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
1R2VR2V
1f11o
b Vf R1
Vo R1 R2
R2 0 R2
1 R1 2 1 R1 2
R11 R1R2
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
1R2VR2V
1f11o
b Vf R1
Vo R1 R2
R2 0 R2 1R1 21R1 2
R11 R1R2
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 Vg2gm2Vg2 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 RidmVtRL
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 = 1010.00625 + [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 2Vo
(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 −2S+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
101 100 101 vdifferentiator 1τ
|A|
“rate of closure” 20 dB/decade 20 dB/decade
Exercise 11–12
102
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
RidVrVtRidV1
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 =μRL ∥[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 1W
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 =10kincluded:
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 R1R2
R1R2
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
R11RE //RF
RF RE 0
RC1
R11
RF 1RE 2
R22RE 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
Afideal= =− 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 VV mV1 r Ridt1Rid
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
RF640
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 RE2100
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 V1RF 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
( b1)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
R11RF
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 =0.1kandAf =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
bVf 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 2Vo
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 QVrV V2
V2 1t
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 2Vo 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:
V1
A≡ o =−(gm1RD1) RF ∥ gm2(RD2 ∥RF)
Is gm1
This figure belongs to Problem 11.69, part (a).
Chapter 11–45
RC
50.356 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 =10kgives
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 RsV11.65V2V3RFRL
RF k 330V 1.65 330V2 1.65 1k 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 fV
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 10A
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 1W 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
IsVs 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 2Vo 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 R1R2 R2
1 R1 2
R22 R1R2 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
R11Rf RE
R22RERf
Figure 2
bIf 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
R11Rf 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
20,0002
20,00022 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 s2
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 ωω2
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
b1
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 103
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
s3
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 s2 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
−1ω180 − 2 tan 105
ω 103
−180 = −3 tan 103 ⇒ω180=103tan60=1732rad/s
104 −1ω180
−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 2.112 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)=ss
√ 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 Ss
ω 2 ω 2
(1K) 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 3.16104 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
=ss 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