[PPT] - Problem for Problems for Discrete Transistor Amplifiers ECE 65, PowerPoint Presentation

SLIDE 1

Problem for Problems for Discrete Transistor Amplifiers

ECE 65, Winter2013, F. Najmabadi

SLIDE 2

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (2/42)

Exercise 1: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a common collector amplifier

(emitter follower) .

Input at the base, output at the emitter.
It has a emitter-degeneration bias with a

voltage divider.

When Rsig & vsig are not given, it implies that vsig = vi and Rsig = 0

SLIDE 3

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (3/42)

Bias circuit Thevenin form of the Voltage divider

V 95 . 4 9 k 18 k 22 k 22 k 90 . 9 k 18 || k 22 = × + = = =

BB B

V R

Real circuit

Exercise 1 (cont’d): β = 200, VA = 150 V.

SLIDE 4

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (4/42)

Bias Calculations

A 3 . 20 / mA 05 . 4 10 7 . ) 1 /( 10 9 . 9 4.95 4.95 : KVL BE

3 3

µ β β = = ≈ = + + + × = + + = −

C B C E E E E E BE B B

I I I I I I R I V R I

V 7 . and V, 7 . : Active in is BJT Assume ≥ > =

CE C BE

V I V

V 0.7 V 5 10 10 4 9 9 : KVL CE

3 3

= > = × × + = + = −

− D CE CE E E CE

V V V R I V V 95 . 4 k 9 . 9 = =

BB B

V R

k 28 . 1 r k . 37 10 x 4 150 r mA/V 156 10 26 10 4.05

3

3

3

= = = = = ≈ = × × = =

m B T

C

A

T

C m

g I V I V V I g β

π

Thevenin form of the Voltage divider

Exercise 1 (cont’d): β = 200, VA = 150 V.

SLIDE 5

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (5/42)

1 151 1 151 151 965 10 156 ) || || ( 965 k 100 || k 1 || k 8 . 38 ) || || ( ) || || ( 1 ) || || (

3

≈ + = = × × = Ω = = + =

− i

L

E

m

L E

L

E

m

L E

m

i

v

v R R r g R R r R R r g R R r g v v k 28 . 1 k, 8 . 38 mA/V, 156 = = =

π

m

r r g

[ ] [ ]

k 9 . 9 k 194 k 3 . 1 || k 9 . 9 ) || || ( || ≈ + = + =

i L E

B

i

R R R r r R R β

π

Ω = = + = 4 . 6 || || || β β

π π

r R R R r R R

E sig B E

Exercise 1 (cont’d): β = 200, VA = 150 V.

Signal circuit (IVC = 0) Amplifier Parameters Emitter Follower 1 = = × + =

i

i
sig

i i sig

v

v v v R R R v v

SLIDE 6

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (6/42)

Ω = = 4 . 6 k 9 . 9

i

R R k 28 . 1 k 8 . 38 mA/V 156 = = =

π

m

r r g Hz 39 . 3 10 47 . ) 10 100 (6.4 2 1 ) ( 2 1 Hz 2 . 34 10 47 . ) 10 9 . 9 ( 2 1 ) ( 2 1

6 3 2 2 2 6 3 1 1 1

= × × × + = + = = × × + × = + =

− −

π π π π

p c L

p

p c sig i p

f C R R f f C R R f Hz 6 . 37 4 . 3 2 . 34

2 1

= + = + ≈

p p p

f f f

Exercise 1 (cont’d): β = 200, VA = 150 V.

Amplifier Cut-off frequency 2 caps: 2 poles

SLIDE 7

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (7/42)
This is a common emitter amplifier with RE .
Input at the base, output at the collector.
It has a emitter-degeneration bias with a

voltage divider.

Exercise 2: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

SLIDE 8

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (8/42)

Bias calculations

Thevenin form of the Voltage divider

A 2 . 14 / mA 84 . 2 510 7 . ) 1 /( 10 . 5 2.22 2.22 : KVL BE

3

µ β β = = ≈ = + + + × = + + = −

C B C E E E E E BE B B

I I I I I I R I V R I

V 7 . and V, 7 . : Active in is BJT Assume ≥ > =

CE C BE

V I V

V 0.7 V 5 . 10 ) 510 10 ( 10 84 . 2 15 15 : KVL CE

3 3

= > = + × × + = + + = −

− D CE CE E E CE C C

V V V R I V R I

k 83 . 1 r k 8 . 52 10 84 . 2 150 r mA/V 10.9 10 26 10 2.84

3

3

3

= = = = × = ≈ = × × = =

m B T

C

A

T

C m

g I V I V V I g β

π

Exercise 2 (cont’d): β = 200, VA = 150 V

Bias circuit Caps open

V 22 . 2 15 k 9 . 5 k 34 k 9 . 5 k . 5 k 9 . 5 || k 34 = × + = = =

BB B

V R

SLIDE 9

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (9/42)

k 83 . 1 k, 8 . 52 mA/V, 10.9 = = =

π

m

r r g Amplifier Parameters CE amplifier with RE k 8 . 4 104k || k . 5 ) / 1 ]( / ) || [( 1 || = =       + + + =

i E

L

C E B i

R r R r R R R r R R

π π

β 64 . 1 0.990k k 100 || k 1 || ) / 1 ]( / ) || [( 1 ) || ( − = = = + + + − =

i

L

C E

L

C E m L C m i

v

v R R r R r R R R g R R g v v

π

k . 1 || 1 || = ≈                 + + + ≈

C

sig

B E E

C
R

R R R R r R r R R

π

β 59 . 1 ) 64 . 1 ( 100 800 , 4 800 , 4 − = − × + = × + =

i

sig

i i sig

v

v R R R v v

Exercise 2 (cont’d): β = 200, VA = 150 V

Signal circuit (IVC = 0)

SLIDE 10

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (10/42)

k . 1 k 8 . 4 = =

i

R R

Amplifier Cut-off frequency (2 caps: 2 poles)

Hz 8 . 15 10 100 ) 10 100 10 ( 2 1 ) ( 2 1 Hz 91 . 6 10 7 . 4 ) 100 10 8 . 4 ( 2 1 ) ( 2 1

9 3 3 2 2 2 6 3 1 1 1

= × × × + = + = = × × + × = + =

− −

π π π π

p c L

p

p c sig i p

f C R R f f C R R f Hz 7 . 22 8 . 15 9 . 6

2 1

= + = + ≈

p p p

f f f k 83 . 1 r k 8 . 52 r mA/V 10.9

=

= =

π m

g

Exercise 2 (cont’d): β = 200, VA = 150 V

SLIDE 11

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (11/42)

Exercise 3: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a PNP common emitter amplifier (no RE ).
Input at the base, output at the collector (RE is shorted
ut for signal because of the by-pass capacitor)
It has a emitter-degeneration bias with two voltage

sources (RB = ∞)

SLIDE 12

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (12/42)

Exercise 3 (cont’d). β = 200, VA = 150 V.

Bias circuit (Signal = 0, Caps open)

A . 5 / mA . 1 100 10 3 . 2 3 : KVL BE

3

µ β = = ≈ = + + × = −

C B C E B EB E

I I I I I V I

V 7 . and V, 7 . : Active in is BJT Assume ≥ > =

EC C EB

V I V

V 0.7 V 4 . 1 10 10 6 . 4 6 3 10 3 . 2 10 3 . 2 3 : KVL CE

3 3 3 3

= > = × × − = − × + + × = −

− D EC EC C EC E

V V V I V I

k 26 . 5 r k 150 10 150 r mA/V 38.5 10 26 10

3

3
3

= = = = = ≈ = × = =

m B T C A T C m

g I V I V V I g β

π

SLIDE 13

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (13/42)

3 . 85 ) k 100 || k 3 . 2 || k 150 ( 10 38.5 ) || || (

3

− = × − = − =

− i

L

C

m

i

v

v R R r g v v k 26 . 5 k, 150 mA/V, 38.5 = = =

π

m

r r g Amplifier Parameters 1) CE amplifier with no RE 2) No voltage divider biasing: no RB RB is open circuit, RB = ∞ k 27 . 2 k 3 . 2 || k 150 || = = =

C
R

r R R k 26 . 5 || = = =

π π

r r R R

B i

Exercise 3 (cont’d). β = 200, VA = 150 V.

3 . 85 − = ≈ × + =

i

i
sig

i i sig

v

v v v R R R v v

i sig

R R << Signal circuit (IVC = 0)

SLIDE 14

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (14/42)

Signal circuit (IVC = 0) k 26 . 5 r k 150 r mA/V 38.5

=

= =

π m

g Hz 145 129 8 . 15

3 2 1

= + + = + + ≈

p p p p

f f f f

Exercise 3 (cont’d). β = 200, VA = 150 V.

k 27 . 2 k 26 . 5 = =

i

R R

Amplifier Cut-off frequency 1) 2 caps: 2 poles 2) no coupling capacitors at the input

) ( 2 1

1 1

= + =

c sig i p

C R R f π Hz 129 10 47 ) 5 . 26 || 10 (2.3 2 1 ]} / ) || ( / 1 [ || { 2 1

6 3 3 3

= × × × = + =

−

π β π

p E sig B m E p

f C R R g R f Hz 8 . 15 10 100 ) 10 10 (2.3 2 1 ) ( 2 1

9 5 3 2 2 2

= × × + × = + =

−

π π

p c L

p

f C R R f

SLIDE 15

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (15/42)
This is a MOS common source amplifier (no RS ).
Input at the gate, output at the drain (RS is shorted
ut for the signal because of the by-pass capacitor)
It has a source-degeneration bias with a voltage

divider.

Exercise 4: Find the amplifier parameters of this circuit. (µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1. Assume capacitors are large and ignore channel width modulation in biasing. )

SLIDE 16

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (16/42)

V 353 . 1 8 . 1 100 33 100 V

G

= + = ⇒ = k k k IG Assume Saturation

2 3 2

10 x 3 5 .

OV OV

x

n D

V V L W C I

−

= = µ V 803 . = + =

t OV GS

V V V Saturation ⇒ >

OV DS

V V mA 275 . 10 x 3

2 3

= =

− OV D

V I V 550 . = − =

GS G S

V V V V 249 . 1 8 . 1 = − =

D D D

I R V V 699 . = − =

S D DS

V V V V 303 . =

OV

V

D S t OV D S GS G

I R V V I R V V + + = + =

2 3 3

10 3 10 2 5 . 353 . 1

OV OV

V V

−

× × × + + = 853 . 6

2

= − +

OV OV

V V GS-KVL:

Exercise 4 (cont’d): µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1.

Bias circuit (Signal = 0, Caps open)

SLIDE 17

k 3 . 36 10 75 2 . 1 . 1 1 r mA/V 1.82 0.303 10 0.275 2 2

3

3

= × × = = = × × = =

D
OV

D m

I V I g λ

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (17/42)

32 . 3 ) k 50 || k 2 || k 3 . 36 ( 10 1.82 ) || || (

3

− = × − = − =

− i

L

D

m

i

v

v R R r g v v Amplifier Parameters This is a CS amplifier with no RS k 90 . 1 k 2 || k 3 . 36 || = = =

D
R

R r R k 8 . 24 = =

G i

R R k 8 . 24 k 33 || k 100 = =

G

R

Exercise 4 (cont’d): µnCox = 100 µA/V2, (W/L) = 6/0.1, Vt = 0.5 V, λ = 0.1 V-1.

Signal circuit (IVC = 0) 19 . 3 k 1 k 8 . 24 k 8 . 24 − = × + = × + =

i

i
sig

i i sig

v

v v v R R R v v

SLIDE 18

Exercise 4 (cont’d):

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (18/42)

Hz 168 4 . 75 7 . 30 7 . 61

3 2 1

= + + = + + ≈

p p p p

f f f f k 9 . 1 k 8 . 24 = =

i

R R

Amplifier Cut-off frequency 1) 3 caps: 3 poles

Hz 7 . 61 10 100 ) 10 10 (24.8 2 1 ) ( 2 1

9 3 3 1 1 1

= × × + × = = + =

−

π π

p c sig i p

f C R R f Hz 4 . 75 10 7 . 4 ] 579 || 10 [2 2 1 )]} /( ) || [( || { 2 1

6 3 3 3

= × × × = + =

−

π π

p S

m

L D

S

p

f C r g R R r R f Hz 7 . 30 10 100 ) 10 50 10 (1.9 2 1 ) ( 2 1

9 4 3 2 2 2

= × × × + × = + =

−

π π

p c L

p

f C R R f k 3 . 36 mA/V 1.82 = =

m

r g

SLIDE 19

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (19/42)

Exercise 5: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a common collector amplifier (emitter follower).
Input at the base, output at the emitter.
It is biased with a current source!
No emitter degeneration: no RE (RE = ∞) and no RB (RB = ∞)

SLIDE 20

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (20/42)

Exercise 5 (cont’d): β = 200, VA = 150 V.

V 72 . 10

3

− = + + =

E E BE B

V V V I

V 7 . and V, 7 . : Active in is BJT Assume ≥ > =

CE C BE

V I V

k 21 . 1 r k 9 . 34 10 4.3 150 r mA/V 165 10 26 10 4.3

3

3

3

= = = = × = ≈ = × × = =

m B T

C

A

T

C m

g I V I V V I g β

π

A 5 . 21 / mA 3 . 4 µ β = = ≈ =

C B C E

I I I I V 0.7 V 72 . 4 4

0 =

> = − =

D E CE

V V V

Bias circuit (Signal = 0, Caps open)

E

V

SLIDE 21

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (21/42)

k 21 . 1 k, 9 . 34 mA/V, 165 = = =

π

m

r r g 1 10 4.28 1 10 4.28 10 4.28 ) || || ( k 9 . 25 k 100 || || k 9 . 34 || || ) || || ( 1 ) || || (

3 3 3

≈ × + × = × = = ∞ = + =

i

L

E

m

L E

L

E

m

L E

m

i

v

v R R r g R R r R R r g R R r g v v Ω = + ≈ 111 || || β

π sig B E

R

R r R R

[ ]

M 2 . 5 k 9 . 25 201 k 2 . 1 ) || || )( 1 ( || = × + = + + =

i L E

B

i

R R R r r R R β

π

Exercise 5 (cont’d): β = 200, VA = 150 V.

Signal circuit (IVC short, ICS open)) Amplifier Parameters 1) Emitter Follower 2) Biased with an ICS (RE = ∞ ) 3) No voltage divider biasing: RB is open circuit, RB = ∞ 1 = ≈ × + =

i

i
sig

i i sig

v

v v v R R R v v

SLIDE 22

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (22/42)

Signal circuit k 21 . 1 r k 9 . 34 r mA/V 165

=

= =

π m

g Hz 34 . 3 10 47 . ) 111 10 100 ( 2 1 ] [ 2 1

6 3 2 2 2

= × × + × = + =

−

π π

p c

L

p

f C R R f Ω = = 111 M 2 . 5

i

R R

Exercise 5 (cont’d): β = 200, VA = 150 V.

Hz 34 . 3

2 =

=

p p

f f

Amplifier Cut-off frequency 1) 1 cap: 1 pole 2) no coupling capacitors at the input

) ( 2 1

1 1

= + =

c sig i p

C R R f π

SLIDE 23

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (23/42)

Exercise 6: Find the bias point and the amplifier parameters of the circuit

below. (µpCox (W/L) = 400 µA/V2, Vtp = − 4 V, λ = 0.01 V-1. Ignore

channel width modulation in biasing).

This is a PMOS common drain amplifier.
Input at the gate, output at the source.
It has a source-degeneration bias with two

voltage sources.

SLIDE 24

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (24/42)

Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1

Assume Saturation: 9 2 9 10 400 5 . 10 | | 10 10 13

2 2 6 4 4 4

= − + = − + × × × + + × = + × =

− OV OV OV OV tp OV D SG D

V V V V V V I V I SG-KVL: 5 .

2 OV

x

p D

V L W C I µ = V 9 . 5 = + =

t OV SG

V V V Saturation ⇒ >

OV DS

V V mA 71 . 10 400 5 .

2 6

= × × =

− OV D

V I V 9 . 5 = → − =

S S SG

V V V V 9 . 10 ) 5 ( = − − =

S SD

V V V 9 . 1 =

OV

V k 141 10 0.71 01 . 1 1 mA/V 0.747 0.303 10 0.71 2 2

3 3

= × × = = = × × = =

− − D

OV

D m

I r V I g λ

S

V

Bias circuit (Signal = 0, Caps open)

SLIDE 25

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (25/42)

Signal circuit 87 . 38 . 6 1 38 . 6 6.38 ) || || ( k 54 . 8 k 100 || k 10 || k 141 || || 88 . ) || || ( 1 ) || || ( = + = = = = = + =

i

L

S

m

L S

L

S

m

L S

m

i

v

v R R r g R R r R R r g R R r g v v Amplifier Parameters Common collector Amp.

Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1

k 2 . 1 k 34 . 1 || k 10 1 || = = = ∞ = =

m S

G

i

g R R R R 87 . = = × + =

i

i
sig

i i sig

v

v v v R R R v v k 141 mA/V, 0.747 = =

m

r g

SLIDE 26

Amplifier Cut-off frequency 1) 1 cap: 1 pole 2) no coupling capacitors at the input

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (26/42)

Signal circuit

Exercise 6 (Cont’d): µpCox (W/L) = 400 µA/V2, Vtp = − 4 V,λ = 0.01 V-1

k 141 mA/V 0.747 = =

m

r g k 2 . 1 = ∞ =

i

R R Hz 39 . 3 10 47 . ) 10 2 . 1 10 ( 2 1 ] [ 2 1

6 3 5 2

= × × × + = + =

−

π π

p c

L

p

f C R R f

SLIDE 27

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (27/42)

Exercise 7: Find the bias point and the amplifier parameters of the circuit

below. (µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1. Ignore

channel width modulation in biasing).

This is a NMOS common gate amplifier.
Input at the source, output at the drain.
It has a source-degeneration bias with a

voltage divider.

SLIDE 28

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (28/42)

Bias circuit 5 4 5 10 800 5 . 10 10 10 6

2 2 6 4 4 4

= − + = − + × × × × + + = × + = =

− OV OV OV OV D t OV D GS G

V V V V I V V I V V GS-KVL: Assume Saturation: 5 .

2 OV

x

n D

V L W C I µ = V . 2 = + =

t OV GS

V V V Saturation ⇒ >

OV DS

V V mA 40 . 10 800 5 .

2 6

= × × =

− OV D

V I V 7 4 4 15 10 10 15

4 4

= − − = × + + × =

DS D DS D

V I V I V . 1 =

OV

V

Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.

G

V V 6 15 M 8 . 1 M 2 . 1 M 2 . 1 = × + =

G

V DS-KVL: k 250 10 0.4 01 . 1 1 r mA/V 0.800 1 10 0.4 2 2

3

3

= × × = = = × × = =

− − D OV D m

I V I g λ

SLIDE 29

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (29/42)

69 . 7 ) || k 10 || k 250 ( 10 8 . ) || || (

3

= ∞ × = =

− i

L

S

m

i

v

v R R r g v v

Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.

Signal circuit (IVC short) Amplifier Parameters 1) CG amplifier 2) Biased with two voltage sources (RG = 0) 11 . 4 69 . 7 k 1 k 15 . 1 k 15 . 1 = × + = × + =

i

sig

i i sig

v

v R R R v v k 250 mA/V 0.800 = =

m

r g k 15 . 1 k 1.30 || k 10 / ) || ( 1 || = =       + =

i m

L

D S i

R g r R R R R k 77 . 9 k 432 || k 10 )] || ( 1 ( [ || = = + =

sig

S m

D
R

R R g r R R

SLIDE 30

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (30/42)

Hz 740

2 1

= + ≈

p p p

f f f k 250 mA/V 0.80 = =

m

r g

Exercise 7 (Cont’d): µnCox (W/L) = 800 µA/V2, Vt = 1 V, λ = 0.01 V-1.

Signal circuit (IVC short)

Amplifier Cut-off frequency 2 caps: 2 poles

Hz 740 10 100 ) 10 10 (1.15 2 1 ) ( 2 1

9 3 3 1 1 1

= × × + × = = + =

−

π π

p c sig i p

f C R R f 10 100 ) 10 (9.77 2 1 ) ( 2 1

9 3 2 2 2

= × × ∞ + × = + =

−

π π

p c L

p

f C R R f k 77 . 9 k 15 . 1 = =

i

R R

SLIDE 31

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (31/42)

Exercise 8: Design a common emitter amplifier with RE with vo / vi = 5 when driving a 100 kΩ load and a bias point of IC = 4 mA and VCE = 6 V. The circuit should have a cut-off frequency of 50~Hz for Rsig = 100 Ω. A 16 V power supply is available. Use a Si BJT with β = 200, β min = 100, VA = 150 V, ignore Early effect in bias calculations.

Prototype of the circuit is shown.
Problem specification gives
We should find RC , RE , RB1 , RB2

, Cc1 , and Cc2

k 100 , 100 V, 16 = Ω = =

L sig CC

R R V

* See Exercise 4 of BJT lecture slides for bias design

SLIDE 32

mA, 4 V, 6 k, 100 , 100 V, 16 = = = Ω = =

C CE L sig CC

I V R R V

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (32/42)

Exercise 8 (Cont’d):

k 30 . 1 k 5 . 37 10 4 150 mA/V 154 10 26 10 4

3 3 3

= = = = × = ≈ = × × = =

m B T π

C

A

T

C m

g I V r I V r V I g β

1. Bias point (IC = 4 mA & VCE = 6 V)

gives the SSM parameters as well as a relationship between RC and RE through CE-KVL:

k 5 . 2 10 4 6 16 ) (

3 =

× − = + + ≈ − + + =

− E C E C C CE CC E E CE C C CC

R R R R I V V I R V I R V

2. Amp gain of vo / vi = 5 also gives another

relationship between RC and RE: CE-KVL: 5 ) / 1 ]( / ) || [( 1 ) || ( = + + + =

π

r R r R R R g R R g v v

E

L

C E m L C m i

Solution can be simplified by noting RC + RE =

2.5k and thus RC < 2.5k: 1) Since RC << RL = 100 k, then RC || RL ≈ RC 2) Since RC << ro = 37.5 k, we can drop the 3rd term in the denominator of the gain formula:

5 . 32 5 5 5 5 1 + = + = = + ≈

E m E C E m C m i

R

g R R R g R g v v k 09 . 2 411 k 5 . 2 5 . 32 5 = Ω = = + +

C E E E

R R R R

A . 20 / µ β = =

C B

I I

SLIDE 33

Exercise 8 (Cont’d):

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (33/42)

3

10 2 . 7 3 . 63 1 316 ) / 1 ]( / ) || [( 1 ) || (

−

× + + = + + + =

π

r R r R R R g R R g v v

E

L

C E m L C m i

In fact, because gm terms are large:
Note that the (open-loop gain) is independent of BJT parameters!

E C E m C m E m C m i

R

R R g R g R g R g v v = ≈ + ≈ 1

For discrete BJT amplifiers (RC << ro ), dropping the third term is a

very good approximation because gm terms are actually large:

k 09 . 2 , 411 = Ω =

C E

R R k 30 . 1 k, 5 . 37 mA/V, 154 = = =

π

m

r r g

A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ = = = = Ω = =

B C CE L sig CC

I I V R R V

SLIDE 34

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (34/42)

Exercise 8 (Cont’d):

A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ = = = = Ω = =

B C CE L sig CC

I I V R R V

1. Bias point (IC = 4 mA & VCE = 6 V)

and requirements of good bias give RB1 and RB2 through BE-KVL. However, it is easier to find RB and VBB (Thevenin equivalent) first.

2. Resistors RB1 and RB2 are found from:

BE-KVL: Dividing the above equations gives RB1 : k 89 . 4 k 4 . 27

2 1

= =

B B

R R k 09 . 2 , 411 = Ω =

C E

R R

k 4.15 411 101 1 . 411 101 1 . ) 1 ( 0.1 ) 1 (

min min

= × × = × × = + ≤ + <<

B E B E B

R R R R R β β V 43 . 2 11 4 10 4 0.7 10 4.15 10 20

3 3 6

= × × + + × × × = + + =

BB

BB

E E BE B B BB

V V R I V R I V 152 . 16 2.43 k 15 . 4 ||

2 1 2 2 1 2 1 2 1

= = + = = + = =

B B B CC BB B B B B B B B

R R R V V R R R R R R R

The 2nd condition for stable bias is RE IE ≥ 1 V:

V 1 64 . 1 11 4 10 4

3

> = × × =

E

ER

I

SLIDE 35

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (35/42)

Exercise 8 (Cont’d):

Coupling capacitors Cc1 and Cc2 are found from specification of cut-off frequency (50 Hz). We need to compute Ri and Ro first. Can choose either capacitor and compute the other one. A good design guideline is to keep capacitors small. k 15 . 4 , k 89 . 4 , k 4 . 27 , k 09 . 2 , 411

2 1

= = = = Ω =

B B B C E

R R R R R k 95 . 3 k 81.6 || k 15 . 4 ) / 1 ]( / ) || [( 1 || = =       + + + =

i E

L

C E B i

R r R r R R R r R R

π π

β k 09 . 2 M 1.74 || || 1 || = ≈ =                 + + + =

C C

sig

B E E

C
R

R R R R R r R r R R

π

β k 30 . 1 k, 5 . 37 mA/V, 154 = = =

π

m

r r g

Amplifier Cut-off frequency (2 caps: 2 poles)

2 3 2 2 1 3 1 1

10 102 2 1 ) ( 2 1 10 05 . 4 2 1 ) ( 2 1

c c L

p

c c sig i p

C C R R f C C R R f × × = + = × × = + = π π π π 50 10 102 2 1 10 05 . 4 2 1 Hz 50

2 3 1 3 2 1

= × × + × × = + ≈

c c p p p

C C f f f π π

A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ = = = = Ω = =

B C CE L sig CC

I I V R R V

SLIDE 36

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (36/42)

Exercise 8 (Cont’d):

We can choose either capacitor and compute the other one. A good design guideline is to keep capacitors small. Since the coefficient of CC1 is 25 times smaller than that CC1, a reasonable choice is to assign 80% of fp to CC1 : k 15 . 4 , k 89 . 4 , k 4 . 27 , k 09 . 2 , 411

2 1

= = = = Ω =

B B B C E

R R R R R k 30 . 1 k, 5 . 37 mA/V, 154 = = =

π

m

r r g nF 156 50 2 . 10 102 2 1 F 982 . 50 8 . 10 05 . 4 2 1 Hz 50

1 2 3 1 1 3 2 1

= → × = × × = → × = × × = + ≈

c c c c p p p

C C C C f f f π µ π Design Commercial Value Value (5%) nF 156 F 982 . 411 k 09 . 2 k 89 . 4 k 4 . 27

2 1 2 1

= = Ω = = = =

C C E C B B

C C R R R R µ nF 160 F 1 390 k 2 k 7 . 4 k 27

1 2 1

= = Ω = = = =

Cw C E C B B

C C R R R R µ

A. 20 mA, 4 V, 6 k, 100 , 100 V, 16 µ = = = = Ω = =

B C CE L sig CC

I I V R R V

SLIDE 37

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (37/42)

Exercise 9: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a two-stage amplifier.
Stage 1 is a common emitter amplifier with RE
Stage 2 is a common collector amplifier (emitter follower).
Both stages have source-degeneration bias with a voltage divider.
Each stage is independently biased because of the 470 nF

coupling capacitor between stages.

SLIDE 38

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (38/42)

Stage 1 Bias

Thevenin form of the Voltage divider

V 37 . 2 15 k 2 . 6 k 33 k 2 . 6 k 22 . 5 k 2 . 6 || k 33

1 1

= × + = = =

BB B

V R

Exercise 9 (cont’d) : β = 200, VA = 150 V

Bias circuit Caps open

A 9 . 15 / mA 18 . 3 500 7 . ) 1 /( 10 22 . 5 2.37 : KVL BE

1 1 1 1 1 1 1 1 3 1 1 1 1 1 1

µ β β = = ≈ = + + + × = + + = −

C B C E E E E E BE B B BB

I I I I I I R I V R I V

V 7 . and V, 7 . : Active in is BJT Assume

1 1 1

≥ > =

CE C BE

V I V

V 0.7 V 05 . 7 ) 500 10 2 ( 10 18 . 3 15 15 : KVL CE

1 3 3 1 1 1 1 1 1

= > = + × × × + = + + = −

− D CE CE E E CE C C

V V V R I V R I

k 64 . 1 k 2 . 47 10 18 . 3 150 mA/V 122 10 26 10 3.18

1 1 1 1 3 1 1 1 3 3 1 1

= = = = × = = = × × = =

m B T π

C

A

T

C m

g I V r I V r V I g β

SLIDE 39

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (39/42)

Stage 2 Bias

Thevenin form of the Voltage divider

V 25 . 8 15 k 18 k 22 k 22 k 90 . 9 k 22 || k 18

2 2

= × + = = =

BB B

V R

Exercise 9 (cont’d) : β = 200, VA = 150 V

Bias circuit Caps open

A 36 / mA 20 . 7 10 7 . ) 1 /( 10 9 . 9 8.25 8.25 : KVL BE

2 2 2 2 2 2 3 2 2 3 2 2 2 2 2

µ β β = = ≈ = + + + × = + + = −

C B C E E E E E BE B B

I I I I I I R I V R I

V 7 . and V, 7 . : Active in is BJT Assume

2 2 2

≥ > =

CE C BE

V I V

V 0.7 V 80 . 7 10 10 20 . 7 15 15 : KVL CE

2 3 3 2 2 2 2

= > = × × + = + = −

− D CE CE E E CE

V V V R I V

k 722 . k 8 . 20 10 20 . 7 150 mA/V 277 10 26 10 7.20

2 2 2 2 3 2 2 2 3 3 2 2

= = = = × = = = × × = =

m B T π

C

A

T

C m

g I V r I V r V I g β

SLIDE 40

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (40/42)

Exercise 9 (cont’d) : β = 200, VA = 150 V

1 262 1 262 262 945 10 277 ) || || ( ) || || ( 1 ) || || (

2 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2

≈ + = = × × = + =

− i

L

E

m

L E

m

L E

m

i

v

v R R r g R R r g R R r g v v

[ ] [ ]

k 41 . 9 k 189 722 || k 90 . 9 ) || || ( ||

2 2 2 2 2 2 2 2

= + = + =

i L E

B

i

R R R r r R R β

π

Stage 2: Emitter Follower

For Gain & Ri , start from the load side because we need to know RL

k 41 . 9

2 1

= =

i L

R R k 100

2

= =

L L

R R k 90 . 9 k 722 . k 8 . 20 mA/V 277

2 2 2 2

= = = =

B π

m

R r r g

SLIDE 41

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (41/42)

Exercise 9 (cont’d) : β = 200, VA = 150 V

k 95 . 4 97.2k || k 22 . 5 ) / 1 ]( / ) || [( 1 ||

1 1 1 1 1 1 1 1 1 1 1

= =       + + + =

i E

L

C E B i

R r R r R R R r R R

π π

β 25 . 3 . 62 201 046 . 500 10 122 1 10 65 . 1 10 122 046 . ) / 1 ]( / ) || [( k 1.65 k 41 . 9 || k 2 || ) / 1 ]( / ) || [( 1 ) || (

3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

− = − = + × × + × × × − = = + = = + + + − =

− − i

E
L

C L C E

L

C E m L C m i

v

v r R r R R R R r R r R R R g R R g v v

π π

Stage 1: CE amplifier with RE k 41 . 9

2 1

= =

i L

R R k 22 . 5 k 64 . 1 k 2 . 47 mA/V 122

1 1 1 1

= = = =

B π

m

R r r g

SLIDE 42

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (42/42)

Exercise 9 (cont’d) : β = 200, VA = 150 V

Stage 2: Emitter Follower

For Ro , start from the source side because we need to know Rsig

k . 2

1 2

= =

sig

R R 100

1

Ω = =

sig sig

R R k 2 || 1 ||

1 1 1 1 1 1 1 1 1 1 1

= ≈                 + + + =

C

sig

B E E

C
R

R R R R r R r R R

π

β Stage 1: CE amplifier with RE Ω = = + = 8 . 11 9 . 11 || k 1 || ||

2 2 2 2 2 2

β

π sig B E

R

R r R R k 22 . 5 k 64 . 1 k 2 . 47 mA/V 122

1 1 1 1

= = = =

B π

m

R r r g k 90 . 9 k 722 . k 8 . 20 mA/V 277

2 2 2 2

= = = =

B π

m

R r r g

SLIDE 43

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (43/42)

Exercise 9 (cont’d) : β = 200, VA = 150 V

k . 2 k 95 . 4 25 . 3 / k 41 . 9 100

1 1 1 1 2 1 1

= = − = = = Ω = =

i

i

i

L sig sig

R R v v R R R R k 22 . 5 k 64 . 1 k 2 . 47 mA/V 122

1 1 1 1

= = = =

B π

m

R r r g k 90 . 9 k 722 . k 8 . 20 mA/V 277

2 2 2 2

= = = =

B π

m

R r r g Ω = = = = = = = 8 . 11 k 41 . 9 1 / k 100 k . 2

2 2 2 2 2 1 2

i

i

L

L

sig

R R v v R R R R 19 . 3 1 ) 25 . 3 ( 98 . 11.8 k, 4.95

2 2 1 1 2 1

− = × − × = × × + = Ω = = = =

i

i
sig

i i sig

i

i

v v v v R R R v v R R R R

SLIDE 44

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (44/42)

Exercise 9 (cont’d) : β = 200, VA = 150 V

k . 2 k 41 . 9 11.8 k 4.95

1 2

= = Ω = =

i
i

R R R R Hz 3 . 52 7 . 29 9 . 15 71 . 6

3 2 1

= + + = + + ≈

p p p p

f f f f

Amplifier Cut-off frequency 3 caps: 3 poles

Hz 71 . 6 ) ( 2 1

1 1

= + =

c sig i p

C R R f π Hz 9 . 15 ) ( 2 1

2 2

= + =

c L

p

C R R f π Hz 7 . 29 10 470 ) ( 2 1 ) ( 2 1

9 1 2 1 , ,

= × × + = + =

− −

i

pj cj j

j

i pj

R R f C R R f π π

Coupling cap at the input: Coupling cap at the output: Coupling cap between stages:

SLIDE 45

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (45/42)

Exercise 10: Find the amplifier parameters of the circuit below. (Si BJT with β = 200, VA = 150 V, ignore Early effect in bias calculations).

This is a two-stage amplifier. Both stages are common emitter

amplifier with RE.

Both stages have source-degeneration bias with two power

supplies (no voltage divider).

There are NO coupling capacitors and the second stage is

biased from VC1.

SLIDE 46

A 3 . 33 / mA 65 . 6 510 ) 1 /( 10 6 . 3 510 10 6 . 3 51 . 3 3 510 7 . 10 6 . 3 10 83 . 3 10 6 . 3 15 3 510 ) ( 10 6 . 3 15

2 2 2 2 2 2 2 3 2 2 3 2 2 3 3 3 2 2 2 1 3

µ β β = = ≈ = + + × = + × = − + + × + × × × = − + + + × =

− C B C E E E E B E B E BE B C

I I I I I I I I I I I V I I

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (46/42)

Exercise 10 (cont’d) : β = 200, VA = 150 V

A 2 . 19 / mA 83 . 3 3 600 7 . ) 1 /( 100 3 600 100

1 1 1 1 1 1 1 1 1 1 1

µ β β = = ≈ = − + + + = − + + =

C B C E E E E BE B

I I I I I I I V I

V 7 . and V, 7 . : Active in are s BJT both Assume

1 1 1

≥ > =

CE C BE

V I V

BE1-KVL: BE2-KVL: CE2-KVL:

0.7 V 63 . 4 3 510 10 5 . 1 15

2 2 2 2 3

= > = − + + × =

D CE E CE C

V V I V I

CE1-KVL:

0.7 V 79 . 1 3 600 ) ( 10 6 . 3 15

1 1 1 2 1 3

= > = − + + + × =

D CE E CE B C

V V I V I I

SLIDE 47

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (47/42)

Exercise 10 (cont’d) : β = 200, VA = 150 V

k 36 . 1 k 2 . 39 10 83 . 3 150 mA/V 147 10 26 10 3.83

1 1 1 1 3 1 1 1 3 3 1 1

= = = = × = = = × × = =

m B T π

C

A

T

C m

g I V r I V r V I g β

V 63 . 4 A 2 . 19 mA 83 . 3

1 1 1

= = =

CE B C

V I I µ V 79 . 1 A 3 . 33 mA 65 . 6

2 2 2

= = =

CE B C

V I I µ

k 782 . k 6 . 22 10 65 . 6 150 mA/V 256 10 26 10 65 . 6

2 2 2 2 3 2 2 2 3 3 2 2

= = = = × = = = × × = =

m B T π

C

A

T

C m

g I V r I V r V I g β

SLIDE 48

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (48/42)

Exercise 10 (cont’d) : β = 200, VA = 150 V

For Gain & Ri , start from the load side because we need to know RL

k 8 . 92

2 1

= =

i L

R R k 100

2

= =

L L

R R ∞ = = = =

2 2 2 2

k 782 . k 6 . 22 mA/V 256

B π

m

R r r g k 8 . 92 92.8k || ) / 1 ]( / ) || [( 1 ||

2 2 2 2 2 2 2 2 2 2 2

= ∞ =       + + + =

i E

L

C E B i

R r R r R R R r R R

π π

β 88 . 2 132 379 108 . 510 10 56 2 1 10 48 . 1 10 56 2 108 . ) / 1 ]( / ) || [( k 1.48 k 100 || k 5 . 1 || ) / 1 ]( / ) || [( 1 ) || (

3 3 3 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

− = − = + × × + × × × − = = + = = + + + − =

− − i

E
L

C L C E

L

C E m L C m i

v

v r R r R R R R r R r R R R g R R g v v

π π

Stage 2: CE amplifier with RE

SLIDE 49

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (49/42)

Exercise 10 (cont’d) : β = 200, VA = 150 V

k 108 k 108 || ) / 1 ]( / ) || [( 1 ||

1 1 1 1 1 1 1 1 1 1 1

= ∞ =       + + + =

i E

L

C E B i

R r R r R R R r R R

π π

β 71 . 5 3 . 89 510 127 . 600 10 147 1 10 47 . 3 10 147 127 . ) / 1 ]( / ) || [( k 47 . 3 k 8 . 92 || k 6 . 3 || ) / 1 ]( / ) || [( 1 ) || (

3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

− = − = + × × + × × × − = = + = = + + + − =

− − i

E
L

C L C E

L

C E m L C m i

v

v r R r R R R R r R r R R R g R R g v v

π π

Stage 1: CE amplifier with RE k 8 . 92

2 1

= =

i L

R R ∞ = = = =

1 1 1 1

k 36 . 1 k 2 . 39 mA/V 147

B π

m

R r r g

SLIDE 50

∞ = = = =

2 2 2 2

k 782 . k 6 . 22 mA/V 256

B π

m

R r r g ∞ = = = =

1 1 1 1

k 36 . 1 k 2 . 39 mA/V 147

B π

m

R r r g

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (50/42)

Exercise 10 (cont’d) : β = 200, VA = 150 V

For Ro , start from the source side because we need to know Rsig

k 6 . 3

1 2

= =

sig

R R 100

1

Ω = =

sig sig

R R k 6 . 3 || 1 ||

1 1 1 1 1 1 1 1 1 1 1

= ≈                 + + + =

C

sig

B E E

C
R

R R R R r R r R R

π

β Stage 1: CE amplifier with RE k 5 . 1 || 1 ||

2 1 2 2 2 2 2 2 2 2 2

= ≈                 + + + =

C

sig

B E E

C
R

R R R R r R r R R

π

β Stage 2: CE amplifier with RE

SLIDE 51

∞ = = = =

2 2 2 2

k 782 . k 6 . 22 mA/V 256

B π

m

R r r g ∞ = = = =

1 1 1 1

k 36 . 1 k 2 . 39 mA/V 147

B π

m

R r r g

F. Najmabadi, ECE65, Winter 2013, Problems for Discrete Amplifiers (51/42)

Exercise 10 (cont’d) : β = 200, VA = 150 V

k 6 . 3 k 108 71 . 5 / k 8 . 92 100

1 1 1 1 2 1 1

= = − = = = Ω = =

i

i

i

L sig sig

R R v v R R R R k 5 . 1 k 8 . 92 88 . 2 / k 100 k 6 . 3

2 2 2 2 2 1 2

= = − = = = = =

i

i

L

L

sig

R R v v R R R R 4 . 16 ) 88 . 2 ( ) 71 . 5 ( 1 k 5 . 1 k, 108

2 2 1 1 2 1

= − × − × = × × + = = = = =

i

i
sig

i i sig

i

i

v v v v R R R v v R R R R Amplifier Cut-off frequency:

nly Coupling cap at the output:

Hz 7 . 15 ) ( 2 1

2

= + =

c L

p