eeeb273 1011s1 mt qna
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College of EngineeringDepartment of Electronics and Communication Engineering
Midterm Test Answer
SEMESTER 1, ACADEMIC YEAR 2010/2011
Subject Code : EEEB273
Course Title : Electronics Analysis & Design II
Date : 21 August 2010
Time Allowed : 2 hours
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Question 1 [40 marks]
(a) Consider the circuit in Figure 1a, the bias voltages are V +=5V, V
¯ =-5V. Let
I S1= I S2=5x10-15
A, V T =0.026 V, I REF=200µA and I o=50µA. Assume V A=∞ and neglect
base currents.
i. Calculate V BE1 and V BE2.[6 marks]
ii. Design the circuit that provides the bias current ( I o) and the reference current ( I REF).
[6 marks]
iii. List the advantages of a Widlar current source as compared to a basic two-transistor
current source.
[2 marks]
Figure 1a
1
2
/
/
1 15
2 15
200ln( ) (0.026) ln 0.6347
5 10
50ln( ) (0.026) ln 0.5987
5 10
BE T
BE T
V V
REF S
V V
O S
REF BE T
S
O BE T
S
I I e
I I e
I V V V
I
I V V V
I
µ
µ
−
−
=
=
= = =
×
= = =
×
1
1
2
2
1 2
1 2
1
1
0.6347 0.5987720
50
(5) ( 5) 0.634746.826
200
BE BE o E
BE BE E
o
REF BE
BE
REF
V V I R
V V R
I
I R V V V
V V V R k
I
µ
µ
+ −
+ −
− =
− −= = = Ω
+ = −
− − − − −= = = Ω
1
2
1
2
The advantage of a Widlar current source is that:
• It requires small resistance values to produce the
desired bias currents
• A difference in V BE voltages of the two transistors can
produce a large magnitude difference between I REF
and I o.
1
1
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(b) Consider the circuit shown in Figure 1b, derive the equations that gives the relationship
between I o and I REF. Assume all transistors are identical and >>1.
[6 marks]
Figure 1b
(c) In the NMOS cascode current source as shown in Figure 1c, I REF=0.1mA. All transistors
are matched with parameters V TN
=0.5V, K n
=100µA/V2
and λλλλ=0.02V-1
.
i. Draw the ac equivalent circuit if the gates voltages of all transistors are constant.
[3 marks]
Grounds – 1 mark
M2 and M4 correctly connected – 1 marks Labels – 1 mark
Figure 1c
)ln(
)ln(
2
1
/
2
/
1
2
1
S
OT BE
S
REF T BE
V V
S C O
V V
S C REF
I
I V V
I
I V V
e I I I
e I I I
T BE
T BE
=
=
==
=≅
)ln(
)ln(
112
12
REF
OT E REF
E REF E E BE BE
REF
OT BE BE
I
I V R I
R I R I V V
I
I V V V
=
≅=−
=−
Label
1
1
1
1
1
1
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ii. Draw the small-signal equivalent circuit for the circuit obtained in part i.
[5 marks]
Each transistor - 1mark = 2marks
r o2 connection – 1 mark
Node labels – 2 marks.
iii. Calculate the V GS of each transistors and the output resistance R o looking into M 4.
[10 marks]
iv. Calculate the changes on the output current if V D4 changes from -2V to +2V.
[2 marks]
1
1( 2 ( 2 ) 8 0
5 0
o o
o
o
d I d V R
d I n A M
=
= − − =
Matched transistors:
( )
V V V
I I
V V V K
I V V
V V K I
GS GS
REF o
GS GS
n
REF TN GS
TN GS n REF
5.1
5.11.0
1.05.0
24
213
2
3
==
=
===+=+=
−=
( ) [ ]
Ω===
Ω=++=++=
====
Ω==
Ω====
M k k mr r g R
or
M k mk k r gr r R
V mAm I K I K g
k r r
k m I I
r
oomo
omooO
REF nonm
oo
REF o
o
50)500)(500)(2.0(
51)500)(2.0(15005001
/ 2.0)1.0)(100(222
500
500)1.0)(02.0(
111
42
424
24
2
µ
λ λ
1
1
2
1
2
1
2
1
1
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Question 2 [30 marks]
(a) Sketch the equivalent AC circuit of a BJT differential amplifier with differential-mode
input signal.
[5 marks]
Ac circuit equivalent with (a) two-sided output and (b) one-sided output
(b) With aid of circuit diagram, Design a BJT differential amplifier with a three-transistor
current source to establish the desired bias current I Q=0.2 mA. The supply voltages are
V +=10 V and V
-=-10 V. The transistor parameters for the BJT diff-amp pair are =100,
V A=, V O2=8 V, V T =26 mV and A cm=0.5. The circuit parameters of the current source are
V BE(on)=0.7 V, V A=, and =150. Neglect the base currents and consider one-side output.
i. Draw the designed circuit. Label the circuit clearly.
[5 marks]
Any circuit (a) or (b) is correct
1
0.5
0.5
11
0.5
0.5
1
0.5
0.5
1
1
0.50.5
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ii. Determine the value of RC and R ref .
[6 marks]
2
0.20.1 ,
2 2
10 8 20 ,0.1
Q
o
I Ic mA
V v Rc k Ic mA
+
= = =
− −= = = Ω
3 1
0.2
10 1.4 1093
0.2
REF
BE BE REF
REF
I Io mA
V V V V R k
I mA
+ −
≅ =
− − − − += = = Ω
iii. Determine the differential voltage gain ( A d ) and the CMRR.
[6 marks]
3
3 3
2
0.13.846 10
26
3.846 10 20 1038.4615
2
38.461576.923
0.5
20 (76.923) 37.72
m C d
CQ
m
T
d
d
cm
dB
g R A
I mAg
V mV
A
ACMRR
A
CMRR Log
−
−
=
= = = ×
× × ×= =
= = =
= =
iv. Determine the differential-mode input resistance ( Rid ).
[4 marks]
3
22
2 100 0.02652
0.1 10 ^
id
CQ
id
VT R r
I
R K
π
β
−
= =
× ×= = Ω
×
v. Find the output voltage of the differential amplifier when a differential-mode inputvoltage of v d =0.3 sin t V is applied.
[4 marks]
,
38.461 0.3 11.538
o d d
o
V A V
V Sin t Sin t V ω ω
=
= × =
1
2
1
2
1
1
2
1
1
2
2
2
2
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Question 3 [30 marks]
Figure 2 shows a MOSFET differential amplifier biased using a MOSFET current source. M 1
and M 2 are always in saturation. Study Figure 2 carefully. Transistor parameters are V TN = 2 V,
K n = 0.25 mA/V2, and = 0. Given that V GS3 is 3.16 V. Show all your calculations clearly
when answering the following questions.
(a) Find I 1, I Q, and I D1.
[12 marks]
Figure 2
(b) Determine V GS1, V DS1, and V DS4.
[9 marks]
I 1 = (V +
- V GS3 - V -) / ( R1) [2]
= (10 - 3.16 - (-10))/(50k) [2]
= 0.337 mA
I Q = I REF [2]
= 0.337 mA [2]
I D1 = I Q / 2 [2]
= 0.168 mA [2]
I D1 = K n (V GS1 – V TN )2
[1]
0.168m = (0.25m)(V GS1 – 2)2
[1]
V GS1 = 2.821 V [1]
V DS1 = V D1 - V S1 [1]
= (V +
– I D1 R D) – (V G1 – V GS1) [1]
= (10 – (0.168m)(24k)) – (0 – 2.82) [1]
= 8.779 V
V DS4 = V S1 – V -
[1]
= (V G1 – V GS1) – (V -) [1]
= (0 – 2.82) – (-10) [1]
= 7.179 V
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(c) What is the output resistance ( RO) of the current source if for M 4 is 0.01 V-1
?
[3 marks]
(d) A one-sided output voltage at the drain of M 2 can be derived using small-signal equivalent
circuit to produce
1
RO is the output resistance of the current source. Using Equation 1 above, find thedifferential-mode voltage gain ( A d ), the common-mode voltage gain ( A cm), and the common
mode rejection ratio (CMRR) of the MOSFET differential amplifier in Figure 2. Use the
output resistance value obtained from question (c) above.
[6 marks]
cm
om
Dmd
Dmo V
Rg
RgV
RgV
212 +
−=
RO = rO4 [1]
= 1 / ( I Q) [1]
= 1 / (0.01)(0.337m) [1]
= 296.9 k
A d = ( g m R D) / 2
= R D SQRT( K n I Q / 2) [1]
= (24k) SQRT((0.25m)(0.337m) /2) [1]
= 4.92
A cm = (- g m R D) / (1+2 g m R o)= (- R D SQRT(2 K n I Q)) / (1+2 R o SQRT(2 K n I Q))
i.e. Equation 11.82(b) in textbook [1]
= Put in all relevant values [1]
= -0.04
CMRR = | A d / A cm | [1]
= |4.92 / 0.04| [1]
= 122.34
CMRR dB = 20 log10 | A d / A cm |
= 41.75 dB