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Final 2013 Answer

1. Active loads can provide high resistance and relax matching requirements compared to using large passive resistors in integrated circuits. They improve the biasing. 2. The circuit remains perfectly symmetric so the voltage at the common source connection is virtually ground. 3. To improve the common-mode rejection ratio (CMRR), we can raise the source resistance by increasing the length of transistor Q5. However, this increases the size and power consumption of the transistor.

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40 views2 pages

Final 2013 Answer

1. Active loads can provide high resistance and relax matching requirements compared to using large passive resistors in integrated circuits. They improve the biasing. 2. The circuit remains perfectly symmetric so the voltage at the common source connection is virtually ground. 3. To improve the common-mode rejection ratio (CMRR), we can raise the source resistance by increasing the length of transistor Q5. However, this increases the size and power consumption of the transistor.

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LLA
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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1. a.) Large resistance is expensive in IC.

b.) Actively loaded can improved biasing


c.) Active load can provide high resistance
d.) Active load will relax matching requirements

2. The circuit remains perfectly symmetric, as a result the voltage at the common source connection will
be zero. Therefore, the source can be regarded as virtually ground.

3. a.) CMRR=¿ A d∨ ¿ ¿
¿ A cm∨¿=gm 1 ,2 ¿ ¿
In order to improve CMRR, we can raise R SS by increasing L of Q5

b.) The price to make bigger transistor

4. Provide the op amp with a dominate pole.

5. In low frequency, bypass and coupling capacitor cause gain roll-offs.


In high frequency, stray, and transistor capacitor cause gain roll-offs.
In the midband range the gain is almost constant because coupling and bypass capacitors act as short
circuits, while stray and transistor capacitors act as open circuits.

6. By teaching of Dr. Yeh, Ting-Jen, you can handle it fucking easily.

7. Miller’s Approximation “misses” the zero introduced by the feedback capacitor

gm gm
8. a.) MOSFET: f T = BJT: f T =
2 π (C gs+ C gd) 2 π (C π +C μ )

b.) Because BJT has higher gm than MOSFET.

c.) To improve f T , we can raise the gm ( i.e., increase drain/collector current). However, higher
drain/collector current will cause higher power consumption.

9.
RB rπ 1
AM= [ ]
−g m ( r o||R C||R L ) f H =
Rsig + RB r x + r π + Rsig ∨¿ R B
( )
2 π [C π +C μ 1+ gm (r o|| RC|| R L ) ](r π ∨¿ (r x + R sig∨¿ R B ))

10. Because CG amplifier has higher f H than CS amplifier

11. Because C m acts as a short circuit causing V gs=0 and hence I d 4 =0 at high frequency.
12. The corner frequency of CS amplifier with degeneration is
1
f H=
R sig + R S gm R L Rsig
2 π [C gs +C gd (R sig + R L + )]
1+ g m R s 1+ gm Rs
By substituting R S=0 , 100 Ω into equation of f H , we can find CS amplifier with degeneration has higher
f H than that of CS amplifier.

13. Because large R sig will reduce the dc gain of the cascade. While the dc gain of the cascade will be the
same as that achieved in a CS amplifier( A o ¿, the bandwidth f H will be greater for the cascade amplifier.

R sig + r π 2
R¿ =( β 1 +1 ) ( r π 2+ r e1 ) , R μ 1=R¿ ∨¿ R sig , R π 1=
14. R sig r π 2
1+ +
rπ 1 re 1

(
R μ 2=R L + r e1 +
R sig
β 1+1 ) (
+ g m 2 R L r e1 +
R sig
β 1+1 )
, R π 2=r π 2∨¿ r e1 +
(Rsig
β1 +1 )
vo vo vb 2 vb 1 r π2 R¿
AV = = × × =(−gm 2 RL )
v sig v b 2 v b 1 v sig r π 2+ r e1 R¿ + R sig
1
τ H =C π 1 R π 1 +C μ 1 R μ 1 +C π 2 Rπ 2+ C μ 2 R μ 2 ∴ f H=
2π τH

The purpose of CC-CE cascade configuration is to provide both high gain(from CE) and great high
frequency response(By CC reduces Miller’s effect)

R¿
Gain of CE is AV =( −g m R L ) ≅ AV → high gain
CE
R¿ + Rsig
Time constant of CE isτ H =C π R π +C μ Rμ > τ H → f H ↑
CE

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