5.1: Solution:
(a) For the dc analysis, refer to the figure:
VCC=IERC+IBRB+VBE →1.5=IE×1+ IE ×47+0.7→IE=1.5−0.7=0.546mA β+1 1+ 47
IC = αIE = 0.99×0.546 = 0.54mA
(b)gm=IC =0.54mA=21.6mA VT 0.025V V
rπ = β = 100 = 4.63 kΩ gm 21.6
(c) Vo = −gm(RC||RL) = −21.6(1||1) = −10.8 V Vb
(d) Using Miller’s theorem, the component of Rin due to RB can be found as: Rin1 = RB = 47 kΩ = 4 kΩ
101
1 − (Vo) 1 − (−10.8) Vb
Rin = Rin1||rπ = 4||4.63 = 2.14 kΩ
(e) GV = Rin ×Vo = 2.14 ×−10.8=−7.4 Rin +Rsig Vb 1+2.14
1
(f) Cin=Cπ+(1+|Vo|)Cμ Vb
Where:
gm
21.6 × 10−3
Cπ +Cμ =2πf =2π×600×10−6 =5.73pF T
Cπ =5.73−0.8=4.93pF
Cin =4.93+(1+10.8)×0.8=14.37pF
(g) R′ =R ||R =2.14kΩ||1kΩ=0.68kΩ sig in sig
f= 1 = 1
=16.3MHz
H 2πC R′ 2π×14.37×10−12 ×0.68×103 in 𝑠𝑖𝑔
5.2: Solution:
2
Since the total resistance at the drain is 𝑟 , we have: 𝑜
𝐴 =−𝑔 𝑟 𝑀𝑚𝑜
𝜏 =𝐶𝑅 =𝐶𝑅 𝑔𝑠 𝑔𝑠 𝑔𝑠 𝑔𝑠 𝑠𝑖𝑔
𝑅 =R_sig (1+g_m R_L^’ )+R_L^’= 𝑅 (1 + 𝑔 𝑟 ) + 𝑟 𝑔𝑑 𝑠𝑖𝑔 𝑚𝑜 𝑜
𝜏 =𝐶 𝑅 =𝐶 [𝑅 (1+𝑔 𝑟)+𝑟] 𝑔𝑑 𝑔𝑑𝑔𝑑 𝑔𝑑𝑠𝑖𝑔 𝑚𝑜 𝑜
𝜏 =𝐶𝑅′=𝐶𝑟 𝑐𝐿 𝐿𝐿 𝐿𝑜
Thus,
𝜏 =𝜏 +𝜏 +𝜏 =𝐶 𝑅 +𝐶 [𝑅 (1+𝑔 𝑟)+𝑟]+𝐶𝑟
𝐻𝑐𝐿𝑔𝑑𝑐𝑠𝑔𝑠𝑠𝑖𝑔𝑔𝑑𝑠𝑖𝑔 𝑚𝑜𝑜𝐿𝑜
For the given numerical values:
𝐴𝑀 =−1×20=−20
𝜏𝐻 =20×20+5[20(1+1×20)+20]+10×20=400+2200+200=2800𝑝𝑠
(b)
We see that:
𝑓= 1 = 1 =56.8𝑀𝐻𝑧
𝐻
2𝜋𝜏𝐻 2𝜋 × 2.8 × 10−9
𝐺𝐵 = 20 × 56.8 = 1.14 𝐺𝐻𝑧
𝑉𝑉𝑟𝑉
𝑔1=1, 𝑔2= 𝑜1 , 𝑜=−𝑔𝑟
𝑉𝑉1𝑉 𝑠𝑖𝑔 𝑔1+𝑟𝑔2
𝑚2 𝑜2
𝑔𝑚1
𝑜1
𝑟𝑟
𝐴 =1× 𝑜1 ×−𝑔 𝑟 =− 𝑜1 (𝑔 𝑟 ) 𝑀 1+𝑟 𝑚2𝑜2 1+𝑟𝑚2𝑜2
𝑔𝑚1 𝑜1 𝑔𝑚1 𝑜1
Next, we evaluate the open-circuit time constants. Refer to Figure 2:
𝐶 : Capacitor 𝐶 is between 𝐺 and ground and thus sees the resistance 𝑅 ,
𝑅𝑔𝑑1 = 𝑅𝑠𝑖𝑔
𝑔𝑑1 𝑔𝑑1 1
𝑠𝑖𝑔
3
𝐶 : To find the resistance 𝑅 𝑔𝑠1 𝑔𝑠1
circuit with 𝑉 set to zero, 𝐶 𝑠𝑖𝑔 𝑔𝑑1
circuit is shown in the figure 3:
𝜏=𝐶𝑅 𝑔𝑑1 𝑔𝑑1 𝑠𝑖𝑔
seen by capacitor 𝐶 , we replace 𝑄 with its hybrid-𝜋 equivalent 𝑔𝑠1 1
= 0 and 𝐶 𝑔𝑠1
replaced by a test voltage 𝑉 . The resulting equivalent 𝑥
Analysis of the circuit in figure 3 proceeds as follows:
𝑉=𝐼𝑅 𝑔1 𝑋𝑠𝑖𝑔
𝑉=𝑉−𝑉=𝐼𝑅−𝑉 𝑠1 𝑔1 𝑋 𝑋𝑠𝑖𝑔 𝑋
At𝑆1:
𝐼=𝑔𝑉−𝑆1=𝑔𝑉−𝑋𝑠𝑖𝑔 𝑋 →𝐼(1+𝑠𝑖𝑔)=𝑉(𝑔 + )
Figure 2
Figure 3
𝑉𝐼𝑅−𝑉𝑅1
𝑋𝑚1𝑋𝑟𝑚1𝑋𝑟 𝑋𝑟𝑋𝑚1𝑟 𝑜1 𝑜1 𝑜1 𝑜1
𝑉𝑅+𝑟
𝑅𝑔𝑠1=𝑋=𝑠𝑖𝑔 𝑜1 𝐼1+𝑔𝑟
𝑋
𝑚1 𝑜1
4
𝜏=𝐶
𝑔𝑠1 𝑔𝑠11+𝑔 𝑟
𝑅+𝑟 𝑠𝑖𝑔 𝑜1
𝐶 𝑔𝑠2
: Capacitor 𝐶 𝑔𝑠2
𝑚1 𝑜1
sees the resistance between 𝐺 and ground, which is the output resistance of source 2
𝑅 = 1 ||𝑟 𝑔𝑠2 𝑔𝑚1 𝑜1
𝜏 =𝐶 (1||𝑟) 𝑔𝑠2 𝑔𝑠2 𝑔𝑚1 𝑜1
follower 𝑄1
Thus,
𝐶 : Transistor 𝑄 operates as a CS amplifier with an equivalent signal-source resistance equal to the 𝑔𝑠2 2
output resistance of the source follower 𝑄 , that is 1 ||𝑟 and with a gain from gate to drain of 𝑔 𝑟
.
1 𝑔𝑚1𝑜1
Thus, the formula for 𝑅𝑔𝑑 in a CS amplifier can be adapted as follows:
𝑚2𝑜2
And thus,
𝑅 2=( 1 ||𝑟 )(1+𝑔 𝑟 )+𝑟 𝑔𝑑 𝑔𝑚1 𝑜1 𝑚2 𝑜2 𝑜2
𝜏 =𝐶 [(1||𝑟)(1+𝑔 𝑟)+𝑟] 𝑔𝑑2 𝑔𝑑2𝑔𝑚1𝑜1 𝑚2𝑜2 𝑜2
𝐶 : Capacitor 𝐶 sees the resistance between 𝐷 and ground which is 𝑟 , 𝐿 𝐿 2 𝑜2
𝜏=𝐶𝑟 𝐶𝐿 𝐿𝑜2
Summing 𝜏𝑔𝑑1, 𝜏𝑔𝑠1 , 𝜏𝑔𝑠2 and 𝜏𝐶𝐿 gives 𝜏𝐻 in the problem statement. For the given numerical values:
𝐴𝑀 =− 20 (1×20)=−19 1+20
𝜏 𝑔𝑑1
𝜏 𝑔𝑠1
= 𝐶 𝑅 = 5 × 20 = 100 𝑝𝑠 𝑔𝑑1 𝑠𝑖𝑔
𝑅 +𝑟 = 𝐶 𝑠𝑖𝑔 𝑜1
𝑔𝑠11+𝑔 𝑟 𝑚1 𝑜1
= 20
20+20 1+1×20
= 38 𝑝𝑠
𝜏 =𝐶 [( 1 ||𝑟 )(1+𝑔 𝑟 )+𝑟 ]=5[(1||20)(1+20)+20]=200ps 𝑔𝑑2 𝑔𝑑2𝑔𝑚1 𝑜1 𝑚2𝑜2 𝑜2
𝜏 =𝐶 𝑟 =10×20=200𝑝𝑠 𝐶𝐿 𝐿𝑜2
𝜏𝐻 =100+38+19+200+200=557𝑝𝑠
𝑓= 1 = 1 =286𝑀𝐻𝑧
𝐻
2𝜋𝜏𝐻 2𝜋 × 557 × 10−12
5
GB = 19× 286 = 5.43 𝐺𝐻𝑧
6