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

7/29/2019 Eeeb273 1011s1 Mt Qna


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

7/29/2019 Eeeb273 1011s1 Mt Qna


(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

7/29/2019 Eeeb273 1011s1 Mt Qna


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

7/29/2019 Eeeb273 1011s1 Mt Qna



(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

7/29/2019 Eeeb273 1011s1 Mt Qna


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

7/29/2019 Eeeb273 1011s1 Mt Qna



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

7/29/2019 Eeeb273 1011s1 Mt Qna


(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

eeeb273 1011s1 mt qna

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