=========================preview======================
(ELEC304)Midterm_2002Spring.pdf
Back to ELEC304 Login to download
======================================================

The Hong Kong University of Science and Technology
Department of Electrical and Electronic Engineering

ELEC304 Midterm Exam 15th April 2003
Name:
Student Number:

1-hour closed-book examination Do all the problems in the answer book and the provided graph paper
Q1. It is known that a common-source amplifier has effective transconductance gm and the input small-signal voltage (vin) must meet the condition v << V .V .
in GS THN

The circuit shown below is a common-source amplifier with source degeneration. Based on the notations used in the figure,
VDD

gm

(a)
Prove that the effective transconductance is given by . (15 marks)1+ gmRS

(b)
Find the condition of the input small-signal voltage. (15 marks)

(b)
Comparing to a common-source amplifier and based on (a) and (b), state one advantage and one disadvantage of the common-source amplifier with source degeneration. (10 marks)


Notes: Show all evaluation steps.
Hints:
1. V = V . IR

GS INDS
.W .
2. gm =nCox . ..(VGS .VTHN ) where VTHN is the threshold voltage of the NMOS transistor. . L .
VDD
RD1 RD2 vo
2RC
+vin/2-vin/2

CC
M1 M2
2RS
IB IB





v ()
s
(a) Neglecting ro of M1 and M2, find the transfer function of the amplifier o . (15 marks)
vin ()
s
.....

.....

s
1

+


v ()
s
z
(b) For
()
As
=

1
=

Ao
.

, find Ao, p1 and z1 in terms of circuit parameters.
o
v
in
()
s s
1

+

.

p
1
.

(5 marks)
(c) Given that Ao = -100V/V, p1 = 100rad/s and z1 = 100krad/s, plot the Bode plot. (10 marks) 20log(|A(s)|)
log()0dB
A(s)
0o
log()
VDD
io

Based on the circuit diagram and including all ro in your calculation,
(a)
What type of feedback is applied? (5 marks)

(b)
Evaluate o . (10 marks)


i
v
in
(c)
Evaluate Rocl. (10 marks)

(d)
Explain briefly on why the above circuit is a good current-signal source. (5 marks) Notes: You must show all your evaluation steps.


END OF MIDTERM PAPER
(a) From the figure,
nCox .W . 2
I + i = ..[V + v .(I + i ). R .V ]
Dd INin Dd STHN
2 . L .
nCox .W . 2
=. V D THN + id R )]
. [( IN . I RS .V )( vin . S
2 . L .
nCox .W . 2
= [( GS .V )( in . i RS
.. V THN + vd )]
2 . L .
nCox ..W .. ( GS .VTHN )2 + 2 VGS .VTHN )( in . id RS + vin . id RS )2 ]



=[V ( v )(
2 . L .
For v . iR << V .V ,

indS GS THN
ID + id nCox ..W ..[(VGS .VTHN )2 + 2(VGS .VTHN )( vin . id RS )]
2 . L .
nCox .W . 2 .W .
= ..(V .V ) + C . .(V .V )( v . i RS )
GS THN nox GS THNin d
2 . L .. L .
Therefore,

.W .
id =nCox . .(VGS .VTHN )( vin . id RS )= gm .(vin . id RS ). L .
. gm .
. id =.. ... vin
.1+ gmRS .

gm
As id is the output current, the effective transconductance is given by , and so the1+ gmRS
bandwidth of a common-source amplifier with source degeneration is reduced by 1+ gmRS times when comparing with that of a common-source amplifier.
(b) The above