the figure on next page. 11.26 Because the op amp has an infinite input resistance and a zero output resistance, this circuit is a direct implementation of the ideal feedback structure and thus A = 1000 V/V and β= R1 R1 + R2 The ideal closed-loop gain is Af =1=1+R2 Af2 = 1+100×0.089 0 􏰀7 mV 􏰒7 mV Slope 􏰔 0.99 V/V Figure 2 vS Slope 􏰔 0.99 V/V RefertoFig.1.ForvI =+0.7V,wehavevO =0 and ve = vI = +0.7 = +7 mV μ 100 Similarly, for vI = −0.7 V, we obtain vO = 0 and vI −0.7 ve = μ = 100 =−7mV Thus, the limits of the deadband are now ±7 mV. Outside the deadband, the gain of the feedback amplifier, that is, vO/vS, can be determined by noting that the open-loop gain A ≡ vO/ve = 100 V/V and the feedback factor β = 1, thus A≡vO= A f vS 1 + Aβ = 100 1+100×1 = 0.99 V/V The transfer characteristic is depicted in Fig. 2. 11.25 The closed-loop gain for the first (high-gain) segment is Af1 = 1000 1+1000β (1) β R1 This figure belongs to Problem 11.25. Chapter 11–9 Thus, 10 = 1 + R2 10 ⇒R2 =90k􏱹 β = 10 = 0.1 V/V 10+90 Aβ = 1000 × 0.1 = 100 Af= A (b) From Example 11.3, we obtain RL ∥[R2 +R1 ∥(Rid +Rs)] Aβ =μ􏲛RL ∥[R2 +R1 ∥(Rid +Rs)]􏲜+ro × R1 ∥(Rid +Rs) × Rid [R1 ∥(Rid +Rs)]+R2 Rid +Rs = 1 + Aβ 10 ∥ [90 + 10 ∥ (100 + 100)] Aβ = 1000 􏲛10 ∥ [90 + 10 ∥ (100 + 100)]􏲜 + 1 × 100 100 + 100 (c) To obtain Af = 9.9 V/V, we use 9.9= A 1 + Aβ =A 1 + A × 0.1 ⇒ A = 1010 V/V Thusμmustbeincreasedbythefactor 1010 = 2.343 to become 431.1 μ = 2343 V/V 10 ∥ (100 + 100) [10 ∥ (100 + 100)] + 90 1000 1+100 × = 1000 × 0.9009 × 0.0957 × 0.5 = 9.9 V/V that is exactly 10, we use 1 + Aβ ⇒ Aβ = 99 To obtain Af 10 = 1000 = 43.11 Aβ 43.11 β = 0.099 0.099 = R1 A= β = 0.1 =431.1V/V A 431.1 Af = 1+Aβ = 1+43.11 =9.77V/V R1 + R2 0.099 = 10 10+R2 ⇒R2 =91k􏱹 11.27 Refer to Fig. 11.11. (a) The ideal closed-loop gain is given by Af = 1 = R1 +R2 =1+ R2 β R1 R1 10=1+ R2 10 ⇒R2 =90k􏱹 11.28 Refer to Fig. 11.10. (a) β = R1 The dc analysis is shown in Fig. 1 from which we see that IE1 ≃ 0.1 mA IE2 ≃ 0.3 mA VE2 =+7.7V (c) Setting Vs = 0 and eliminating dc sources, the feedback amplifier circuit simplifies to that shown in Fig. 2. Chapter 11–10 R1 + R2 A 􏲀􏲀 = 1 = 1 + R 2 fideal β R1 5 = 1 + R2 1 ⇒R2 =4k􏱹 (b) From Example 11.2, we have Aβ = (gm1RD1)(gm2RD2) 1 × 1 + gm1R1 􏱾1􏱿 R1 RD2+R2+ R1∥g Q2 Q1 Rs R1 RL Rib Figure 2 Now, breaking the feedback loop at the base of Q1 while terminating the right-hand side of the circuit (behind the break) in the resistance Rib, Rib =(β1 +1)(re1 +Rs) results in the circuit in Fig. 3 which we can use to determine the loop gain Aβ as follows: m1 11 B1R2 = (4 × 10)(4 × 10) 1 + 4 × 1 × 10 + 4 + (1 ∥ 0.25) = 22.54 A= Aβ = 22.54 =112.7V/V β 0.2 Af= A = 1 + Aβ 112.7 = 4.79 V/V 11.29 (a) The feedback network consists of the 1 + 22.54 voltage divider (R1R2), thus β= R1 R1 + R2 If the loop gain is large, the closed-loop gain approaches the ideal value Af =1=1+R2 β R1 = 1 + 10 = 11 V/V 1 (b) 0.1 mA 0.1 mA Rs 100 k􏰵 Q 􏰀8.4 V Q2 0.003 mA 0 􏰀0.7 V R2 10 k􏰵 0.7 mA R1 1 k􏰵 Figure 1 0.3 mA 􏰀7.7 V 1 mA 1 Figure 3 Ie1= Vt (1) re1 + Rs Ic1 = α1Ie1 (2) 􏲛􏲜 Ve2 =−(β2 +1)Ic1 RL ∥[R2 +(R1 ∥Rib)] (3) V = V R1 ∥ Rib r e2(R1 ∥Rib)+R2 Combining (1) to (4), we can determine Aβ as Aβ≡−Vr Vt (β +1)􏲛R ∥[R +(R ∥R )]􏲜 r +R e1 s R ∥R × 1 ib (R1 ∥Rib)+R2 Substituting α1(β2 + 1) = α(β + 1) = β = 100 re1 = VT = 25mV =250􏱹 (4) Thus, A = 4× 10×1000 = 39.6 V/V Chapter 11–11 10 + 1000 Af= A =α2 L21ib 5= ⇒β=0.175V/V 1 + Aβ 39.6 1 1+39.6β R1 0.175= R1 +R2 ⇒ R1 = 0.175×1000 = 175 k􏱹 IE1 Rs =100􏱹 RL = 1 k􏱹 R1=1k􏱹 R2 =10k􏱹 0.1 mA 11.31 (a) The feedback network consists of the voltage divider (RF , RS1). Thus, β= RS1 RS1 +RF and the ideal value of the closed-loop gain is Af =1=1+RF R =101(0.25+0.1)=35.35k􏱹 ib we obtain 􏲛􏲜 Aβ= 100 1∥[10+(1∥35.35)] 0.25 + 0.1 × 1 ∥ 35.35 (1 ∥ 35.35) + 10 = 23.2 (d) A= Aβ = 23.2 =255.2V/V β RS1 10=1+ RF 0.1 ⇒RF =0.9k􏱹 (b) RD1 Id1 RS1 Vd2 RD2 Id3 Q3 Q2 Id3 RS2 Vr Vt β Af=A 1 + Aβ (1/11) Q 1 1/gm1 Id1 R F Figure 1 = 255.2 = 10.5 V/V 1 + 23.2 If 11.30 Refer to Fig. 11.8(c) and to the expressions for β, Aβ, and A given in the answer section of Exercise 11.6. A=g RD(R1+R2) m RD + R1 + R2 where gm = 4 mA/V RD =10k􏱹 R1 + R2 = 1 M􏱹 (the potentiometer resistance) Figure 1 shows the circuit for determining the loop gain. Observe that we have broken the loop at the gate of Q2 where the input resistance is infinite, obviating the need for adding a termination resistance. Also, observe that as usual we have set Vs = 0. To determine the loop gain Aβ ≡ −Vr Vt we write the following equations: Vd2 = −gm2RD2Vt (1) 0.0968 = 0.1 0.1+RF ⇒RF =933􏱹 (an increase of 33 􏱹). 11.32 (a) The feedback circuit consists of the voltage divider (RF , RE ). Thus, β=RE RE +RF Id3=1 􏲝 􏱼Vd2􏱾 1􏱿􏱽􏲞 g + RS2∥ RF+ RS1∥g (2) (3) (4) (5) Chapter 11–12 m3 If=Id3􏱼 􏱾 RS2 􏱿􏱽 m1 RF+ RS1∥ 1 gm1 +RS2 RS1 Id1 =If 1 RS1 + gm1 Vr =Id1RD1 and, A 􏲀􏲀 = 1 = 1 + R F Substituting the numerical values in (1)–(5), we obtain fideal β RE Thus, Vd2 = −4 × 10Vt = −40Vt Id3=1 􏲝 􏱼Vd2􏱾 1􏱿􏱽􏲞 4+ 0.1∥ 0.9+ 0.1∥4 Id3 = 2.935Vd2 (6) (7) (8) (9) (10) 25=1+ RF 0.05 ⇒RF =1.2k􏱹 (b) Figure 1 on next page shows the feedback amplifier circuit prepared for determining the loop gain Aβ. Observe that we have eliminated all dc sources, set Vs = 0, and broken the loop at the base of Q2. We have terminated the broken loop in a resistance rπ 2 . To determine the loop gain If =Id3􏱼 􏱾 0.1 􏱿􏱽 1 0.9+ 0.1∥4 +0.1 If = 0.0933Id3 Aβ≡Vr Vt we write the following equations: Ic2 = gm2Vt I =I 0.1 =0.286I d1 f 1 f 0.1 + 4 Vr = 10Id1 Combining (6)–(10) gives Vr = −31.33Vt ⇒ Aβ = 31.33 A= Aβ = 31.33 =313.3V/V (1) β Af= A 1 + Aβ = 313.3 1 + 31.33 0.1 = 9.7 V/V RE Ie1=Ie3R+r (4) E e1 Vr = −α1Ie1(RC1 ∥ rπ2) (5) Substituting α1 = 0.99 RC1=2k􏱹 IC2 2 mA gm2 = V = 0.025V =80mA/V T β2 100 rπ2=g =80=1.25k􏱹 m2 RE=0.05k􏱹 re1 = VT ≃ 25mV =25􏱹=0.025k􏱹 IE1 1 mA β3 = 100 RC2=1k􏱹 RF=1.2k􏱹 I=I RC2 b3 c2R +(β +1)[R +(R ∥r )] (2) Ie3 = (β3 + 1)Ib2 (3) C2 3 F E e1 Thus, Af is 0.3 V/V lower than the ideal value of 10 V/V, a difference of −3%. The circuit could be adjusted to make Af exactly 10 by changing β through varying RF . Specifically, 10 = 313.3 1 + 313.3β ⇒ β = 0.0968 But, β= RS1 RS1 +RF This figure belongs to Problem 11.32, part (b). Chapter 11–13 RC1 re1 RE RC2 Ib3 Q3 Ic2 Ic1 􏰀 Q2 rp2 VrV􏰀 Q1 􏰒t􏰒 Ie3 Ie1 Ie3 RF we obtain Ic2 = 80Vt (a) Figure 1 I=I 1 b3 c2 1 + 101[1.2 + (0.05 ∥ 0.025)] = 8.072 × 10−3 Ic2 Ie3 = 101Ib3 50 Ie1 = Ie3 50 + 25 = 0.667Ie3 Vr = −0.99(2 ∥ 1.25)Ie1 Vr = −0.7615Ie1 Combining (6)–(10) results in Aβ = 33.13 Aβ 33.13 A= β = 1/25 =828.2V/V Af= A 1 + Aβ = (6) (7) (8) (9) (10) Q3 Q4 Rs Vo R1 􏰀􏰒 V t 􏰀 Vr R2 􏰒 ID =100μA=0.1mAand|VOV|=0.2V,thus gm1,2 = 2×0.1 = 1 mA/V 0.2 All devices have ro = |VA| = 10 =100k􏱹 ID 0.1 5 = 47.62 1 + 47.62β ⇒ β = 0.179 V/V Q1 Q2 Figure 1 Figure 1 shows the circuit prepared for determining the loop gain Aβ. Vo = −gm1,2[ro2 ∥ ro4 ∥ (R1 + R2)]Vt (1) Vr=R2Vo=βVo R1 + R2 Thus, Aβ≡−V =gm1,2[ro2 ∥ro4 ∥(R1 +R2)]β Vr t 828.2 1 + 33.13 = 1(100 ∥ 100 ∥ 1000)β = 47.62β Thus, A = 47.62 V/V (b)Af= A 1 + Aβ = 24.3 V/V 11.33 All MOSFETs are operating at R2 = 0.179 R1 + R2 ⇒R2 =179k􏱹 R1 = 821 k􏱹 11.34 Ri = 2 k􏱹 Ro =2k􏱹 A=1000V/V β = 0.1 V/V Loop Gain ≡ Aβ = 1000 × 0.1 = 100 1 + Aβ = 101 Af = A 1 + Aβ = 1000 = 9.9 V/V 101 Rif =Ri(1+Aβ) = 2 × 101 = 202 k􏱹 Rof = Ro 1+Aβ = 2 = 19.8 􏱹 101 11.35 Since the output voltage is sampled, the resistance-with-feedback is lower. The reduction is by the factor (1 + Aβ), thus 1 + Aβ = 200 Aβ = 199 Rof = Ro 200 ⇒ Ro = 200×100 = 20,000 􏱹 = 20 k􏱹 11.36 A= 1+Aβ=1+ s 1 + ωH Zif =Ri(1+Aβ) =R+ A0βRi Figure1 Chapter 11–14 Zof = Ro 1 + Aβ = Ro 1+ A0β 1+s ωH A0 1+s Thus, the output admittance Yof is Yof≡1=1+ 􏱾A0β 􏱿 Zof Ro Ro1+s ωH which consists of a resistance Ro in parallel with an impedance Z given by Ro Ro Z = A0β +sA0βωH which consists of a resistance (Ro/A0β) in series with an inductance L = Ro/A0βωH . The equivalent circuit of Zof is shown in Fig. 2. Figure 2 11.37 A = 1000 V/V Ri = 1 k􏱹 Rif =10k􏱹 Thus, the connection at the input is a series one, and 1+Aβ= 10 =10 1 Af = A 1 + Aβ ωH A0β 1000 is ==100V/V 1 + ωH Thus, Zif consists of a resistance Ri in series with 10 To implement a unity-gain voltage follower, we use β = 1. Thus the amount of feedback is 1 + Aβ = 1 + 1000 = 1001 and the input resistance becomes Rif = (1 + Aβ)Ri = 1001 × 1 = 1001 k􏱹 = 1.001 M􏱹 an admittance Y , Y=1+s A0βRi A0βRiωH which is a resistance (A0βRi) in parallel with a capacitance 1/A0βRiωH . The equivalent circuit is shown in Fig. 1. 11.38 (a) β = 1 A 􏲀􏲀 = 1 V/V Chapter 11–15 f ideal (b) Substituting R1 = ∞ and R2 = 0 in the expression for A in Example 11.4, we obtain A=μ RL Rid RL+ro Rid+Rs Aβ = A × 1 = A (c)A=104× 2 × 100 2+1 100+10 = 6060.6 V/V Aβ = 6060.6 Af= A 1 + Aβ = 6060.6 = 0.9998 V/V 1 + 6060.6 From Example 11.4 with R1 = ∞ and RL = 0, we have Ri =Rs +Rid =10+100=110k􏱹 Rif =Ri(1+Aβ) = 110 × 6061.6 = 667 M􏱹 Rin =Rif −Rs ≃667M􏱹 Figure 1 (c) The A circuit is shown in Fig. 1. Ro =ro ∥RL =1∥2=0.67k􏱹 Rof= Ro =0.67k􏱹=0.11􏱹 I= e1 Vi R1 ∥R2 Rs + re1 + β + 1 1+Aβ Rof = Rout ∥ RL 6061.6 Rout ≃ 0.11 􏱹 11.39 Refer to the solution to Problem 11.29. (a) β = R1 R1 + R2 Af=1=1+R2 β R1 10 =1+ 1 =11V/V (b) From the solution to Problem 10.29, we have IE1 ≃0.1mA IE2 ≃ 0.3 mA VE2 = +7.7 V 1 Vo =Ie2[RL ∥(R1 +R2)] = (β + 1)α V RL ∥ (R1 + R2) 2 1i R1∥R2 Rs + re1 + β1 + 1 Sinceβ1 =β2 =βandα= β ,wehave β+1 A≡Vo=β RL∥(R1+R2) Vi Rs + re1 + R1 ∥ R2 β1 + 1 Substitutingβ=100,RL =1k􏱹,R1 =1k􏱹, R2 =10k􏱹,Rs =0.1k􏱹,andre1 =0.25k􏱹 gives A=100 1∥11 0.1+0.25+ 1 ∥ 10 101 = 255.3 V/V (d) β = 1 R1 + R2 ID = μpCox |VOV3|2 2L R1 ∥R2 β1+1 1 􏱾W􏱿 ID= μnCox V2 Ri = Rs + re1 + = 0.1+0.25+ 1 ∥ 10 = 0.359 k􏱹 101 Ro = RL ∥ (R1 + R2) = 1 ∥ 11 = 0.917 k􏱹 2 LOV 100 = 1 × 120 × 20 V 2 R 1􏱾W􏱿 Chapter 11–16 2 1 OV1,2 ⇒ VOV1,2 = 0.29 V For Q3 , use 100=1×60×40|VOV3|2 1+1011 21 =1=1 (e) Vo =Af = A 11 Af =255.3=10.5V/V 24.21 Rif =Ri(1+Aβ) = 0.359 × 24.21 = 8.69 k􏱹 Rin = Rif − Rs = 8.69−0.1 = 8.59 k􏱹 ⇒ |VOV3| = 0.29 V Since VSG4 = VSG3 , we have |VOV4| = |VOV3| = 0.29 V Finally for Q5, use 1 20 2 800=2×120× 1 ×VOV5 ⇒VOV5 =0.82V If perfect matching pertains, then VD4 = VD3 = VDD −VSG3 = 2.5 − |Vt| − |VOV3| = 2.5−0.7−0.29 = 1.51 V VO = VD4 − VGS5 = VD4 − Vt − VOV 5 = 1.51 − 0.7 − 0.82 = −0.01 V which is approximately zero, as stated in the Problem statement. (c) gm1 =gm2 =gm3 = 2ID |VOV | = 2×0.1 = 0.7 mA/V 0.29 2×0.3 gm4 = 0.29 ≃ 2 mA/V gm5 = 2×0.8 ≃2mA/V 0.82 ro1 =ro2 =ro3 = |VA| = |VA′|×L ID ID = 24×1 =240k􏱹 0.1 ro4 = 24 =80k􏱹 0.3 ro5 = 24 =30k􏱹 0.8 (d) Figure 2 on the next page shows the A circuit. Observe that since the β network is simply a wire connecting the output node to the gate of Q2, we have R11 = 0 and R22 = ∞. To determine A, we write Vs 1 + Aβ 1 + Aβ = 1 + 255.3 = 24.21 Rof = Ro = 0.917 k􏱹 = 37.9 􏱹 1 + Aβ Rof = Rout ∥ RL 37.9 = Rout ∥ 1000 ⇒ Rout = 39.4 􏱹 24.21 The value of Af (10.5 V/V) is 0.5 less than the ideal value of 11, which is 4.5%. 11.40 (a) Refer to Fig. P11.40. Assume that for some reason vS increases. This will increase the differential input signal (vS − vO) applied to the differential amplifier. The drain current of Q1 will increase, and this increase will be mirrored in the drain current of Q4. The increase in iD4 will cause the voltage at the gate of Q5 to rise. Since Q5 is operating as a source follower, the voltage at its source, vO, will follow and increase. This will cause the differential input signal (vS − vO) to decrease, thus counteracting the originally assumed change. Thus, the feedback is negative. (b) Figure 1 on the next page shows the circuit prepared for dc analysis. We see that ID1 =ID2 =100μA ID3 = 100 μA ID4 = 300 μA ID5 = 0.8 mA For Q1 and Q2, use This figure belongs to Problem 11.40, part (b). Chapter 11–17 􏰀2.5 V Q3 (40/1) Q4 (120/1) Q5 (20/1) 300 􏱌A Q1Q2 VO (20/1) (20/1) This figure belongs to Problem 11.40, part (d). Q3 Id1 Vi 􏰀􏰒 Q5 Vo 200 􏱌A Figure 1 Q4 3Id1 Vd4 0.8 mA Q1 Q2 0 Id1, 2 1 2 1 2 Vi = 1gm1,2Vi 2/gm1,2 2 Finally, Vo is related to Vd4 as Vo ro5 Id1,2 = 􏱾􏱿􏱾􏱿 = R11 􏰔 0 Figure 2 R22 􏰔 􏰿 Since W =3 W ,thedraincurrentofQ Vd4 ro5 + 1 LL 4 gm5 43 will be Id4 =3Id1 = 3gm1,2Vi 2 The voltage at the drain of Q4 will be Vd4 = Id4ro4 = 3gm1,2ro4Vi 2 Thus, A≡ Vo = 3g r ro5 Vi2m1,2o4 1 ro5 + g m5 Substituting numerical values, we obtain A = 3 × 0.7 × 80 × 30 2 30+0.5 = 82.6 V/V The output resistance Ro is 1 Ro = ro5 ∥ g m5 = 30 ∥ 0.5 = 0.492 k􏱹 source follower; thus the signal at its source (which is the output voltage) will follow that at its gate and thus will be positive. The end result is that we are feeding back through the voltage divider (R2 , R1 ) a positive incremental signal that will appear across R1 and thus at the source of the Q1. This signal, being of the same polarity as the originally assumed change in the signal at the gate of Q1(Vs), will subtract from the original change, causing a smaller signal to appear across the gate-source terminals of Q1. Hence, the feedback is negative. (b) β = R1 R1 + R2 A 􏲀􏲀 = 1 = 1 + R 2 fideal β R1 Thus, Chapter 11–18 = 492 􏱹 (e) Af = (f) To obtain a closed-loop gain of 5 V/V, we connect a voltage divider in the feedback loop, as shown in Fig. 3. A 1 + Aβ = 82.6 1+82.6 = 0.988 V/V Rof=Ro =492=5.9􏱹 1 + Aβ 1 + 82.6 Rout =Rof =5.9􏱹 Thus, β = 2 = 0.1 V/V If the loop gain is large, the closed-loop gain 2+18 approaches the ideal value Q5 R2 􏲀 Q2 β= R1 R1 + R2 Af= A 1 + Aβ 82.6 5= 1+82.6β R1 Figure 3 Af 􏲀ideal = 1 + 18 = 10 V/V 2 (c) VG1 =0.9V VS1 = VG1 − VGS1 =VG1 −Vt1 −VOV1 = 0.9−0.5−0.2 = 0.2 V VG2=VDD−VSG2 =VDD−|Vt2|−|VOV2| = 1.80−0.5−0.2 = 1.1 V Thus, current source I1 will have 0.7-V drop across it, more than sufficient for its proper operation. Since VS1 = 0.2 V the dc current through R1 will be IR1 = VS1 = 0.2V =0.1mA R1 2 k􏱹 Now, a node equation at S1 reveals that because ID1 =0.1mAandIR1 =0.1mA,thedccurrentin R2 will be zero. Thus, it will have a zero voltage drop across it and VS3 =VS1 =0.2V Thus, current source I3 will have across it, the minimum voltage required to keep it operating properly. Finally, VG3 = VS3 + VGS3 =VS3 +Vt3 +VOV3 = 0.2+0.5+0.2 = 0.9 V ⇒ β = 0.188 Selecting R1 = 1 M􏱹, we obtain 0.188 = 1 1 + R2 ⇒ R2 = 4.319 M􏱹 Note that by selecting large values for R1 and R2, we have ensured that their loading effect on the A circuit would be negligible. 11.41 (a) Let Vs increase by a small increment. Since Q1 is operating in effect as a CS amplifier, a negative incremental voltage will appear at its drain. Transistor Q2 is also operating as a CS amplifier; thus a positive incremental voltage will appear at its drain. Transistor Q3 is operating as a Thus, current source I2 will have across it a voltage more than sufficient to keep it operating properly. (d) rocs3 = |VA| = 10 =100k􏱹 I3 0.1 Refer to the A circuit. Transistor Q1 is a CS amplifier with a resistance R11 in its source: Rs =R11 =R1 ∥R2 =2∥18=1.8k􏱹 Transistor Q1 will have an effective transconductance: Gm1 = gm1 = 2 =0.43mA/V 1+gmRs 1+2×1.8 The output resistance of Q1 will be Ro1 = (1 + gm1Rs)ro1 = (1 + 2 × 1.8) × 100 = 460 k􏱹 The total resistance at the drain of Q1 is Rd1 = rocs1 ∥ Ro1 = 100 ∥ 460 = 82.1 k􏱹 Thus, the voltage gain of the first stage is A1 = −Gm1Rd1 = −0.43 × 82.1 = −35.3 V/V The gain of the second stage is A2 = −gm2(rocs2 ∥ ro2) = −1(100 ∥ 100) = −50 V/V To determine the gain of the third stage, we first determine the total resistance between the source of Q3 and ground: Rs3 =rocs3 ∥ro3 ∥(R1 +R2) Rs3 =100∥100∥20 = 14.3 k􏱹 Thus, A= Rs3 3 1 Rs3 + gm3 14.3 Chapter 11–19 Figure 1 Figure 1 shows the A circuit as well as the β circuit and how the loading-effect resistances R11 and R22 are determined. To determine A, let’s first determine the small-signal parameters of all transistors as well as ro of each of the three current sources. gm1 = 2ID1 = 2×0.1 =1mA/V VOV1 0.2 |VA| ro1 = I D1 10 = 0.1 = 100 k􏱹 1 = 0.935 V/V 11 |VA| 10 rocs1 = I = 0.1 =100k􏱹 = gm2 = 2ID2 = 2×0.1 =1mA/V VOV 2 0.2 ro2 = |VA| = 10 =100k􏱹 ID2 0.1 rocs2 = |VA| = 10 =100k􏱹 14.3+ TheoverallvoltagegainAcannowbefoundas A = A1A2A3 = −35.3 × −50 × 0.935 = 1650 V/V (e) We already found β in (b) as β = 0.1 V/V (f) 1+Aβ = 1+1650×0.1 = 166 Af = A = 1650 =9.94V/V 1 + Aβ 166 I2 gm3 = 2ID3 VOV 3 0.1 = 2×0.1 = 1 mA/V 0.2 r =|VA|=10=100k􏱹 o3 ID3 0.1 which is lower by 0.06 or 0.6% than the ideal value obtained in (b). Ro (g) Rof = 1 + Aβ To obtain Ro refer to the output part of the A circuit. gate of Q2. The increase in the voltage of the gate of Q2 will subtract from the initially assumed increase of the voltage of the gate of Q1, resulting in a smaller increase in the differential voltage applied to the (Q1, Q2) pair. Thus, the feedback counter acts the originally assumed change, verifying that it is negative. (b) The negative feedback will cause the dc voltage at the gate of Q2 to be approximately equal to the dc voltage at the gate of Q1, that is, zero. Now, with VG2 ≃ 0, the dc current in R2 will be zero and similarly the dc current in R1 will be zero, resulting in VO = 0 V dc. (c) Figure 1 shows the A circuit. It also shows how the loading effect of the β network on the A circuit, namely R11 and R22, are found. The gain of the A circuit can be written by inspection as A = gm1,2(ro2 ∥ ro4 ∥ R22) where gm1,2 = 2ID1,2 VOV 1,2 = 2 × 0.1 = 1 mA/V 0.2 Vo R1 R22 R2 R1 R2 Chapter 11–20 Ro = (R1 + R2) ∥ rocs3 ∥ ro3 ∥ = 20 ∥ 100 ∥ 100 ∥ 1 = 935 􏱹 Rof = 935 =5.6􏱹 166 1 gm3 Note: This problem, though long, is extremely valuable as it exercises the student’s knowledge in many aspects of amplifier design. 11.42 (a) Refer to Fig. P11.33. Let Vs increase by a positive increment. This will cause the drain current of Q1 to increase. The increase in Id 1 will be fed to the Q3 − Q4 mirror, which will provide a corresponding increase in the drain current of Q4 . The latter current will cause the voltage at the output node to rise. A fraction of the increase in Vo is applied through the divider (R1, R2) to the This figure belongs to Problem 11.42, part (c). Q3 Q4 Rs Vi Q1 Q2 R11 Ro R1 0R1 R11 1R2 21R2 2 Figure 1 R22 ro2 =ro4 = |VA| = 10 =100k􏱹 ID3,4 0.1 R22 =R1 +R2 =1M􏱹 A=1(100∥100∥1000)=47.62V/V This is identical to the value found in the solution to Problem 11.33. (f) WithRL =10k􏱹, Vo =5× RL Vs RL +Rout 5× 10 =3.33V/V (g) As an alternative to (f), we shall redo the analysis of the A circuit in (c) above with RL =10k􏱹included: A=gm1,2(ro2 ∥ro4 ∥R22 ∥RL) = 1(100 ∥ 100 ∥ 1000 ∥ 10) = 8.26 V/V Using β = 0.179, we obtain A = 8.26 =3.33V/V f 1+8.26×0.179 which is identical to the value found in (f) above. 11.43 All transistors have L = 1 μm, thus all have |VA| = |VA′ | × L = 10 × 1 = 10 V. Also, all have |Vt | = 0.75 V. (a) Figure 1 shows the circuit prepared for dc design. We have also indicated some of the current and voltage values. We now find the (W/L) ratios utilizing 􏱾􏱿 I=1μC WV2 D 2 n ox L OV for the NMOS transistors, and 1 􏱾W 􏱿 Chapter 11–21 (d) Vo = A = V f s 5= 47.62 1+47.62β ⇒ β = 0.179 Thus, A 1+Aβ Q4 5 10 + 5 R2 =0.179 R1 +R2 R2 =0.179M􏱹=179k􏱹 R1 = 1000−179 = 821 k􏱹 Again, these values are identical to those found in Problem 11.33. (e) Refer to Fig. 1. Ro =R22 ∥ro2 ∥ro4 = 1000 ∥ 100 ∥ 100 = 47.62 k􏱹 R Rout=Rof= o 1+Aβ = 47.62 1 + 47.62 × 0.179 = 5 k􏱹 This value cannot be found using the loop-gain analysis method of Problem 11.33. This figure belongs to Problem 11.43, part (a). ID = 2μpCox L |VOV|2 for the PMOS devices. 50 􏱌A Q1 Q2 􏰀1.5 V 􏰒 50 􏱌A 100 􏱌A Q7 2.5􏰒(􏰒1.5) 80 􏰔 50 􏱌A Q6 VDD 􏰔 􏰀2.5 V 􏰀1.5 V Q3 80 k􏰵 􏰀 50 􏱌A 􏰒1.5 V 􏰒VSS 􏰔 􏰒2.5 V Figure 1 0 􏱌A Q5 250 􏱌A VO 􏰔 0 V 250 􏱌A Q8 For Q6, 1 􏱾W􏱿 50= ×100× ×0.252 Thus, −1.25 V ≤ VICM ≤ +2.25 V (c) gm1,2 = 2ID1,2 VOV 1,2 For Q7, (W/L)7 =100μA=2 = 2×0.05 0.2 = 0.5 mA/V 2ID 2 × 0.25 Chapter 11–22 ⇒ 􏱾W􏱿 L6 = 16 2L6 (W/L)6 50 μA ⇒(W/L)7 =2×16=32 For Q8, (W/L)8 =250μA=5 gm5 = |VOV 5| = 0.75 = 0.67 mA/V (d) ro1 =ro2 =ro3 =ro4 =ro6 = |VA| = 10 = 200 k􏱹 0.05 ro7 = 10 =100k􏱹 0.01 ro5 =ro8 = 10 =40k􏱹 0.25 ID (W/L)6 50 μA = 5 × 16 = 80 ⇒ 􏱾W􏱿 L8 For Q1 and Q2, 1 􏱾W􏱿 50= 2 ×100× L 􏱾W􏱿 􏱾W􏱿 ⇒ L = L 12 ×0.252 = 16 (e) Figure2onthenextpageshowstheAcircuit, the β circuit, and how the loading effects of the β circuit on the A circuit, namely R11 and R22, are determined. Vg5 = gm1,2(ro2 ∥ ro4) Vi = 0.5(200 ∥ 200) = 50 V/V Vo = Rs Vg5 R+1 34 sgm5 1,2 For Q3 and Q4, 1 􏱾W􏱿 × 0.252 ⇒ L = L =32 50 = 2 × 50 × L 􏱾W􏱿 􏱾W􏱿 3,4 Finally, since VG3 = VD4 = VD3 = 1.5 V and we require VO = 0 V, we have VGS5 =1.5V VOV5 = 1.5−0.75 = 0.75 V where Rs =ro8 ∥ro5 ∥(R1 +R2)∥RL = 40 ∥ 40 ∥ 100 ∥ 100 = 14.3 k􏱹 Thus, 1 􏱾W􏱿 250= ×100× 2L5 leaving the saturation region, VICM max = VD1 + Vt = 1.5+0.75 = 2.25 V ×0.752 􏱾W􏱿 L5 Vo V 14.3 = 14.3 + (1/0.67) = 0.905 V/V = 8.9 (b) ThemaximumvalueofV g5 A= V = V × V ⇒ ICM islimitedbyQ 1 Vo Vg5 Vo i i g5 = 50 × 0.905 = 45.3 V/V A TheminimumvalueofVICM islimitedbythe need to keep Q7 in saturation. This is achieved by keeping VD7 at a minimum voltage of −2.5 + |VOV7| = −2.5 + 0.25 = −2.25 V Thus, VICM min = −2.25 + VGS1 = −2.25 + 1 = −1.25 V Af =10=1+Aβ 10= 45.3 1 + 45.3β ⇒ β = 0.078 R2 = 0.078 R1 + R2 R2 = 7.8 k􏱹 R1 = 100−7.8 = 92.2 k􏱹 This figure belongs to Problem 11.43, part (e). Chapter 11–23 Q3 Q4 Vg5 Q5 Vo R2 Ro V􏰀 R1 i􏰒 RRro8 RL 21 R11 R22 A Circuit R1 R1 0R1 1R2 21R221R2 2 b Circuit b􏰔 R2 (f) Refer to Fig. 2. Ro =RL ∥(R1 +R2)∥ro8 ∥ro5 ∥ 1 R11 􏰔 R2 􏰐􏰐R1 Figure 2 R22 􏰔 R1􏰀R2 R1􏰀R2 11.44 (a) Figure1onthenextpageshowstheA circuit and the circuit for determining β as well as the determination of the loading effects of the β circuit. Q1 Q2 (c)25=1+ RF 50 􏱹 gm5 =Rs ∥ 1 =14.3∥(1/0.67) gm5 = 1.36 k􏱹 ⇒RF =1.2k􏱹 (d) Refer to the A circuit in Fig. 1. The voltage gain of Q1 is given by Vc1 =−α RC1 ∥rπ2 R V1r+R Rof=o i e111 1 + Aβ = 1.36 k􏱹 1+45.3×0.078 Rout ∥RL =Rof ⇒Rout ≃300􏱹 ≃ 300 􏱹 where re1=VT =25mV=25􏱹 IE1 1 mA R11 =RE ∥RF =50􏱹∥1200􏱹=48􏱹 gm2 = IC2 ≃ IE2 = 2mA VT VT 0.025 mA rπ2 = β2 = 100 =1.5k􏱹 80 80 α1 =0.99≃1 Vc1 =−10=− RC1 ∥1.5 Vi 0.025 + 0.048 ⇒RC1=1.42k􏱹 =80mA/V (b) If Aβ is large, then Af≡Vo≃1 Vs β Since Next consider the second stage composed of the CE transistor Q2. The load resistance of the second stage is composed of RC2 in parallel with the input resistance of emitter-follower Q3. The latter resistance is given by Ri3 = (β3 + 1)(re3 + R22) R β=E RF +RE we have Af =RF+RE RE Q.E.D. This figure belongs to Problem 11.44, part (a). Chapter 11–24 RF RF 􏰀􏰀 Vf1RE 2Vo 1 RE 2 􏰒􏰒 b Circuit b􏰔 RE R11􏰔RE //RF RF 􏰀 RE 0 RC1 R11 RF 1RE 2 R22􏰔RE 􏰀RF RC2 Q3 Q2 A Circuit Figure 1 = 23.8 V/V Vi Q1 Vo R22 Ro A ≡ −10 × −50 × 0.996 =498V/V Ri where re3 = VT = 25mV =5􏱹 IE3 5mA R22 =RF +RE =1.2+0.05=1.25k􏱹 Thus, Ri3 =101×1.25=126.3k􏱹 Vc2 A2 ≡ V =−gm2(RC2 ∥Ri3) b2 −50 = −80(RC2 ∥ 126.3) Vo 498 Af ≡ V = s 1+498× 50 1250 ⇒RC2 =628􏱹 (e) A = A1A2A3 where (f) Refer to the A circuit in Fig. 1. R =(β +1)(r +R ) i 1 e1 11 Ri = 101(0.025 + 0.048) = 7.37 k􏱹 Rif =Ri(1+Aβ) where 1 + Aβ = 1 + 498 = 20.92 25 A3 = R22 = R22 + re3 1.25 1.25 + 0.005 = 0.996 V/V Rif = 7.37 × 20.92 = 154 k􏱹 (d) Figure 2 on the next page shows the A circuit and the β circuit together with the determination of its loading effects, R11, and R22. We can write 􏱼 Ro = R22 ∥ re3 + 􏱼 R 􏱽 C2 β3 + 1 0.628􏱽 101 20.92 11.45 (a) Refer to Fig. P11.45. If Vs increases, the output of A1 will decrease and this will cause the output of A2 to increase. This, in turn, causes the output of A3, which is Vo, to increase. A portion of the positive increment in Vo is fed back to the positive input terminal of A1 through the voltage divider (R2, R1), The increased voltage at the positive input terminal of A1 counteracts the originally assumed increase at the negative input terminal, verifying that the feedback is negative. 􏲀 1 (b) Af 􏲀ideal = β V1 =− 82 Vi 82+9+16 V2 = 20V1 × 5 3.2+5 =−0.766V/V = 12.195V1 Chapter 11–25 = 1.25 ∥ = 11.1 􏱹 Rout = Rof = 0.005+ Ro V3 = −20V2(20 ∥ 20) = −200V2 1 ∥ 100 1 + Aβ = 11.1 =0.53􏱹 Vo =V3(1∥100)+1 =0.497V3 Thus, A ≡ Vo = 0.497 × −200 × 12.195 × −0.766 Vi = 928.5 V/V 20 (e) β = 20 + 80 = 0.2 V/V where β=1 1 + Aβ = 1 + 928.5 × 0.2 = 186.7 (f) Af ≡ Vo = A Vs 1 + Aβ = 928.5 =4.97V/V 186.7 which is nearly equal to the ideal value of 5 V/V. (g) From the A circuit, Ri = 9+82+16 = 107 k􏱹 Rif = Ri(1 + Aβ) = 107 × 186.7 = 19.98 M􏱹 Rin = Rif − Rs ≃ 19.98 M􏱹 (h) From the A circuit, Ro = RL ∥ R22 ∥ 1 k􏱹 = 1 ∥ 100 ∥ 1 = 497.5 􏱹 R R1 +R2 Thus, to obtain an ideal closed-loop gain of 5 V/V we need β = 0.2: 0.2 = 20 20+R2 ⇒R2 =80k􏱹 (c) Figure 1 shows the small-signal equivalent circuit of the feedback amplifier. This figure belongs to Problem 11.45, part (c). Figure 1 This figure belongs to Problem 11.45, part (d). Chapter 11–26 Rof = Ro 1 + Aβ R ∥R =R out L of = 497.5 = 2.66 􏱹 186.7 = 10, 000 × 0.1658 =1.658A/V Figure 2 Rout ∥1000=2.66􏱹 Rout ≃ 2.66 􏱹 (i) fHf =fH(1+Aβ) = 100 × 186.7 = 18.67 kHz (j) If A1 drops to half its nominal value, A will drop to half its nominal value: A = 1 × 928.5 = 464.25 A Af = 1+Aβ = 1658 1+1658×0.1 11.47 (a) IF = 9.94 mA/V RF R I 2Mo and Af becomes 464.25 = 4.947 V/V 1+464.25×0.2 Af = Thus, the percentage change in Af is Figure 1 Figure 1 shows the β network with the input port short-circuited. Thus, β≡If =− RM Io RM +RF = 4.947 − 4.97 = −0.47% 4.97 I 11.46 ToobtainAf ≡ o ≃10mA/V,weselect 1 􏱾 R 􏱿 􏲀 Af􏲀ideal= =− 1+ F RF =β= A =100􏱹 f (b) Figure2onthenextpageshowsthecircuit for determining the loop gain Aβ, Aβ=−Vr Vt First, we express Id 2 in terms of Vt : Id2 = −gm2Vt (1) Vs 1 βRM From Example 11.6, we obtain μ gm(RF ∥Rid ∥ro2) A= RF 1+gm(RF ∥Rid ∥ro2) ≡ 1000 2(0.1 ∥ 100 ∥ 20) 0.1 k􏱹 1+2(0.1 ∥ 100 ∥ 20) Chapter 11–27 For A 􏲀􏲀 f ideal RF =1k􏱹 (b) 􏰒 􏰀􏰀 RD Vt Q2 =−1k􏱹,wehave RF V V􏰀V 􏰀􏰒mV1 r Ridt􏰒1Rid􏰒 􏰒􏰒 Figure 1 Figure 1 shows the circuit for determining the loop gain Aβ: Vr Id1 Q1 Id2 RL RM ro RF Id1 Figure 2 Then we determine Id1: I =I RM d1d21 Vt (2) The returned voltage Vr can now be obtained as RM +RF + gm1 Aβ≡−Vr WritingVr intermsofV1 =Vt yields Rid Vr =−μVt Rid +RF +ro Thus, Aβ≡−Vr =μ Rid Vr =Id1RD Combining Eqs. (1)–(3), we find Vr /Vt : Vr =− gm2RDRM (3) Vt Thus, RM +RF + 1 gm1 Vt (c) Aβ = 1000 = 980.4 980.4 Rid +RF +ro 100 100 + 1 + 1 980.4 Q.E.D. gm2RDRM R +R + 1 Aβ = Dividing the expression for Aβ by A=β=−1/RF = −980.4 k􏱹 M F gm1 β=− RM yields A Af = 1 + Aβ RM + RF A = − gm2RD Q.E.D. 980.4 1 + 980.4 1+1/[gm1(RM +RF)] (c) A=− 4×10 =− =−0.999k􏱹 1+1/[4×1] = −32 A/A 11.49 Af =−5=− 32 1−32×β β = −0.169 A/A − RM = −0.169 RM +RF RM = 0.169×1 = 0.169 k􏱹 = 169 􏱹 11.48 (a) Refer to Fig. P11.48(b). β≡If =−1 Vo RF A 􏲀􏲀ideal = 1 = −R fβF Figure 1 Figure 1 shows the A circuit where Ri1 = 10 k􏱹 Ro1 = 100 k􏱹 β = 200 􏱹 R22 = 200 􏱹 R11 =10k􏱹 Rs =10k􏱹 RL =10k􏱹 To determine A, A ≡ Io Vi 11.50 (a) Figure 1 Figure 1 shows the β circuit from which we obtain Chapter 11–28 RF􏰔640􏰵 1RE1 2 100 􏰵 we write V =V Ri1 RE1 β = RE1 + RF = 100 = 0.135 V/V 100 + 640 (b) ForAβ≫1, 1 i Ri1 + Rs + R11 =V 10 =1V (1) (2) i10+10+10 3 i I =G V Ro1 o m 1Ro1+RL+R22 100 Ve3 Vs (c) 􏲀􏲀 1 =Af ideal= =7.4V/V = 0.6 × 100 + 10 + 0.2 Vi = 0.544 Vi β 5 k􏰵 Combining (1) and (2), we obtain Io = 0.544 × 1 Vi = 0.1815 Vi 3 RC2 RC3 􏰔 600 􏰵 Q3 Vo RE2􏰔100􏰵 A = 0.1815 A/V Af=Io= A Vs 1 + Aβ 0.1815 0.1815 Q2 = 1+0.1815×200 = 1+36.2 =4.88mA/V Rif =Ri(1+Aβ) R22 Figure 2 Ri is obtained from the A circuit as Ri = Rs + Ri1 + R11 = 10+10+10 = 30 k􏱹 Thus, Rif =30×37.2=1.116M􏱹 R =R −R Ro Figure 2 shows the portion of the A circuit relevant for calculating Ro: 􏱼 R 􏱽 C2 Ro=RE2∥R22∥ re3+ where RE2 = 100 􏱹, R22 (from β circuit) in if s β3 + 1 =740􏱹, re3 =5􏱹, RC2 =5 k􏱹, β3 =100; = 1.116 − 0.010 = 1.006 M􏱹 ≃ 1 M􏱹 R =R(1+Aβ) of o whereRo isobtainedfromtheAcircuitas Ro =RL +Ro1 +R22 = 10+100+0.2 = 110.2 k􏱹 Rof = 110.2 × 37.2 = 4.1 M􏱹 Rout = Rof −RL = 4.1−0.01 = 4.09 M􏱹 thus, Ro =100∥740∥ 5+ 101 =33.7􏱹 Rof = Ro 33.7 = 1+246.3 = 0.14 􏱹 􏱼 5000􏱽 1 + Aβ 11.51 (a) 􏰀 Vf 􏰒 RS1 RF Figure 1 RS2 Io R11 = RS1 ∥ (RF + RS2) = 100 ∥ (800 + 100) = 80 􏱹 R22 =RS2 ∥(RF +RS1)=80􏱹 The value of A is determined as follows: Chapter 11–29 Vd1 =− Vi RD1 (1/gm1) + R11 Figure1showstheβnetwork.Thevalueofβcan be obtained from β≡Vf Io = − Vd2 =−g R =−4×10=−40V/V 10 = −30.3 V/V (1/4) + 0.08 V m2 D2 d1 Io= 1 Vd2 1/gm5 +R22 = RS1RS2 RS2 +RF +RS1 = 1 ≃ 3 mA/V 0.25 + 0.08 If Aβ ≫ 1, then Af≃1=1+1+ RF (1) β RS1 RS2 RS1RS2 For Af ≃ 100 mA/V, Thus, A = Io = 3×−40×−30.3 = 3636 mA/V Vi (c) β = 0.01 k􏱹 1 + Aβ = 1 + 3636 × 0.01 = 37.36 Io 3636 Af = V = 37.36 = 97.3 mA/V i Difference from design value = 97.3 − 100 × 100 100 = −2.7% 1 1 R 100= + + F 0.1 0.1 0.1×0.1 ⇒RF =0.8k􏱹 (b) Figure 2 shows the A circuit and the determination of the loading effects of the β circuit, namely R11 and R22, This figure belongs to Problem 11.51, part (b). Figure 2 To make Af exactly 100 mA/V, we can increase RF (see Eq. (1) to appreciate why we need to increaseRF). (d) From the A circuit in Fig. 2, we have R =r +R +g r R Figure 4 shows the A circuit for this case. To determine A, we write Vd1 =− RD1 Vi 1/gm1 + R11 Vd1 =− 10 Vi 0.25+0.0889 Chapter 11–30 o o3 22 m3 o3 22 = 20+0.08+4×20×0.08 = 26.48 k􏱹 Rout =Rof =Ro(1+Aβ) = 26.48 × 37.36 = 989.3 k􏱹 (e) =−29.5V/V Vd2 =−gm2RD2 =−4×10=−40V/V Vd1 Vo = Vd2 Thus, A ≡ Vo Vi R22 R22 + 1 = 88.9 88.9+250 =0.26V/V RF 1 RS1 2 R11 0 RF 1 RS1 R22 Figure 3 Figure 3 shows the β circuit for the case the output is Vo. β= RS1 RS1 +RF = 100 =1 100 + 800 9 Also shown is how R11 and R22 are determined in this case: R11 =RS1 ∥RF =100∥800=88.9􏱹 R22 =RF +RS1 =800+100=900􏱹 gm5 = 0.26×−40×−29.5 = 306.9 V/V 1 RF RS1 1+Aβ=1+306.9×1=35.1 9 which is a little lower than the value (37.36) found when we analyzed the amplifier as a transconductance amplifier. Af= A 1 + Aβ = 306.9 = 8.74 V/V 35.1 (f) From the A circuit in Figure 4, we have Ro=R22∥ 1 gm3 Ro =900􏱹∥250􏱹 = 195.7 􏱹 Rout2 = Rof = Ro 2 1 + Aβ = 35.1 =5.6􏱹 195.7 11.52 (a) RD1 Vd2 Vd1 Q1 Vi􏰀􏰒 R11 Figure 4 RD2 Q3 R22 Figure 1 Vo Ro Figure 2 Figure 1 shows the small-signal equivalent circuit of the feedback amplifier. Observe that the resistance RF senses the output current Io and provides a voltage IoRF that is subtracted from Vs. Thus the feedback network is composed of the resistanceRF,asshowninFig.2.Becausethe feedback is of the series-series type, the loading resistances R11 and R22 are determined as indicated in Fig. 2, R11 = RF R22 = RF (b) The β circuit is shown in Fig. 2 and β = RF Figure 3 shows the A circuit. (c) A ≡ Io = g ro V mr+R Chapter 11–31 i o F 1 + Aβ = 1 + gmroRF ro + RF Io A V =Af = 1+Aβ s = = gm ro /(ro + RF ) 1 + gmroRF /(ro + RF ) gm 1+gmRF +RF ro From the A circuit in Fig. 3, we have Ro = ro + RF Rof = (1 + Aβ)Ro 􏱾grR􏱿 = 1+ m o F (ro+RF) 􏰀 Vgs 􏰀􏰒 􏰒 Vi Io Ro ro + RF =ro +RF +gmroRF gmVgs R11 ro which is a familiar relationship! 11.53 (a) The β circuit is shown in Fig. 1: β = RF For Aβ ≫ 1, Af ≡ Io/Vs approaches the ideal value Io RF Figure 3 To determine A = Io/Vi, we write Af 􏲀􏲀 1 1 ideal = β = RF Vgs = Vi I=gV ro To obtain Af ≃ 5 mA/V, we use RF = 1 =0.2k􏱹=200􏱹 5 0 􏰀 o m gsro+RF This figure belongs to Problem 11.53, part (a). 1 RF 2 1V1 RF 2 Io 􏰒 b V1􏰔RF Io 1 RF 2 1 RF 2 00 R11 􏰔RF R22 􏰔RF Figure 1 This figure belongs to Problem 11.53, part (b). Chapter 11–32 􏰀􏰀 Io Ro V1 Vgs 􏰀 􏰒 􏰀 􏰒 gmVgs ro Vi 􏰒 􏰒 mV1 Figure 2 R11􏰔 RF Io R22􏰔 RF (b) Determining the loading effects of the β network is illustrated in Fig. 1: R11 =R22 =RF Figure 2 shows the A circuit. An expression for Ro = 20 + 0.2 + 2 × 20 × 0.2 = 28.2 k􏱹 Rof = 20+0.2+1001×2×20×0.2 = 20+0.2+8008 = 8028.2 k􏱹 ≃ 8 M􏱹 11.54 Figure1onthenextpageshowsthe equivalent circuit with Vs = 0 and a voltage Vx applied to the collector for the purpose of determining the output resistance Ro, Ro ≡ Vx Ix Some of the analysis is displayed on the circuit diagram. Since the current entering the emitter node is equal to Ix , we can write for the emitter voltage Ve =Ix[Re ∥(rπ +Rb)] (1) The base current can be obtained using the A ≡ Io/Vi can be derived as follows: V1 = Vi (1) Vgs =μV1 −IoRF (2) I =g V ro (3) o m gsro +RF Combining Eqs. (1)–(3) yields A ≡ Io = μgmro Vi ro +RF +gmroRF Forμ=1000V/V,gm =2mA/V,ro =20k􏱹, and RF = 0.2 k􏱹, we have A = 1000 × 2 × 20 20 + 0.2 + 2 × 20 × 0.2 = 1418.4 mA/V (c) Aβ = μgmroRF ro +RF +gmroRF Aβ = 283.7 1 + Aβ = 284.7 (d) Af ≡ Io = Vs A 1 + Aβ current-divider rule applied to Re and (rπ + Rb) as I = −I Re (2) = 1418.4 = 4.982 mA/V 284.7 which is very close to the ideal value of 5 mA/V. (e) From the A circuit in Fig. 2, we have Ro =ro +RF +gmroRF 1 + Aβ = 1 + μgmroRF ro +RF +gmroRF b x Re + rπ + Rb The voltage from collector to ground is equal to Vx and can be expressed as the sum of the voltage drop across ro and Ve, Vx =(Ix −βIb)ro +Ve Substituting for Ve from (1) and for Ib from (2), we obtain Vx Ro=I =ro+[Re∥(rπ+Rb)] R =(1+Aβ)R ofo x = ro + RF + gmroRF + μgmroRF + Reβro = r + R + (μ + 1)g r R Re + rπ + Rb oFmoF 􏱼β􏱽 ≃μgmroRF =ro +[Re ∥(rπ +Rb)] 1+rorπ +Rb This figure belongs to Problem 11.54. Chapter 11–33 Ib Ix BC (Ix 2 bIb) rp R Ve 􏰀􏰒V bIb ro bEx Ix Re Figure 1 Since β = gm rπ , we obtain Ro=ro+[Re∥ and thus R ≡Vx =r +(β+1)r 􏱼􏱽oπo (r + R )] 1 + g r rπ π b m orπ +Rb For Rb = 0, Ro = ro +(Re ∥ rπ)(1+gmro) Q.E.D. Ix whichisidenticaltotheresultinEq.(3). The maximum value of Ro will be obtained when Re ≫ rπ . If Re approaches infinity (zero signal current in the emitter), Ro approaches the theoretical maximum: Ib rp 0 E (Ix 2 bIb) bIb Ix 􏰀􏰒 Vx 11.55 Romax =ro +rπ(1+gmro) =ro+rπ+βro ≃ βro Ix Ix BC ( b􏰀1)Ix ro (3) Ix rp R Vx out I bI ro x Figure 2 The situation that pertains in the circuit when Re = ∞ is illustrated in Fig. 2. Observe that since the signal current in the emitter is zero, the base current will be equal to the collector current (Ix ) and in the direction indicated. The controlled-source current will be βIx, and this currentaddstoIx toprovideacurrent(β+1)Ix in the output resistance ro. A loop equation takes the form Vx =(β+1)Ixro +Ixrπ E Ix 􏰀Vx 􏰒x Figure 1 Figure 1 shows the situation that pertains in the transistor when μ is so large that Vb ≃ 0 and Ie ≃ 0. Observe that Ib = −Ix Writing a loop equation for the C-E-B, we obtain Vx =(Ix −βIb)ro −Ibrπ Substituting Ib = −Ix, we obtain Vx Rout = I =rπ +(β+1)ro x or if β is denoted hfe, Rout =rπ +(hfe +1)ro Q.E.D. This figure belongs to Problem 11.56. Chapter 11–34 0 􏰀 1 RF 2 Vf RF Io 􏰒 bCircuit b Vf 􏰔RF Io 00 1 RF 2 1 RF 2 R11 􏰔RF Thus, for large amounts of feedback, Rout is limited to this value, which is approximately hfero independent of the amount of feedback. This phenomenon does not occur in MOSFET circuits where hfe = ∞. 11.56 Figure 1 above shows the β circuit together Figure 1 R22 􏰔RF Combining (1) and (2) results in A≡Io = Vi gm2RD (1/gm1) + RF gm2RDRF (1/gm1) + RF with the determination of β , R11 β = RF R11 =R22 =RF and R22 , Aβ = Io=Af= A Vs 1 + Aβ ⇒Af = gm2RD (1/gm1)+RF +gm2RDRF From the A circuit, breaking the loop at XX gives Ro =RF +RL +ro2 Rof = (1 + Aβ)Ro RD 􏱼g+RR􏱽 =1+m2 DF [RF+RL+ro2] Vd1 Q1 R11􏰔RF Q Ro R22􏰔 RF (1/gm1) + RF For 2 Io RL gm1 =gm2 =4mA/V, RD =20k􏱹, ro2 =20k􏱹, RF =100􏱹,andRL =1k􏱹, we obtain A = 4×20 = 228.6 mA/V 0.25 + 0.1 β = 0.1 k􏱹 Aβ = 22.86 1 + Aβ = 23.86 Af = 228.6 = 9.56 mA/V 23.86 Ro = 0.1+1+20 = 21.1 k􏱹 Rof = 23.86 × 21.1 = 503.4 k􏱹 Vi 􏰀􏰒 Figure 2 Figure 2 shows the A circuit. To determine A ≡ Io/Vi, we write for Q1 Vd1 =− RD (1) Vi (1/gm1) + RF and for Q2 Io = −gm2Vd1 (2) 11.57 Chapter 11–35 R2 􏰀 V1 1 R3 􏰒 R1 2 Io R12 R22 μ (1/gm)+R22 1 R3 R11 R2 0 0 R2 R12 1R3 Figure 1 Thus, A≡Io = Vi Figure 1 shows the feedback network fed with a current Io to determine β: Vf R1R3 β≡I =R+R+R o123 For Aβ ≫ 1, Since β = 0.01, we have Aβ = 0.01μ Af = Io ≃ 1 Vs β Thus, Af = 1 + R2 R3 R1R3 ForR1 =R3 =0.1k􏱹andAf =100mA/V, 100=10+ R2 +10 0.01 ⇒R2 =0.8k􏱹 To obtain the loading effects of the feedback network, refer to Fig. 1. R11 = R3 ∥ (R2 + R1) = 100 􏱹 ∥ (800 + 100) 􏱹 = 90 􏱹 R22 =R1 ∥(R2 +R3) = 100 ∥ (800 + 100) = 90 􏱹 1/gm + R22 = 0.01μ = 9.17 × 10−3μ + 1 R1 1+0.09 For a 60-dB amount of feedback, Vi 􏰀 􏰒 􏰀 􏰒 R11 V Io mg Ro R22 1+Aβ=1000 Aβ = 999 9.17 × 10−3μ = 999 ⇒μ=1.09×105 V/V Rout=Rof =(1+Aβ)Ro=1000Ro where Ro can be obtained from the A circuit as Ro =ro +R22 +gmroR22 = 50+0.09+1×50×0.09 = 54.6 k􏱹 Thus, Rout =1000×54.6=54.6M􏱹 11.58 (a) Since Vs has a zero dc component, the gate of Q1 is at zero dc voltage. The negative feedback will force the gate of Q2 to be approximately at the same dc voltage as that at the gate of Q1, thus VO = 0 VD1 =1.2−VSG3 = 1.2 − |Vt| − |VOV3| = 1.2 − 0.4 − 0.2 = +0.6 V Figure 2 The A circuit is shown in Fig. 2. We can write Vg =μVi (1) Io = Vg (2) (1/gm) + R22 VD2 = VO + VGS5 =0+Vt +VOV5 = 0.6 V (b) 􏰀 Vf 1 􏰒 RF2 Io gm1,2 = 2ID1,2 VOV 1,2 = 2 × 0.1 = 1 mA/V 0.2 ro2 = ro4 = |VA| ID2,4 =20=200k􏱹 0.1 gm5=2ID5 =2×0.8=8mA/V VOV5 0.2 A = 1×(200 ∥ 200) 0.125 + 10 = 9.88 mA/V Io =Af = A Chapter 11–36 b􏰒􏰔Vf 􏰔R I F o Figure 1 The feedback network is shown in Fig. 1, from whichwefind β=RF =10k􏱹 For Aβ ≫ 1, Af ≃ 1 = 1 β RF Af = 1 =0.1mA/V 10 k􏱹 (c) From the β circuit in Fig. 1 and noting that the feedback topology in series-series, the loading effects of the feedback network are Vs 1+Aβ = 9.88 = 0.099 mA/V 1+9.88×10 (d) FromtheAcircuit,wehave R11 =R22 =RF =10k􏱹 Figure 2 shows the A circuit. We can write Vg5 = −gm1,2(ro2 ∥ ro4) Vi Io = Vg5 (1/gm5) + R22 Thus, A≡ Io = gm1,2(ro2 ∥ro4) Vi (1/gm5) + R22 This figure belongs to Problem 11.58, part (b). 10 k􏰵 Vi (1) Ro =ro5 +R22 +gm5ro5R22 where |VA| 20 ro5 = I = 0.8 = 25 k􏱹 D5 Ro = 25+10+8×25×10 = 2035 k􏱹 Rout=Rof =Ro(1+Aβ) = 2.035×(1+9.88×10) = 203 M􏱹 (e) Vo =IoRF = Af VsRF Vo =AfRF =0.099×10=0.99V/V Vs Q Q Io 3 4 Ro Q5 Vg5 Q1 Q2 R22 􏰔 RF 􏰔 10 k􏰵 􏰔 10 k􏰵 R11 􏰔 RF Figure 2 Rout = Output resistance at source of Q5 1 + Aβ ≃ 1/gm5 1+Aβ 125 􏱹 = 1+9.88×10 = 1.25 􏱹 A = −103 × 0.909 × = −818.9 k􏱹 Is (10 ∥ 1) 0.1 + (10 ∥ 1) Chapter 11–37 Af =Vo = A 1 + Aβ 818.9 11.59 Rs = − 1 − 818.9 × −0.1 = − 818.9 = −9.88 k􏱹 Q3 Q4Vg5 1 + 81.89 Rif =Ri/(1+Aβ) = 0.909k􏱹 =11􏱹 1 + 81.89 Rif = Rin ∥ Rs =Rin ∥1k􏱹 Rin= 1 = 1 1−1 1−1 Rif 1000 11 1000 = 11.1 􏱹 From Eq. (11.42), we have Ro =ro ∥RF ∥RL = 0.1 ∥ 10 ∥ 1 = 90.1 􏱹 Rof = Ro 1 + Aβ = 90.1 = 1.1 􏱹 1 + 81.89 Rof = Rout ∥ RL = Rout/1 k􏱹 ⇒ Rout ≃ 1.1 􏱹 Comparison to the values in Example 11.9: Q5 Q1 Q2 Vr First we write for the gain of differential amplifier Vg5 = −gm1,2(ro2 ∥ ro4) (1) V􏰀 RF t􏰒 Figure 1 Figure 1 shows the circuit prepared for determining the loop gain Aβ: Aβ ≡ − Vr Vt Vt Next we write for the source follower, Vr = RF ∥ ro5 Vg5 (RF ∥ro5)+(1/gm5) Combining (1) and (2) yields Aβ=−Vr =gm1,2(ro2∥ro4) RF∥ro5 Vt (RF ∥ro5)+(1/gm5) (2) μ=104 V/V μ = 103 Af −9.99 k􏱹 −9.88 k􏱹 Rin 1.11 􏱹 11.1 􏱹 Rout 0.11 􏱹 1.1 􏱹 Q.E.D. 11.60 μ=103 V/V, Rid =∞, ro =100􏱹, RF =10k􏱹,andRs =RL =1k􏱹.From Example 11.9 Eqs. (11.37) and (11.41), 1 1 β=−R =−10k􏱹=−0.1mA/V F A=−μR (RF ∥RL) iro+RF ∥RL) where Ri =Rid ∥RF ∥Rs Ri =∞∥10∥1=0.909k􏱹 11.61 Comparing the circuit of Fig. E11.19 and that of Fig. 11.24(a), we note the following: μ = gmro,Q ro = ro,Q (This is based on representing the transistor output circuit (gmVgs, ro,Q) by its Thévenin equivalent.) Rid =∞ RL =∞ Using Eq. (11.39), we obtain 11.62 􏰀 Ri =Rid ∥RF ∥Rs =RF ∥Rs Using Eq. (11.44), we obtain gmro(RF∥Rs) RF A = − r o + R F RF g 􏰀 V 􏰀􏰒 V g m V g s r o Chapter 11–38 V r R s f 1 tgs 1+gmro(RF ∥Rs)ro+RF =− (Rs ∥RF)gm(ro ∥RF) 1+(Rs ∥ RF)gm(ro ∥ RF)/RF which is the expression given in the answer to Exercise 11.19(b). Rif = Ri 1 + Aβ R ∥R =sF t 􏰒 􏰒 Vgs 􏰔Vt Figure 1 R = Rs∥RF if 1 + g (R ∥ R )(r ∥ R )/R Figure 1 shows the circuit prepared for the determination of the loop gain Aβ: Vr Aβ=−V 1 + Aβ Substituting for Aβ from Eq. (11.43), we obtain An expression for Vr can be written by inspection as Vr=−gmVt[ro∥(Rs+RF)] Rs Rs + RF msFoFF 1 =1 +1 +gm(ro∥RF) Thus, Aβ=gm[ro ∥(Rs +RF)] Rif Rs RF RF But, Rs (1) Rs + RF RF 1 2􏰀􏰒Vo Figure 2 111 If =+ Rif Rs Rin Thus, 1 = 1[1+gm(ro∥RF)] Rin RF R ⇒ Rin = F The feedback network (β circuit) is shown in Fig. 2fedwithVo atport2andwithport1 short-circuited to determine β: β≡If =−1 (2) Vo RF Equations (1) and (2) can how be used to determine A: A=−gm[ro∥(Rs+RF)](Rs∥RF) Using the numerical values given in Exercise 11.19(c), we obtain A=−5[20∥(1+10)](1∥10) = −32.3 k􏱹 β=− 1 =−0.1mA/V RF [1+gm(ro ∥RF)] which is the answer given to Exercise 11.19(c). Using Eq. (11.42), we obtain Ro =ro ∥RF ∥RL =ro ∥RF Rof = Ro Substituting for Aβ from Eq. (11.43), we obtain Rof = (ro∥RF) 1+gm(Rs ∥RF)(ro ∥RF)/RF 1 =1+1+gm(Rs∥RF) RofroRF RF Rof = ro ∥ RF 1+gm(Rs ∥RF) Rout=Rof =ro∥ RF 1+gm(Rs ∥RF) which is identical to the result given in the answer to Exercise 11.19(d). 1 + Aβ Aβ = 3.23 Af =Vo = Is = − 32.3 1 + 3.23 A 1 + Aβ = −7.63 k􏱹 Compare these results to those found in Exercise 11.19: A = −32.3 k􏱹 (−30.3 k􏱹), β = −0.1 mA/V (−0.1 mA/V), Aβ = −3.23 (−3.03), and Af = −7.63 k􏱹 (−7.52 k􏱹). The slight differences are due to the approximation used in the systematic analysis method. 11.63 (a) Also, observe that β=−1 Rf From the A circuit in Fig. 3, we have Ri = Rs ∥ Rf ∥ rπ (1) Vπ =IiRi (2) Vo =−gmVπ(RC ∥Rf) (3) Combining (1), (2) and (3) gives A≡Vo =−gm(Rs∥Rf ∥rπ)(RC∥Rf) Ii Ri =10k􏱹∥56k􏱹∥1.56k􏱹 = 1.32 k􏱹 A = −64 × 1.32 × (5.6 ∥ 56) = −429 k􏱹 From the A circuit, we have Ro =RC ∥Rf = 5.6 k􏱹 ∥ 56 k􏱹 = 5.1 k􏱹 Chapter 11–39 Figure 1 Figure 1 illustrates the dc analysis. We can express the dc collector voltage VC in two alternativeways: 􏱾I􏱿 VC =+15−RC IC + C +0.07 β (d) β =− 1 =− Rf 429 Aβ= 56 =7.67 1 56 k􏱹 and V =0.7+R C +0.07 􏱾I􏱿 Cfβ 1+Aβ=8.67 (e) Af = Vo = Is Equating these two expressions yields 15 − 5.6(IC + 0.01IC + 0.07) = 0.7 + 56(0.01IC + 0.07) ⇒IC =1.6mA VC ≃5.5V (b) Figure 2 on the next page show the small-signal equivalent circuit of the amplifier where gm = 1.6mA =64mA/V 0.025 V rπ = β = 100 = 1.56 k􏱹 gm 64 (c) Figure 3 on the next page shows the A circuit. It includes the loading effects of the feedback network: R11 =R22 =Rf A 1 + Aβ =−429 =−49.5k􏱹 8.67 Ri Rif =1+Aβ = 1.32k􏱹 =152􏱹 8.67 Rif = Rs ∥ Rin 152 􏱹 = 10 k􏱹 ∥ Rin Rin = 155 􏱹 Rout=Rof= Ro 1 + Aβ 5.1 k􏱹 = 8.67 =588􏱹 This figure belongs to Problem 11.63, part (b). Chapter 11–40 Figure 2 The below two figures belong to Problem 11.63, part (c). Figure 3 (f) Vo = Vo = Af =−49.5 =−4.95V/V Vs IsRs Rs 10 Figure 4 11.64 (a) Refer to Fig. P11.63 and assume the gain of the BJT to be infinite so that the signal voltage at its base is zero (virtual ground). In this case, we have Vo =−Rf =−56=−5.6V/V Vs Rs 10 Thus, the actual gain magnitude (≃ 5 V/V) is only about 12% below the ideal value; not bad for a single transistor inverting op amp! Refer to the feedback network shown in Fig. 11.24(b) and to the determination of β illustrated in Fig. 11.24(c). Thus, β=−1 RF Ro =ro ∥RF =2∥20=1.818k􏱹 1.818 Chapter 11–41 If Aβ ≫ 1, then we have Af = Vo ≃ 1 = −RF Rout = Rof = 82.18 = 22.1 􏱹 (c)fHf =fH(1+Aβ) = 1 × 82.18 = 82.18 kHz Is β and the voltage gain realized will be 11.65 􏰀1.4 V 􏰀0.7 V Q1 Vo = Vo ≃−RF Vs IsRs Rs IfRs =2k􏱹,toobtainVo/Vs ≃−10V/V,we required RF =10×Rs =20k􏱹 (b) Refer to the solution to Example 11.9. 11 β=−R =−20k􏱹=−0.05mA/V F Using Eq. (11.39), we obtain Ri =Rid ∥RF ∥Rs = 100 ∥ 20 ∥ 2 = 1.786 k􏱹 Using Eq. (11.41) with RL = ∞, we get A≡ Vo =−103 ×1.786× 20 Ii 20+2 VDD I Q2 􏰀0.7 V I = −1623.6 k􏱹 Af ≡Vo = A (a) See Figure 1. RF 0 Figure1 Is 1 + Aβ where VG1=VGS1=Vt+VOV = 0.5 + 0.2 = +0.7 V (because the dc voltage across RF is zero) VO =VG1 VO = +0.7 V VD1 =VO +VGS2 = 0.7+0.5+0.2 = +1.4 V (b) gm1,2 = 2I = 2×0.4 =4mA/V VOV 0.2 VA 16 V ro1,2= I =0.4mA=40k􏱹 1 + Aβ = 1 + 1623.6 × 0.05 = 82.18 Af =Vo =−1623.6 Is 82.18 = −19.76 k􏱹 Vo = Vo =Af =−19.76 VsIsRsRs 2 = −9.88 V/V Rif= Ri 1 + Aβ = 1.786 = 21.7 􏱹 (c) Figure 2 on the next page shows the β circuit and the determination of its loading effects, R11 =R22 =RF Figure 2 shows also the A circuit. We can write Vg1 =IiRi (1) where Ri =R11 =RF (2) 82.18 Rif = Rs ∥ Rin 21.7 􏱹 = 2000 􏱹 ∥ Rin Rin ≃ 21.7 􏱹 Rout=Rof= Ro 1 + Aβ where from Eq. (11.42) with RL = ∞ we get This figure belongs to Problem 11.65, part (c). Chapter 11–42 RF RF RF 121212 R11 􏰔 RF R22 􏰔 RF Vd1 Q2 R22 􏰔 RF Vg1 Q1 R11 􏰔 RF Ii Vo Ri Vd1=−gm1ro1 (3) Figure 2 Ro A 1 + Aβ Vo = Vd1 R22 ∥ ro2 (4) (R22 ∥ ro2) + 1 (e)Af ≡Vo = Is gm2 Combining Eqs. (1)–(4) results in Vo RF ∥ro2 A≡ I =−gm1ro1RF(R ∥r )+1/g gm1ro1RF(RF ∥ ro2) = −(RF ∥ ro2) + 1/gm2 + (gm1ro1)(RF ∥ ro2) (f) Ri = RF Rin =Rif =Ri/(1+Aβ) i (d) F o2 m2 􏲈􏱼 R ∥r 􏱽 =R 1+gr F o2 If RF F m1 o1(RF ∥ro2)+1/gm2 Rout=Rof =Ro/(1+Aβ) where from the A circuit we have I 􏰀 V b f 􏰔􏰒1 Ro = RF ∥ ro2 ∥ 1 gm2 􏱾 1􏱿􏲈 Rout= RF∥ro2∥gm2 􏰒o Vo RF Figure 3 􏱼R∥r􏱽 1+gr F o2 FromFig.3weseethat β=− 1 m1 o1 (RF ∥ ro2) + 1/gm2 (g) A=−4×40×10 10∥40 RF Aβ=g r RF∥ro2 = −1551.5 k􏱹 β=−1 =− 1 =−0.1mA/V (10∥40)+0.25 m1 o1 (RF ∥ ro2) + 1/gm2 RF 10 k􏱹 Aβ=155.15 1+Aβ=1+g r RF ∥ro2 m1 o1(RF ∥ro2)+1/gm2 1+Aβ=156.15 Af = − 1551.5 = −9.94 k􏱹 156.15 = 10, 000 􏱹 = 64 􏱹 156.15 gm2 Ro =10∥40∥0.25=242􏱹 Rout=Rof= Ro 1 + Aβ = 242 =1.55􏱹 156.15 Combining Eqs. (1) and (2), we obtain Vr ro2 Aβ≡−V =(gm1ro1)r +1/g (3) t o2 m2 This expression differs from that obtained in Problem 11.65 utilizing the general feedback analysis method. To determine the numerical difference, we evaluate Aβ in Eq. (3) using gm1 =gm2 =4mA/V, ro1 =ro2 =40k􏱹 Aβ = 4 × 40 × 40 = 159 which is larger than the value found in Problem 11.65 (155.15) by about 2.5%. This small difference is a result of the approximations involved in the general method. The more accurate result for Aβ is the one obtained here. However, the loop-gain method does not make it possible to determine the input and output resistances of the feedback amplifier. 11.67 Figure 1 shows the small-signal-equivalent circuit of the feedback amplifier of Fig. E11.19. Analysis to determine Vo/Is proceeds as follows: Ri =RF =10k􏱹 Rin = Rif = RF Chapter 11–43 1 + Aβ Ro = RF ∥ ro2 ∥ 1 40 + 0.25 11.66 Q2 QVr􏰀V V2 􏰀 V2 1􏰒t 􏰒 RF 0 Figure 1 Writing a node equation at the output node provides gmVgs+Vo +Vo−Vgs =0 Figure 1 shows the circuit of Fig. P11.65 prepared ⇒ V for determining the loop gain Aβ: gs ro = −V o RF 1+1 ro RF 1 gm − RF V1 Aβ ≡ −Vr t ⇒ Vgs = −Vo 􏱾 1 􏱿 The gain of the source-follower Q2 can be found gm − R as F (1) V2 = ro2 (1) Vt ro2+1/gm2 The gain of the CS transistor Q1 can be found as Vr = −gm1ro1V2 (2) This figure belongs to Problem 11.67. (ro ∥ RF ) Writing a node equation at node G provides Is−Vgs +Vo−Vgs =0 Is − RF Vgs = 0 (2) Rs (Rs ∥RF) RF + Vo RF (Vo – Vgs)/RF G Vo Vgs/Rs 􏰀 Vo/ro Is Rs Vgs gmVgs ro 􏰒 Figure 1 This figure belongs to Problem 11.68, part (a). If 􏰒If 1 b􏰔 V 􏰔􏰒R oF Chapter 11–44 RF RF RF 1 2􏰀􏰒Vo 1 2 1 2 R11 􏰔 RF R22 􏰔 RF Figure 1 Substituting for Vgs from (1) into (2), we obtain Is+Vo􏱾 􏱿1 +Vo RD1 Vd1 Q1 RD2 gm− 1 RF = 0 ⇒ Vo Is 􏱾1􏱿 gm − R (ro∥RF)(Rs∥RF) RF Vo (ro ∥ RF )(Rs ∥ RF ) 1+gm−1 (ro∥RF)(Rs∥RF)/RF Q2 R22 􏰔 RF R o =􏱾F􏱿 RF For the feedback analysis to be reasonably accurate, we use gm ≫ 1 RF I i R11 􏰔 RF Ri 11.68 Figure 1 shows the feedback network with a voltage Vo applied to port 2 to determine β: β≡If =−1 Vo RF Figure 2 Figure 2 shows the A circuit. For the CG amplifier Q1, we can write 1 Ri = RF ∥ g (1) m1 Vsg =IiRi (2) Vd1 = gm1VsgRD1 (3) Combining (1)–(3) yields 􏱾1􏱿 Vd1=(gm1RD1) RF∥g Ii (4) m1 For the CS stage Q2, we can write Vo =−gm2(RD2 ∥RF) (5) Vd1 Combining (4) and (5), we obtain the open-loop For Aβ ≫ 1, we have Vo ≡ Af ≃ 1 = −RF Is β Thus, for Vo ≃ −10 k􏱹, we select Is RF =10k􏱹 The loading of the feedback network on the A circuit can be determined as shown in Fig. 1: R11 = R22 = RF gain A: V􏱾1􏱿 A≡ o =−(gm1RD1) RF ∥ gm2(RD2 ∥RF) Is gm1 This figure belongs to Problem 11.69, part (a). Chapter 11–45 RC 5􏰒0.356 􏰁10 􏰔 􏰀1.44 V 􏰀0.7 V 0.0035 mA 􏰀5 V 10 k􏰵 0.356 mA 0.35 0.0058 mA mA Q2 Q1 0.577 ~ 0.58 mA VO 􏰔 0.7 􏰀 0.035 􏰔 0.735 V 0.0035 0.735􏰒(􏰒5) 􏰔 0.5735 mA 10 RE 10 k􏰵 RF 􏰔10 k􏰵 􏰒5 V Substituting gm1 = gm2 = 4 mA/V, RD1 =RD2 =10k􏱹,andRF =10k􏱹gives A = −(4 × 10) × (10 ∥ 0.25) × 4 × (10 ∥ 10) A = −195 k􏱹 Aβ = 195 = 19.5 10 (b) Figure 1 1 + Aβ = 20.5 Af≡Vo= A Is 1 + Aβ =−195 =−9.52k􏱹 20.5 Rin=Rif= Ri 1 + Aβ From Eq. (1), we obtain Figure 2 Figure 2 shows the β circuit and the determination of its loading effects on the A circuit: R11 =R22 =RF =10k􏱹 Ri =10∥0.25=244􏱹 Rin = 244 =11.9􏱹 20.5 From the A circuit, Ro =RD2 ∥RF = 10 ∥ 10 = 5 k􏱹 Rout=Rof= Ro 1 + Aβ = 5000 =244􏱹 20.5 11.69 (a) Figure 1 (see figure above) shows the dc analysis. We assumed IC1 = 0.35 mA and found that IC2 = 0.58 mA, thus verifying the given values. The dc voltage at the output is VO = +0.735 V Figure 3 Figure 3 shows the A circuit. The input resistance is given by Ri =RF =RF ∥rπ1 where gm1 = IC1 = 0.35 =14mA/V VT 0.025 rπ1= β =100=7.14k􏱹 gm1 14 β=−1 =− 1 =−0.1mA/V RF 10 k􏱹 Aβ = −567.6 × −0.1 = 56.76 1 + Aβ = 57.76 Chapter 11–46 VA (d) Af ≡ o = Thus, Ri =10k􏱹∥7.14k􏱹=4.17k􏱹 The input voltage Vb1 is given by Vb1 = IiRi = 4.17Ii Is 1 + Aβ Af = − 567.6 = −9.83 k􏱹 (1) 􏲛􏲜 57.76 Rin=Rif= Ri The collector voltage of Q1 is given by Vc1 =−gm1Vb1 RC ∥(β2 +1)[re2 +(RE ∥RF)] where 1 + Aβ = 57.76 = 72.2 􏱹 re2 = VT = 25mV =43.1􏱹 IE2 0.58 mA 4.17 k􏱹 From the A circuit, we have 􏲛􏲜􏱼􏱽 Vc1 = −14Vb1 10 ∥ 101[0.0431 + (10 ∥ 10)] = −137.3Vb1 The gain of the emitter-follower Q2 is given by Ro=RF∥RE∥ re2+ RC β2 + 1 Ro Vo= RE∥RF Vc1 (RE ∥RF)+re2 5 = 5 + 0.0431 = 0.99 V/V Combining (1)–(3) gives A≡ Vo =−0.99×137.3×4.17 Ii = −567.6 k􏱹 (c) The value of β can be obtained from the β circuit as shown in Fig. 4: Figure 4 This figure belongs to Problem 11.70, part (a). = 10 ∥ 10 ∥ = 138.2 􏱹 Rout=Rof =1+Aβ (2) (3) 􏱼 10 􏱽 0.0431+ 101 138.2 = 57.76 = 2.4 􏱹 11.70 (a) Converting the signal source to its Norton’s form, we obtain the circuit shown in Fig. 1(a). This is a shunt–shunt feedback amplifier with the feedback network consisting of the resistor Rf . To determine β, we use the arrangement shown in Fig. 1(b), β=−1 Rf Figure 1 This figure belongs to Problem 11.70, part (b). 􏰀 V1 􏰒 7.5 k􏰵 􏰒􏰀 330V1 7.5 k􏰵 Chapter 11–47 1.65 k􏰵 (a) 7.5 k􏰵 􏰀􏰀􏰀 7.5 k􏰵 Ii RsV11.65􏰒􏰀V2􏰒􏰀V3􏰒􏰀RFRL RF k􏰵 330V 1.65 330V2 1.65 􏰒 1􏰒k􏰵 􏰒k􏰵 330V3 Ri 10 k􏰵 1 M􏰵 1 M􏰵 Ro Now, for Aβ ≫ 1, the closed-loop gain becomes Af ≡ Vo ≃ 1 =−Rf rπ = β = 100 = 2.27 k􏱹 gm 44 Rin =10∥15∥2.27=1.65k􏱹 Rout = 7.5 k􏱹 Avo =−gm ×7.5k􏱹 = −44 × 7.5 = −330 V/V (b) Figure 2 1 M􏰵 Is β The voltage gain Vo is obtained as Vs Vo=Vo =Af Vs Is Rs Rs Thus, approximately −100 V/V, we use −100 = − Rf Rs For Rs = 10 k􏱹, we obtain Rf =1M􏱹 Now consider the amplifier stage shown in Fig. P11.70(b). First, we determine the dc bias point as follows: 10 15 × 10 + 15 − 0.7 Figure 2(a) shows the equivalent circuit of the amplifier stage. Figure 2(b) shows the A circuit of the feedback amplifier made up of the cascade of three stage. Observe that we have included Rs and RL as well as R11 and R22. The overall gain VR o ≃− f Vs Rs Q.E.D. (b) To obtain a closed-loop voltage gain of A ≡ Vo/Is cam be obtained as follows: Ri =Rs ∥RF ∥1.65k􏱹 = 10 ∥ 1000 ∥ 1.65 = 0.623 k􏱹 V1 = Ii Ri = 0.623Ii V2 = −330V1 × V3 = −330V2 × 1.65 1.65 + 7.5 1.65 1.65 + 7.5 (1) = −59.5V1 (2) = −59.5V2 (3) =1.11mA IC = 1.11×0.99 = 1.1 mA 1 ∥ 1000 Vo =−330V3 × (1∥1000)+7.5 = −38.8V3 (4) Combining (1)–(4) gives A≡ Vo =−8.558×104 k􏱹 Ii IE = 10∥15 4.7 + 101 gm = IC = 1.1 =44mA/V VT 0.025 Since β=−1=−1 Rf 1 M􏱹 we have Aβ = 85.58 and 1+Aβ=86.58 Thus, Af ≡ Vo =−8.558×104 Vs 86.58 = −988 k􏱹 and the voltage gain realized is Vo =Af =−988=−98.8V/V Vs Rs 10 Rif= Ri 1 + Aβ = 623􏱹 =7.2􏱹 86.58 Rif = Rs ∥ Rin Rin ≃ 7.2 􏱹 From the A circuit, we have Ro =RL ∥RF ∥7.5k􏱹 Ro =1∥1000∥7.5=881.6􏱹 Ro Rof =1+Aβ The A circuit is shown in Fig. 1. Ri= 1 =1=0.2k􏱹 gm1 5 A≡ Vo =RD =10k􏱹 Ii Ro =RD =10k􏱹 Chapter 11–48 I C1 f􏰀V Qf 􏰒 o C2 Figure 2 The β circuit is shown in Fig. 2. Vo C1 + C2 = 1.8 mA/V β≡If=C1 gmf=0.9×2 Vo A Af ≡V =1+Aβ s 0.9 + 0.1 881.6 = 10 = 0.53 k􏱹 1 + 10 × 1.8 Ri 200 􏱹 Rin =Rif = 1+Aβ = 19 =10.5􏱹 = 10.2 􏱹 Rof =Rout ∥RL = ⇒ Rout = 10.3 􏱹 86.58 Rout = Rof = 11.73 (a) I1 Ro 1 + Aβ = 10 k􏱹 = 526 􏱹 19 11.71 (a) Shunt-Series (b) Series-Series (c) Shunt-Shunt 11.72 VDD 􏰔 3.3 V Q2 Q1 RD Q1 Figure1 ID2 􏰔 1 mA R RVG 10􏱌A 1R RL 1 0.99 mA Figure 1 Figure 1 shows the circuit for the purpose of performing a dc design. ID1 =100μA⇒I1 =100μA 02 Vo VG VO R o Ii Ri 1 􏱾W􏱿 I=μCV2 Chapter 11–49 D1 2 n ox L OV1 1Q 1􏱾W􏱿 2 100 = 2 × 200 × 􏱾W 􏱿 L × 0.22 Q1 R2 1 ⇒ VG1 =Vtn +VOV1 = 25 Vo RL Rof L1 = 0.6+0.2 = 0.8 V Is􏰔Vs Rs Rs R1 Since IR2,R1 = 10 μA, we have R1 = 0.8V =80k􏱹 0.01 mA IRL = ID2 − IR2,R1 =1−0.01=0.99mA VO = 0.99×2 = 1.98 V 1 􏱾W􏱿 I = μC V2 Rif Rin Rout Vo = Vo = Af Vs Is Rs Rs Thus, −6=−118 Figure 3 D2 2 n ox L OV2 2 Rs 1 􏱾W􏱿 1= ×0.2× ×0.22 ⇒R =19.7k􏱹 s ⇒ 2L2 􏱾W􏱿 L2 (d) Ii = 250 R2 = VO − VG 0.01 mA Q2 = 1.98−0.8 = 118 k􏱹 0.01 VGS2 =Vtn +VOV2 =0.8V VD1 = VG2 = 1.98+0.8 = 2.78 V (b) The β circuit consists of resistance R2. The value of β can be determined as shown in Fig. 2. If Rs R1 Ri Q1 R11 R22 Vo RL Ro (1) (2) 􏰔R2 􏰔R2 Figure 4 R2 1 2􏰀􏰒Vo Figure 2 The A circuit is shown in Fig. 4. Ri =Rs ∥R1 ∥R2 = 19.7 ∥ 80 ∥ 118 = 13.92 k􏱹 Vgs1 = IiRi = 13.92Ii Vd1 = −gm1Vgs1ro1 where gm1 = 2ID1 = 2×0.1 =1mA/V VOV1 0.2 ro1 = VA = 20 =200k􏱹 ID1 0.1 Thus, Vd1 = −200Vgs1 Vo = RL ∥ R2 ∥ ro2 Vd1 (RL ∥ R2 ∥ ro2) + 1/gm2 β≡If =−1=− 1 Vo R2 118k􏱹 =−8.47×10−3 mA/V Thus, A 􏲀􏲀 f ideal ≡ 1 = −118 k􏱹 β (c) Converting the signal source to its Norton’s form, the feedback amplifier takes the form shown in Fig. 3. where gm2 = V = 0.2 =10mA/V Vo = Af = − 113 = −5.73 V/V Vs Rs 19.7 Ri 13.92 (f) Rif = 1+Aβ = 23.34 =0.596k􏱹 R=R∥R if s in 0.596 = 19.7 ∥ Rin ⇒Rin =615􏱹 Chapter 11–50 2ID2 2×1 OV2 ro2=VA =20=20k􏱹 ID2 1 Thus, Vo = Vd1 (2 ∥ 118 ∥ 20) (2 ∥ 118 ∥ 20)+0.1 (3) = 0.947 V/V Combining (1)–(3), we obtain Rof = Ro = 94.7 =4.06􏱹 A= Vo =−13.92×200×0.947 Ii = −2636.7 k􏱹 Ro=RL∥R2∥ro2∥ 1 gm2 (e)Af=Vo= A Vs 1 + Aβ 23.34 4.06=Rout ∥2000⇒Rout =4.1􏱹 1 + Aβ Rof = Rout ∥ RL = 2 ∥ 118 ∥ 20 ∥ 0.1 = 94.7 􏱹 11.74 (a) Figure 1 shows the β network as well as the determination of its loading effects on the A circuit: R11 =RF +RM R22=RM∥RF Figure 2 shows the A circuit. Some of the analysis is shown on the diagram. Ri=R11∥ 1 (1) gm1 Vsg1 = IiRi (2) 2636.7 1 + (2636.7/118) = − = − 2636.7 = −113 k􏱹 23.34 These figures belong to Problem 11.74, part (a). RF RF0RF 1 RM 2 1 RM 2 1 RM 2 R11 􏰔 RF 􏰀 RM R22 􏰔 RM // RF RD gm1Vsg1 Q1 Figure 1 􏰒 Vgs2 􏰀 􏰒 Q2 Ii Vsg1 􏰀 R11 Io 􏰔 􏰒gm2Vgs2 RL R22 Ro Ri Figure 2 (3) Io = −gm2Vgs2 (4) Vgs2 = gm1Vsg1RD Combining (1)–(4) gives 11.75 Refer to Fig. 11.27(c), which shows the determination of β, β = If = − R1 (1) Io R1 + R2 Refer to Fig. 11.27(e), which shows the A circuit. The input resistance Ri is given by Ri = Rs ∥ Rid ∥ (R1 + R2) For our case here, Rs = Rid = ∞, thus Ri = R1 + R2 ForRin =Rif =1k􏱹,wehave Rif= Ri 1 + Aβ ⇒Ri =Rif(1+Aβ) Thus R1+R2=1k􏱹×(1+Aβ) Since1+Aβ is40dB,thatis, 1+Aβ=100 wehave R1 +R2 =1×100=100k􏱹 (2) Now, Af= A 1 + Aβ −100= A 100 Chapter 11–51 I 􏱾 A ≡ o = − R11 ∥ (b) If RF Ii 1 􏱿 (gm1RD)gm2 1 RM 2 Io Figure 3 (5) gm1 The β circuit prepared for the determination of β is shown in Fig. 3. β≡If =− RM Io RM +RF (c) From (5)–(6), we obtain (6) RM =10k􏱹, andRF =90k􏱹, thus R11 =RF +RM =90+10=100k􏱹 R22 =RM ∥RF =10∥90=9k􏱹 A = −(100 ∥ 0.2) × (5 × 20) × 5 = −99.8 A/A β=− 10 =−0.1A/A 10+90 R 􏱾 1􏱿 Aβ = M R11 ∥ (gm1RD)gm2 RM +RF gm1 (d) gm1 =gm2 =5mA/V, RD =20k􏱹 Aβ=9.98 1 + Aβ = 10.98 Af ≡Io = A Is 1+Aβ Rin =Rif = Ri 1 + Aβ ⇒A=−104 A/A β= Aβ = 99 =−0.0099 =−99.8 =−9.1A/A 10.98 R1 −0.0099=−R1 +R2 ⇒R1 =0.0099×100=0.99k􏱹 R2 = 100−0.99 = 99.01 k􏱹 Now, using Eq. (11.53) (page 869), we obtain A = −μ Ri ro2 1/gm +(R1 ∥ R2 ∥ ro2) ro2 +(R1 ∥ R2) −104 = where Ri =100k􏱹∥0.2k􏱹≃0.2k􏱹 Rin = 200 􏱹 = 18.2 􏱹 10.98 (e) Breaking the output loop of the A circuit between XX , we find Ro = R22 +RL +ro2 =(RM ∥RF)+RL +ro2 = (10 ∥ 90) + 1 + 20 = 30 k􏱹 Rof =Ro(1+Aβ)=30×10.98 = 329.4 k􏱹 Rout =Rof −RL =328.4k􏱹 −μ 100 20 0.2+(0.99 ∥ 99.01 ∥ 20) 20+(0.99 ∥ 99.01) A −104 Using Eqs. (1) and (2), we obtain μ = 119 V/V From Example 11.10, we have Ro = ro2 +(R1 ∥ R2)+gm2ro2(R1 ∥ R2) Ro =20+(0.99∥99.01)(1+5×20) = 119 k􏱹 Rout =Rof =Ro(1+Aβ)=119×100=11.9M􏱹 11.76 Thus, A = −9709 and Chapter 11–52 Io 􏰀m 􏰒􏰀􏰒 971.9 ro2 Af=−9709=−9.99A/A Vr 􏰒 Vt 􏰀 R2 Io Io 0 R1 which is identical to the value obtained in Example 11.10. Thus while Aβ and A differ slightly for the earlier results, Af is identical; an illustration of the power of negative feedback! Figure 1 11.77 (a) 0.2 mA Figure 1 shows the shunt-series feedback amplifier circuit of Fig. 11.27(a) prepared for determining the loop gain, Aβ ≡ −Vr Vt observe that here Rs = Rid = ∞. Thus Io can be obtained as 􏰀0.7 V Q2 R2 Q1 14 k􏰵 0 R1 3.5 k􏰵 0.2 mA 􏰀0.7 V V r Io=t o2 (1) 0.2 mA 1 +(ro2 ∥R1) ro2 +R1 gm Figure 1 Figure 1 shows the dc analysis. It starts by noting thatID1 =0.2mA.ThusVOV1 =0.2Vand VG1 =VGS1 =Vt +VOV1 =0.5+0.2 =0.7V Since the dc current through R2 is zero, the dc voltage drop across it will be zero, thus VS2 =+0.7V and IR1 = 0.7V =0.2mA 3.5 k􏱹 Thus, Q2 is operating at ID = 0.2 mA Q.E.D. (b) gm1 = gm2 = 2ID VOV = 2 × 0.2 = 2 mA/V 0.2 ro1 =ro2 = 10 =50k􏱹 0.2 The voltage Vr can be obtained as Vr = IoR1 × −μ = −μR1Io (2) Combining Eqs. (1) and (2), we obtain V Aβ≡− r =μ Vt R r 1 o2 1 +(ro2 ∥R1) ro2 +R1 gm For μ=1000V/V, R1 =10k􏱹, gm =5mA/V, andro2 =20k􏱹 we obtain Aβ=1000× 10 20 0.2+(20 ∥ 10) 20+10 = 970.9 which is slightly lower than the value found in Example 11.10 (1076.4), the difference being about −10%. This is a result of the assumptions and approximations made in the general feedback analysis method. From Example 11.10 (or directly from the β circuit) we have β = −0.1 A/A (c) Ro = ro2 + R22 + gm2ro2R22 R2 R2 0 = 50+2.8+2×50×2.8 = 332.8 k􏱹 1R121 R12(d) Chapter 11–53 If Figure 4 Figure 4 shows the determination of the value of R11 􏰔 R1􏰀R2 R2 1 R1 2 R22 􏰔 R1􏰐􏰐R2 Figure 2 Figure 2 shows the β circuit and the determination of its loading effects, R11 and R22, R11 =R1 +R2 =3.5+14=17.5k􏱹 R22 =R1 ∥R2 =3.5∥14=2.8k􏱹 R2 1 R1 2 Io β: β≡If =− R1 Io R1 + R2 Thus, Io Ro β = − 3.5 = −0.2 A/A 3.5 + 14 (e) Aβ = −525.8 × −0.2 = 105.16 1+Aβ=106.16 Af = − 525.8 = −4.95 A/A 106.16 (f) Rin = Rif = Ri 1 + Aβ 17.5 k􏱹 = 106.16 = 164.8 􏱹 Rout =Rof =Ro(1+Aβ) = 332.8 × 106.16 = 35.3 M􏱹 11.78 (a) If μ is a very large, a virtual ground will appear at the input terminal. Thus the input resistance Rin = V−/Ii = 0. Since no current flows in Rs, or into the amplifier input terminal, all the current Is will flow in the transistor source terminal and hence into the drain, thus Io = Is and Io =1 Is (b) This is a shunt-series feedback amplifier in which the feedback circuit consists of a wire, as shown in Fig. 1. As indicated, R11=∞ R22 = 0 The A circuit is shown in Fig. 2, for which we can write Vid = −Ii(Rs ∥ Rid) ≃−IiRs (1) Vgs1 Vd1 Q2 Q1 ro2 Ii R11 R22 Io= 1 gm2 Vd1 +(ro2 ∥R22) o2 22 (gm1ro1) Io Figure 3 Figure 3 shows the A circuit. To obtain A = Io/Ii, we write Vgs1 = IiR11 Vd1 = −gm1ro1Vgs1 (1) (2) (3) ro2 ro2+R22 ro2 r +R Combining (1)–(3) yields A=Io =− Ii R11 1 +(ro2∥R22) A = − 17.5 0.5+(50 ∥ 2.8) × 2 × 50 × 50 50+2.8 gm2 A = −525.8 A/A Ri =R11 =17.5k􏱹 Chapter 11–54 These figures belong to Problem 11.78, part (b). 121212 R22 􏰔 0 R11 􏰔 􏰿 Figure 1 G Io Ro S 􏰀 Vid 􏰒 Rs Rid mVid 􏰀􏰒 􏰀0 Vgs gm2Vgs ro2 Ii 􏰒 short because R22 􏰔 0 Figure 2 Ri (sinceRid isverylarge) Vgs = μVid Io = gm2Vgs Combining (1)–(3), we obtain A≡Io =−μg R (2) (3) Notethatthenegativesignisduetoour assumption that Is flows into the input node (see Fig. 2 for the way Ii is applied). If instead Is is flowing out of the input node, as indicated in Fig. P11.78, then Af ≡Io = μgm2Rs Is 1+μgm2Rs If μ is large so that μgm2Rs ≫ 1, Af ≃1 (e)Rif= Ri 1 + Aβ = Rs μgm2 Rs Rif = Rin ∥ Rs 1=1+1 I i Ri = Rs Ro = ro2 (c) m2 s If 12 Io Figure 3 From Fig. 3 we find I β≡f=−1 Io (d) Aβ = μgm2Rs Af= A 1 + Aβ =− μgm2Rs 1 + μgm2Rs Rif Rin Rs 1+μg =1 +1 R m2 R R s ins ⇒Rin= 1 μgm2 Rout=Rof =Ro(1+Aβ) = ro2(1 + μgm2Rs) (f) gm1Vi Io Figure 1 shows the circuit for determining Rin, Rin ≡ Vx R Ix out Chapter 11–55 􏰀 m 􏰒 Figure 4 Q A node equation at D1 gives Vx = (Ix − gm1Vgs)ro1 ⇒ Vx = Ixro1 − gm1Vgsro1 But, Vgs=μVx, thus Vx =Ixro1 −μgm1ro1Vx ⇒Rin≡Vx= ro1 Ix 1 + μgm1ro1 Since μgm1 ro1 ≫ 1, we have 1 Rin ≃ μgm1 (d) Since the drain of Q2 is outside the feedback loop, we have Rout = ro2 11.80 See figures on the next two pages. 2 ro1 Figure 4 shows the circuit of the cascode amplifier in Fig. P11.78 with VG replaced by signal ground, and Q1 replaced by its equivalent circuit at the drain. This circuit looks identical to that of Fig. 11.78(a). Thus we can write Io ≃ gm1vi Rout = ro2(1 + μgm2ro1) = ro2 + μ(gm2ro2)ro1 ≃ μ(gm2ro2)ro1 Compared to the case of a regular cascode, we see that while Io ≃ gm1Vi as in the regular cascode, utilizing the "super" CG transistor results in increasing the output resistance by the additional factor μ! 11.79 (a) Refer to Fig. P11.79. Let Is increase. This will cause the voltage at the input node, which is the voltage at the positive input of the amplifier μ, to increase. The amplifier output voltage will correspondingly increase. Thus, Vgs of Q1 increases and Io1 increases. This counteracts the originally assumed change of increased current into the input node. Thus, the feedback is negative. (b) The negative feedback will cause the dc voltage at the positive input terminal of the amplifier to be equal to VBIAS. Thus, the voltage at the drain of Q1 will be equal to VBIAS. For Q1 to operate in the active mode, the minimum voltage atthedrainmustbeequaltoVOV whichis0.2V. Thus the minimum value of VBIAS must be 0.2 V. (a) Refer to Fig. (a). (b) Refer to Fig. (b). (c) Routz = ro2 ∥ ro4 11.81 100 􏱌A 􏰀0.7 V 􏰀15 V 10 mA 200 􏱌A RL 􏰔 500 􏰵 V 􏰔 􏰀10 V O Q2 100 􏱌A 􏰀0.7 V 70 􏱌A 0 Q1 Rf 􏰔 10 􏰵 10.07 mA 􏰀1.4 V RE 140 􏰵 70 􏱌A Rs 10 k􏰵 10 mA (c) Figure 1 Figure 1 The dc analysis is shown in Fig. 1, from which we see that IC1 = 0.1 mA IC2=10mA These figures belong to Problem 11.80. Chapter 11–56 Q1 Q2 Vx /R Vx /R QN Vx/R y Vx 􏰀 Vx 􏰒 x Vx 0􏰀 􏰀m 􏰒􏰒z 0 QP R R Iz 0 Vy 􏰔 Vx 􏰔 Vx R 0 (a) Vx positive Q3 Q4 Figure 1 These figures belong to Problem 11.80. Chapter 11–57 Q1 Q2 0 QN Vy 􏰔 0 0y Rin 􏰔 0 Iy x 0 􏰀 􏰀􏰒 m 􏰒 0 z Iy QP Vy 􏰔 0 Iz 􏰔 Iy Iy Q4 Iy Q3 Q1 (b) Iy positive Iy z Iy QN Vy􏰔 0 y Iy x 0 􏰀 􏰀􏰒 m 􏰒 Q2 Iz 􏰔 Iy 0 Q4 Iy 0 QP Vy 􏰔 0 0 Q3 (b) Iy negative Figure 2 These figures belong to Problem 11.81. Rf Rf 0 Chapter 11–58 1 RE 2 1 RE 2 b Circuit Rf IfRf 1 RE 2 1 RE 2 Io R11􏰔Rf 􏰀RE R22􏰔RE􏰐􏰐Rf Figure 2 b􏰔If􏰔􏰒 RE Io Rf 􏰀 RE RL 􏰔500􏰵 b2 gm1Vp1 􏰔 􏰒Io gm1Vp1 􏰀 Ii Rs R11 Vp1 􏰒 Q2 Q1 Thus, gm1 =4mA/V rπ1 =25k􏱹 gm2 = 400 mA/V The β circuit is shown in Fig. 2 together with the determination of its loading effects and of β. R11 =Rf +RE =10.14k􏱹 R22 =Rf ∥RE =10∥0.14=0.138k􏱹 β = − RE = − 0.14 = 0.0138 A/A Rf +RE 10+0.4 The A circuit is shown in Fig. 3. Ri Vp1 􏰔 IiRi Figure 3 R22 Ri =Rs ∥R11 ∥rπ1 =10∥10.14∥25 = 4.19 k􏱹 Vπ1 =IiRi Io = −β2gm1Vπ1 ⇒ A = Io = −β2gm1Ri Ii A = −100 × 4 × 4.19 = −1676 A/A Aβ = −1676 × −0.0138 = 23.13 1 + Aβ = 24.13 Af=Io= A Is 1 + Aβ where Is = Vs Rs Af = − 1676 = −72.5 A/A 24.13 Vo = −IoRL = 72.5× 0.5 = 3.62 V/V 10 The small-signal parameters of Q1 and Q2 can now be obtained as gm1 = 0.86 = 34.4 mA/V 0.025 rπ1 = 100 = 2.91 k􏱹 34.4 0.76 gm2 = = 30.4 mA/V 0.025 100 rπ2 = 30.4 = 3.3 k􏱹 (b) The equivalent circuit of the feedback amplifier is shown in Fig. 1, where Rs =10k􏱹 Is = Vs Rs RB =RB1 ∥RB2 =13k􏱹 (c) See figure on the next page. The determination of the loading effects of the β circuit on the A circuit is shown in Fig. 2: R11 =Rf +RE2 =10+3.4=13.4k􏱹 R22 =RE2 ∥Rf =3.4∥10=2.54k􏱹 The A circuit is shown in Fig. 3 on the next page. Analysis of the A circuit to determine A ≡ Io/Is proceeds as follows: Ri =Rs ∥R11 ∥RB ∥rπ1 = 10 ∥ 13.4 ∥ 13 ∥ 2.91 = 1.68 k􏱹 Vπ1 =IiRi (1) I =−g V RC1 (2) b2 m1 π1RC1 +rπ2 +(β+1)R22 Chapter 11–59 Vs −IsRs R Rif = i 1 + Aβ 4190 􏱹 24.13 Rif = Rs ∥ Rin = = 173.6 􏱹 173.6 􏱹 = 10, 000 􏱹 ∥ Rin ⇒ Rin = 176.7 􏱹 11.82 (a) Refer to the circuit in Fig. P11.82. Observe that the feedback signal is capacitively coupled and so are the signal source and RL; thus, these do not enter into the dc bias calculations and the feedback does not affect the bias. The dc emitter current in Q1 can be determined from 15 12× 100+15 −0.7 =0.865mA IC1 =0.99×0.865=0.86mA IE1 = 100∥15 0.870 + 101 Next consider Q2 and let its emitter current be IE2. The base current of Q2 will be IE2/(β + 1) ≃ 0.01 IE2. The current through RC1 will be (IC1 + IB2) = (0.86 + 0.01 IE2). We can thus write the following equation: 12 = (0.86 + 0.01 IE2) × 10 + 0.7 + 3.4 × IE2 ⇒ IE2 = 12−8.6−0.7 = 2.7 = 0.77 mA 3.4 + 0.1 3.5 IC2 =0.76mA Io =Ie2 =(β+1)Ib2 􏰀 (3) Iout This figure belongs to Problem 11.82, part (b). 􏰀 Ic2 rp2 Vp2 Is Rs R Vp1 RC1 􏰒 gm2Vp2 RC2 RL B rp1 gm1Vp1 Io Rout Rof RE2 􏰒 Rif Rin Rf Figure 1 Chapter 11–60 These figures belong to Problem 11.82, part (c). Rf Rf0Rf 1 RE221 RE221 RE22 b Circuit R22 􏰔 RE2 􏰐􏰐Rf R11􏰔Rf 􏰀RE2 Figure 2 Combining Eqs. (1)–(3) results in A≡Io =− (β+1)Rigm1RC1 Ii RC1 +rπ2 +(β+1)R22 = 101×1.68×34.4×10 10 + 3.3 + 101 × 2.54 = −216.3 A/A Breaking the emitter loop of Q2 at XX gives R=R +rπ2+RC1 Rof = Ro(1 + Aβ) = 2.67 × 55.88 = 149.2 k􏱹 (f) Rif = Rs ∥ Rin 30.1 􏱹 = 10 k􏱹 ∥ Rin Rin = 30.2 􏱹 Rs Iin=IsR+R ≃Is s in RC2 RC2 Iout =IC2R +R =αIoR +R C2 L C2 L Iout ≃ Iout = Io ×α RC2 Iin Is Is RC2 +RL ⇒ Iout =−3.87×0.99×0.99× 8 Figure 3 o 22 3.3+10 (d) If β+1 = 2.54 + = 2.67 k􏱹 Rf 1 RE2 2 Io Figure 4 101 Iin 8+1 = −3.41 A/A β≡If =− RE2 Io RE2 +Rf = −0.254 A/A (e) Aβ = −216.3 × −0.254 = 54.88 1 + Aβ = 55.88 Af =Io =−216.3=−3.87A/A RC1 Rout Q2 Rof Is Rif = Ri 1 + Aβ 55.88 = 1.68 k􏱹 = 30.1 􏱹 55.88 Figure 5 To determine Rout, consider the circuit in Fig. 5. Using the formula given at the end of Example 11.8, adapted to our case here, we get At ω = ω180, the magnitude of A becomes | A(jω180 )| = Rout =ro2 +[Rof ∥􏱼 􏱽 (r +R )] 1+g r rπ2 􏱾20,000􏱿2􏲠􏲟 􏱾20,000􏱿2􏲠2 20,000 Chapter 11–61 􏲌 􏲋􏲋􏲟 105 π2 where C1 m2 o2rπ2+RC1 1+ βcr = 1 = 4×10−3 V/V 250 􏲊 1+ |A(jω180)|= 200×2 =250V/V ro2 = 75V 0.76mA 105 |A(jω180)|βcr =1 100 =98.7k􏱹 = 98.7 + [149.2 ∥ R = 98.7+12.21×743 = 9.17 M􏱹 Check: MaximumpossibleRout ≃βro ≃10M􏱹 So, our result is reasonable. out 􏱼􏱽 (3.3+10)] 1+30.3×98.7× 3.3 3.3 + 10 105 11.83 A(s) = 􏲃 􏲄 􏱾 1+s1+s 100 20,000 −1ω −1ω 􏱿2 11.84 A(s) = 105 􏲃s􏲄􏱾 s􏱿2 1+ 100 1+ 20,000 Correspondingly, Af = 105 1+105 ×4×10−3 105 = 1 + 400 ≃ 250 V/V φ=−tan 100−2tan 20,000 ωω 􏲃ω􏲄􏱾ω􏱿2 A(jω)= 105 1+j20,000 180◦ = tan−1 180 +2 tan−1 180 100 20, 000 Since ω180 will be much greater than 100 rad/s, we can assume that at ω180, tan−1(ω180/100) is approximately 90◦ , thus 1+j100 2tan 􏲄 􏲟 􏱾 −1 ω180 20,000 ◦ =90 φ(ω) = −tan−1 |A(jω)|= 􏲆 􏲃 1 + − 2 tan−1 105 (1) 􏱿2􏲠 (2) 􏲃 ω 􏲄 100 􏱾 ω 20, 000 􏱿 ⇒ tan−1 ω180 20, 000 ω2 ω 100 1 + 20, 000 = 45◦ ⇒ ω180 = 20, 000 rad/s Using Eqs. (1) and (2), we can obtain the data required to construct Nyquist plots for the two cases: β = 1 and β = 10−3. The results are given in the following table. which is indeed much greater than 100 rad/s, justifying our original assumption. This table belongs to Problem 11.84. ω rad/s −tan−1 􏲃 ω 􏲄 100 􏱾ω􏱿 −2 tan−1 20, 000 φ | A| | Aβ| β=1 | Aβ| β = 10−3 0 0 0 0 105 105 100 102 −45◦ −0.6◦ −45.6◦ 0.7 × 105 0.7 × 105 70 103 −84.3◦ −5.7◦ −90◦ 104 104 10 104 −89.4◦ −53.1◦ −142.5◦ 800 800 0.8 2×104 −89.7◦ −90◦ ≃ −180◦ 250 250 0.25 ∞ −90◦ −180◦ −270◦ 0 0 0 This figure belongs to Problem 11.84. Chapter 11–62 􏰒1 b􏰔1 w 􏰔 􏰒45.6o v 􏰔 0 􏱛Ab􏱛 􏰔105 Re Im v 􏰔 2 􏰁 104 􏰔 250 v 􏰔 104 􏰔 800 v 􏰔 103 􏰔 104 Im v 􏰔 102 􏰔 0.7 􏰁 105 v 􏰔 2 􏰁 104 􏰔 0.25 􏰒1 v 􏰔 104 􏰔 0.8 v 􏰔 103 􏰔 10 Not to scale b 􏰔 10􏰒3 w 􏰔 –45.6° v 􏰔 102 􏰔 70 Not to scale Figure 1 v 􏰔 0 􏰔100 Re Using these data, we obtain the two Nyquist plots shown in Fig. 1. We observe that the amplifier with β = 1 will be unstable and that with β = 10−3 will be stable. 104k 􏱿3/2 106 | A(jω180)β(jω180)| = 􏱾 1 + 17322 11.85 A(s)β(s) = ⇒0.125×104k <1 −4 104k 􏲃 s􏲄3 1 + 103 = 0.125 × 104k For stable operation, | A(jω180)β(jω180)| < 1 A(jω)β(jω) = 􏲃 |A(jω)β(jω)|=􏱾 ω2 􏱿3/2 11.86A(s)=􏲃 s􏲄􏲃 s􏲄2 1 + 104 1 + 105 104 ω 􏲄􏲃 ω 􏲄2 k < 8 × 10 1+j103 104 104k 􏲄 ω3 104 k 1 + 106 A(jω) = φ = −tan−1 − 2 tan−1 φ(ω) = −3 tan−1 ◦ −1 ω180 1+j104 1+j105 􏲃ω􏲄 􏲃ω􏲄 105 −1􏲃ω180 􏲄 − 2 tan 105 􏲃ω􏲄 103 􏲃 −180 = −3 tan 103 ⇒ω180=103tan60=1732rad/s 104 −1􏲃ω180 􏲄 −180 = −tan 104 By trial and error we find ω180 = 1.095 × 105 rad/s At this frequency, 104 | A| = √1 + 10.952√1 + 1.0952 = 413.6 For stable operation, |A|βcr <1 βcr < 2.42 × 10−3 Thus, oscillation will commence for −3 11.89 (a) The closed-loop poles become coincident when Q = 0.5. Using Eq. (11.70), we obtain √(1 + A0β)ωP1ωP2 ωP1 + ωP2 √(1 + A0β)ωP1ωP2 ωP1 + ωP2 Chapter 11–63 Q = 0.5 = β≥2.42×10 ⇒ 1 + A0β = 0.52 (ωP1 + ωP2)2 ωP1ωP2 = 0.52 × (2π)2(104 + 105)2 (2π)2 × 104 × 105 2 112 =0.5 × 10 =3.025 11.87 Af (0) = where A0 1+A0β β = 2.025 × 10−4 1 ωc = 2(ωP1 +ωP2) 1 A0 = 1MHz =105 V/V 10 Hz = 2 ×2π(fP1 +fP2) fc = 1 ×(104 +105)=5.5×104 Hz Thus, Af(0)= 1+105 ×0.1 ≃10V/V f3dB = 10(1 + A0β) =10(1+105 ×0.1)≃105 Hz Unity-gain frequency of closed-loop amplifier = Af (0)×f3dB =10×105 =106 Hz=1MHz Thus,thepoleshiftsbyafactorequaltothe amount-of-feedback, (1 + A0β). 11.88 A0 =10V/V fP =100Hz fHf=fP(1+A0β)=10kHz 10×103 ⇒ 1 + A0β = 100 = 100 2 (b) Af(0)= A0 105 1 + A0β 104 = 1+2.205×10−4 ×104 =3306V/V Af (s) = A(s) 1+A(s)β where A0 A(s)= 􏱾 s 􏱿􏱾 s 􏱿 1 + ω P1 = 1 + ω P2 Af (s) = 􏱾 1+ s 􏱿 􏱾 A0 􏱿 1+ s +A0β ωP2 (1+A0β)+s 1 + 1 + s ωP1 􏱾 A0 􏱿 2 ωP1 ωP2 ωP1ωP2 Af (jω) = 104 ωωωω ⇒β= 99 =0.0099V/V A0 􏱿 􏱾 􏱿􏱾 􏱿 ω + ω − ω ω 104 Af(jωc)= j6.05 104 | Af |(jωc) = 6.05 = 1653 V/V (c) Q=0.5. 􏱾 A0 1 + A0β A (s) = 100 f 1+s/2π×104 (1 + A0β) + j Af (jωc) = 3.025 + j(5.5 + 0.55) − 5.5 × 0.55 Af (0) = P1 P2 P1 P2 104 104 100 = 100 V/V A (s)= Af(0) = fs 1 + ωHf (d) If β = 2.025 × 10−3 V/V. Using Eq. (11.68), 11.91 we obtain Chapter 11–64 10R V C I3 12 s = −2 × 2π(104 + 105) ±1× 2􏲡 2π (104 +105)2 −4(1+104 ×2.025×10−3)×104 ×105 T(s)= Vr V1 can be determined as indicated in Fig. 1. We start at the left-hand side where the voltage is Vr . The current through (C/10) will be I1 =sCVr 10 This is the same current that flows through (10R); thus V2 = I1 × 10R + Vr =sC ×10RVr +Vr 10 = Vr (1 + sCR) The current I2 through R can now be found as I2 = V2 = Vr(1+sCR) RR The input current I3 is the sum of I and I2: I3 = s C Vr + Vr (1+sCR) 10 R V􏱾 C􏱿 I3= r 1+sCR+s R R 10 = Vr (1 + 1.1sCR) R TheinputvoltageV1 cannowbefoundas V1 =V2 + 1 I3 sC 1 (1+1.1 sCR) r r sCR 􏰀 I1 I1 I2 VCV 􏰀 r 10 R 1 􏰒􏰒 s = −5.5 × 104 Figure 1 Figure 1 shows the feedback circuit modified according to the specifications in this problem. This circuit replaces that in the feedback path in Figs. 13.34 (a) and (b). Its transfer function T (s), 2π 􏲇 ±0.5 121 × 108 − 4 × 21.25 × 109 Q = =1.325 4 4√ ± 0.5 × 10 121 − 40 × 21.25 = −5.5 × 10 =−5.5×104 ±j0.5×104 ×27 =(−5.5±j13.25)×104 Hz Using Eq. (11.70), we obtain 􏲇(1 + 104 × 2.025 × 10−3 )104 × 105 104 + 105 11.90 Af (0) = 10= 1000 A0 1 + A0β 1 + 1000β ⇒ β = 0.099 To obtain a maximally flat response, Q=0.707 Using Eq. (11.70), we obtain 􏲇 0.707 = 100×1×fP2 1 + fP2 1= 100fP2 2 (1 + fP2)2 f2 +2f +1=200f P2P2 P2 f2 −198f +1=0 P2 P2 fP2 ≃198kHz (the other solution is a very low frequency which obviously does not make physical sense). The 3-dB frequency of the closed-loop amplifier is f0, which can be obtained from Eq. (11.68) and the graphical construction of Fig. 11.32: f0 = 1(fP1 +fP2) 2Q 2 1 f0 = √2 (1 + 198) = 140.7 kHz = V (1+sCR)+V Finally, T (s) can be obtained as T(s) ≡ Vr V1 = s(1/CR) 2.1 􏱾 1 􏱿2 s2+s + CR CR Thus, L(s) = s2 + s(2.1/CR) + (1/CR)2 −s(K /CR) The characteristic equation is 1 + L(s) = 0 that is, 2 2.1􏱾1􏱿2 K s+s+−s=0 10 = K2 1+K2β = K2 2 ⇒K2 =20⇒K =√20=4.47V/V β= 1 =0.05V/V 20 UsingEq.(11.68),weobtain ω0 = 1(ωP1 +ωP2) Q2 f0 1 √ =2×2fP 1/ 2 ⇒fP =√2f0 =√2×1=1.414kHz A (s) = 10ω02 f ω0 Af (s) = + (2π × 103)2 1/ 2 11.93 Let each stage have the transfer function Chapter 11–65 CR CR CR 2 2.1−K 􏱾 1 􏱿2 s +s CR + CR =0 Thus, ω0 = 1 CR 1 Q = 2.1 − K For the poles to coincide, Q = 0.5, thus 0.5 = 1 s 2 + s Q + ω 02 10(2π × 103)2 2.1−K ⇒ K = 0.1 2π × 103 s2 + s √ For the response to become maximally flat, Q = 0.707: 0.707 = 1 2.1−K ⇒ K = 0.686 The circuit oscillates for K = 2.1. 11.92 Af = 10 Maximally flat response with fsdB = f0 = 1 kHz. Let each stage in the cascade have a dc gain K and a 3-dB frequency fP , thus √ andsubstitutingQ=1/√2,ω =ω =ω , P1 P2 P T(s) = K 1+s K2 1 + ωP A(s) = 􏱾 Using the expression for Q in Eq. (11.70), we get s 􏱿2 Q= (1+A0β)ωP1ωP2 ωP1 + ωP2 ωP whereωP =2π×100×103 rad/s ⎛ ⎞3 A(s)=⎜⎝ K ⎟⎠ β=1 Thus the characteristic equation is given by 1 + A(s)β = 0 K3 1 + 􏱾 s 􏱿3 = 0 1+ωP Tosimplifymatters,let s =S,whereSisa ωP normalized frequency variable, thus (1+S)3 +K3 =0 (1) This equation has three roots, which are the poles of the feedback amplifier. One of the roots will be real and the other two can be complex conjugate depending on the value of K. The real pole can be directly obtained from Eq. (1) as (1+S1)3 =−K3 (1 + S1) = −K S1 =−(1+K) (2) 1+s ωP and A0 = K2, we obtain 1 􏲇1+K2β √2= 2 1 + K2β = 2 and K2β = 1 Now, Af0= A0 1 + A0β Now we need to obtain the two other poles. The characteristic equation in (1) can be written as S3 +3S2 +3S+(1+K3)=0 (3) Equivalently it can be written as (S+1+K)(S2 +aS+b)=0 (4) Equating the coefficients of corresponding terms in (3) and (4), we can find a and b and thus obtain the quadratic factor S2 +(2−K)S+(1−K+K2)=0 (5) This equation can now be easily solved to obtain the pair of complex conjugate poles as Since the loop gain rolls off at a uniform slope of −20 dB/decade, it will reach the 0 dB line (| Aβ| = 1) five decades beyond 10 Hz. Thus the unity-gain frequency will be f1 =105 ×10=106 Hz=1MHz The phase shift will be that resulting from a single pole, −90◦, resulting in a phase margin: Phase margin = 180◦ − 90◦ = 90◦ 60° Chapter 11–66 11.95 105 A(s)=􏱾s􏱿􏱾s􏱿 √ S2,3=−1+K±j 3K 22 Im 1+ 1+ 2π × 10 β=0.01 Aβ(jω) = 2π × 103 3 Normalized s-plane 1000 􏱾 ω􏱿􏱾 ω􏱿 (1) 1+j2π ×10 1+j2π ×103 |Aβ|= K2K vP 􏲌 1000 (2) 3 S􏰔s 􏲋􏲋􏲟 􏱾 ω 􏱿2􏲠􏲟 􏱾 ω 􏱿2􏲠 􏰒(1􏰀K) 􏰒1 0 Re 􏲊 1+ 2π×10 1+ 2π×103 K 2 3 Figure 1 Figure 1 shows the root locus of the three poles in the normalized s plane (normalized relative to ωP = 2π × 105). Observe that as K increases the real pole S1 moves outwardly on the negative real axis. The pair of conjugate poles move on straight lines with 60◦ angles to the horizontal. These two poles reach the jω axis (which is the boundary for stable operation) at K = 2 at which point √ S2,3=±j 3 Thus the minimum value of K from which oscillations occur is K = 2. Oscillations will be at √ ω= 3×2πfP or √ f= 3×100kHz = 173.2 kHz 11.94 A0 = 105 with a single pole at fP =10Hz For a unity-gain buffer, β = 1, thus A0β=105 andfP =10Hz Figure 1 Figure 1 shows a sketch of the Bode plot for | Aβ|. Observe that the unity-gain frequency will occur on the −40 dB/decade line. The value of f1 can be obtained from the Bode plot as follows: The −40 dB/decade line represents gain drop proportional to 103/f 2. For a drop by only 20 dB (a factor of 10) the change in frequency is 103 1 f12 =10 f1 =√10×103 Hz=3.16×103 Hz However, the Bode plot results are usually approximate. If we require a more exact value for f1 we need to iterate a couple of times using the exact equation in (2). The result is f1 =3.085×103 Hz The phase angle can now be obtained using (1) as Peaking, P ≡ | Af (jω)| 1/β Thus, P = 1/|1 + e−jθ | = 1/|1+cos θ −j sin θ| 􏲡 =1/ (1+cosθ)2+sin2θ √ =1/ 2+2cosθ = 1/􏲇2 + 2 cos(180◦ − φ) 􏲇 P=1/ 2(1−cosφ) 􏱾1􏱿 ⇒ φ = cos−1 1 − 2P2 We use this equation to obtain the following results: follows: φ = −tan−1 = −tan−1 f1 10 ω1 2π × 103 ω1 2π × 10 − tan−1 − tan−1 f1 Chapter 11–67 103 = −tan−1(3.085 × 102) − tan−1(3.085) = −89.81◦ − 72.03◦ = −161.84 Thus, Phase margin = 180◦ − 161.84 = 18.15◦ To obtain a phase margin of 45◦: The phase shift due to the first pole will be ≃ 90◦. Thus, the phase shift due to the second pole must be ≃ −45◦, thus −45◦ = −tan−1 f1 103 f1 ≃ 103 rad/s. Since f1 is two decades above fP we need A0β to be 100. Thus, β will now be β = 100/105 = 10−3 Figure 2 shows a sketch of the Bode plot for | Aβ| in this case. 11.97 Figure 1 on the next page shows magnitude and phase Bode plots for the amplifier specified in this problem. From the phase plot we find that θ = −135◦ (which corresponds to a phase margin of 45◦) occurs at f = 3.16×105 Hz At this frequency, | A| is 70 dB. The β horizontal straight line drawn at 70-dB level gives 􏱾1􏱿 20 log β = 70 dB ⇒ β = 3.16 × 10−4 Correspondingly, 104 Af = 1+104 ×3.16×10−4 = 2.4×103 V/V or 67.6 dB 11.98 Figure 1 on the next page shows the Bode plot for the amplifier gain and for a differentiator. Observe that following the rate-of-closure rule the intersection of the two graphs is arranged so that the maximum difference in slopes is 20 dB/decade. P φ 1.05 56.9◦ 1.10 54.1◦ 0.1 dB ≡ 1.0115 59.2◦ 1.0 dB ≡ 1.122 52.9◦ 3 dB ≡ 1.414 41.4◦ 40 20 􏰒20 dB/decade b 􏰔 0.001 0 1 10 102 103 f Figure 2 f (Hz) 􏰒40 dB/decade 11.96 Using Eq. (11.82), we obtain | Af (jω1)| = 1/β 1 |1+e−jθ| θ = 180◦ − φ φ ≡ Phase margin where This figure belongs to Problem 11.97. Chapter 11–68 80 70 60 50 40 30 20 10 0 􏱔° 0 􏰒45° 􏰒90° 􏰒135° 􏰒180° 􏰒225° 􏰒270° 20 log 􏰒20 dB/decade 20 log(1/b) 􏰔 70 dB 􏰒40 dB/decade 􏰒60 dB/decade 104 Pole fP1 104 105 3.16 􏰁 105 106 107 fP1 fP2 fP3 f, Hz f, Hz Pole fP2 105 Pole fP3 3.16 􏰁 105 106 107 1 2pCR and the corresponding closed-loop gain is Af=A0= 105 1+A0β 1+105 ×3.16×10−5 = 2.4 × 104 V/Vor 87.6 dB and for PM = 45◦, we have 1 20logβ =80dB ⇒β=10−4 V/V Total phase This figure belongs to Problem 11.98. 11.99 Figure 1 is a replica of Fig. 11.37 except here we locate on the phase plot the points at which the phase margin is 90◦ and 135◦, respectively. Drawing a vertical line from each of those points and locating the intersection with the | A| line enables us to determine the maximum β that can be used in each case. Thus, for dB 60 􏰒20 dB/decade 􏰒40 dB/ decade 100 f (MHz) Figure 1 20 dB/decade 40 Bode plot for differentiator 20 PM = 90◦, we have 20log1 =90dB 0 0.001 0.01 0.1 1 10 β ⇒ β = 3.16 × 10−5 2π CR ⇒CR≥ 2π×103 =0.159ms Figure 1 Thus, to guarantee stability, 1 ≤ 0.001 MHz or 1 kHz 1 This figure belongs to Problem 11.99. dB 100 90 80 70 60 50 40 30 20 10 􏰒20 dB/decade X1 Chapter 11–69 20 log (a) 90o PM 45o PM 20 log 1/b 􏰔 85 dB (stable) 20 log 1/b for zero margins 20 log 1/b 􏰔 50 dB (unstable) f180o 10 102 10 102 0 103 104 105 w 0 􏰒45o 􏰒90o 􏰒135o 􏰒180o 􏰒225o 􏰒270o 25 dB gain margin 􏰒40 dB/decade X2 (b) 􏰒60 dB/decade 107 108 f(Hz) 107 108 f(Hz) 72o phase margin 106 f180o 104 105 90o PM 106 􏰒108o and the corresponding closed-loop gain is 11.101 We must move the 1 MHz pole to a new location, fD = 20MHz =2kHz 104 This reduction in frequency by a factor of 1 MHz = 500 will require that the total 2 kHz capacitance at the controlling node must become 500 times what it originally was. 11.102 Refer to Fig. 11.38. (a) For β = 0.001, 1 20 log β = 60 dB A horizontal line at the 60-dB level will intersect the vertical line at fP2 = 106 Hz at a point Z1. Drawing a line with a slope of −20 dB/decade A = A0 = f 1+A0β = 9.09 × 103 V/V or 79.2 dB 105 1+105 ×10−4 45o phase margin Figure 1 11.100 The new pole must be placed at fD = 1MHz =100Hz 104 In this way the modified open-loop gain will decrease at the uniform rate of −20 dB/decade, thus reaching 0 dB in four decades, that is at 1 MHz where the original pole exists. At 1 MHz, the slope changes to −40 dB/decade, but our amplifier will be guaranteed to be stable with a closed-loop gain at low as unity. from Z1 will intersect the 100-dB horizontal line at a frequency two decade lower than fP 2 , thus the frequency to which the 1st pole must be moved is ′ fP2 106 fD = 100 = 100 =10kHz (b) For β = 0.1, 20log1 =20dB β Following a process similar to that for (a) above, the first pole must be lowered to ′ 106 fD = 104 = 100 kHz 11.103 R1 =R2 =R C1 = 10C C2 = C Cf ≫C gm1R = 100 ωP1= ωP2 = gC ω′= mf jv log scale 10 1 0.1 0.01 0.01 s CR CR CR CR CfR Chapter 11–70 s-plane v P' 2 v P 2 v P 1 Figure 1 v ' P 1 11.104 The fourth, dominant pole must be at fP1 fD=A0 106 =105 =10Hz C 1 =0.1 10CR CR R 􏰔 1 M􏰵 􏰒 􏰀 1 M􏰵 Figure 1 (1) 1 (2) Refer to Fig. 1. fD=1 ω′= P1 R CR gm RCf R 100Cf 2π RC 10= 1 1=1 ω′ =0.01 Thus, P1 Cf R 2π × 1 × 106 × C ⇒ C = 15.9 nF (3) P2 C1C2 +Cf (C1 +C2) 1 11.105 fP1 = 2πC1R1 105= 1 2π ×150×10−12 ×R1 gmCf = gmCf 10C2 + 11CCf C(10C + 11Cf ) = Since Cf ≫ C, we have ⇒R1=10.61k􏱹 fP2=1 ω′ ≃ gmCf = gm P2 11CCf 11C Substituting gm = 100/R, we obtain ′ 100 10 2πC2R2 106 = 2π ×5×10−12 ×R2 1 ⇒ R2 = 31.83 k􏱹 ωP2 = 11CR ≃ CR Equations (1)–(4) provide a summary of pole First, we determine an approximate value of f ′ P2 from Eq. (11.94) (4) splitting: The two initial poles with frequencies gm Cf P2 2π[C1C2 + Cf (C1 + C2)] 0.1 1 CR and CR, a decade apart in frequency, are split further apart. The lower frequency pole is moved 0.01 to a frequency Cf R which is more than a decade lower (because Cf ≫ C) and the higher frequency pole is moved to a frequency 10 which is a CR decade higher. This is further illustrated in Fig. 1. f′= Assume that Cf ≫ C2, then gm P2 2π(C1 +C2) f′≃ = 2π(150 + 5) × 10−12 40 × 10−3 = 41.1 MHz which is much greater than fP3. Thus, we use fP3 to determine the new location of fP1, ′ 2×106 fP1 = 104 = 200 Hz Using Eq. (11.93), we obtain A(s)β(s) = −Vr Vt = A(s) 1/sC R+1/sC 105 1 Chapter 11–71 A(s)β(s) = CR = 0.01×10−6 ×100×103 = 10−3 s A(s)β(s)=􏲃 s􏲄􏲃 s􏲄 1+ 10 1+ 103 (a) Bode plots for the magnitude and phase of Aβ are shown in Fig. 2. From the magnitude plot we find the frequency f1 at which | Aβ| = 1 is f1 = 3.16×104 Hz (b) From the phase plot we see that the phase at f1 is 180◦ and thus the phase margin is zero. A more exact value for the phase margin can be obtained as follows: 3.16 × 104 103 f′= P1 1 2πgmR2Cf R1 1+ s 1+sCR 10 200 = 2π ×40×10−3 ×31.83×103 ×Cf ×10.61×103 105 1 Since Cf is indeed much greater than C2, the pole at the output will have the frequency already calculated: ⇒Cf =58.9pF ′ fP2 ≃ 41.1 MHz 11.106 Rs 􏰔 100 k􏰵 −1 3.16 × 104 −1 10 − tan 􏰀 θ(f1) = −tan = −89.98 − 88.19 = −178.2 A (s) 􏰒 􏰀􏰀 R 􏰔 100 k􏰵 C 􏰔 0.01 􏱌F Thus the phase margin is 1.8◦ . A(s) Vt 􏰒 Vr 􏰒 (c) Af (s) = 1 + A(s)β(s) 􏲃s􏲄 = 105/ 1+10 105 1+􏲃 s􏲄􏲃 s􏲄 Figure 1 This figure belongs to Problem 11.106, part (a). 1 + 10 1 + 103 100 80 60 40 20 0 0o 􏰒45o 􏰒90o 􏰒135o 􏰒180o 􏰒20 dB/decade 􏰒40 dB/decade f1 􏰔3.16􏰁104 Hz 1 10 fP1 100 1000 fP2 10,000 f (Hz) 100,000 Total phase Phase of Phase f (Hz) offP2 fP1 Figure 2 􏲃s􏲄 105 1 + 103 =􏲃s􏲄􏲃s􏲄 105 + 1+ 1+ Chapter 11–72 The pair of complex-conjugate poles have ω0 ≃31.62krad/s Q=31.3 Thus, the response is very peaky, as shown in 1 + 10−5(1 + 0.101s + 0.0001s2) Fig. 4. 􏲃10 1+ s 􏲄 103 = 103 At s = 0, Af ≃1 The transmission zero is sZ = −103 rad/s 2 The poles are the roots of 10−9s2 +1.01×10−6s+1=0 which are s=(−0.505±j31.62)×103 rad/s The poles and zero are shown in Fig. 3. Af 1000 1000 1 0 1.01 krad/s v0 􏰔 31.62 krad/s krad/s Figure 4 v Figure 3