CS计算机代考程序代写 Ex:10.1AM =− RG

Ex:10.1AM =− RG
RG +Rsig
10
=−10+0.1 ×2(10∥10)
fL= 1 􏱼1 + 1 + 1􏱽 2π τC1 τCE τC2
1􏱼1 1 1􏱽 = 2π 7.44 + 0.071 + 13
= 2.28 kHz fZ=1
3.45×10−11 F/m 10×10−9 m
gm(RD∥RL)
2π ×1×10−6(0.1+10)×106 = 0.016 Hz
1
fP3 = 2πCC2(RD + RL)
= 1
2π ×1×10−6(10+10)×103
= 8 Hz fZ= 1
×103
1
Since fZ is much lower than fL it will have a
Exercise 10–1
= −9.9 V/V fP1= 1
2πCC1(Rsig + RG) = 1
=
2πCERE
2π ×1×10−6 ×5×103
= 31.8 Hz
g + 1/R
negligible effect on fL. εox
fP2 = m (2+0.1)×10−3
S
2πCS
= 2π×1×10−6 =334.2Hz
Ex:10.3 Cox = t = ox
= 3.45 × 10−3 F/m2 = 3.45 fF/μm2
Cov =WLovCox
= 10×0.05×3.45 = 1.72 fF
Cgs = 2 WLCox + Cov 3
2πCSRS =13
2π ×1×10−6 ×10×103 = 15.9 Hz
Since the highest-frequency pole is fP2 = 334.2 and the next highest-frequency singularity is fZ at 15.9 Hz, the lower 3-dB frequency fL will be
= 2 ×10×1×3.45+1.72 = 24.72 fF
Cgd =Cov =1.72fF
= 􏲆 10
1 + 0.6
Csb = 􏲆 Csb0 VSB
1
= 6.1 fF
fL ≃fP2 =334.2Hz
Ex: 10.2 Refer to Fig. 10.10.
τC1 =CC1[Rsig +(RB ∥rπ)]
= 1×10−6[5+(100 ∥ 2.5)]×103
= 7.44 ms
􏱼 􏱾 R∥R􏱿􏱽
τCE=CE RE∥ re+ B sig β+1
β=gmrπ =40×2.5=100 re ≃1/gm =25􏱹
= 􏲆 10
1 + 2 + 1
0.6
1 + V 0
Cdb = 􏲆 Cdb0
1 + VDB
V0
􏲡
Ex: 10.4 gm = 2k′n(W/L)ID 􏲇
=4.1fF
= 2×0.16×(10/1)×0.1 = 0.566 mA/V fT= gm
100 ∥ 5􏱿􏱽 τCE =1×10−6 5∥ 0.025+ 101
τCE = 0.071 ms
τC2 =CC2(RC +RL)
= 1×10−6(8+5)×103 = 13 ms
2π(Cgs+Cgd) = 0.566×10−3
2π(24.72 + 1.72) × 10−15 fT =3.4GHz
Ex:10.5 Cde =τFgm where
τF =20ps
IC 1 mA
gm = V = 0.025V =40mA/V
T
􏱼 􏱾
×103

Thus,
Cde = 20×10−12 ×40×10−3 = 0.8 pF
fH= 1 2πCin(Rsig ∥ RG)
=1
2π × 4.26 × 10−12(0.01 ∥ 4.7) × 106
Exercise 10–2
Cje ≃ 2Cje0
= 2 × 20 = 40 f F
Cπ = Cde + Cje =0.8+0.04=0.84pF
C Cμ=􏱾 μ0 􏱿m
VCB 1 + V0c
= 3.7 MHz Ex:10.9 fH =
1 2πCin(Rsig ∥ RG)
1
2πCin(0.1 ∥ 4.7) × 106
=􏱾 1+
20
2􏱿0.33 =12fF
0.5
1×106 =
⇒ Cin = 1.625 pF
But,
Cin =Cgs +Cgd(1+gmR′L) 1.625 = 1+Cgd(1+7.14) ⇒ Cgd = 0.08 pF
Ex: 10.10 To reduce the midband gain to half the value found, we reduce R′L by the same factor, thus
R′L = 1.5 k􏱹
But,
R′L =ro ∥RC ∥RL
1.5 = 100 ∥ 8 ∥ RL
⇒RL =1.9k􏱹
Cin =Cπ +Cμ(1+gmR′L) = 7+1(1+40×1.5)
= 68 pF
fH = 1
2π Cin R′sig
1
= 2π × 68 × 10−12 × 1.65 × 103
= 1.42 MHz
Thus, by accepting a reduction in gain by a factor of 2, the bandwidth is increased by a factor of 1.42/0.754 = 1.9, approximately the same factor as the reduction in gain.
gm 2π(Cπ +Cμ)
fT =
= 2π(0.84 + 0.012) × 10−12
40×10−3
= 7.47 GHz
Ex: 10.6 | hfe | = 10 at f = 50 MHz Thus,
fT = 10×50 = 500 MHz
Cπ+Cμ= gm 2πfT
= 40 × 10−3 2π ×500×106
= 12.7 pF
Cπ = 12.7−2 = 10.7 pF
Ex: 10.7 Cπ = Cde + Cje
10.7 = Cde + 2
⇒ Cde = 8.7 pF
Since Cde is proportional to gm and hence IC, at IC =0.1mA,
Cde = 0.87 pF
and
Cπ = 0.87+2 = 2.87 pF
4×10−3
fT = 2π(2.87 + 2) × 10−12 = 130.7 MHz
Ex: 10.11 ft = |AM |fH 2×109 = 12.5
=−
=−4.7+0.01 ×7.14=−7.12V/V
Ex:10.8 A M
2π(CL+Cgd)×10×103 ⇒CL+Cgd =99.5fF
RG g R′ R +R m L
4.7
G sig
CL =99.5−5=94.5fF

Ex:10.12 T(s)=
GB = 1000×100×103 = 108 Hz
ωP1 ωP1 ⇒ωH =0.84ωP1
TheapproximatevalueofωH isobtainedfrom Eq. (10.77),
􏲈􏲉
11 ωH≃1 ω2+4ω2
P1 P1 ⇒ωH =0.89ωP1
Fork=4,theexactvalueofωH isobtainedfrom 􏲟 􏱾ω􏱿2􏲠􏲟 􏱾ω􏱿2􏲠
2=1+H1+H ωP1 4ωP1
Exercise 10–3
1000 􏱾ω 􏱿4 􏱾ω 􏱿2
s H+5H−4=0
1 + 2π × 105 AM
ω 􏱿 ωP2
Ex:10.13 T(jω)=􏱾 ω 􏱿􏱾 1+j 1+j
ωP1
􏲋􏲋 􏲟 􏱾 ω 􏱿 2 􏲠 􏲟 􏱾 ω 􏱿 2 􏲠
|T|=􏲌 |AM| 􏲊1+ω 1+ω
P1 P2
|T|= 􏲌 |AM|
􏲋􏲟 􏱾 􏱿􏲠􏲟 􏱾 􏱿􏲠
􏱾 􏱿2 􏱾 􏱿4 17ωH1ωH
􏲋ω2 ω2
1+ H kωP1
2=(1+0.81) 1+ ⇒k =2.78
ForωH =0.99ωP1, 􏱾
􏲊1+ 1+ ωP1
=1+16 ω +16 ω
kωP1 􏱾ω􏱿2
P1
H +17H −16=0 ωP1 ωP1
P1 􏲟􏲠􏲟􏲠 􏱾ω􏱿4􏱾ω􏱿2
􏱾ω􏱿2 2=1+ H
ωP1 For ωH = 0.9ωP1,
⇒ωH =0.95ωP1
The approximate value of ω
Eq. (10.77):
􏲈􏲉
ωH≃1 1+1 ω2 16ω2
= 0.97ωP1
is found from
􏱾 0.81􏱿
H
k2
0.992 􏱿 2=(1+0.992) 1+ k2
P1 P1
⇒k =9.88
􏲟 􏱾ω􏱿2􏲠􏲟 􏱾ω􏱿2􏲠 Ex:10.142= 1+ H 1+ H
Ex: 10.15 Cin = Cgs + Cgd (1 + gmR′L) = 20+5(1+1.25×10)
= 87.5 fF
fH= 1
2π Cin R′sig
=1
2π ×87.5×10−15 ×10×103
= 181.9 MHz
This is greater than the value obtained in Example10.8,fH =135.5MHz,by34%.The value obtained in Example 10.8 is a better estimateoffH asittakesintoaccountthe effect of CL.
Ex: 10.16
|AM|=gmR′L =1.25×10=12.5V/V GB = | AM | fH
= 12.5 × 135.5 = 1.69 GHz
1
Ex:10.17 |AM|= 2 ×12.5=6.25V/V
Rgs = 10 k􏱹
ωP1
􏲟 􏱾 􏱿2􏲠􏲟 􏱾 􏱿2􏲠
kωP1
Fork =1,
2= 1+ ωH
=1+H ωP1
1+ ωH ωP1
ωP1
􏲟 􏱾ω􏱿2􏲠2
⇒ωH =0.64ωP1
The approximate value using Eq. (10.77) is
􏲈􏲉
2 ωP1 ωH≃1 ω2=√2
P1
= 0.71ωP1
Fork=2,theexactvalueofωH isobtainedfrom 􏲟 􏱾ω􏱿2􏲠􏲟 􏱾ω􏱿2􏲠
2=1+H1+H ωP1 2ωP1
5􏱾ω 􏱿2 1􏱾ω 􏱿4
=1+H+H 4 ωP1 4 ωP1

Rgd =R′sig(1+gmR′L)+R′L
= 10(1+6.25)+5 = 77.5 k􏱹
RCL = R′L = 5 k􏱹
τgs = CgsRgs = 20×10−15 ×10×103 = 200 ps τgd = Cgd Rgd = 5×10−15 ×77.5×103 = 387.5 ps τCL = CLRCL = 25×10−15 ×5×103 = 125 ps τH = τgs + τgd + τCL
= 200 + 387.5 + 125
= 712.5 ps
fH =1 2π τH
Ex: 10.19 (a) gm = 40 mA/V rπ = 200 = 5 k􏱹
40
ron = VAn = 130 =130k􏱹
I1
rop = |VAp| = 50 =50k􏱹
I1
R′L =ron ∥rop =130∥50=36.1k􏱹
AM =− rπ gmR′L rπ +rx +Rsig
5
=−5+0.2+36 ×40×36.1
= −175 V/V
(b) Cin =Cπ +Cμ(1+gmR′L) = 16+0.3(1+40×36.1)
= 450 pF R′sig=rπ∥(rx+Rsig)
= 5 ∥ (0.2 + 36) = 4.39 k􏱹
fH= 1
2π Cin R′sig
=1 2π×450×10−12 ×4.39×103
= 80.6 kHz
(c) Rπ =R′sig =4.39k􏱹
Rμ = R′sig(1 + gmR′L) + R′L
= 4.39(1 + 40 × 36.1) + 36.1
= 6.38 M􏱹
RCL = R′L = 36.1 k􏱹
τH = Cπ + CμRμ + CLRCL
= 16×4.39+0.3×6.38×103 +5×36.1 = 70.2 + 1914 + 180.5
= 2164.7 ns
fH= 1
2π × 2164.7 × 10−9
Exercise 10–4
= 1
2π × 712.5 × 10−12
= 223.4 MHz GB = 6.25 × 223.4 = 1.4 GHz
􏲆
Ex:10.18g = 2μC WI m n ox L D
Since ID is increased by a factor of 4, gm doubles: gm = 2×1.25 = 2.5 mA/V
′
R′L = 1 × 10 = 2.5 k􏱹 4
|AM|=gmR′L =2.5×2.5=6.25V/V Rgs = R′sig = 10 k􏱹
Rgd =R′sig(1+gmR′L)+R′L
= 10(1 + 6.25) + 2.5
= 75 k􏱹
RCL = R′L = 2.5 k􏱹
τH = τgs + τgd + τCL
= CgsR′sig + Cgd Rgd + CLRCL
= 20×10−15 ×10×103 +5×10−15 ×75
×103 +25×10−15 ×2.5×103 = 200+375+62.5
=637.5ps
1
fH = 2π ×637.5×10−12 = 250 MHz
GB = | AM | fH
= 6.25×250 = 1.56 GHz
Since RL is ro/2, increasing ID by a factor of four results in ro and hence R′L decreasing by a factor of4,thus
= 73.5 kHz (d) fZ = gm
2πCμ −3
40 × 10
2π × 0.3 × 10−12
=
(e) GB = 175×73.5 = 12.9 MHz
= 21.2 GHz

Ex:10.20 Rin = RL +ro 1+gmro
τH =CgsRsig +Cgd ×21Rsig = Cgs Rsig + 0.25Cgs × 21Rsig = 6.25CgsRsig
fH = 1
2π × 6.25Cgs Rsig
For the cascode amplifier,
τH ≃ Rsig[Cgs1 + Cgd1(1 + gm1Rd1)] where
Rd1 =ro1 ∥Rin2 =ro ∥ ro +ro gm ro
500 + 20 = 1+25
Gv = RL Rsig + Rin
= 20 k􏱹
= 500 10 + 20
=16.7V/V
Exercise 10–5
Rgs = Rsig ∥ Rin = 10 ∥ 20 = 6.7 k􏱹 Rgd = RL ∥ Ro
= 500 ∥ 280 = 179.5 k􏱹
τH =CgsRgs +(Cgd +CL)Rgd
= 20×10−15 ×6.7×103 +(5+25)×10−15 ×179.5×103
=134+5385 = 5519 ps
has been substantially lowered (by a factor of 9.4). Thus, the high-frequency advantage of the CG amplifier is completely lost!
Ex: 10.21 (a) ACS = −gm(RL ∥ ro) = −gm(ro ∥ ro) = −1gmro
2
1
=− ×40=−20V/V
2
Acascode = −gm(RL ∥ Ro) = −gm(ro ∥ gmroro)
≃ −gmro = −40 V/V Thus,
τH = CgsRgs + Cgd Rgd where
2
2 g ro
=ro∥ = m
gm 2 +ro
gm
2ro ro 􏲃
1
2π × 5519 × 10−12
2ro
=2+gr =2+40=21
fH =
substantially (by a factor of 9.7), the bandwidth
m o = CgsRsig
􏱼
=28.8MHz
Thus, while the midband gain has been increased
gmro 􏲄􏲦 21
fH(CS) 1.73
(c) ft (cascode) = 2 × 3.6 = 7.2
ft (CS)
Ex: 10.22 gm = 40 mA/V rπ = β = 200 =5k􏱹
gm 40
Rin =rπ +rx =5+0.2=5.2k􏱹
A0 =gmro
= 40×130 = 5200 Ro1 =ro1 =130k􏱹
τH
= Cgs Rsig 1 + 0.25
fH= 1
􏲥
1 + 0.25 􏱾
1 +
40 􏱿􏱽
1 + 21 2π × 1.73 CgsRsig
Thus,
fH (cascode) = 6.25 = 3.6
= 1.73CgsRsig
Acascode ACS
=2
(b) For the CS amplifier,
Rgs = Rsig
R =R (1+g R′)+R′
ro2 + RL ro2+RL/(β2+1)
gd sig m L ≃Rsig(1+gmR′L)
􏱾 1 =Rsig 1+2gmro
L
=21Rsig
Rin2 =re2
=25 130+50
􏱿
130+ 50 201
= 35 􏱹
Ro ≃β2ro2 =200×130=26M􏱹
􏱾1􏱿 =Rsig 1+ 2 ×40

AM = − rπ gm(Ro ∥ RL) rπ +rx +Rsig
5
= − 5 + 0.2 + 36 40(26000 ∥ 50)
AM =−242V/V
R′sig =rπ1 ∥(rx1 +Rsig)
= 5 ∥ (0.2 + 36) = 4.39 k􏱹
Rπ1 = R′sig = 4.39 k􏱹
Rc1 = ro1 ∥ Rin2
= 130 k􏱹 ∥ 35 􏱹 ≃ 35 􏱹
R =R′ (1+g R )+R μ1 sig m1 c1 c1
= 4.39(1 + 40 × 0.035) + 0.035
=10.6k􏱹
τH = Cπ1Rπ1 +Cμ1Rμ1 +Cπ2Rc1 +(CL +Cμ2)(RL ∥Ro)
= 16×4.39+0.3×10.6+16×0.035 + (5 + 0.3)(50 ∥ 26,000)
= 70.24 + 3.18 + 0.56 + 264.5 = 338.5 ns
fH = 1 =470kHz 2π ×338.5×10−9
ft =|AM|fH =242×470=113.8MHz
Thus, in comparison to the CE amplifier of Exercise 10.19, we see that | AM | has increased from175V/Vto242V/V,fH hasincreasedfrom 73.5 kHz to 470 kHz, and ft has increased from 12.9 MHz to 113.8 MHz.
Ex: 10.24 From Example 10.11, we get τH =b1 =104ps
1 fH = 2πτ
H
= 1 =1.53GHz 2π × 104 × 10−12
This is lower than the exact value found in Example 10.11 (i.e., 1.86 GHz) by about 18%, still not a bad estimate!
Ex:10.25 gm =40mA/V re = 25 􏱹
rπ = β = 100 =2.5k􏱹 gm 40
R′sig = Rsig +rx = Rsig = 1 k􏱹
R′L =RL ∥ro =1∥100=0.99k􏱹
τH=2πf =2π×1×106=159.2ns H
πe
Exercise 10–6
AM =
R′L
R ′L + r e + β + 1
R′sig 0.99 + 0.025 + (1/101)
=
Cπ +Cμ = 2πfT
40×10−3
= 2π × 400 × 106
= 15.9 pF
Cμ = 2 pF Cπ = 13.9 pF
0.99
= 0.97 V/V
gm
TohavefH equalto1MHz, 1 1
1
fZ=2πCr =2π×13.9×10−12×25
= 458 MHz
􏲟 􏲣R′􏲤􏲠 􏲟 􏲣R′􏲤􏲠
1
Thus,
159.2 = 70.24 + 3.18 + 0.56
+(CL +Cμ)(50∥26000) ⇒ CL + Cμ = 1.71 pF
Thus,CL mustbereducedto1.41pF. Ex: 10.23 From Eq. (10.120), we obtain
R′sig+ Cπ +CL 1+ sig R′L rπ
′ R′ 1+ RL + sig
= 0.99 1 1++
R = Rsig
gs gmR′L +1
R′L gmR′L +1
= Rsig + R′L gmR′L +1
= 2.66 × 10−9 s
Rgd =Rsig R= L
+
Cπ +Cμ 1+ L re
b1=
􏱼 􏱾
re rπ
13.9+2 1+ 0.025 ×1+(13.9+0)0.99
0.99 􏱿􏱽
0.025 2.5
b = CπCμR′LR′sig = 13.9×2×0.99×1
2 R′R′ 0.991 1+ L + sig 1+0.025+2.5
re rπ
−18 =0.671×10
R′
CL gmR′L +1

ωP1 and ωP2 are the roots of the equation 1+b1s+b2s2 =0
Solving we obtain,
fP1 = 67 MHz
Ex: 10.28 Ad = gm1,2(ro2 ∥ ro4) where
gm1,2 = 0.5 = 20 mA/V 0.025
100
ro2 =ro4 = 0.5 =200k􏱹
Ad = 20(200 ∥ 200) = 2000 V/V
The dominant high-frequency pole is that
fP2 =563MHz Since fP1 ≪ fP2,
fH ≃fP1 =67MHz
0.4 =
⇒VOV =0.2V
gm=2ID =2×0.4=4mA/V VOV 0.2
(b) Ad =gm(RD ∥ro) where
ro = VA = 20 =50k􏱹 ID 0.4
Ad =4(5∥50)=4×4.545 = 18.2 V/V
(c) fH = 1
2π(CL + Cgd + Cdb)(RD ∥ ro)
=1 2π(100+10+10)×10−15 ×4.545×103
= 292 MHz
(d) τgs = CgsRsig = 50×10 = 500 ps
τgd = CgdRgd = Cgd[Rsig(1+gmR′L)+R′L]
= 10[10(1 + 18.2) + 4.545]
= 1965.5 ps
τCL =(CL +Cdb)R′L =110×4.545=500ps τH = τgs + τgd + τCL
= 500+1965.5+500 = 2965.5 ps
fH = 1
2π × 2965.5 × 10−12
Exercise 10–7
introduced at the output node, 1􏱾W􏱿 1
Ex: 10.26 (a) I D1,2
=
μ C V 2 fH =
2 n ox L 1=1
1,2
OV
2πCL(ro2 ∥ro4)
2π ×2×10−12 ×100×103
= 0.8 MHz
Ex: 10.29 (a) AM = −gmR′L
where
R′L =RL ∥ro =20∥20=10k􏱹
AM =−2×10=−20V/V
τH =CgsRgs +CgdRgd +CLR′L
= CgsRsig + Cgd [Rsig(1 + gmR′L) + R′L] + CLR′L = 20×20+5[20(1+20)+10]+5×10
= 400+2150+50
= 2600 ps
×0.2×100V2 2 OV
1 fH = 2πτH
= 1
2π × 2600 × 10−12
= 53.7 MHz Ex:10.27 fZ = 1
Ro ≃ ro(1 + gmRs)
= 20 × 3 = 60 k􏱹
R′L =RL ∥Ro =20∥60=15k􏱹 AM =−GmR′L
= −0.67 × 15 = −10 V/V
Rgd =Rsig(1+GmR′L)+R′L
= 20(1+10)+15
= 235 k􏱹
= 61.2 MHz GB = | AM | fH = 20 × 61.2 = 1.22 GHz
(b) Gm = gm 1+gmRs
= 2 1+2
= 0.67 mA/V
2πRSSCSS =1
2π ×75×103 ×0.4×10−12 = 5.3 MHz
Thus, the 3-dB frequency of the CMRR is 5.3 MHz.

RCL = R′L = 15 k􏱹
Rsig + Rs + RsigRs/(ro + RL)
where
Rs = 2 = 1 k􏱹
gm
fP2 =
1 2πCμRL
Exercise 10–8
Rgs= 􏱾r􏱿 1+gRo
1 2π×2×10−12 ×10×103
m
s
ro +RL
=
= 8 MHz
T(s)=􏱾 s 􏱿􏱾 s 􏱿
20 × 1 20 + 1 + 20 + 20
1 + ω P2
50
P1 √
Atω=ωH,|T|=50/ 2,thus
􏲟 􏱾ω 􏱿2􏲠􏲟 􏱾ω 􏱿2􏲠
2=1+H 1+H ωP1 ωP2
1 + ω P1
|T(jω)|=􏲌􏲋􏲋􏲟
􏲊1+ ω 1+ ω
Rgs = 20 1+2×20+20
=10.75k􏱹
τH = CgsRgs + Cgd Rgd + CLRCL
=20×10.75+5×235+5×15 =215+1175+75=1465ps
1
fH = 2π×1465×10−12 =109MHz
GB = 10 × 109 = 1.1 GHz
Ex: 10.30 Refer to Fig. 10.42(b).
2rπ 1 AM=2r+R ×2×gmRL
π sig where
gm = 20 mA/V
50 􏱾 ω 􏱿2􏲠􏲟
􏱾 ω 􏱿2􏲠 P2
= 1 + ω4
ωH ωP1
+
􏱾 1
ωH + ωH ωH ωP2 ωP1 ωP2
1 􏱿
􏱾 􏱿2 􏱾 􏱿2 􏱾 􏱿2􏱾 􏱿2
H +ω2 + −1=0 ω2ω2 Hω2 ω2
P1 P2 P1 P2 4 􏱾 􏱿
fH +f2 1 + 1 −1=0 f2f2Hf2f2
􏱿
100 20
f411
H +f2 + −1=0
P1 P2 P1 P2
􏱾
rπ = AM =
fP1 =
= 6.4 MHz
= 5 k􏱹
10 ×1×20×10=50V/V
(2rπ ∥Rsig)
6.42 × 82 H 6.42 82
⇒ fH = 4.6 MHz (Exact value)
Using Eq. (10.164), an approximate value for fH can be obtained:
􏲉
10+10 2 1
2π
= 􏱾6 􏱿
􏱾C 􏱿 π +Cμ
211
1
2π 2+2 ×10−12(10∥10)×103
fH≃1/ f2 +f2 P1 P2
􏲆 1 1
=1/ 6.42 +82 =5MHz

10.1 Refer to Fig. 10.3(b). Is = gmVgYS V R YS +gm
g=G 􏱾1􏱿
Chapter 10–1
Vsig RG+Rsig+ 1 sCC1
where
RG =RG1 ∥RG2 =2M􏱹∥1M􏱹=667k􏱹 and
Rsig = 200 k􏱹
gm +sCS Is= RS
Vg
= gm
gm + 1 + sCS RS
s+1/CSRS
s+ Thus,
gm + 1/RS CS
Vg = RG
Vsig RG+Rsig s+
Thus,
f= 1
s
1 CC1(RG + Rsig)
fP2=gm+1/RS 2πCS
fZ = 1 2πCSRS
where
gm =5mA/VandRS =1.8k􏱹 To make fP2 ≤ 100 Hz,
gm + 1/RS ≤ 100 2π CS
5×10−3 +(1/1.8×103)
⇒ CS ≥ 2π ×100 = 8.8 μF
SelectCS =10μF.
Thus,
fP2 = 5×10−3 +(1/1.8×103) = 88.4 Hz 2π ×10×10−6
=8.84Hz
P1 2πCC1(RG + Rsig) We required
fP1 ≤ 10 Hz
thus we select CC1 so that
1 2πCC1(RG +Rsig)
≤10 1
CC1 ≥
⇒CC1 =20nF
10.2 Refer to Fig. 10.3(b). V =−I RD
=18.4nF
2π ×10×(667+200)×103
o d 1 RD+sC +RL
×R L
and
f = 1
Vo = − RDRL fP3 = 1
s Id RD+RLs+ 1
AM =− RG ×gm(RD ∥RL) RG +Rsig
where
C2
Z 2π × 10 × 10−6 × 1.8 × 103 10.4 Refer to Fig. 10.3.
CC2(RD +RL) 2πCC2(RD + RL)
where
RD =10k􏱹andRL =10k􏱹 TomakefP3 ≤10Hz,
1
2πCC2(RD + RL) ≤ 10
⇒CC2 ≥ 1
2π ×10×(10+10)×103
Select, CC2 = 0.8 μF. 10.3 Refer to Fig. 10.3(b).
RG =RG1 ∥RG2 =47M􏱹∥10M􏱹
=8.246M􏱹
Rsig =100k􏱹, gm =5mA/V, RD =4.7k􏱹and RL = 10 k􏱹.
=0.8μF
Thus,
A =− 8.426 ×5(4.7∥10)
M 8.426+0.1 = −15.8 V/V
Vg
Is = 1
g +ZS m
fP1 = 1 2πCC1(RG + Rsig)
=1
2π ×0.01×10−6(8.426+0.1)×106
=1.9Hz

fP2 = gm +1/RS 2π CS
5 × 10−3 + 0.5 × 10−3 = 2π × 10 × 10−6
fZ= 1 2πCSRS
(e) The Bode plot for the gain is shown in Fig. 2.
Chapter 10–2
= 87.5 Hz
Gain, dB
20 dB/decade
= 1 =8Hz 2π ×10×10−6 ×2×103
26 dB
6 dB
20 log 20 􏰔 26 dB
f 􏰔 100 Hz Figure 2
fP3 = 1 2πCC2(RD + RL)
f 􏰔 10 Hz
f (log scale)
=
1
2π × 1 × 10−6(4.7 + 10) × 103
= 10.8 Hz
Vo RD
Observe that the dc gain is 6 dB, i.e. 2 V/V. This makes perfect sense since from Fig. 1 we see that at dc, capacitor CS behaves as open circuit and the gain becomes
DCgain=− = −2 V/V
10.6 See figure on next page. Replacing the MOSFET with its T model results in the circuit shown in the figure.
AM =− RG ×gm(RD ∥RL) RG +Rsig
= − 2 × 3(20 ∥ 10) 2+0.5
= −16 V/V
To minimize the total capacitance we select CS so as to place fP2 (usually the highest-frequency low-frequency pole) at 100 Hz. Thus,
100= gm 2πCS
Since
fP2 ≫ fP1, fP3, fZ , fL ≃fP2 =87.5Hz
10.5
Is
Vi 􏰀􏰒
Figure 1
RD =−􏱾10k􏱹􏱿 1 +R 1+4.5
RS
CS
gS2 m
Id 􏰔 Is
1
gm
Replacing the MOSFET with its T model results in the circuit shown in Fig. 1.
3 × 10−3 2πCS
V= (a)A ≡ o=−gR
MVmD i
⇒CS =4.8μF Select CS = 5 μF.
Of the two remaining poles, the one caused by CC2 has associated relatively low-valued resistances (RD and RL are much lower than RG), thus to minimize the total capacitance we place fP3 at10HzandfP1 at1Hz.Thus,
10= 1 2πCC2(RD + RL)
1
≡ 2πCC2(20 + 10) × 103
⇒C =0.53μF C2
−20 = −2 × RD
⇒RD =10k􏱹
(b) fP = gm + 1/RS 2πCS
2 × 10−3 + (1/4.5 × 103) 2πC
100=
⇒CS =3.53μF
(c) fZ = 1 2πCS RS
S
=
2π ×3.53×10−6 ×4.5×103
1 (d) SincefP ≫fZ,
f ≃f =100Hz L P
=10Hz
Select CC2 = 1 μF. 1= 1
2πCC1(RG + Rsig)

This figure belongs to Problem 10.6.
Chapter 10–3
Id 􏰔 Is
CC2
Vo RD Io RL
Rsig CC1
Vsig 􏰀􏰒
1
0
RG
Is
1/gm CS
1 =
To make CE responsible for 80% of fL, we use 1 =0.8ωL =0.8×2πfL
2πCC1(2 + 0.5) × 106 ⇒ CC1 = 63.7 nF
τCE
⇒ τCE =
Select CC1 = 100 nF = 0.1 μF.
With the selected capacitor values, we obtain
1 ≃ 2 ms 0.8×2π ×100
1
fP1 = 2π ×0.1×10−6 ×2.5×106 = 0.64 Hz
3×10−3
fP2 = 2π×5×10−6 =95.5Hz
Thus,
􏱼
CE 250+
(200 ∥ 20)×103 􏱽
101 =2×10−3
fZ =0(dc)
fP3 = 1
= 5.3 Hz
⇒CE =4.65μF Select, CE = 5 μF.
Using the information in Fig. 10.10(a), we determine τC1 as
τC1 =CC1[(RB ∥rπ)+Rsig]
To make the contribution of CC1 to the
determination of fL equal to 10%, we use 1 =0.1ωL =0.1×2πfL
1 =15.92ms 0.1×2π ×100
2π ×1×10−6(20+10)×103 Since fP2 ≫ fP1 and fP3, we have
fL ≃ fP2 = 95.5 Hz
10.7 The amplifier in Fig. P10.7 will have the equivalent circuit in Fig. 10.9 except with
RE = ∞ (i.e. omitted). Also, the equivalent circuits in Fig. 10.10 can be used to determine the three short-circuit time constants, again with
RE = ∞. Since the amplifier is operating at IC ≃IE =100μA=0.1mAandβ=100,
re = 25mV =250􏱹 0.1 mA
gm = 0.1mA =4mA/V 0.025 V
β 100
rπ=g = 4 =25k􏱹
m
Using the equivalent circuit in Fig. 10.10(b), we get
􏱾R∥R􏱿 τCE=CE re+ B sig
τC1 ⇒τC1 =
β+1
CC1[(200 ∥ 25)×103 +20×103] = 15.92×10−3
⇒CC1=0.38μF
Select CC1 = 0.5 μF.
For CC2 we use the information in Fig. 10.10(c) to determine τC 2 :
τC2=CC2(RC+RL)
To make the contribution of CC2 to the
determination of fL equal to 10%, we use 1 = 0.1ωL = 0.1 × 2π fL
τC2
⇒τC2 = 1 =15.92ms 0.1×2π×100
Thus,

Thus,
CC2(20 + 10) × 103 = 15.92 × 10−3
⇒CC2 =0.53μF
Although, to be conservative we should select CC2 = 1 μF; in this case we can select
CC2 =0.5μF
because the required value is very close to 0.5 μF and because we have selected CC1 and CE larger than the required values. The resulting fL will be
fL = 1 􏱼 1 + 1 + 1 􏱽 2π τCE τC1 τC2
Using the method of short-circuit time constants and the information in Fig. 10.10, we obtain
τC1 =CC1[(RB ∥rπ)+Rsig] = CC1(Rin + Rsig)
= 1×10−6(5.7+5)×103 = 10.7 ms 􏱼 􏱾 R∥R􏱿􏱽
τCE=CE RE∥ re+ B sig β+1
Chapter 10–4
= 20 × 􏱼
10−6 3.9×103 ∥ = 2.2 ms
􏱾
83.3+
(33 ∥ 22 ∥ 5) × 103 􏱿􏱽 121
􏱼
τCE = 5 × 10−6 × 250 + = 2.15 ms
(200 ∥ 20)×103 􏱽 101
τC2 =CC2(RC +RL)
= 1×10−6(4.7+5.6)×103 = 10 ms
τC1 = 0.5×10−6[(200 ∥ 25)×103 +20×103] =21.1ms
τC2 = 0.5×10−6(20+10)×103 = 15 ms
1 􏱼103 103 103􏱽 fL =2π 2.15+21.1+ 15
= 92.2 Hz
which is lower (hence more conservative) than
the required value of 100 Hz. Ctotal =5+0.5+0.5=6.0μF
10.8 Refer to Fig. 10.9. In the midband,
Rin =RB1 ∥RB2 ∥rπ
where
RB1 =33k􏱹, RB2 =22k􏱹
gm = IC = 0.3mA =12mA/V VT 0.025 V
re = VT = 25mV =83.3􏱹 IE 0.3 mA
rπ = β =120=10k􏱹 gm 12
Thus,
Rin =33∥22∥10=5.7k􏱹
AM =− Rin gm(RC ∥RL) Rin + Rsig
where
Rsig =5k􏱹, RC =4.7k􏱹, andRL =5.6k􏱹 Thus,
AM =− 5.7 ×12(4.7∥5.6) 5.7+5
= −16.3 V/V
fL ≃ 1 2π
􏱾􏱿
1 + 1 + 1 τC1 τCE τC2
1􏱾 1
= 2π 10.7×10−3 + 2.2×10−3 + 10.3×10−3
1 1 􏱿 10.9 Refer to the data given in the statement for
Problem 10.8.
RB =RB1 ∥RB2 =33k􏱹∥22k􏱹=13.2k􏱹 IC ≃IE ≃0.3mA
gm = IC = 0.3mA =12mA/V VT 0.025 V
re = VT = 25mV =83.3􏱹 IE 0.3 mA
β 120 rπ=g=12=10k􏱹
m
From Fig. 10.10, we have τC1=CC1[(RB∥rπ)+Rsig]
For CC 1 to contribute 10% of fL , we use
1
τ =0.1ωL =0.1×2πfL
C1
= 0.1×2π ×50
⇒τC1 =31.8ms
Thus, CC1[(13.2∥10)+5]×103=31.8×10−3 ⇒CC1 =3μF
τC2 =CC2(RC +RL)
For CC2 to contribute 10% of fL, we use
1 = 0.1ωL = 0.1 × 2πfL τC2
= 0.1×2π ×50 ⇒τC2 =31.8ms
= 102.7 Hz

Thus,
CC2(4.7 + 5.6) × 103 = 31.8 × 10−3 ⇒CC2 =3.09μF≃3μF
Finally,
􏱼 􏱾 R∥R􏱿􏱽 τCE=CE RE∥ re+ B sig
For CC1 and CE to contribute equally to fL,
τC1 = τCE
Thus,
CC1 × 5.91 × 103 = CE × 41.8
Chapter 10–5
⇒CE =141.4 β+1 CC1
For CE to contribute 80% of fL, we use 1 = 0.8ω = 0.8 × 2π f
τ L L CE
10.11 Replacing the BJT with its T model results in the equivalent circuit shown in Fig. 1 below.
(a) At midband, CE and CC act as short circuits. Thus
Rin =(β+1)re
Vo =− (β+1)re g(R ∥R)
V (β+1)r+R m C L 121 sig esig
= 0.8×2π ×50 ⇒ τCE = 3.98 ms Thus,
􏱼 􏱾 (13.2∥5)×1000􏱿􏱽
CE3900∥83.3+ =3.98×10−3
⇒CE =36.2μF
10.10 Using the information in Fig. 10.10, we get
τC1 =CC1[(RB ∥rπ)+Rsig]
=− β(RC ∥RL) (β + 1)re + Rsig
(b) Because the controlled current source αIe is ideal, it effectively separates the input circuit from the output circuit. The result is that the poles caused by CE and CC do not interact. The pole duetoCE willhavefrequencyωPE:
1
= CC1[(10 ∥ 1)+5]×103 3
ωPE=􏱼 R􏱽 CE re+ sig
=CC1×5.91×10
􏱼 􏱾 R∥R􏱿􏱽
β+1
and the pole due to CC will have a frequency ωPC
τCE=CE RE∥ re+ B sig β + 1
where
re = rπ
β + 1
τ = CE􏱼
= 1000 ≃10􏱹 101
(10 ∥ 5) × 1000􏱿􏱽 101
ωPC = 1 CC(RC +RL)
(c) The overall voltage transfer function can be expressed as
CE 1.5×103 ∥ = CE × 41.8
􏱾
10+
Vo =A s s V Ms+ω s+ω
This figure belongs to Problem 10.11, part (a).
aIe 􏰀
VbIe re CE
􏰒
Zin
CC
sig
RC
PE PC
Vo
Rsig
Vsig 􏰀 􏰒
RL
Figure 1

(d)re=VT =25mV=25􏱹 IE 1 mA
Replacing the BJT with its T model results in the circuit shown in Fig. 1.
A = − M
100(10 k􏱹 ∥ 10 k􏱹) 101×25×10−3 k􏱹+10 k􏱹
Vsig
re + Re + sCE
Chapter 10–6
(a) Ie =
Vo =−αIeRC
1
= −40 V/V
(e) To minimize the total capacitance we choose to make the pole caused by CE the dominant one and make its frequency equal to fL = 100 Hz,
Thus,
Vo =− αRC s (1)
Vsig re+Re s+ 1 CE(re +Re)
From this expression we obtain
αRC
AM =−r +R (2)
ee
and
1
fL =fP = 2πCE(re +Re) (3)
(b) From Eq. (2) we see that
| AM | = αRC 1
re 1 + Re re
Thus,includingR reducesthegainmagnitudeby 􏱾e􏱿
the factor 1 + Re . re
2π × 100 = 􏱼 1
CE 25+10,000
101
Placing the pole due to CC at 10 Hz, we obtain
2π × 10 = 1
CC(10 + 10) × 103
⇒CC =0.8μF
(f) A Bode plot for the gain magnitude is shown
in Fig. 2.
􏱽
⇒ CE = 12.83 μF
(c) From Eq. (3), we obtain fL=1 1
2πCEre 1+Re re
Thus, including R reduces f by the factor
Figure 2
The gain at fP2 (10 Hz) is 12 dB. Since the gain decreases by 40 dB/decade or equivalently
12 dB/octave, it reaches 0 dB (unity magnitude) atf =fPC/2=5Hz.
1 + Re . This is the same factor by which the re
magnitude of the gain is reduced. Thus, Re can be used to tradeoff gain for decreasing fL (that is, increasing the amplifier bandwidth).
(d) I =0.25mA, RC =10k􏱹, CE =10μF re=VT = 25mV =100􏱹
I 0.25 mA For Re = 0:
|AM|= αRC ≃ 10k􏱹 =100V/V re 100 􏱹
􏱾􏱿
eL
10.12
Vsig 􏰀􏰒
Vo
aIe
re Re CE
RC
fL =
1 = 159.2 Hz 2π ×10×10−6 ×100
Ie
To lower fL by a factor of 10, we use
1+Re=10 re
⇒Re =900􏱹
The gain now becomes
|AM|= 100 = 100 =10V/V 1 + Re 10
re
Figure 1
See Fig. 2 for the Bode plot.

Gain (dB)
Cdb = 􏲆 Cdb0
|VDB| V0
Chapter 10–7
40 dB
20 dB
Re 􏰔 0
15.9 Hz 159 Hz
1+ =􏲆 20
Re 􏰔 900 􏰵
=9.4fF gm
1 + 2.5 0.7
f (Hz) (log scale)
fT =
2π(Cgs + Cgd ) 1.3 × 10−3
Figure 2
= 2π(61.6 + 4.3) × 10−15 = 3.1 GHz 10.14 gm = 2ID = 2×0.2 = 1.33 mA/V
10.13 Cox = εox tox
= 3.45×10−11 F/m = 0.43×10−2 F/m2 8×10−9 m
=0.43×10−2 ×10−12 F/μm2 = 4.3 fF/μm2
k′n =μnCox
= 450 × 108 (μm2/V·s)
×4.3×10−15 F/μm2 = 193.5 μA/V2
1 􏱾W􏱿
I = k′ V2 (1+λV )
VOV 0.3 fT = gm
2π(Cgs + Cgd ) 1.33 × 10−3
= 2π×(25+5)×10−15 =7.1GHz gm
D2nLOV DS
12 gs
10.15 fT = 2π(C +C ) gs gd
ForCgs≫Cgd gm
(1)
Cgs = 3WLCox +WLovCox
If the overlap component (WLov Cox ) is small, we
200= 2 ×193.5×20×VOV(1+0.05×1.5)
⇒VOV =0.31V
gm = 2ID = 2 × 0.2 = 1.3 mA/V VOV 0.31
fT ≃ 2πC 2
get χ=􏲇γ 2
2 2φf+VSB
= √0.5 =0.19
2 0.65+1
gmb =χgm =0.25mA/V
ro = |VA| = 1 = 1
ID | λ|ID 0.05 × 0.2
C =2WLC +WL C gs3ox ovox
= 100 k􏱹
Cgs≃3WLCox (2) The transconductance gm is given by
􏲉 􏱾W􏱿
gm = 2μnCox L ID (3)
Substituting from (2) and (3) into (1), we get
􏲉 􏱾 􏱿
= 2 ×20×1×4.3+20×0.05×4.3 3
= 57.3+4.3 = 61.6 fF
Cgd = WLovCox = 20×0.05×4.3 = 4.3 fF
C Csb = 􏲆 sb0
fT = = 1.5
2μCWI n ox L D
2
2π × 3WLCox
􏲉
μnID Q.E.D. 2Cox WL
π L Weobservethatforagivendevice,fT is
􏲇
1 + | VSB | V0
ID ; thus to obtain faster
Also, we observe that f is inversely proportional
proportional to
operation the MOSFET is operated at a higher ID.
20
= 􏲆
= 12.8 fF
√T
to L WL; thus faster operation is obtained from
1+1 0.7
smaller devices.

10.16 fT = gm 2π(Cgs + Cgd )
10.18 fT = gm 2π(Cπ + Cμ)
where
gm=IC =0.5mA=20mA/V
Chapter 10–8
For Cgs ≫ Cgd fT≃ gm
(1)
(2)
2πCgs 2
VT Cπ = 8 pF
Cμ =1pF Thus,
0.025 V
Cgs = 3WLCox +WLovCox
If the overlap component is small, we get
Cgs ≃ 2 WLCox 3
The transconductance gm can be expressed as 􏱾W􏱿
20 × 10−3
fT = 2π ×(8+1)×10−12 = 353.7 MHz
fβ=fT =353.7=3.54MHz β 100
(3) Substituting from (2) and (3) into (1), we obtain
gm =μnCox L VOV 􏱾􏱿
10.19 See figure on next page. Cπ = Cde + Cje whereC isproportionaltoI .
μ C W V
n ox L OV
de C AtIC =0.5mA,
fT =
= 3μnVOV
2
2π × 3WLCox
8=Cde +2⇒Cde =6pF 1
4πL2
Wenotethatforagivenchannellength,fT canbe increased by operating the MOSFET at a higher VOV .
For L = 0.5 μm and μn = 450 cm2/V·s,
wehave
VOV =0.2V⇒fT = 3×450×108 ×0.2
At IC = 0.25 mA, Cde = 2 × 6 = 3 pF, and Cπ =3+2=5pF.
= 5.73 GHz
VOV =0.4V⇒fT = =11.46GHz
1 mA
= 0.025 V = 40 mA/V
β
rπ = g = 40 =2.5k􏱹
m
ro=VA =50=50k􏱹
IC 1
Cde =τFgm =30×10−12 ×40×10−3 =1.2pF Cje0 = 20 pF
Cπ = Cde +2Cje0 = 1.2+2×0.02 = 1.24 pF
Cje0 Cμ=􏱾 V􏱿m
1 + CB V0c
4π × 0.52 3×450×108 ×0.4
4π × 0.52
Also,atIC =0.25mA,gm =10mA/V.ThusfT at IC =0.25mAis
10×10−3
fT = 2π(5+1)×10−12 =265.3MHz
10.20 rx = 100 􏱹
IC gm = V
T
100
2V 10.17 A0 = A =
2V′L A
VOV
A0 = 2×5×L =50L, V/V(Linμm)
0.2
fT ≃ 3μnVOV = 3×400×108 ×0.2
VOV
4π L2 4π L2 1.91
fT = L2 , GHz (L in μm)
The expressions for A0 and fT can be used to
20
2 􏱿0.5 = 10.4 fF
0.75
fT= gm 2π(Cπ +Cμ)
= 40 × 10−3 2π(1.24 + 0.01) × 10−12
= 5.1 GHz
obtain their values for different values of L. The results are given in the following table.
Cμ = 􏱾
1 +
L
Lmin 0.13 μm
2Lmin 0.26 μm
3Lmin 0.39 μm
4Lmin 0.52 μm
5Lmin 0.65 μm
A0 (V/V)
6.5
13
19.5
26
32.5
fT (GHz)
113
28.3
12.6
7.1
4.5

This figure belongs to Problem 10.20.
Chapter 10–9
rx B’ Cm BC
􏰀
Vp 􏰒
rp
gm Vp ro
gm 2π(Cπ +Cμ)
Cp
E
10.21 Forf ≫fβ, |hfe| = fT
10.22 fT = IC
f
Atf =50MHzandIC =0.2mA,
f |hfe|=10= T
50 ⇒fT =500MHz
Atf =50MHzandIC =1.0mA,
|hfe|=12=
50
⇒fT =600MHz Now,
gm
fT = 2π(Cπ +Cμ)
where
Cπ =Cde +Cje =τFgm +Cje Cμ = 0.1 pF
1 mA
gm = V = 0.025V =40mA/V
T
10×109 = ⇒Cπ =0.54pF
−12
=83.3MHz
=
10.23 For f ≫ fβ,
fβ =
β0
40×10−3 2π(Cπ +0.1)×10
β 120
rπ= = =3k􏱹
fT
gm fT
|hfe| ≃ fT f
40
10 × 109 120
40 = 2000 MHz f
⇒f =50MHz
fβ = fT = 2000MHz =10MHz
AtIC =0.2mA,gm = 0.2 =8mA/V,thus 0.025
500×106 = 8×10−3 2π(Cπ + 0.1) × 10−12
⇒ Cπ = 2.45 pF
τF ×8×10−3 +Cje = 2.45×10−12
AtIC =1mA,gm = 1 =40mA/V,thus 0.025
600×106 = 40×10−3 2π(Cπ + 0.1) × 10−12
⇒ Cπ = 10.51 pF
τF ×40×10−3 +Cje = 10.51×10−12 Solving (1) together with (2) yields
τF =252ps
Cje = 0.43 pF
β0 10.24
200
rp Cp
B
rx
B’
C, E
Cm
(1)
(2)
Zin
With the emitter and the collector grounded, the equivalent circuit takes the form shown in the figure, and the input impedance becomes
Zin = rx +
1
1 +jω(Cπ +Cμ) rπ
= rx +
rπ 1+jω(Cπ +Cμ)rπ

Zin = rx +
rπ =
fT = 2π(Cπ + Cμ)
􏱾 ω 􏱿 ωβ
1−j ωβ
0 , k􏱹 gm (mA/V)
1 fβ=2π(C +C)r
Chapter 10–10
Sinceωβ= 1 ,then gm=IC =αIE ≃IE =IE(mA) (Cπ + Cμ)rπ VT VT VT 0.025 V
rπ β
1+j
􏱾 ω 􏱿
gm
=rx+rπ
􏱾ω􏱿2
1+ πμπ
ωβ
ein x 􏱾ω􏱿2
fT fβ = β0
10.26 Cin =Cgs +Cgd(1+gmR′L) = 1 + 0.1(1 + 39)
= 5 pF
fsdB= 1 2πCinRsig
R (Z ) = r +
rπ 1+ ωβ
For the real part to be an estimate of rx accurate to within 10%, we require
rπ
􏱾 ω 􏱿2
≤ 0.1r x
1+ωβ
1 ≤0.1rx
􏱾 ω 􏱿2 1+ ω
=
1 2π×5×10 Rsig
􏱾 􏱿
rπ Butrx≤10,thusr ≤0.1,
−12 For fsdB > 1 MHz,
β
rπ
1 ≤ 0.1 × 0.1
rx
π
1
Rsig < 2π ×5×10−12 ×1×106 = 31.8 k􏱹 10.27 (a) Vo =−AVi If the current flowing through Rsig is denoted Ii , we obtain Ii sC(Vi −Vo) Yin = V = V ii 􏱾 V 􏱿 =sC 1− o Vi = sC(1 + A) Thus, Cin =C(1+A) 􏱾 ω 􏱿2 1+ ωβ or, equivalently, β 􏱾 ω 􏱿2 1+ ω ≥100 ⇒ ω ≥ 10ωβ 10.25 To complete the table we use the following relationships: re = VT = 25 mV IE IE (mA) This table belongs to Problem 10.25. Transistor IE (mA) re (􏱹) gm (mA/V) rπ (k􏱹) β0 fT (MHz) Cμ (pF) Cπ (pF) fβ (MHz) (a) 2 12.5 80 12.5 100 500 2 23.5 5 (b) 1 25 40 3.13 125 500 2 10.7 4 (c) 1 25 40 2.5 100 500 2 10.7 5 (d) 10 2.5 400 0.25 100 500 2 125.3 5 (e) 0.1 250 4 25 100 150 2 2.2 1.5 (f) 1 25 40 0.25 10 500 2 10.7 50 (g) 1.25 20 50 0.2 10 800 1 9 80 (b) Vi(s) = 1/sCin Vsig(s) Rsig + 1 10.29 fH = 1 2πCinRsig sCin =111 Chapter 10–11 1+sCinRsig Vo(s) =− A Vsig(s) 1 + sCinRsig (c) DCgain=40dB=100V/V, ForfH ≥6MHz Cin≤2πf R =2π×6×106×1×103 H sig Cin ≤26.5pF ⇒ A = 100 V/V 1 But, Cin =Cgs +(1+gmR′L)Cgd =5+(1+gmR′L)×1, pF =6+gmR′L,pF ForCin ≤26.5pFwehave gmR′L ≤20.5 ′ 20.5 RL≤ 5 =4.1k􏱹 ′ Corresponding to RL = 4.1 k􏱹, we have |AM|=gmR′L =20.5V/V GB = | AM | fH = 20.5×6 = 123 MHz IffH =2MHz,weobtain Cin = 26.5×3 = 79.5 pF gmR′L =79.5−6=73.5 Thus, | AM | = 73.5 V/V GB = 73.5×2 = 147 MHz 10.30 Refer to Example 10.3. If the transistor is replaced with another whose W is half that of the original transistor, we obtain W2 = 1W1 2 Since Cgs = 2WLCox +WLovCox 3 f3dB = 2π Cin Rsig 100×103 = 1 2πCin ×1×103 ⇒ Cin = 1591.5 pF Cin 1591.5 C=A+1= 101 =15.8pF (d) The Bode plot is shown in the figure below. From the figure we see that the gain reduces to unity two decades higher than f3dB , that is at 10 MHz. 10.28 Cin = 0.2(1 + 1000) = 200.2 pF Vo (s) 1000 V (s)=−1+sCR sig in sig f3dB = 1 2π Cin Rsig =1 then 2π ×200.2×10−12 ×1×103 = 795 kHz The gain falls off at the rate of 20 dB/decade. For thegaintoreach0dB(unity),thegainhastofall by 60 dB. This requires three decades or a factor of 1000, thus funitygain =795×1000=795MHz 1 Cgs2 = 2Cgs1 = 0.5 pF Also, Cgd =WLovCox thus, Cgd2 = 1Cgd1 =0.2pF 2 Since 􏲆 AM =− 100 ×3×4.1 100 + 100 = −6.1 V/V Cin =Cgs +Cgd(1+gmR′L) (1) = 1+0.2(1+3×4.1) = 3.66 pF fH= 1 (2) 2πCinR′sig where R′sig = Rsig ∥ Rin g = 2μC WI m n ox L D gm1 = 0.71 mA/V 􏱾 􏱿 1 Chapter 10–12 then gm2 = √ 2 Since I=1μC WV2 D 2 n ox L OV then √ VOV2 = Finally, ro2 =ro1 =150k􏱹 Thus, R′L2 = R′L1 = 7.14 k􏱹 Thus, Cin2 = Cgs2 + (gm2R′L2 + 1)Cgd2 = 0.5 + (0.71 × 7.14 + 1) × 0.2 = 1.71 pF This should be compared to Cin1 = 4.26 pF. Thus, 1 fH2 = 2πCin2(Rsig ∥ RG) =1 2π × 1.71 × 10−12(0.1 ∥ 4.7) × 106 =952MHz in comparison to fH 1 = 398 MHz 4.8 = 5 V/V in comparison to | AM 1| = 7 V/V. GB2 = 5×952 = 4.73 GHz in comparison to GB1 = 7 × 398 = 2.79 GHz. 10.31 A = − Rin g R′ MR+RmL where 2VOV1 = 100 k􏱹 ∥ 100 k􏱹 = 50 k􏱹 fH= 1 2π ×3.66×10−12 ×50×103 = 870 kHz 4.7 TodoublefH bychangingRin,Eq.(2)indicates that R′sig must be halved: R′sig=25k􏱹 which requires Rin to be changed to Rin2, 25 k􏱹 = 100 ∥ Rin2 ⇒ Rin2 = 33.3 k􏱹 This change will cause | AM | to become | AM 2| = 33.3 × 3 × 4.1 33.3 + 100 = 3.1 V/V which is about half the original value. TodoublefH bychangingRL,Eq.(2)indicates that Cin must be halved: 1 Cin2 = × 3.66 = 1.83 pF 2 Using Eq. (1), we obtain 1.83 = 1+0.2(1+gmR′L2) ⇒ gmR′L2 = 3.15 Thus, R′L2 = 1.05 k􏱹 and RL2 can be found from 1.05=RL ∥8k􏱹∥50k􏱹 ⇒ RL = 1.24 k􏱹 and the midband gain becomes | AM 2| = 100 × 3.15 = 1.6 V/V 100 + 100 which is about a quarter of the original gain. Clearly, changing Rin is the preferred course of action! |A |= ×g R′ M 2 4.7 + 0.1 = 4.7 ×0.71×7.14 m2 L2 in sig R′L =RD ∥RL ∥ro = 8 k􏱹 ∥ 10 k􏱹 ∥ 50 k􏱹 = 4.1 k􏱹 10.32 (a) AM =− RG RG + Rsig where R′L =RD ∥RL ∥ro = 20 k􏱹 ∥ 20 k􏱹 ∥ 100 k􏱹 = 9.1 k􏱹 gmR′L where Cin =Cgs +Cgd(1+gmR′L) = 1+0.2(1+3×3.1) = 3.06 pF and R′sig = Rsig ∥ RG =100k􏱹∥8.25M􏱹 = 99 k􏱹 Thus, fH = 1 2π ×3.06×10−12 ×99×103 ×5×9.1 Cin =Cgs +Cgd(1+gmR′L) = 3 + 0.5(1 + 5 × 9.1) = 26.25 pF and R′sig =Rsig ∥RG = 500 k􏱹 ∥ 2000 k􏱹 = 400 k􏱹 Thus, Chapter 10–13 AM =− 2M􏱹 2 M􏱹+0.5 M􏱹 = −36.4 V/V (b) fH = 1 2πCinR′sig where = 525 kHz 􏲇 1 2π ×26.25×10−12 ×400×103 10.34 gm = 2μnCox(W/L)1ID1 √ = 2×0.09×100×0.1 = 1.34 mA/V ro1 = |VA1| = 12.8 =128k􏱹 ID1 0.1 ro2 = |VA2| = 19.2 = 192 k􏱹 ID2 0.1 The total resistance at the output node, R′L, is given by R′L =ro1 ∥ro2 =128k􏱹∥192k􏱹 = 76.8 k􏱹 AM=−gm1R′L fH = = 15.2 kHz (c)fZ= gm 2π Cgd = −1.34 × 76.8 = −103 V/V 5×10−3 1 =2π×0.5×10−12 = 1.6 GHz 10.33 RG =RG1 ∥RG2 =47M􏱹∥10M􏱹 = 8.25 M􏱹 fH =2πCinRsig where Cin =Cgs +Cgd(1+gm1R′L) = 0.2 + 0.015(1 + 103) = 1.76 pF Thus, MR+RmL 1 A=−RG gR′ G sig where R′L =RL ∥RD ∥ro =10k􏱹∥4.7k􏱹∥100k􏱹 = 3.1 k􏱹 AM =− 8.25 ×3×3.1 8.25+0.1 =−9.2V/V fH = 1 2πCinR′sig fH = 2π ×1.76×10−12 ×200×103 = 452 kHz 1.34×10−3 = 2π×0.015×10−12 fZ = 2πC gd gm = 14.2 GHz 10.35 g R′ =50 m L Cin =Cπ +Cμ(1+gmR′L) =10+1(1+50) = 61 pF fH= 1 2π Cin R′sig =1 2π ×61×10−12 ×5×103 Also, now RB =50k􏱹 RC = 4 k􏱹 ThenewvalueofAM is AM =− RB rπ (gmR′L) RB +Rsig rπ +rx +(RB ∥Rsig) where R′L =ro ∥RC ∥RL = 50 ∥ 4 ∥ 5 = 2.13 k􏱹 Thus, gmR′L =80×2.13=170V/V and 50 1.25 AM =−50+5 × 1.25+0.05+(50∥5) ×170 = −33 V/V and 20 log|AM | = 30.4 dB This should be compared to the previous value of 39 V/V (32 dB). To determine fH , we first find Cin, Cin =Cπ +Cμ(1+gmR′L) = 15+1(1+170) = 186 pF and the effective source resistance R′sig, R′sig =rπ ∥[rx +(RB ∥Rsig)] = 1.25 ∥ [0.05 + (50 ∥ 5)] = 0.98 k􏱹 = 522 kHz 10.36 R AM=− B where R′L =ro ∥RC ∥RL = 100 k􏱹 ∥ 10 k􏱹 ∥ 10 k􏱹 = 4.76 k􏱹 and rπ = β/gm = 100/40 = 2.5 k􏱹 π RB +Rsig rπ +rx +(Rsig ∥RB) gmR′L Chapter 10–14 r AM=− 100 × 100+10 × 40 × 4.76 = −37 V/V fH= 1 2π Cin R′sig 2.5 2.5+0.1+(10 ∥ 100) where Cin =Cπ +Cμ(1+gmR′L) = 10+1×(1+40×4.76) = 201.4 pF and R′sig =rπ ∥[rx +(RB ∥Rsig)] = 2.5 ∥ [0.1 + (100 ∥ 10)] = 2 k􏱹 Thus fH=1 1 2π ×201.4×10−12 ×2×103 = 395 kHz 10.37 Refer to Example 10.4. Since IE is doubled to 2 mA, we have gm = IC = 2mA =80mA/V fH = 2πCinR′sig =1 2π ×186×10−12 ×0.98×103 = 873 kHz This should be compared to the previous value of 754 kHz. The gain-bandwidth product becomes GB=|AM|fH =33×873=28.8MHz This should be compared to the previous value of 39 × 754 = 29.4 MHz. Thus, increasing the bias current by a factor of 2 results in an increase in fH by a factor of 1.16—that is, by about 16%. However, because of the attendant reduction in input resistance, the overall gain decreased by about the same factor and GB remained nearly constant. The price paid for the slight increase in fH isanincreaseinpowerdissipationbyafactor of about two. VT rπ = β = gm 25 mV 100 =1.25k􏱹 80 mA/V ro = VA = 100 V = 50 k􏱹 IC 2 mA gm 80 × 10−3 Cπ+Cμ=ω =2π×800×106 =16pF T Cμ = 1 pF Cπ =15pF rx = 50 􏱹 10.38 (a) AM = − RB rπ gmR′L The Bode plots are shown in the figure. If the midband gain is unity, fH =GB=6.37MHz This is obtained when R′L is R′ 1=100× L 25 ⇒R′L =0.25k􏱹=250􏱹 10.39 RB =RB1 ∥RB2 =68k􏱹∥27k􏱹 = 19.3 k􏱹 gm = IC = 0.8mA =32mA/V VT 0.025 V β 200 rπ=g =32=6.25k􏱹 m R′sig = rπ ∥ RB ∥ Rsig =6.25k􏱹∥19.3k􏱹∥10k􏱹 = 3.2 k􏱹 R′L =RC ∥RL =4.7k􏱹∥10k􏱹 = 3.2 k􏱹 AM=− RB rπ gmR′L RB +Rsig rπ +(Rsig ∥RB) 19.3 6.25 =−19.3+10 6.25+(10∥19.3)×32×3.2 Chapter 10–15 RB +Rsig rπ +rx +(Rsig ∥RB) ForRB ≫Rsig,rx ≪Rsig,Rsig ≫rπ, rπ AM ≃ −R sig (b) Cin =Cπ +(gmR′L +1)Cμ ForgmR′L ≫1andgmR′LCμ ≫Cπ, Cin ≃gmR′LCμ f=1 H 2πCinR′sig where R′sig =rπ ∥[rx +(RB ∥Rsig)] ≃rπ ∥Rsig ≃rπ Thus, ′ ′ gmRL = −βRL/Rsig Q.E.D. fH ≃ fH = 1 2πgmR′LCμrπ 1 2 π C μ β R ′L Q.E.D. GB = 1 2π ×1×10−12 ×25×103 (c) GB=|AM|fH R′ =β L Rsig 1 2π Cμ β R′L 1 2π Cμ Rsig = ForRsig =25k􏱹andCμ =1pF, Q.E.D. = 6.37 MHz | AM | 100 V/V (ii) R′L = 2.5 k􏱹: 2.5 AM =−100× 25 =−10V/V fH = GB =6.37MHz=637kHz 10 V/V = −32.8 V/V gm ForIC =1mAandβ=100, (i) R′ =25k􏱹: fT = 2π(Cπ +Cμ) 32×10−3 L AM =−100×25=−100V/V 1×109=2π(C +C) π μ 25 fH = GB = 6.37MHz =63.7kHz ⇒Cπ +Cμ =5.1pF Cπ =5.1−0.8=4.3pF Cin =Cπ +(gmR′L +1)Cμ = 4.3+(32×3.2+1)×0.8 =87pF fH = 1 2π Cin R′sig =1 2π ×87×10−12 ×3.2×103 = 572 kHz 10.40 Rin = R 1−K = 100k􏱹 =1000k􏱹=1M􏱹 1−0.9 | AM | 1−A 1−1 A Chapter 10–16 10.41 Using Miller’s theorem, we obtain Zin= Z , Zout= Z For Z=1 jωC Zin = Zout = 1 jωC(1 − A) ⇒Cin =C(1−A) 􏱾 1􏱿 􏱾 1􏱿 ⇒Cout =C 1− A Figure2 (a) RefertoFig.1. Ii=Vi−Vo =Vi−2Vi =−Vi RRR Thus, Rin ≡ Vi = −R Ii (b) Replacing the signal source with its equivalent Norton’s form results in the circuit in Fig.2.ObservethatReq =∞whenRsig =R.In this case, IL = Vsig = Vsig Rsig R (c) If ZL = 1 , sC Vi =ILZL =Vsig × 1 R sC = 1 Vsig sCR and 2 Vo =2Vi = sCRVsig Thus, Vo = 2 Vsig sCR which is the transfer function of an ideal 1 jωC 1− A (a) A=−1000V/V, C=1pF Cin = 1(1 + 1000) = 1001 pF 􏱾1􏱿 Cout =1 1+1000 =1.001pF (b) A=−10V/V, C=10pF Cin =10(1+10)=110pF 􏱾1􏱿 Cout =10 1+10 =11pF (c) A=−1V/V, C=10pF Cin =10(1+1)=20pF Cout =10(1+1)=20pF (d) A=+1V/V, C=10pF Cin =C(1−1)=0 Cout =C(1−1)=0 (e) A=+10V/V, C=10pF Cin =10(1−10)=−90pF 􏱾 1􏱿 Cout=101−10 =9pF The −90 pF input capacitance can be used to cancel an equal (+90 pF) capacitance between the input node and ground. 10.42 This figure belongs to Problem 10.42. Rsig Ii noninverting integrator. 0 􏰀 Vi 􏰒 Rin Figure 1 􏰀2 Vo IL Vsig 􏰀 􏰒L Z (Vi􏰒Vo)/R R 10.43 (a) For the dc analysis, refer to the figure. Ii R2 􏰔 10 k􏰵 R1 􏰔 1 k􏰵 Ii 0 􏰒􏰒 VCC =IERC +IBRB +VBE 1.5=IE ×1+ IE ×47+0.7 Chapter 10–17 β+1 ⇒ IE = 1.5−0.7 = 0.546 mA 1+ 47 V􏰀 􏰀 􏰀 101 sig 􏰒 Vi 􏰀A Rin 􏰔 􏰒Vi Vo 􏰒 I =αI =0.99×0.546=0.54mA CE (b) gm = IC = 0.54mA =21.6mA/V VT 0.025 V β 200 rπ = g = 21.6 =4.63k􏱹 m (c) Vo = −gm(RC ∥ RL) Vb = −21.6(1 ∥ 1) = −10.8 V/V (d) Using Miller’s theorem, the component of Rin due to RB can be found as Rin1 = RB Ii From the figure we see that Vo = AVi (1) From Miller’s theorem, we have R2 R2 R2 Rin=􏱾􏱿= = (2) 1−Vo 1+Vo1+A −Vi Vi Using the voltage divider rule at the input, we get −V = V Rin 1 − (Vo/Vb) i sig Rin + R1 ⇒V =−V Rin (3) = 47 k􏱹 1 − (−10.8) = 4 k􏱹 i sig Rin + R1 Rin = Rin1 ∥ rπ = 4 ∥ 4.63 = 2.14 k􏱹 For each value of A we use Eq. (2) to determine Rin,Eq.(3)todetermineVi (forVsig =1V),Eq. (1) to determine Vo, and finally we calculate the value of Vo/Vsig. The results are given in the table below. (e) Gv = = 2.14 Rin × Vo Rin + Rsig Vb × −10.8 = −7.4 V/V 􏱾 􏲀􏲀V􏲀􏲀􏱿 2.14+1 (f) Cin = Cπ + 1 + 􏲀􏲀 o 􏲀􏲀 Cμ A (V/V) Rin (k􏱹) Vi (V) Vo (V) Vo /Vsig (V/V) 10 9.091 × 10−1 −0.476 −4.76 −4.76 100 9.901 × 10−2 −0.0901 −9.01 −9.01 1000 9.990 × 10−3 −9.89 × 10−3 −9.89 −9.89 10,000 9.999 × 10−4 −9.989 × 10−4 −9.99 −9.99 Vb Cπ +Cμ = gm = 21.6×10−3 2πfT 2π × 600 × 106 Cπ + Cμ = 5.73 pF Cπ = 5.73−0.8 = 4.93 pF Cin = 4.93+(1+10.8)×0.8 = 14.37 pF (g) R′sig = Rin ∥ Rsig where 10.44 VCC 􏰔 􏰀1.5 V IE RC RB IIC B IE = 2.14 k􏱹 ∥ 1 k􏱹 = 0.68 k􏱹 1 􏰀 VBE 􏰒 fH = 2πCinR′sig =1 2π ×14.37×10−12 ×0.68×103 = 16.3 MHz 10.45 Chapter 10–18 BB C 􏰀 Vp rp 􏰒 gm Vp re Cp Cm Cp Zi (s) From the figure we see that the controlled current-sourcegmVπ appearsacrossitscontrol voltage Vπ , thus we can replace the current source with a resistance 1/gm. Now, the parallel equivalent of rπ and 1/gm is rπ(1/gm) = rπ = rπ =re rπ + 1 gmrπ +1 β+1 gm Thus, the equivalent circuit simplifies to that of re re 1+sCπre Zi (s) fT≃ 1 EE 4πCπre It follows that in this case, in parallel with Cπ , Zi(s) = 1 = f45◦ = 1fT =200MHz 2 10.46 AM =−gmR′L = −4 × 20 = −80 V/V fsdB = fH = 1 2π(CL + Cgd )R′L =1 2π(2+0.1)×10−12 ×20×103 1 +sCπ re Zi(jω) = re 1+jωCπre = 3.8 MHz gm 4×10−3 Zi(jω) will have a 45◦ phase at ω45Cπre =1 ⇒ω45= 1 Cπ re Now, fZ = 2πC = 2π×0.1×10−12 =6.4GHz gd ft =|AM|fH = 80×3.8 = 304 MHz gm 2π(CL +Cgd) g fT= m 10.47 ft = CL+Cgd=2πft 2π(Cπ +Cμ) At high bias currents, gm Cπ ≫Cμ and f ≃ gm T 2πCπ 2×10−3 = 2π ×2×109 = 0.159 pF To reduce ft to 1 GHz, an additional capacitance of 0.159 pF must be connected to the output node. (Doubling the effective capacitance at the output node reduces ft by a factor of 2.) 10.48 Refer to Fig. P10.48. To determine gm1 weuse􏲉 􏱾 􏱿 g = 2μC W I Since gm ≃ 1/re, we have fT≃ 1 2πCπre f45◦ ≃ fT = 400 MHz Thus, If the bias current is reduced to the value that results in Cπ ≃ Cμ, m1 􏲆 n ox L D1 1 gg fT= m = m 100 = 2×0.090× ×0.1 1.6 = 1.06 mA/V ro1 = VA1 = 12.8 =128k􏱹 ID1 0.1 2π × 2Cπ 4πCπ Again, gm ≃ 1 , thus re ro2 = |VA2| = 19.2 =192k􏱹 ID2 0.1 R′L =ro1 ∥ro2 =128∥192=76.8k􏱹 AM =−gm1R′L =−1.06×76.8=−81.4V/V CL =Cdb1 +Cdb2 +Cgd2 f3dB = 1 2π(CL + Cμ)ro =1 2π(1+0.2)×10−12 ×500×103 Chapter 10–19 = 265.3 kHz fZ = gm = 20+36+15 = 71 fF fH = 1 = 2π ×0.2×10−12 = 6.37 GHz ft =|Adc|f3dB 2π(CL + Cgd1)R′L fH = 1 = 4000 × 265.3 = 1.06 GHz TheBodeplotisshowninFigure2. 2π Cμ 8×10−3 2π(71+15)×10−15 ×76.8×103 = 24.1 MHz fZ = gm1 = 1.06×10−3 2π Cgd 1 2π × 0.015 × 10−12 = 11.2 GHz 10.49 Figure 1 shows the amplifier high-frequency equivalent circuit. A node equation at the output provides 􏱾1􏱿 r +sCL Vo+gmVπ+sCμ(Vo−Vπ)=0 o Replacing Vπ by Vi and collecting terms results in 􏱼1􏱽 Figure 2 10.50 The equivalent circuit is shown in the figure. gm=IC = 2mA =80mA/V VT 0.025 V r = β = 120 =1.5k􏱹 π g 80 Vo r +s(CL+Cμ) =−Vi(gm−sCμ) o ⇒ Vo =−g r 1−s(Cμ/gm) V m o 1 + s(C + C )r Q.E.D. iLμom ForIC =200μA=0.2mAandVA =100V, gm = IC = 0.2mA =8mA/V VT 0.025V ro = VA = 100 =500k􏱹 IC 0.2 DCgain=−gmro =8×500=−4000V/V This figure belongs to Problem 10.49, part (a). Cm 􏰀􏰀 Vi􏰀􏰒rpVpCpgmVp ro CLVo 􏰒􏰒 (a) AM =− rπ gmR′L rπ +rx 1.5 −10=−1.5+0.1 ×gmRL ⇒gmR′L =10.7V/V R′L = 0.133 k􏱹 ′ Figure 1 This figure belongs to Problem 10.50. Chapter 10–20 Cπ+Cμ= gm 2π fT 80 × 10−3 = 2π ×2×109 = 6.37 pF Cπ = 6.37−1 = 5.37 pF Cin =Cπ +(gmR′L +1)Cμ Cin = 5.37+(10.7+1)×1 = 17.07 pF R′sig =rπ ∥rx =1.5k􏱹∥0.1k􏱹 = 0.094 k􏱹 fH= 1 2πCinR′sig =1 2π × 17.07 × 10−12 × 0.094 × 103 = 99.2 MHz (b) If|AM|isreducedto1,weobtain 1 = 1.5 × gmR′L 1.6 ⇒gmR′L =1.07 Cin =Cπ +(gmR′L +1)Cμ = 5.37+(1.07+1)×1 = 7.44 pF fH = 1 2π ×7.44×10−12 ×0.094×103 = 227.6 MHz 10.51 AM =40dB⇒100V/V A(s) = 100􏱾 s 􏱿􏱾 s 􏱿 1 + 2π × 2 × 106 1 + 2π × 20 × 106 SincefP1 ≪fP2 ≪fZ,wehave f3dB ≃ fP1 = 2 MHz The Bode plot is shown in the figure. Gain (dB) 50 40 30 20 10 0 􏰒10 􏰒20 􏰒30 􏰒20 dB/decade 􏰒40 dB/decade 2 200 f (MHz) 􏰒20 dB/decade 1 1 + s/(2π × 105) 20 10.52 (a) A(s) = 1000 (b) |A|, dB 60 40 20 0 3 dB 1 MHz 1 MHz Figure 1 􏰒20 dB/decade f f 􏱖 􏰒5.7° 􏰒45° 􏰒90° 5.7° 10 kHz 100 kHz 0 10 kHz 100 kHz 10 MHz 100 MHz 1+s/(2π ×200×106) Figure 1 shows the Bode plot for the gain magnitude and phase. (c) GB=1000×100kHz=100MHz (d) The unity-gain frequency ft is ft = 100 MHz (e) If the frequency of the finite zero is lowered to 103 rad/s the zero will cancel the pole at 103 rad/s and the transfer function becomes Chapter 10–21 A(s) = −103 1 1+s 105 The 3-dB frequency now becomes Figure 2 ω3dB = 105 rad/s 10.55 If at ω = 107 rad/s the excess phase due to the 3 coincident poles (at frequency ωP) is 30◦, then each pole is contributing 10◦. Thus, P 10.56 τH = CgsRgs + Cgd Rgd + CLRCL where Cgs = 30 fF Rgs = R′sig = 10 k􏱹 Cgd =5fF Rgd =R′sig(1+gmR′L)+R′L = 10(1+2×20)+20 = 430 k􏱹 CL =30fF RCL =R′L =20k􏱹 Thus, τH = 30×10+5×430+30×20 = 3050 ps fH= 1 2πτH 1 = 2π × 3050 × 10−12 = 52.2 MHz gm 2 × 10−3 Figure 2 shows the magnitude response when a second pole at 1 MHz appears in the transfer function. The unity-gain frequency ft now is ft = 10 MHz which is different from the gain-bandwidth product, GB = 100 MHz 10.53 Using the dominant-pole approximation, ωH ≃ωP1 Using the root-sum-of-squares formula, we get ω≃􏲉1 =􏲉ωP1 −1 107 ◦ ω = 10 tan ωP = tan10◦ =5.67×107 rad/s 1 + 1 ω2 ω2 ωP2 107 H 􏱾 􏱿2 1+ ωP1 P1 P2 (a) For a difference of 10%, ω 􏲉 P1 􏱾ω 􏱿2 1+ P1 ωP2 ⇒ωP2 =4.26 ωP1 (b) For a difference of 1%, = 0.9ω P1 􏲉 ωP1 􏱾ω 􏱿2 1+ P1 ωP2 ⇒ωP2 =49.3 ωP1 1 + 103 (a) ω ≃ 103 rad/s = 0.99ω P1 s fZ=2πC =2π×5×10−15=63.7GHz gd 10.54 A(s)=−10 􏲃 s 􏲄􏲃 3 10 1+4 s 􏲄 1 + 105 RG gmR′L RG +Rsig H 10.57 AM =− 0.65 􏲈􏲉􏱾1 1􏱿 2 (b) ωH =1 106 +1010 −108 = 1010 Hz =−0.65+0.15 ×5×10 = −40.6 V/V τgs = CgsRgs = CgsR′sig = Cgs(Rsig ∥ RG) = 2×10−12 ×(150 ∥ 650)×103 = 243.8 ns τgd =CgdRgd = Cgd [R′sig(1 + gmR′L) + R′L] = 0.5×10−12[(150 ∥ 650)(1+5×10)+10]×103 = 3112.8 ns τ = C R′ CL LL = 30×10−12 ×10×103 = 300 ns τH = τgs + τgd + τCL = 243.8 + 3112.8 + 300 = 3656.6 ns = 1.59 MHz At node 2: Req2 =Ro1 ∥Rin2 =2k􏱹∥10k􏱹 = 1.67 k􏱹 Ceq2 = Co1 + Cin2 = 2 + 10 = 12 pF fP2 = 1 2π Ceq2 Req2 1 = 2π ×12×10−12 ×1.67×103 = 7.94 MHz At node 3: Req3 =Ro2 ∥RL =2k􏱹∥1k􏱹=0.67k􏱹 Ceq3 =Co2 +CL =2+7=9pF 1 fP3 = 2πCeq3Req3 fH = 1 2π τH = 2π ×3656.6×10−9 = 43.5 MHz 10.58 The figure shows the equivalent circuit of the two-stage amplifier where we have modeled each stage as a transconductance amplifier. At node 1: Req1 = Rsig ∥ Rin1 = 10 k􏱹 ∥ 10 k􏱹 = 5 k􏱹 Ceq1 =Cin1 +Csig =10+10=20pF Thus, 1 fP1=2πC R eq1 eq1 fP1 = 1 2π ×20×10−12 ×5×103 This figure belongs to Problem 10.58. Chapter 10–22 1 fP3=1 2π ×9×10−12 ×0.67×103 = 26.4 MHz Thus, the three poles have frequencies 1.59 MHz, 7.94 MHz, and 26.4 MHz. Since the frequency of the second pole is more than two octaves higher than that of the first pole, the 3-dB frequency will be mostly determined by fP1, f3dB ≃ fP1 = 1.59 MHz A slightly better estimate of f3dB can be determined using the root-sum-of-squares formula, 􏲉􏱾 1 􏱿2 􏱾 1 􏱿2 􏱾 1 􏱿2 f3dB = 1/ 􏲆 f + f + f P1 P2 P3 =1/ = 1.56 MHz 1 + 1 + 1 1.592 7.942 26.42 10.59 AM =−gmR′L = −4 × 20 = −80 V/V Cin =Cgs +Cgd(gmR′L +1) = 2+0.1(80+1) = 10.1 pF Using the Miller approximation, we obtain fH=1 (1) 2π τH 1 = 2π ×620×10−9 = 257 kHz The interaction of R′sig with the input capacitance contributes all of τgs (50 ns) and a significant part of τgd , namely Cgd [R′sig(1 + gmR′L)] = 1 × 10(1 + 50) = 510 ns Thus, the total contribution of R′ is sig 50+510 = 560 ns whichis 560 =90.3%ofτH.TodoublefH,we 620 mustreduceτH tohalfofitsvalue: τH =1×620=310ns 2 Now, τH = R′sig[Cgs + Cgd (1 + gmR′L)] +CgdR′L +CLR′L 310 = R′sig[5+1(1+50)]+1×10+5×10 ⇒ R′sig = 4.46 k􏱹 10.61 To lower fH from 135.5 MHz (see Example 10.8)to100MHz,τH mustbeincreasedto τH=1= 1 2πfH 2π × 100 × 106 = 1591.5 ps Now, τH = τgs + τgd + τCL 1591.5 = 200 + 725 + τCL ⇒ τCL = 666.5 But, τCL = CL′ R′L 665.5 = CL′ × 10 ⇒CL′ =66.6fF Thus, the original CL of 25 fF must be increased by 66.6 − 25 = 41.6 fF 10.62 We will assume that the value given in the problem statement is for Rsig (not R′sig): Rsig = 5 k􏱹 rπ = β =100=5k􏱹 gm 20 fH ≃ 1 2πC R′ Chapter 10–23 in sig =1 2π ×10.1×10−12 ×20×103 =788kHz Using the open-circuit time constants, we get τgs = CgsRgs = CgsR′sig = 2 × 20 = 40 ns Rgd =R′sig(1+gmR′L)+R′L = 20(1+80)+20 = 1640 k􏱹 τgd = CgdRgd = 0.1×1640 = 164 ns τCL =CLR′L = 2 × 20 = 40 ns τH = τgs + τgd + τCL = 40+164+40 = 244 ns (1) fH= 1 2π τH = 1 2π × 244 × 10−9 = 652 kHz The estimate obtained using the open-circuit time constants is more appropriate as it takes into account the effect of CL. We note from Eq. (1) that τCL is 16.4% of τH , thus CL has a significant effect on the determination of fH . 10.60 τgs = CgsRgs = CgsR′sig = 5 × 10 = 50 ns Rgd =R′sig(1+gmR′L)+R′L = 10(1+5×10)+10 = 520 k􏱹 τgd = CgdRgd = 1×520 = 520 ns RCL =CLRL = 5 × 10 = 50 ns τH =τgs+τgd +τCL = 50+520+50 = 620 ns R′sig = rπ ∥ Rsig = 5 k􏱹 ∥ 5 k􏱹 = 2.5 k􏱹 τH =CπRπ =CπR′sig = 10×2.5 = 25 ns τμ = CμRμ = Cμ[R′sig(1 + gmR′L) + R′L] = 1×[2.5(1+20×5)+5] = 257.5 ns τCL = CLR′L =10×5=50ns τ =τ +τ +τ H π μ CL Using the open-circuit time constants, we get τπ = CπRπ = CπR′sig =10×0.76=7.6ns τμ = Cμ[R′sig(gmR′L +1)+R′L] = 0.3[0.76(40×5+1)+5] = 47.3 ns τCL = CLR′L = 3×5 = 15 ns τH = τπ + τμ + τCL Chapter 10–24 = 7.6+47.3+15 = 69.9 ns fH = 1 = 25+257.5+50 = 332.5 ns f= 1 = 1 2πτH = 1 = 2.28 MHz =479kHz 10.63 Refer to Fig. 10.19(a). Since RB is not specified, we assume that its value is very large. 2π × 69.9 × 10−9 ThisisamorerealisticestimateoffH asittakes H 2πτH 2π×332.5×10−9 AM=−rπ gmR′L into account the effect of CL. 10.64 CSamplifierwith: R′sig =ro/2 R ′L = r o / 2 Cgs =Cgd =0.1pF Using the Miller approximation, we get Cin =Cgs +Cgd(gmR′L +1) 􏱾1 􏱿 rπ +Rsig =− 5 ×20×5 5+5 = −50 V/V AM =− rπ rπ +rx +Rsig gmR′L rπ = β =100=2.5k􏱹 where gm 40 =0.1+0.1 2 gmro +1 􏱾1 􏱿 =0.1 2gmro+2 f= 1 Thus, 2.5 AM =−2.5+0.1+1×40×5 =−138.9V/V Using the Miller approximation, we obtain Cin =Cπ +Cμ(1+gmR′L) = 10 + 0.3(1 + 40 × 5) = 70.3 pF R′sig =rπ ∥(rx +Rsig) = 2.5 ∥ (0.1 + 1) = 0.76 k􏱹 fH = 1 2πCinR′sig H 2πCinR′sig = 􏱾 1􏱿 = 􏱿 1 gmro + 2 2 (ro/2) × 10−9 2π × 0.1 􏱾 1 wherero isink􏱹. For initial design, gm = 2 mA/V, ro = 20 k􏱹 2 1 0.1π 2 gmro +2 ro ×10−9 􏱾 1 􏱿 0.1π 1×40+2 ×20×10−9 1 2π ×70.3×10−12 ×0.76×103 fH = = 7.23 MHz = = 2.98 MHz (i) Forthecase,I isreducedbyafactorof4: τH =τgs +τgd +τCL = 20+246.5+75 = 341.5 ns f = 1 H2πτH = 1 = 466 kHz 2π × 341.5 × 10−9 (c) CL =50pF τCL =CLR′L =50×7.5=375ns τH = τgs + τgd + τCL = 20+246.5+375 = 641.5 ns fH= 1 2π τH = 1 = 248 kHz 2π × 641.5 × 10−9 Using the Miller approximation, since CL is not taken into account, then for all three cases we obtain Cin =Cgs +Cgd(gmR′L +1) = 0.2 + 0.2(1.5 × 7.5 + 1) =2.65pF fH = 1 2πCinR′sig =1 2π ×2.65×10−12 ×100×103 = 600 kHz which is very close to the estimate obtained using the method of open-circuit time constants for the case CL = 0. However, as CL is increased, the estimate obtained using the Miller approximation becomes less and less realistic, which is due to the fact that it does not take CL into account. Since 􏲆 g = 2μC WI m noxLD reducing ID by a factor of 4 reduces gm by a factor of 2, gm =1mA/V Since ro = VA , ID reducing ID by a factor of 4, increases ro by a factor of 4, Chapter 10–25 ro =80k􏱹 Thus, 􏱾 1 􏱿 1 × 80 + 2 fH = 0.1π × 80 × 10−9 2 = 0.95 MHz (ii) For the case, I is increased by a factor of 4, gm increases by a factor of 2, gm =4mA/V and ro decreases by a factor of 4, ro = 5 k􏱹 Thus, fH= 􏱾 1􏱿 0.1π 1 ×20+2 2 = 53.1 MHz ×5×10−9 10.65 R′L = ro ∥ RL = 20 k􏱹 ∥ 12 k􏱹 = 7.5 k􏱹 τgs = CgsRgs = CgsR′sig 10.66 Refer to Fig. 10.26(c). = 0.2 × 100 = 20 ns Vo = τ =C [R′ (g R′ +1)+R′] Vsig gdgdsigmL L 1/gm gmRL 1 +Rsig gm = 0.2[100(1.5 × 7.5 + 1) + 7.5] = 246.5 ns (a) CL =0 τCL = 0 τH =τgs +τgd =20+246.5=266.5ns = 1/5 ×5×10 1+1 5 = 8.3 V/V fP1 = 2πCgs 􏱾 1 􏱿 Rsig ∥ 1 gm fH = 1 2π×266.5×10−9 =597kHz 1 􏱿 ×103 fP1 = =239MHz 􏱾 (b) CL =10pF τCL =CLR′L =10×7.5=75ns 2π×4×10−12 1∥ 1 5 This figure belongs to Problem 10.67. Rsig E Vsig 􏰀􏰒 fP2 = 1 2π(CL + Cgd )RL C Vo RL (CL􏰀Cm) Chapter 10–26 re = 7.23 MHz Since fP1 ≫ fP2, fP2 will be dominant and fH ≃ fP2 = 7.23 MHz Cp 􏰒 Vp 􏰀 ≃ 2ro = 2 gmro gm Rgs = Rsig ∥ Rin = ro ∥ 2 gmVp B =1 2π(2+0.2)×10−12 ×10×103 Now, Rin=ro+RL =ro+ro 1+gmro 1+gmro 10.67 See figure above. Replacing the BJT with its high-frequency T model while neglecting ro and rx results in the equivalent circuit shown in the figure. (a) There are two separate poles, one at the input given by fP1 = 1 2πCπ (Rsig ∥ re) and the other at the output, given by 1 fP2 = 2π(C +C )R Q.E.D. L μ L (b) IC =1mA, IC 1 mA gm = V = 0.025V =40mA/V T re ≃ 1 = 25 􏱹 gm fP1 = 1 2π × 10 × 10−12(1 ∥ 0.025) × 103 2 gm ro×2 2 gm = ro ro+2 1gr+2 = ≃ τgs = CgsRgs ≃ 2Cgs gm 2gm 2mo ro =2 1grgm 2mo Ro = ro + Rsig + gmroRsig 1 1 Ro = ro + 2ro + 2gmroro ≃ 1 g m r o2 2 = ro ∥ 2gmro ≃ ro τgd = (CL + Cgd )Rgd = (Cgs + Cgd )ro τH = τgs + τgd =2Cgs +(Cgs+Cgd)ro gm Since gmro ≫ 1, we obtain τH ≃(Cgs+Cgd)ro and fH= 1 2π τH 1 fH = 2π(Cgs +Cgd)ro Rgd = RL ∥ Ro 12 = 652.5 MHz fP2 = 1 2π(1+1)×10−12 ×10×103 = 7.96 MHz Since fP2 ≪ fP1, fP2 will be dominant and fH ≃ fP2 = 7.96 MHz 10.68 Refer to Fig. 10.26 with RL = ro Rsig =ro/2 CL =Cgs Since fT= gm 2π(Cgs + Cgd ) τH =τgs +τgd =0.26+6.42 fH = 2πτH = 1 = 23.8 MHz 10.71 Ro = ro2 + ro1 + (gm2ro2)ro1 = 2 r o + g m r o2 = 2×20+2×20×20 = 840 k􏱹 Av = −gm1(Ro ∥ RL) = −2(840 ∥ 1000) = −913 V/V Using Eq. (10.109), we obtain τH = Rsig[Cgs1 + Cgd1(1 + gm1Rd1)] + Rd1(Cgd1 + Cdb1 + Cgs2) + (RL ∥ Ro)(CL + Cgd2) where Rd1 =ro1 ∥Rin2 ro2 + RL Rin2 = 1+gm2ro2 20 + 1000 = 1+2×20 = 24.9 k􏱹 Rd1 =20∥24.9=11.1k􏱹 Thus, τH = 100[20+5(1+2×11.1)] +11.1(5+5+20) + (1000 ∥ 840)(20 + 5) = 13587 + 333 + 11413 = 25, 333 ps = 25.33 ns fH = 1 =6.28MHz 2π × 25.33 × 10−9 10.72 (a) AM = −gmR′L where R′L =RL ∥ro =20∥20=10k􏱹 Thus, AM =−4×10=−40V/V τgs = CgsRgs =CgsRsig =2×20=40ns then fH = fT gm ro Q.E.D. 10.69 Refer to Example 10.9. To reduce fH to 200 MHz,τH mustbecome τH=1= 1 2πfH 2π × 200 × 106 = 795.8 ps Since τgs remains constant at 26.6 ps, τCL must be increased to τCL = 795.8 − 26.6 = 769.2 ps But, τCL = (Cgd + CL)Rgd thus, 769.2 = (5 + CL) × 18.7 ⇒CL +5=41.1fF CL = 36.1 fF Thus, the amount of additional capacitance to be connected at the output is 36.1 − 25 = 11.1 fF 10.70 ro = VA = 10 =100k􏱹 ID 0.1 Rsig =ro/2=50k􏱹 RL =ro =100k􏱹 gmro = 1.5×100 = 150 ro +RL 100+100 Rin = 1+g r = 1+150 =1.32k􏱹 mo Ro = Rsig + ro + gmroRsig = 50+100+150×50 = 7650 k􏱹 τgs = Cgs(Rin ∥ Rsig) τgs = 0.2(1.32 ∥ 50) = 0.26 ns Rgd =RL ∥Ro = 100 ∥ 7650 = 98.7 k􏱹 τgd =(Cgd +CL +Cdb)Rgd = (0.015 + 0.03 + 0.02) × 98.7 = 6.42 ns Chapter 10–27 = 6.68 ns 1 2π × 6.68 × 10−9 Rgd =Rsig(1+gmR′L)+R′L = 20(1+4×10)+10 = 830 k􏱹 τgd = CgdRgd = 0.2×830 = 166 ns τCL =CLR′L =1×10=10ns τ =τ +τ +τ H gs gd CL |AM|=gm(RL ∥Ro) 5000 = 1gmRo Chapter 10–28 2 = 2(gmro) 1 ⇒gmro =100 2 = 40+166+10 = 216 ns f = 1 2VA =100 VOV 2×10 ⇒VOV = 100 =0.2V H 2πτH 1 1 􏱾W 􏱿 I=μC V2 = 2π ×216×10−9 = 737 kHz D 2 n ox L OV = 1 ×0.2×50×0.22 GB≡|AM|fH =40×737=29.5MHz (b) AM = −gm1(Ro ∥ RL) where Ro = ro1 + ro2 + gm2ro2ro1 =2ro +gmro2 = 2×20+4×20×20 = 1640 k􏱹 AM =−4(1640∥20)=−79V/V Rin2 = ro2 + RL 1 + gm2ro2 = 20+20 ≃ 0.49 k􏱹 1+4×20 Rd 1 = ro1 ∥ Rin2 = 20 ∥ 0.49 = 0.48 k􏱹 Using Eq. (10.109), we obtain τH = Rsig[Cgs1 + Cgd1(1 + gm1Rd1)] + Rd1(Cgd1 + Cdb1 + Cgs2) + (RL ∥ Ro)(CL + Cgd2) = 20[2 + 0.2(1 + 4 × 0.48)] +0.48(0.2+0.2+2) +(20 ∥ 1640)(1+0.2) τH =51.7+1.15+23.7=76.6ns 2 = 0.2 mA ft = gm 2π(CL +Cgd) where 2ID gm = V OV 2×0.2 = 0.2 = 2 mA/V 2×10−3 2π(1 + 0.1) × 10−12 ft = = 289.4 MHz f3dB = ft = 289.4 = 57.9 kHz | AM | 5000 If the cascode transistor is removed, AM = −gm(ro ∥RL) ≃ −gmro = −100 V/V 10.74 (a) For the CS amplifier, |AM|=gm(ro ∥ro)= 1gmro fH = 1 2πCinRsig 2 11 􏱿􏱽 fH =2π×76.6×10−9 =2.08MHz GB≡|AM|fH =79×2.08=164MHz Note the increase in bandwidth and in GB. 10.73 20log|AM|=74dB ⇒ | AM | = 5000 Ro ≃ (gmro)ro RL = Ro = 􏱼 􏱾1 2π Cgs +Cgd 2gmro +1 (1) Rsig For the cascode amplifier, |AM|=gm(Ro ∥ro) = gm[(gmro)ro ∥ ro] ≃ gmro Thus, the gain increases by a factor of 2. fH = 1 2π Cin Rsig where Cin =Cgs +Cgd(1+gmRd1) Rd1 =ro ∥Rin2 = ro ∥ RL + ro gm ro ro + ro Rd1=ro∥ gr mo = ro ∥ 2 ≃ 2 gm gm 􏱾2􏱿 Cin=Cgs+Cgd 1+gm×gm = Cgs + 3Cgd fH = 1 (2) From (1) and (2), the ratio N of fH of the cascode amplifiertofH oftheCSamplifieris = 1 ×0.4×10×0.22 2 = 0.08 mA = 80 μA 10.75 (a) From Fig. 10.29, we have ft = gm 2π(CL +Cgd) √2μnCox(W/L) 􏲇 = 2π(CL +Cgd) ID Q.E.D. (1) (2) (3) (4) (5) (6) Chapter 10–29 (b) gm = 􏲇2μnCox(W/L) 􏲇ID 􏲌􏲋􏲋 ID 2π(Cgs + 3Cgd )Rsig ro = VA ID Ro = (gmro)ro AM = −gm(Ro ∥ RL) = −gm(Ro ∥ Ro) = −1gmRo 2 Substituting VOV =􏲊1 2μnCox(W/L) N = Thus, 􏱾1􏱿 Cgs +Cgd 2gmro +1 Cgs + 3Cgd Cgs + 1(gmro)Cgd N≃ 2 Cgs + 3Cgd (b) 50= 1gmro 2 ⇒gmro =100 Cgs + 1 ×100×0.1Cgs N=2 Cgs +3×0.1Cgs 1+5 = 1+0.3 =4.6 2V (c) gmro = A VOV 100= 2×10 VOV Q.E.D. μnCox = 0.4 mA/V2, W/L = 20, CL = 20 fF, obtain the following results in the table below. 10.76 Refer to Fig. 10.30. IC 1mA gm = V = 0.025V =40mA/V T rπ = β =100=2.5k􏱹 gm 40 rπ AM=−r+r+R gm(βro∥RL) Cgd =5fFandVA =10VinEqs.(1)–(6),we ⇒VOV =0.2V 1 􏱾W􏱿 π x sig = − 2.5 40(100 × 100 ∥ 2) I=μC V2 D 2 n ox L OV 2.5+0.05+5 ≃ −26.5 (V/V) This table belongs to Problem 10.75, part (b). ID (mA) ft (GHz) VOV (V) gm (mA/V) ro (k􏱹) Ro (M􏱹) AM (V/V) fH (MHz) 0.1 8 0.16 1.26 100 12.6 −7938 1 0.2 11.5 0.22 1.80 50 4.5 −4050 2.8 0.5 18 0.35 2.83 20 1.13 −1600 11.3 R′sig = rπ ∥ (rx + Rsig) = 2.5 ∥ (0.05 + 5) = 1.67 k􏱹 Rπ1 =R′sig =1.67k􏱹 rπ = β = 100 = 2.5 k􏱹 gm 40 (i) Rsig = 1 k􏱹 R′sig =rπ ∥Rsig =2.5∥1=0.71k􏱹 1 fP1 = 2π ×0.71×103(10+2×2)×10−12 ≃ 40 MHz 􏲈􏲉 =1 162 +402 =14.9MHz (ii) Rsig =10k􏱹 R′sig =rπ ∥Rsig =2.5∥10=2k􏱹 1 fP1 = 2π ×2×103(10+4)×10−12 = 5.7 MHz fP2 = 40 MHz 􏲈􏲆1 1 fH=1 2+ 2=5.6MHz 5.7 40 10.78 Refer to Fig. 10.30. IC =0.1mA gm = 0.1 mA = 4 mA/V 0.025 V rπ = β = 100 =25k􏱹 gm 4 ro = VA = 100 V = 1000 k􏱹 Chapter 10–30 􏱼r+R􏱽 R=r∥r o2 L c1 o1 e2 ro2 +RL/(β2 +1) ⎡⎤ =16MHz P2 2π(0+0+2)×10−12 ×2×103 ⎢ 100+2 ⎥ =100∥⎣0.025 2⎦ f= 1 100+ = 25.5 􏱹 Rμ1 =R′sig(1+gm1Rc1)+Rc1 = 1.67(1 + 40 × 0.0255) + 0.0255 = 3.4 k􏱹 Ro =β2ro2 =100×100=10,000k􏱹 τH =Cπ1Rπ1 +Cμ1Rμ1 +(Ccs1 +Cπ2)Rc1 +(CL +Ccs2 +Cμ2)(RL ∥Ro) = 10×1.67+2×3.4+(0+10)×0.0255 +(0+0+2)(2 ∥ 10,000) 101 fH =1 􏲈􏲆1 1 1 + 1 f2 f2 p1 p2 = 16.7+6.8+0.255+4 = 27.8 ns fH= 1 = 1 2πτH 2π × 27.8 × 10−9 = 5.7 MHz 10.77 (a) Gain from base to collector of Q1 = −1. Thus, Cin = Cπ1 +Cμ1(1+1) =Cπ1 +2Cμ1 fP1= 1 2πR′sigCin =1Q.E.D. 1 ID 0.1 mA = 0.25 k􏱹 2πR′sig(Cπ1 + 2Cμ) At the output node, the total capacitance is (CL +Cc2 +Cμ2)andsincero islarge,Ro willbe very large, thus the total resistance will be RL. Thus the pole introduced at the output node will have a frequency fP2, re ≃ g m fP2 = 1 2π(CL + Cc2 + Cμ2)RL (b) I = 1 mA gm = 1mA =40mA/V 0.025 V Q.E.D. = − =−1βgmro rπ AM =−r +r +R gm(βro ∥RL) π x sig gm(βro ∥ βro) rπ rπ + rπ 4 =−1 ×100×4×1000 4 =−100,000V/V R′sig =rπ ∥Rsig =rπ ∥rπ = 1rπ =12.5k􏱹 2 Rπ1 = R′sig = 12.5 k􏱹 ⎛⎞ ⎜ro+RL ⎟ Rc1=ro∥re⎝ro+ RL ⎠ β+1 ⎛⎞ ⎜r+βr⎟ =1000∥0.25⎜ o o ⎟ 2 = 1000 ∥ 12.5 = 12.3 k􏱹 Rμ1 = R′sig(1 + gm1Rc1) + Rc1 = 12.5(1+4×12.3)+12.3 = 639.8 k􏱹 Ro =βro =100×1000=100M􏱹 τH = Cπ1Rπ1 + Cμ1Rμ1 + (Ccs1 + Cπ2)Rc1 +(CL +Cc2 +Cμ2)(RL ∥Ro) To determine Cπ , we use fT= gm 2π(Cπ +Cμ) 4×10−3 1×109 = 2π(Cπ +Cμ) ⇒Cπ +Cμ =0.64pF Cπ = 0.64−0.1 = 0.54 pF τH = 0.54×12.5+0.1×639.8+0.54×12.3 +0.1× 1 ×100×1000 2 = 6.8+64+6.6+5000 ns Obviously the last term, which is due to the pole at the output node, is dominant. The frequency of the output pole is This estimate of ft is not very good (too high!). The other three poles have frequencies much lower than 3.18 GHz and will cause the gain to decrease faster, reaching the 0 dB value at a frequency lower than 3.18 GHz. Also note that fT of the BJTs is 1 GHz and the models we use for the BJT do not hold at frequencies approaching fT . ⎝r+βr⎠ ≃ 1000 ∥ 0.25 × β + 1 1 Chapter 10–31 10.79 AM = oβ+1o m R′L R′L + g where 1 1.82 AM = 1 =0.91V/V 1.82 + 5 Ro =ro ∥ 1 =20∥ 1 ≃0.2k􏱹=200􏱹 R′L =RL ∥ro ∥ gmb = 2 ∥ 20 = 1.82 k􏱹 gm 5 fZ = 2πCgs 5×10−3 = 2π ×2×10−12 = 398 MHz Next, we evaluate b and b : 12 gm 􏱾C􏱿􏱾C+C􏱿 b1=Cgd+gsRsig+gsLR′L gmR′L +1 gmR′L +1 􏱾2􏱿􏱾2+1􏱿 = 0.1+ 5×1.82+1 20+ 5×1.82+1 ×1.82 1 fP = −9 = 31.8 kHz = 5.96+0.54 = 6.50×10−9 s b2 = (Cgs + Cgd )CL + CgsCgd RsigR′ g m R ′L + 1 L (2+0.1)×1+2×0.1 = 5×1.82+1 ×20×1.82 = 8.3 × 10−18 √b √8.3 Q= 2= =0.44 b1 6.5 Thus, the poles are real and their frequencies can be obtained by finding the roots of the polynomial (1 + b1s + b2s2) = 1 + 6.5 × 10−9s + 8.3 × 10−18s2 which are ωP1 = 0.21 × 109 rad/s and 2π ×5000×10 fH ≃fP =31.8MHz Because the other poles are at much higher frequencies, an estimate of the unity-gain frequency can be found as ft =|AM|fP =105 ×31.8×103 =3.18GHz ωP 2 = 0.57 × 109 rad/s Thus, fP1=ωP1=33.4MHz 2π fP2=ωP2=90.7MHz 2π Since the two poles are relatively close to each other,anestimateoffH canbeobtainedusing 10.81 Refer to Fig. 10.31(c). Replacing Cgs with an input capacitance between G and ground, we get Ceq=Cgs(1−K) where g m R ′L K = 1 + g m R ′L then f=11+1 ′ Chapter 10–32 􏲈􏲉 Hf2f2 P1 P2 Ceq = Cgs/(1 + gmRL) and the total input capacitance becomes Cin =Cgd +Ceq =Cgd+ Cgs 1+gmR′L = 31.6 MHz 10.80 fH ≃fP1 ≃ 1 2πb1 where 􏱾C􏱿􏱾C+C􏱿 b=C+gs R+gs LR′ 1 gd gmR′L+1 sig gmR′L+1 L The frequency of the input pole is 1 1 gd sig gmR′L+1 sig Cgs (Rsig +R′L) g m R ′L + 1 fH ≃fP1 10.82 For a maximally flat response we have Q= 1 √ 􏱾C􏱿2 fP1= 􏱾 C 􏱿 2πRsig Cgd + gs ForCL =0, b1 =CgdRsig + For Rsig ≫ R′L, b≃CR+Cgs R 1+gmR′L = and Cgd + gs Rsig gmR′L +1 = 2π × 106 rad/s Thus, the transfer function will be Vo(s)= dcgain Vi(s) s2 +sω0 +ω2 􏱾 1 2πRsig Cgd + Cgs 􏱿 fH = For the given numerical values, Q.E.D. ω = ω 0 3dB g m R ′L + 1 Q 0 0.8 = √ fH = 2π ×100×103 = 156 kHz s2 +s 􏱼1 􏱽 = 0.8 (Note: An error was made in the first printing of the book and the values of Cgs and Cgd were exchanged.TheabovevalueoffH correspondsto the numbers in the first printing.) ForCgs =10pFandCgd =2pF, fH = 􏱼1􏱽 2π×100×103 2+ 10 5×(2 ∥ 20)+1 = 532 kHz ×10−12 For AM = 0.9, 0.9= gmRL 2×2π ×106 +(2π ×106)2 s2 +s 8.886×106s+39.48×1012 10.83 With gmb = 0 and ro large, we obtain R′L ≃ RL and AM = gmRL gmRL +1 10+ 2 ×10−12 5×(2 ∥ 20)+1 gmRL +1 ⇒gmRL =9 Now, for a maximally-flat response, Q = 1/√2. Using the expression for Q in Eq. (10.129), we get √ 􏲡 gmRL + 1 [(Cgs + Cgd )CL + CgsCgd ]RsigRL 2 × 109 = 40 × 10−3 2π(Cπ + Cμ) ⇒ Cπ + Cμ = 3.2 pF Cπ = 3.2−0.1 = 3.1 pF R′L = RL ∥ ro = 1 ∥ 20 = 0.95 k􏱹 R′sig = Rsig +rx = 1+0.1 = 1.1 k􏱹 R′ Chapter 10–33 Q = [Cgs +Cgd(gmRL +1)]Rsig +(Cgs +CL)RL 1 √9+1√[(10+1)10+10×1]×100×RL √2 = [10+1(9+1)]×100+(10+10)RL √√ 1 3RL √=􏱾􏱿 A=L 2RL MR′ 10 1+100 􏱾R 􏱿2 􏱾R 􏱿 ⇒ L −4 L +1=0 100 100 This equation results in two solutions, RL =27k􏱹andRL =373k􏱹 The second answer is not very practical as it implies the transistor is operating at gm = 9/373 = 0.024 mA/V, a very small transconductance!. We will pursue only the first answer. Thus, RL =27k􏱹 gm=0.33mA/V and the 3-dB frequency is found using Eq. (10.127): R′L+re+ sig β+1 AM= = 0.96 V/V 0.95 0.95 + 0.025 + 1.1 101 fZ= 1 2πCπre 1 2π × 3.1 × 10−12 × 25 = ≃ 2 GHz b1 = 􏲟 􏲣R′􏲤􏲠 􏲟 􏲣R′􏲤􏲠 Cπ+Cμ 1+ L R′sig+ Cμ+CL 1+ sig R′L re R′L rπ R′sig f3dB = f0 = 2π√b2 􏲉 = = 0.27 × 10−9 b2 = 0.025 1+ 0.95 + 1.1 0.025 2.5 1 􏱼􏱾􏱿􏱽 1++ re rπ 3.1+0.1 1+ 0.95 ×1.1+(3.1+0)×0.95 ω3dB = 􏲉 gmRL +1 RsigRL[(Cgs +Cgd)CL +CgsCgd] 9 + 1 100×27[(10+1)×10+10×1]×106 ×10−24 [(Cπ + Cμ)CL + Cπ Cμ]R′LR′sig R′L R′sig = = 5.55 Mrad/s f3dB = 884 kHz 10.84 Refer to Fig. 10.33. IC =1mA gm =40mA/V, re =25􏱹 rπ = 100 =2.5k􏱹 40 ro = VA = 20 = 20 k􏱹 IC 1 fT= gm 2π(Cπ +Cμ) 1++ re rπ = (0+3.1×0.1)×0.95×1.1 1+ 0.95 +1.1 0.025 2.5 = 8.2 × 10−21 √b Q= 2= b1 √8.2 × 10−21 0.27 × 10−9 =0.335 Thus, the poles are real and their frequencies can be found as the roots of the polynomial (1 + b1s + b2se) = 1+0.27×10−9s+8.2×10−21s2 􏱾 s􏱿􏱾 s􏱿 = 1+ω 1+ω P1 P2 ⇒ωP1 =4.25×109 rad/s ωP2 = 28.6 × 109 rad/s Thus, fP1 = 676 MHz fP2 = 4.6 GHz Thus, f3dB ≃ fP1 = 676 MHz = 40×10−15 ×40×103 = 1.6 ns Rgd = Rsig(gmR′L + 1) + R′L = 40(1.6×9.1+1)+9.1 = 631.5 k􏱹 τgd = CgdRgd = 5×631.5 = 3.16 ns τCL =(CL +Cdb)R′L = (100 + 5) × 9.1 = 955.5 ps = 0.96 ns τH = τgs + τgd + τCL = 1.6+3.16+0.96 = 5.72 ns fH= 1 2π τH =1 2π × 5.72 × 10−9 = 27.8 MHz 10.86 The common-mode gain will have a zero at fZ= 1 2π RSS CSS =1 2π ×100×103 ×1×10−12 = 1.59 MHz Thus, the CMRR will have two poles, one at fZ , i.e. at 1.59 MHz, and the other at the dominant pole of Ad , 20 MHz. Thus, the 3-dB frequency of CMRR will be approximately equal to fZ , f3dB = 1.59 MHz 10.87 At low frequencies, Ad =100V/V Acm = 0.1 V/V CMRR= Ad =1000or60dB Acm The first pole of CMRR is coincident with the zero of the common-mode gain, fP1 = 1 MHz 10.85 I = 0.4 mA 1 􏱾W􏱿 (a)I = μC V2 D 2 n ox L OV Chapter 10–34 0.2 = 1 × 0.4 × 16V 2 2 OV ⇒VOV =0.25V gm=2ID =2×0.2 VOV 0.25 = 1.6 mA/V (b)ro=VA =20 ID 0.2 = 100 k􏱹 RD ∥ro =10∥100 = 9.1 k􏱹 Ad =gm(RD ∥ro) = 1.6×9.1 = 14.5 V/V (c) Rsig small and the frequency response is determined by the output pole: fP2 = 1 2π(CL + Cgd + Cdb)(RD ∥ ro) =1 2π(100+5+5)×10−15 ×9.1×103 = 159 MHz fH ≃159MHz (d) Rsig = 40 k􏱹 τgs = CgsRgs = CgsRsig The second pole is coincident with the dominant pole of the differential gain, fP2 = 10 MHz A sketch for the Bode plot for the gain magnitude CL = 100 fF 􏱾 􏱿 (a)I=1k′ WV2 D 2 n L OV isshowninthefigure. CMRR (dB) 60 40 20 0 10.88 RSS = VA = 40 I 0.1 = 400 k􏱹 CSS =100fF fZ = 1 2πCSSRSS 2 Chapter 10–35 1 0.040= 2 ×0.2×100×VOV ⇒VOV =0.063V gm = 2ID = 2×0.04 =1.27mA/V VOV 0.063 (b) ro = VA = 20 = 500 k􏱹 ID 0.04 Ad =gm(RD ∥ro) 􏰒20 dB/decade 􏰒40 dB/decade 0.1 1 fP1 fP2 100 1000 f, MHz = 1.27(20 ∥ 500) = 1.27 × 19.2 = 24.4 V/V (c) fH ≃ fP2 = 1 10 2π(CL + Cgd + Cdb)(RD ∥ ro) =1 = 1 2π ×100×10−15 ×400×103 = 4 MHz 2π(100+10+10)×10−15 ×19.2×103 = 69.1 MHz (d) Rsig = 100 k􏱹 τgs =CgsRgs =CgsRsig =50×100=5ns Rgd =Rsig(1+gmR′L)+R′L = 100(1+1.27×19.2)+19.2 = 2559 k􏱹 τgd = CgdRgd = 10×2559 = 25.6 ns τCL =(CL +Cdb)R′L =(100+10)×19.2=2.1ns τH = τgs + τgd + τCL = 5+25.6+2.1 = 32.7 ns fH= 1 2πτH =1 2π ×32.7×10−9 =4.9MHz IfVOV ofthecurrentsourceisreducedbyafactor of 2 while I remains unchanged, (W/L) must be increased by a factor of 4. Assume L remains unchanged, W must be increased by a factor of 4. SinceCSS isproportionaltoW,itsvaluewillbe quadrupled: CSS =400fF The output resistance RSS will remain unchanged. Thus,fZ willdecreasebyafactorof4tobecome fZ =1MHz 10.89 I = 80 μA W =100, k′n =0.2mA/V2, VA =20V, L Cgs = 50 fF, Cgd = 10 fF, Cdb = 10 fF, RD =20k􏱹, 10.90 gm = IC = 0.25mA =10mA/V VT 0.025 V where R′sig =rπ ∥(Rsig +rx) = 10 ∥ 10.1 ≃ 5 k􏱹 fH = 1 =598kHz 2π ×53.2×10−12 ×5×103 GB=|Ad|fH =49.8×598=29.8MHz 10.91 The common-mode gain will have a zero at 1 fZ = 2π ×1×106 ×1×10−12 = 159 kHz Thus, the CMRR will have two poles: The first will be coincident with the zero of Acm, fP1 = 159 kHz and the second will be coincident with the pole of Ad, Chapter 10–36 rπ = β =100=10k􏱹 gm 10 gm 2π(Cπ +Cμ) fT = Cπ +Cμ = gm 2π fT = 10 × 10−3 2π × 500 × 106 = 3.2 pF Cπ = 3.2−0.5 = 2.7 pF (a) RC Vo /2 fP2 = 2 MHz 10.92 gm1,2 = 2ID1,2 = VOV 1,2 = 0.2mA =1mA/V Rsig 2 Vid I VOV 1,2 The figure shows the differential half-circuit and its high-frequency equivalent circuit. 0.2 V ro2=ro4=|VA|= 10V =100k􏱹 ID 0.1 mA Ad = gm1,2(ro2 ∥ ro4) (b) A ≡ Vo =− rπ g R dVr+r+RmC 1 =1(100∥100)=50V/V id π x sig fP1 = 2πC R =− 10 ×10×10 10+0.1+10 = −49.8 V/V (c) Cin =Cπ +Cμ(1+gmRC) = 2.7+0.5(1+10×10) = 53.2 pF fH = 1 2π Cin R′sig This figure belongs to Problem 10.90. L o = 1 2πCL(ro2 ∥ ro4) =1 2π × 0.2 × 10−12(100 ∥ 100) × 103 = 1 2π ×0.2×50×10−9 fP2 = gm3 2π Cm = 15.9 MHz Rsig rx Cm 􏰀 Vprp Cp 􏰒 Vo/2 RC Vid 􏰀 2􏰒 gmVp where gm3=2ID = I =1mA/V For the numerical values given, we have Ad = 20 =100V/V 0.2 I 0.2 gm = V = 0.2 = 1 mA/V Chapter 10–37 | VOV | | VOV | 1×10−3 fP2 = 2π×0.1×10−12 =1.59GHz fZ = 2gm3 =2fP2 =2×1.59=3.18GHz 2πCm 10.93 Ad = gm1,2(ro2 ∥ ro4) gm1,2 = 2ID = I OV f=I P1 2πCL|VA| | VOV | | VOV | ro2 =ro4 = |VA| = 2|VA| = 0.2 × 10−3 2π ×100×10−15 ×20 = 15.9 MHz ft = 15.9×100 = 1.59 GHz fP2 = 4AdfP1 = 4×100×15.9 = 6.37 GHz fZ = 2fP2 = 12.7 GHz A sketch of the Bode plot for |Ad| is shown in the figure. Ad = I |VOV | Ad = |VA| | VOV | fP1= 1 2πCLRo where Ro =ro2 ∥ro4 = |VA| I/2 I 􏱾2|VA| ∥ 2|VA|􏱿 I I I 2πCL| VA| fP1 = I fP2 = gm3 (1) (2) (3) 2π Cm where gm3 = 2ID = I | VOV | Cm = CL 4 fP2 = 4I | VOV | 10.94 See figure on next page. The mirror high-frequency equivalent circuit is shown in the figure. Note that we have neglected rx and ro. The model of the diode-connected transistor Q1 reduces to re1 in parallel with Cπ1. To obtain the current-transfer function Io (s)/Ii (s), wefirstdetermineVπ intermsofIi.Observethat the short-circuit at the output causes Cμ2 to appear in parallel with Cπ1 and Cπ2. Thus, 2π CL | VOV | fZ = 2gm3 = 2fP2 = 8I 2π CL | VOV | 2π Cm Dividing (2) by (1), we obtain fP2 =4|VA| =4Ad Q.E.D. fP1 | VOV | Vπ = Ii(s)􏱾 1 1 1 􏱿 (1) +s(Cπ1 +Cπ2 +Cμ2) Since f = 4 A f P2 dP1 is equal to GB, thus ft = Ad fP1 = |VA| I , the unity-gain frequency f t + re1 At the output node we have Io(s)=gm2Vπ −sCμ2Vπ (2) Combining Eqs. (1) and (2) gives rπ2 |VOV| 2πCL|VA| ft = I /| VOV | 2πCL gm 2πCL Io(s) Ii(s) = 􏱾 1 re1 gm2 − sCμ2 +s(Cπ1 +Cπ2 +Cμ2) = Q.E.D. + 1 􏱿 rπ2 This figure belongs to Problem 10.94. C1, B1, B2 Cm2 Io (s) sCm2Vp gm2Vp 1 2π [(2 × 10.7 + 2) × 10−12 × 25/1.01] Chapter 10–38 􏰀 Ii(s) Vp re1 Cp1 􏰒 Since the two transistors are operating at approximately equal dc bias currents, their small-signal parameters will be equal, thus Cp2 rp2 fP = =2π×23.4×25 I(s) g −sC o=􏱾m􏱿μ 1.01 × 1012 =274.8MHz fZ= gm 2πCμ Ii(s) 1 1+ 1 +s(2Cπ +Cμ) re β+1 = gmre 1−s(Cμ/gm) 1􏱼 􏱾1􏱿􏱽 40×10−3 1+ = 1+s (2Cπ +Cμ)re/ 1+ β+1 β+1 α 1 − s(Cμ/gm) 1 􏱼 􏱾 1 = 2π×2×10−12 =3.18GHz 10.95 Refer to Eqs. (10.146)–(10.153). For our 􏱿􏱽 1+β+1 1+s (2Cπ +Cμ)re/ 1+β+1 case, Gm= gm (1) = 1+2/β 􏱼 1 − s(Cμ/gm) 􏱾 1 + gmRs Ro=verylarge 1 1 􏱿􏱽 1+s (2Cπ +Cμ)re/ 1+β+1 Thus we see that the low-frequency transmission is Io(0)= 1 Ii 1+ 2 β R′L =RL ∥Ro =RL −gmRL Am =−GmRL = 1+g R (2) m s Rgd =Rsig(1+GmRL)+RL (3) Rgs = Rsig + Rs (4) 1 + gmRs as expected. The pole is at fP, fP ≃ 􏱼 1 􏱾 τH =CgsRgs +CgdRgd fH= 1 1􏱿􏱽 2π(2Cπ+Cμ)re/1+β 2π τH For the numerical values given: and the zero is at fZ= gm 2π Cμ For the numerical values given, IC 1mA gm = V = 0.025V =40mA/V T re ≃ 25 􏱹 Cπ+Cμ= gm 2πfT = 40 × 10−3 2π × 500 × 1006 = 12.7 pF Cπ =12.7−2=10.7pF (a)Rs=0 Gm =gm =5mA/V AM =−gmRL =−5×5=−25V/V Rgd = 100(1+5×5)+5 = 2605 k􏱹 τH = 10×100+2×2605 = 6.21 ns 1 fH = 2π×6.21×10−9 =26.6MHz GB = 26.6×25 = 641 MHz (b) Rs =100􏱹 Gm = 5 =3.33mA/V 1+5×0.1 AM =−3.33×5=−16.7V/V Rgd = 100(1+3.33×5)+5 = 1771.7 k􏱹 Rgs = 100+0.1 = 66.7 k􏱹 1+5×0.1 τgd = CgdRgd = 0.1×2040 = 204 ns τCL = CLR′L = 1 × 20 = 20 ns τH = τgs + τgd + τCL = 40+204+20 = 264 ns 1 fH = 2π ×264×10−9 = 603 kHz GB = 100×603 = 60.3 MHz fH = 2π×4.21×10−9 =37.8MHz (c) Rs =200􏱹 5 (b) WithRs =400􏱹, gm Gm = 1+5×0.2 = 2.5 mA/V AM =−2.5×5=−12.5V/V Gm = 1+gmRs = 5 = 1.67 mA/V Chapter 10–39 τH = 10×66.7+2×1771.7 = 4.21 ns 1 GB = 631 MHz Rgd = 100(1+2.5×5)+5 = 1355 k􏱹 Rgs = 100+0.2 = 50.1 k􏱹 1+5×0.4 Ro = ro(1 + gmRs) 1+5×0.2 τH = 10×50.1+2×1355 ns = 40(1 + 5 × 0.4) = 120 k􏱹 R′L =RL ∥Ro =40∥120=30k􏱹 AM =−GmR′L = −1.67 × 30 = −50 V/V Rgd =Rsig(1+GmR′L)+R′L = 20(1+1.67×30)+30 = 1050 k􏱹 τgd = CgdRgd = 0.1×1050 = 105 ns τCL = CLRCL = C L R ′L = 1 × 30 = 30 ns Rsig + Rs + RsigRs/(ro + RL) = = 10.25 k􏱹 τgs = CgsRgs = 2×10.25 = 20.5 ns τH = τgs + τgd + τCL = 20.5+105+30 = 155.5 ns 1 fH = 2πτH = 1 = 1.02 MHz 2π × 155.5 × 10−9 GB = 51.2 MHz 1 fH = 2π ×3.21×10−9 = 49.6 MHz GB = 49.6×12.5 = 620 MHz A summary of the results is provided in the following table: Observe that increasing Rs trades off gain for bandwidth while GB remains approximately constant. 10.96(a)A =−gR′ MmL where R′L = RL ∥ ro = 40 ∥ 40 = 20 k􏱹 AM =−5×20=−100V/V τgs = CgsRgs = CgsRsig = 2 × 20 = 40 ns Rgd =Rsig(1+gmR′L)+R′L = 20(1+5×20)+20 = 2040 k􏱹 Rgs= 􏱾r􏱿 1+gR o Rs =0 Rs = 100 􏱹 Rs = 200 􏱹 | AM | (V/V) 25 16.7 12.5 fH | (MHz) 26.6 37.8 49.6 GB (MHz) 641 631 620 m s ro+RL 20+0.4+20×0.4/(40+40) 􏱾 40 􏱿 1+5×0.4 40+40 = 2π ×0.2×10−12 ×100×103 = 7.96 MHz (b) |AM|=20V/V fH = 7.96 = 398 kHz 1+ 2gmRs For Rsig ≫ Rs, Rgs ≃ Rsig(1+Rs/2ro) 1 + (k/2) 1=1 Chapter 10–40 Rsig + Rs + RsigRs/(ro + RL) MH Rgs= 􏱾r􏱿 10.97 (a) GB=|A |f =1 msro+RL 1+gR o 2πCgdRsig Rsig +Rs +RsigRs/2ro 20 (c) A = g r Forro ≫Rs, Rgs ≃ Rsig 0 mo 100 = 5×ro 1+(k/2) τgs = CgsRgs = CgsRsig ⇒ro =20k􏱹 G=gm=5 ′′ m 1 + gmRs 1 + gmRs Utilizing the expressions for RL and GmRL derived earlier, we obtain 1 + (k/2) Rgd =Rsig(1+GmR′L)+RL 􏱼 A 􏱽 􏱾1+k􏱿 Rgd=Rsig1+0 +ro M mL m s 20+20(1+gmRs) ⇒1+gmRs =4 ⇒Rs = 3 =0.6k􏱹=600􏱹 gm 10.98Gm=gm =gm 1+gmRs 1+k R′L=RL∥Ro =ro ∥ro(1+gmRs) =ro ∥ro(1+k) = ro ×ro(1+k) ro +ro(1+k) = r 1 + k o 2 + k A M = − G m R ′L = − g m r o 2+k Ro =ro(1+gmRs)=20(1+gmRs) 2+k 2+k ′ 􏱾􏱿􏱾􏱿 R′L =RL ∥Ro =20∥20(1+gmRs) τ=CR= A =−GR gd gs gd 20 = 5 1+gmRs gd sig 2+k gd o 2+k [20 ∥ 20(1 + gmRs)] 4(1+g R)= 20×20(1+gmRs) τCL =CLR′L =C r 1+k L o2+k Thus, τH = τgs + τgd + τCL 1+(k/2) 2+k 􏱾1+k􏱿 􏱾1+k􏱿 CR1+A0 +Cr1+k CR􏱾A􏱿 = gs sig +CgdRsig 1+ 0 +Cgdro 2+k +CLro 2+k CR 􏱾 A􏱿 = gs sig +CgdRsig 1+ 0 1+(k/2) 2+k 􏱾1+k􏱿 +(CL +Cgd)ro 2+k Q.E.D. Thus, AM = −A0 Q.E.D. 10.99 Substituting the given numerical values in the expressions for AM and τH given in the statement for Problem 10.98 and noting that A0 =gmro =5×40=200,weobtain fH= 1 2π τH and GB = |AM |fH 2+k we obtain the results in the following table. τgs = CgsRgs = CgsRsig Rgd =Rsig(1+gmR′L)+R′L = Rsig(1 + gmro) + ro τgd =CgdRgd =Cgd[Rsig(1+gmro)+ro] τ C L = C L R ′L = C L r o Thus, τH = τgs + τgd + τCL = CgsRsig + Cgd [Rsig(1 + gmro) + ro] + CLro Q.E.D. For the given numerical values, AM =−1×20=−20V/V τH = 20×20+5[20(1+1×20)+20]+10×20 = 400+2200+200 = 2800 ps = 2.8 ns fH=1= 1 2πτH 2π × 2.8 × 10−9 = 56.8 MHz GB = 20×56.8 = 1.14 GHz (b) From Fig. 1 we see that Chapter 10–41 k |AM|, V/V τH ns fH (MHz) GB (MHz) 0 100 264 0.603 60.3 1 66.7 191.3 0.832 55.6 2 50 155 1.03 51.5 3 40 133.2 1.19 47.6 4 33.3 118.7 1.34 44.6 5 28.6 108.3 1.47 42.0 6 25 100.5 1.58 39.5 7 22.2 94.4 1.69 37.5 8 20 89.6 1.78 35.6 9 18.2 85.7 1.86 33.9 10 16.7 82.3 1.93 32.2 11 15.4 79.6 2.00 30.8 12 14.3 77.2 2.06 29.5 13 13.3 75.1 2.12 28.2 14 12.5 73.3 2.17 27.1 15 11.8 71.6 2.22 26.2 ToobtainfH =2MHz,weseefromthetablethat k = 11 Thus, 1+gmRs =11 ⇒ Rs = 10 = 5 k􏱹 2 The gain achieved is | AM | = 15.4 V/V 10.100 (a) Refer to Fig. P10.100(a). Since the total resistance at the drain is ro, we have AM = −gmro Q.E.D. This figure belongs to Problem 10.100, part (b). Rsig 􏰀 Vg1 Vsig =1 Vg2 = ro1 1 +ro1 gm1 Vg1 Vo = −gm2ro2 Thus, Vg2 AM =1× ro1 ×−gm2ro2 1 +ro1 gm1 = − ro1 1/gm1 + ro1 (gm2ro2) Q.E.D. Vo 􏰒 􏰒gm1 􏰀 Q2 ro2 Vg2 􏰒 Q1 ro1 Vsig 􏰀 Vg1 1 􏰿 Figure 1 Next we evaluate the open-circuit time constants. Refer to Fig. 2. τgs1 = Cgs1Rgs1 = C Rsig + ro1 gs1 1 + gm1ro1 Cgs2: Capacitor Cgs2 sees the resistance between G2 and ground, which is the output resistance of source follower Q1, Rgs2= 1 ∥ro1 gm1 Chapter 10–42 Cgd1 Rsig G 1 Cgs1 Q C gd2 1 G Q2 CL 2 Cgs2 Thus, τgs2 = Cgs2 􏱾1􏱿 g ∥ ro1 m1 Figure 2 Cgd1: Capacitor Cgd1 is between G1 and ground and thus sees the resistance Rsig, Rgd1 = Rsig τ=CR gd1 gd1 sig Cgs1: To find the resistance Rgs1 seen by capacitor Cgs1, we replace Q1 with its hybrid-π equivalent circuit with Vsig set to zero, Cgd1 = 0, and Cgs1 replaced by a test voltage Vx . The resulting equivalent circuit is shown in Fig. 3. Cgd 2 : Transistor Q2 operates as a CS amplifier with an equivalent signal-source resistance equal to the output resistance of the source follower Q , 􏱾1􏱿2 that is, g ∥ ro1 and with a gain from gate to m1 drain of gm2ro2. Thus, the formula for Rgd in a CS amplifier can be adapted as follows: 􏱾1􏱿 Rgd2 = g ∥ro1 (1+gm2ro2)+ro2 m1 􏱼􏱾 􏱿 􏱽 and thus, 1 ∥r (1+g r )+r m1 CL: Capacitor CL sees the resistance between D2, and ground which is ro2, τCL = CLro2 Summing τgd1, τgs1, τgs2, τgd2 and τCL gives τH in the problem statement. Q.E.D. For the given numerical values: AM =− 20 (1×20) 1+20 = −19 V/V τgd1 = Cgd1Rsig = 5×20 = 100 ps τ =C Rsig+ro1 gs1 gs1 1 + gmro1 =20 20+20 =38ps 1+1×20 τ =C gd2gd2go1 m2o2o2 Rsig Vg1G1 D1 Ix Vx􏰀􏰒gV ro1 m1 x S1 Ix Vs1 Figure 3 Analysis of the circuit in Fig. 3 proceeds as follows: Vg1 = IxRsig V=V−V=IR−V s1 g1 x xsig x Node equation at S1, Vs1 Ix = gm1Vx − r o1 = gm1Vx − IxRsig − Vx 􏱾 􏱿ro1􏱾 􏱿 Ix 1+Rsig =Vx gm1+ 1 􏱾1􏱿 τgs2 = Cgs2 ∥ ro1 gm1 = 20 × (1 ∥ 20) = 19 ps ro1 VR+r Rgs1 ≡ x = sig o1 Ix 1 + gm1ro1 ro1 􏱼􏱾1􏱿 􏱽 τgd2 =Cgd2 g ∥ro1 (1+gmro2)+ro2 m1 = 5[(1 ∥ 20)(1+20)+20] = 200 ps Thus, τCL =CLro2 =10×20=200ps τH = 100+38+19+200+200 = 557 ps fH= 1 2π τH = 1 2π × 557 × 10−12 = 286 MHz GB = 19 × 286 = 5.43 GHz Thus, while the dc gain remained approximately thesamebothfH andGBincreasedbyafactorof about5! 10.101 At an emitter bias current of 0.1 mA, Q1 and Q2 have gm =4mA/V re =250􏱹 β 100 rπ=g=4=25k􏱹 showsthecircuitwithVsig =0andthefour capacitances indicated. Again, recall that here RL = ro2. Also, in our present circuit there is a capacitance CL at the output. Chapter 10–43 Capacitance Cμ1 sees a resistance Rμ1, R=R∥R μ1 sig in = 10 k􏱹 ∥ 2.5 M􏱹 ≃ 10 k􏱹 To find the resistance Rπ 1 we refer to the circuit in Fig. 10.40(c) where Rin2 is considered to include ro2 , Rin2 =25k􏱹∥1000k􏱹=24.4k􏱹 We use the formula for Rπ1 given in Example 10.13: Rπ1 = Rπ1 = Rsig + Rin2 1+Rsig +Rin2 rπ1 re1 10 24.4 =347􏱹 1+ + 10+24.4 m ro = VA = 100 IC 0.1 C+C=gm = 1000 k􏱹 25 0.25 Capacitance Cπ 2 sees a resistance Rπ 2 : Rπ2 = Rin2 ∥ Rout1 π μ 2πfT 􏱼 R􏱽 re1 + sig 4×10−3 = 2π×200×106 =3.2pF Cμ = 0.2 pF Cπ = 3 pF To determine Rin and the voltage gain AM , refer to the circuit in Fig. 10.40(a). Here, however, RL is ro2 . β1 + 1 􏱼 􏱽 Rin2 =rπ2 =25k􏱹 Rin = (β1 + 1)[re1 + (ro1 ∥ Rin2)] = 101[0.25 + (1000 ∥ 25)] = (1 + 4 × 1000) × 0.344 + 1000 = 2376 k􏱹 WecandetermineτH from τH = Cμ1Rμ1 + Cπ1Rπ1 + Cμ2Rμ2 + Cπ2Rπ2 + CLro = 0.2×10+3×0.347+0.2×2376 +3×0.344+1×1000 τH = 2+1+475.2+1+1000 = 1479.2 ns Observe that there are two dominant capacitances: the most significant is CL and the second most significant is Cμ2. ≃ 2.5 M􏱹 Vb1 = Vsig Vb2 = Vb1 Rin = Rin +Rsig 2.5M􏱹 2.5 M􏱹+10 k􏱹 ≃1V/V = rπ2 ∥ ro1 ∥ =25∥1000∥ 0.25+ 10 101 Capacitance Cμ2 sees a resistance Rμ2: = 344 􏱹 Rμ2 = (1 + gm2ro2)(Rin2 ∥ Rout1) + ro2 (Rin2 ∥ro1) (Rin2 ∥ ro1) + re1 = 25 ∥ 1000 = 0.99 ≃ 1 V/V (25 ∥ 1000) + 0.25 Vo V = −gm2ro2 = −4 × 1000 = −4000 V/V b2 Thus, Vo AM=V =−4000V/V sig TodeterminefH weusethemethodof open-circuit time constants. Figure 10.40(b) fH= 1 2πτH = 1 2π × 1479.2 × 10−9 =107.6kHz 10.102 fP1 = =􏱾1􏱿 gm = 2ID VOV 0.2 mA 0.2 V = 2(I /2) VOV = 1 mA/V 􏱾 1 􏱿 2πR Cπ+C Chapter 10–44 I VOV sig 2 μ = Vo = RD = Vsig 2/gm 11 2π ×12×103 = 2 MHz 12.2 +0.5 2 ×10−12 =2gmRD =2×1×50=25V/V 1 The high-frequency analysis can be performed in an analogous manner to that used in the text for the bipolar circuit. Refer to Fig. 10.42(b) and adapt the circuit for the MOS case. Thus, fP2 = 2πR C Lμ fP1 = 􏱾 1 2πR Cgs +C = 31.8 MHz Thus, fP1 is the dominant pole and 􏱿 fH ≃ fP1 = 2 MHz 10.104 Using an approach analogous to that utilized for the BJT circuit (Fig. 10.42), we see that there is a pole at the input with frequency fP1: = 1 2π ×10×103 ×0.5×10−12 2π × 100 × 103 = 637 kHz 2 + 0.5 sig 2 gd 1 =􏱾4􏱿 × 10−12 fP1 = f=1 1 􏱾 1 Cgs +C 􏱿 +0.1 ×10−12 and 2πR sig 2 gd P2 2πRDCμ fP1 = 􏱾2 1 2π×20×103 =2 􏱿 2π ×50×103 ×0.5×10−12 = 6.37 MHz Since fP2 ≃ 10fP1, the pole at fP1 will dominate and fH ≃fP1 =637kHz 10.103 gm = IC ≃ 1mA =40mA/V = 7.2 MHz, and a pole at the output with frequency fP2, 1 fP2 = 2π(Cgd + CL)RL =1 2π ×(0.1+1)×10−12 ×20×103 = 7.2 MHz Thus, fP1=fP2=7.2MHz The midband gain AM is obtained as AM=RL=1gmRL 2/gm 2 =1×5×20=50V/V 2 Thus, the amplifier transfer function is Vo(s)=􏱾 50 􏱿 Vsig(s) 1+ s 2 2π ×7.2×106 VT 0.025 V rπ = β =120=3k􏱹 re ≃ 25 􏱹 gm 40 Rin =2rπ =6k􏱹 Vo=Rin αRL Vsig Rin + Rsig 2re ≃ 6 6 + 12 × 10 2 × 0.025 = 66.7 V/V Cπ+Cμ= gm 2πfT 40×10−3 = 2π ×500×106 = 12.7 pF Cμ = 0.5 pF Cπ = 12.2 pF 􏲀􏲀􏲀V 􏲀􏲀􏲀 50 􏲀o􏲀=􏱾 􏱿2 Vsig 1+ ω 2π ×7.2×106 􏲀􏲀􏲀V 􏲀􏲀􏲀 50 Denoting Atω=ω3dB,􏲀 o􏲀= √ ,thus Chapter 10–45 Vi 2 gm1 =G, √􏱾􏱿2 g0 2=1+ ω3dB m2 2π × 7.2 × 106 then, 􏲡√ VoG0 f3dB = 2−1×7.2MHz = 4.6 MHz 10.105 (a) A CS amplifier for which the gain is low so that the Miller effect is negligible and for which Cgd ≪ Cgs has an input capacitance that is approximately given by V=−s The dc gain G0 can be related to W1 and W2 as follows: The bias current I divides between Q1 and Q2 as, 􏱾􏱿 I=1k′ W1 V2 1 2 n L OV i1+ ωT /(G0 + 1) C≃C 1􏱾W􏱿 in Now, gs I=k′ 2V2 2 2n L OV where we have utilized the fact that Q1 and Q2 are operating at the same value of VOV . Thus, ωT = thus, gm + Cgd ≃ gm Cgs Cgs Cgs≃gm I1 = W1 I2 W2 Now, 2I1 gm1 = VOV and gm2 = 2I2 VOV Thus, ωT and Cin≃gm ωT If ro can be neglected, the hybrid-π equivalent circuit reduces to that shown in Fig. P10.105(a). (b) Replacing Q1 by its equivalent circuit and replacing the diode-connected Q2 by its equivalent circuit which consists of the input capacitance (gm2/ωT ) and the resistance 1/gm2 and including the input capacitance of the subsequent stage (gm1/ωT ) gives the equivalent circuit shown in the figure above. The gain can easily be determined as gm1 =I1 =W1 gm2 I2 W2 and G0=gm1 =W1 gm2 W2 (c) G0 = 3 W1 ⇒W=3 2 W1 =3×25=75μm Vo = − Vi = − gm1 gm2 gm1 gm2 +sgm1 +gm2 ωT 1 + s1 + gm1/gm2 ωT 1 This figure belongs to Problem 10.105, part (b). Q1 Q2 Subsequent stage 􏰀 V gm1 i vT 􏰒 􏰀 gm2 vT gm1 V vT o 􏰒 gm1Vi 1 gm1 I =1k′W1V2 1 2nLOV = 1 × 0.2 × 75 × 0.32 = 1.35 mA 2 0.5 I2 = 1 ×0.2× 25 ×0.32 =0.45mA 2 0.5 I =I1 +I2 =1.35+0.45=1.8mA 10.107 (a) VG1 􏰔 2 V 􏰀5 V 5􏰒2 􏰔 1 mA 0 Chapter 10–46 RG 􏰔 10 M􏰵 3 3 k􏰵 f3dB = fT = G0 + 1 12 = 3 GHz 3 + 1 􏰀2 V 1 mA Q2 6.8 k􏰵 􏰀 Q1 􏰒0 10.106 (a) For each of Q1 and Q2, gm= 2ID =2×0.1=1mA/V VGS 􏰔 1.3 V 􏰀0.7 V 0.1 mA 0.1 mA | VOV | 0.2 ro = |VA| = 10 = 100 k􏱹 ID 0.1 gmro = 100 Vo = (gmro)2 = 10,000 V/V Vsig (b) τgs1 = CgsRsig = 20 × 10 = 200 ps Rgd1 = Rsig(1 + gm1ro1) + ro1 = 10(1 + 100) + 100 = 1110 k􏱹 τgd1 =Cgd1Rgd1 =5×1110=5550ps At the drain of Q1 we have (Cdb1 + Cgs2) and the resistance seen is ro: τd1 =(Cdb1 +Cgs2)ro = (5 + 20) × 100 = 2500 ps Rgd2 = ro1(1 + gm2ro2) + ro2 = 100(1+100)+100 = 1110 k􏱹 τgd2 =Cgd2Rgd2 =5×1110=5550ps τd2 = Cdb2ro2 Figure 1 The dc analysis is shown in Fig. 1. It is based on VS1 = VBE2 = 0.7 V. Neglecting IB2, we obtain rπ2 = τH = τgs1 +τgd1 +τd1 +τgd2 +τd2 gm2 0.7 V ID1 = 6.8 k􏱹 ≃ 0.1 mA I =1k′(W/L)V2 D12n OV 0.1 = 1 × 2 × V 2 2 OV Q.E.D. ⇒VOV ≃0.3V VGS =Vt +VOV =1+0.3=1.3V VG1 =0.7+1.3=2V VC1 =VG1 =2V IC2 = 5 − 2 = 1 mA 3 2 × 0.1 0.3 Q.E.D. = 0.67 mA/V = Cgs =Cgd =1pF (b) gm1 = 2ID1 VOV gm2 = IC = 1mA =40mA/V = 5 × 100 = 500 ps β 200 40 VT 0.025 V 40 × 10−3 = 2π ×600×106 = 10.6 pF Cμ2 = 0.8 pF C =9.8pF π2 = Cπ2+Cμ2= gm2 = 200+5550+2500+5550+500 = 14,300 ps = 14.3 ns =5k􏱹 2πfT2 fH =1 2πτH = 1 2π × 14.3 × 10−9 = 11.1 MHz (c) Q1 acts as a source follower, thus Vb2 = 6.8k􏱹∥rπ2 The pole due to C2 has a frequency f2: f2 = 1 Chapter 10–47 Vi 1 +(6.8k􏱹∥rπ2) gm1 = 2πC2(3+1)×103 1 = 40 Hz (6.8 ∥ 5) = 1.5+(6.8 ∥ 5) = 0.66 V/V Neglecting RG , we obtain Vo = −gm2(3 k􏱹 ∥ 1 k􏱹) Vb2 = −40(3 ∥ 1) = −30 V/V Thus, Vo = 0.66 × −30 Vi ≃ −20 V/V Using Miller’s theorem, the input resistance Rin is found as Rin=RG =10M􏱹 2π ×1×10−6 ×4×103 Since f2 ≫ f1, the lower 3-dB frequency fL will be fL ≃f2 =40Hz (d) τgd1 = Cgd1(Rin ∥ Rsig) = 1 × 10−12(476 ∥ 100) × 103 = 82.6 ns To determine the resistance Rgs seen by Cgs , refer to Fig. 2. We can show that Rgs ≡ Vx = Rsig + Rs 1 − Vo Vi = 476 k􏱹 Vi = Rin Vsig Rin + Rsig where Ix 1 + gm1Rs Rs =6.8k􏱹∥rπ2 = 6.8 ∥ 5 = 2.88 k􏱹 Rgs = 100+2.88 = 35.1 k􏱹 τgs = CgsRgs = 1×10−12 ×35.1×103 = 35.1 ns τπ2 = Cπ2(rπ1 ∥ 6.8 k􏱹) = 9.8×10−12 ×2.88×103 = 28.2 ns = 476 = 0.83 V/V 476 + 100 1 − (−20) 1+0.67×2.88 Vo V = 0.83×20 = 16.5 V/V sig (c) The pole due to C1 has a frequency f1: f=1 􏱾1􏱿 1 2πC1(Rsig + Rin) Rμ2 = g ∥ 6.8 k􏱹 [1+gm2(3 ∥ 1)]+(3 ∥ 1) m1 =1􏱾3􏱿 2π×0.1×10−6(100+476)×103 = 2.8 Hz This figure belongs to Problem 10.107, part (d). =(1.5∥6.8) 1+40×4 +0.75 = 38.8 k􏱹 Rsig Ix Rsig 􏰔 100 k􏰵 Cgs 􏲚 􏰀􏰒 Vx Ix Rs gmVx Rs 􏰔 6.8 k􏰵 // rp2 Figure 2 τμ2 =Cμ2Rμ2 =0.8×38.8=31.1ns τH =τgd +τgs +τπ2 +τμ2 = 82.6+35.1+38.8+31.1 =187.6ns R′sig =rπ ∥Rsig =5∥10=3.33k􏱹 Rπ1 =R′sig τπ1 = Cπ1Rπ1 = 6×3.33 = 20 ns Rc1 =re2 =50􏱹 Rμ1 = R′sig(1 + gm1Rc1) + Rc1 = 3.33(1+40×0.05)+0.05 = 3.33(1+2)+0.05 = 10.05 k􏱹 τμ1 = Cμ1Rμ1 = 2×10.05 = 20.1 ns τc1 = Cπ2Rc1 = 6 × 0.05 = 0.3 ns τμ2 =Cμ2RL =2×10=20ns τH = 20+20.1+0.3+20 = 60.4 ns fH= 1 = 1 2πτH 2π×60.4×10−9 = 2.6 MHz (c) This is a CC-CB cascade similar to the circuit analyzed in Fig. 10.42. There are two poles: one at the input, fH = 1 2π τH 1 = 2π ×187.6×10−9 = 848 kHz 10.108 All transistors are operating at IE = 0.5 mA. Thus, gm ≃ 20 mA/V re≃50􏱹 rπ = β = 100 =5k􏱹 gm 20 ro = very high (neglect) rx = very small (neglect) gm 20 × 10−3 Cπ+Cμ=2πf =2π×400×106 1 2π ×10×103 ×408×10−12 fH=1/ 1+1 f2 f2 Chapter 10–48 f = 1􏱾 T P1Cπ 􏱿 =8pF Cμ = 2 pF Cπ = 6 pF (a) CE amplifier AM=−rπ gmRL rπ +Rsig = − 5 × 20 × 10 = −66.7 V/V 5+10 Since CL = 0, we can obtain a good estimate of fH usingtheMillerapproximation: Cin = Cπ +Cμ(gmRL +1) = 6+2(20×10+1) = 408 pF fH = 1 2π Rsig Cin 2π(Rsig ∥2rπ) 2 +Cμ 1 fP1 = 2π(10 ∥ 10)×103(3+2)×10−12 =1 2π ×5×5×10−9 = 6.4 MHz and one at the output, 1 = 8 MHz Since the two poles are relatively close to each other, we use the root-sum-of-the-squares formula toobtainanestimateforfH: 􏲉 1 2πRLCμ fP2 = = 2π ×10×103 ×2×10−12 = =39kHz P1 P2 􏲆1 1 (b) This is a cascode amplifier. Refer to Fig. 10.30 for the analysis equations. AM = − rπ gm(βro ∥ RL) rπ + Rsig ≃ − rπ gmRL rπ +Rsig = −66.7 V(V) (same as the CE in (a)) =1/ 6.42 +82 =5MHz AM = = RL 2re + Rsig β + 1 10 ≃ 50 V/V 2×0.05+ 10 101 (d) This is a CC-CE cascade similar to the circuit analyzed in Example 10.13. Rin =(β1 +1)(re1 +rπ2) = 101(0.05 + 5) = 510 k􏱹 Rμ2 = (1 + gm2RL)(Rin2 ∥ Rout1) + RL 􏱼 􏱾 10 􏱿􏱽 =(1+20×10) 5∥ 101 +0.05 +10 = 39.1 k􏱹 τμ2 =Cμ2Rμ2 =2×39.1=78.2ns τH = 19.6+0.9+0.9+78.2 = 99.6 ns 1 fH =2π×99.6×10−9 =1.6MHz (e) This is a folded cascode amplifier. The analysis is identified to that for (b) above. AM = −66.7 V/V fH =2.6MHz (f) This is a CE-CB cascade. The analysis is identical to that for case (c) above. AM = 50 V/V fH =5MHz Summary of Results Chapter 10–49 Vb1 = Rin Vsig Rin +Rsig = 0.98 V/V = 510 510+10 rπ2 = r +r 5 = 5+0.05 =0.99V/V π2 e1 = −gm2RL = −20 × 10 = −200 V/V AM = Vo =−0.98×0.99×200=−194V/V Vsig Rμ1 = Rsig ∥ Rin = 10 ∥ 510 = 9.81 k􏱹 τμ1 = Cμ1Rμ1 = 2×9.81 = 19.6 ns Rπ1 = Rsig +Rin2 1+Rsig +Rin2 rπ1 re1 Vb2 V b1 Vo Vb2 10+5 = 0.15 k􏱹 1+10+ 5 = τπ1 = Cπ1Rπ1 = 6×0.15 = 0.9 ns Case Configuration AM (V/V) fH (MHz) a CE −66.7 0.039 b Cascode −66.7 2.6 c CC-CB 50 5 d CC-CE −194 1.6 e Folded Cascode −66.7 2.6 f CC-CB 50 5 5 0.05 Rπ2 = Rin2 ∥ Rout1 􏱾R􏱿 =rπ2 ∥ sig +re1 β1 + 1 􏱾10 􏱿 +0.05 =0.15k􏱹 τπ2 = Cπ2Rπ2 = 6×0.15 = 0.9 ns =5∥ 101