Lec2 Frequency Response

7/27/2019 Lec2 Frequency Response

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Frequency Response

BJT/MOSFET



Quiz No 1Group BGroup A

If R1=R2=2kΩ and C=1μFFind ωz & ωp.

ωz = 1/CR1 = 500 rad/sec

ωp = 1/C(R1||R2) = 1000 rad/sec

ωz = 1/CR2 = 500 rad/sec

ωp = 1/C(R1+R2) = 250 rad/sec

15-06-09



High-Pass Filter

Low Frequency Band

2121

12

1

12

2

12

2 1

1

1

||

)(

)(

R sCR R R

sCR R

sCR

R R

R

sC R R

R

sV

sV

i

o

21

2

21

1

21

2

||1

1

)(

)(

R R

Rk

s

sk

R R sC

sCR

R R

R

sV

sV

p

z

i

o



Low-Pass Filter

12

21

2

||1

1

)(

)(

R RC

s R R

R

sV

sV

i

o

121

2

1

1

1||

1

||

)(

)(

pi

o

sk

sC R R

sC R

sV

sV



Low-Pass Filter

p

z

i

o

s

s

k

R R sC

sCR

R R

R

sC R R

sC R

sV

sV sT

1

1

)(1

1

1

1

)(

)()(

21

2

21

2

21

2

Find ‘k’ k = Vo(s)/Vi(s) while C is short circuited

ωZ while Vo(s)=0 s = - 1/CR2

ωp while Vi(s)=0 s = - 1/C(R1+R2)



Complete Analysis (Mid-band Frequency)

D C Analysis.

– Suppress ac independent sources – External & Internal Capacitors be Open circuited

– Determine DC operating Point I C /I D

– Calculate small signal model parameters gm, r pi, r e

Mid-Band Analysis AM.

– Suppress DC independent sources

– External Capacitors be short circuited – Internal Capacitors be Open circuited

– Replace BJT with small signal Model

– Analyze the resulting circuit to find voltage gainsGv Ai, input & output resistance

t



omp ete na ys s ow-FL, g requency FH

Low Frequency Analysis FL.


– Internal Capacitors be Open circuited


– Keep the External Capacitors in the circuited – Analyze the resulting circuit to find Req for each capacitor by

using Short Circuit Time Constant Method.

High Frequency Analysis FH.


– External Capacitors be Short circuited


– Keep the Internal Capacitors in the circuited – Analyze the resulting circuit to find Req for each capacitor by

using Open Circuit Time Constant Method after using Miller theorem.



Gain Function A(s)

A(s) = AMFL(s)FH(s)

• FL(s) & FH(s) are gain function in low & high

frequency bands and are dependent on

frequency

• At mid-band, the gain for ω L<< ω <<ωH

A(s) = AM



Gain Function A(s)• Mid-band gain is determined by analyzing amplifier

equivalent circuit with the assumption• Coupling and bypass capacitors are acting as a perfect short

circuit• Internal capacitors of the amplifier are acting as a perfect open

circuit

• The low frequency function FL(s) is determined by

analyzing amplifier equivalent circuit • Including Coupling and bypass capacitor but assuming

Internal capacitors of the amplifier as a perfect open circuit

• The High frequency function FH(s) is determined byanalyzing amplifier equivalent circuit

• Including Internal capacitors of the amplifier but assumingCoupling and bypass capacitor as a perfect short circuit

121

21

.......

....... )(

p p p

z z L

s

s

s s

s s s F

1

1

.......11

.......11

)(

121

21

p p p

z z

H

s s s

s s

s F



Dominant-Pole Approximation

ωP1 >> ωP2 , ωP3 ……..ωZ1 ,ωZ2

• Dominant-Pole Approximation can bemade if the highest frequency pole is

separated from the nearest pole or zero by

at least two octave ( a factor of four)



Complete Analysis Amplifier

AMFL(s)FH(s)

DC Analysis Mid-band Analysis Low frequency Analysis High frequency Analysis

AM FL(s) FH(s)Suppress Independent AC

sources

Suppress Independent DC

sources


sources


sources

Open Circuit all capacitors Suppress all capacitors Open Circuit Internal capacitors Short Circuit External capacitors

Calculate Node Voltages & IC Draw Small Signal Model Draw Low Frequency Model Draw High Frequency Model

Find out Mode of operation &

SSM parameters

Find out Gain, Input & output

Impedance

Using Short Circuit Time

Constant, Find out Dominant

pole Frequency

Using Open Circuit Time

Constant, Find out Dominant pole

Frequency

......2...

.......

....... )(

22

12

22

12

121

21

z z p p L

p p p

z z L

s

s

s s

s s s F

.....11

2.........11

1)(

1

1

.......11

.......11 )(

22

12

22

12

121

21

z z p p

H

p p p

z z

H

s

s s s

s s s F



Normalized High frequency Response



Low Frequency Response

fL(s)



Reading Material


• Microelectronics Sedra & Smith

– Section 4.9 Ed 5

– Section 5.9 Ed 5 – Appendix ‘E’ Ed5

– Section 7.1 thru 7.3 Ed 4

• Microelectronics Millman

• Electric Cct Franco




k r

r

o

X

100

50

100



DC analysis

External & Internal Capacitors open Circuit



DC analysis

mA

R R

I

E B

E 1

1

7.04

100

k r r

I

V r

AmV I

V g

e

E

T e

E

T m

5.21

25

/25

V R I V V C C CC C 6

V R I V V B B BB B 4

4.0 BC V V Mode Active for Check



Mid-band (dc) Gain AM

Suppress DC independent sources

External Capacitors be short circuitedInternal Capacitors be Open circuited

Replace BJT with small signal Model



Mid-band (dc) Gain AM

sin

in

x

LC om

sig

o

R R

R

r r

r R Rr g

v

v

||||

sig

B

B

o

sig

o

v

v

v

v

v

v

v

v

V V v

v

sig

o /5.22

)(|| r r R R x Bin

+

-

vಗ

vB

Rin



Low Frequency Response : Equivalent CircuitSuppress DC independent sources

Internal Capacitors be Open circuited

Replace BJT with small signal ModelKeep the External Capacitors in the circuitedAnalyze the resulting circuit to find Req for each capacitor by using Short Circuit TimeConstant Method.

Short Circuit Time Constant Analysis : C



sec/1891

11

1

111

rad

RC

RC

C C

p

C C p

k r r R R R x BS eqC 3.5)(||1

Find Req for the CC1 by inspection

Suppress the independent voltage source.

Short-Circuit Time Constant Analysis : CC1

Keep CC1 and short cct all other Capacitors

01 zCC

Short Circuit Time Constant Analysis : C



Short-Circuit Time Constant Analysis : CC2

Keep CC2 and short cct all other Capacitors

k R Rr R LC oC 66.9||2

sec/1031

22

2

222

rad

RC

RC

C C

p

C C p

Find Req for the CC2 by inspection

Suppress the independent voltage source.

02 zCC

or rcu me ons an na ys s :



or - rcu me ons an na ys s : E

(Zero Frequency )

sec/4.4

1

011

1||

3 Rad RC

R sC R sC

RC

E E

z

E E

E E

E E

or rcu me ons an na ys s : E



or - rcu me ons an na ys s : E

Keep CE and short cct all other Capacitors

5.40

1

||||

s B xe E CE

R Rr r R R

sec/2470

13

3

rad RC

RC

CE CE

p

CE CE p

Find Req for the CE by inspectionSuppress the independent voltage source.

r o is neglected to simplify the analysis

Ie /(β+1) Ie(β+1)/β

Ie



Low frequency Response

sin

in

x

LC om M R R

Rr r

r R Rr g A

||||

11

1

1

C C

p RC

01 zCC )(||1 r r R R R x BS eqC

E E

z RC

13

1

||||

s B x E CE

R Rr r R R

CE CE

p RC

13

LC oC R Rr R ||2

22

2

1

C C

p RC

02 zCC

.......

....... )( 1)( )()(

21

21

p p

z z L H L M

s s

s s s F s F s F A s A



Low Frequency : Dominant Pole

Hz rad

Hz rad

Hz rad

Hz Rad Hz rad

Hz rad

p L

z

z z

p

p

p L

393sec/2470PoleDominanatApprox

394sec/24792470103189PoleDominantExact

5.0sec/4.4

0

393sec/247015sec/103

30sec/189

3

222

3

21

3

2

1

Hz f frequencylow L 393 3



Low Frequency Response CE : CE, CC1 & CC2 Included

393 hZ30 hZ

16 hZ

30dB



Designing of Coupling Capacitors

Design the coupling capacitors so that a dominant low

frequency pole at 100 Hz is obtained.Design so that the contribution of the capacitors are in

the ratio of 8:1:1.

pF C C

antpoledo f E

E

L 39.05.40

1min2

10

8

pF C C

poleother f c

C

L 30103.5

1 2

10

113

1

L L L f Hz f 2100

pF C C poleother f cC

L 5.161066.9

1

210

1

232

k RC eqC

3.51

5.40 E C

eqR k RC

C eq 66.92



Low Frequency Response : CC Configuration

β =120



DC Anaysis : CC Configuration

β =120

Mode ActieveV V

V R I V V V

mA R R

V V I

BC

E E BC

B E

BE CC E

4.0

3,5

595.0

1

k r r

I

V r

e

E

T e

084.51

42

Mid b d G i A



Mid-band Gain AM

Configuration is CC so ‘T’ Model

V V v

v

k R Rr RR

R R

R

R Rr

R R

R R

R

R Rr

R Rr g

v

v

R R

R

R Rr

r R R g

v

v

v

v

v

v

v

v

sig

o

L E e Bin

sig in

in

L E e

L E

sig in

in

L E e

L E em

sig

o

sig in

in

L E e

e L E m

sig

i

i

be

be

o

sig

o

/922.0

4.147||1||

||

||

||

||

||

||

Low Frequency Response : C



Low Frequency Response : Cc1

(β+1)IB

A B

k Rr R R R

R R R

Le B sig eq

in sig eq

4.157)')(1(||1

1

3111

1 104.157

11

ceqc p C RC

IB

01

C zC

βIB

Low Frequency Response : C



I

Low Frequency Response : Cc2

k R R

r R R Rsig B

e E Leq 72.5

)1(

||||2

3

222

2

1072.5

11

ceqc

p

C RC

I/β+1

Iβ /β+1

02

C zC



Designing of Coupling Capacitors

Design the coupling capacitors so that a dominant low

frequency pole at 10 Hz is obtained.Design so that the contribution of the capacitors are in

the ratio of 10:1 and the total capacitance is minimized

F C C

antpoledo f c

C

L 06.31072.5

1min211

1023

2

F C C poleother f cC

L 11.11042.157

1

211

1

131

L L Lf Hz f 210

k RC eqC 72.5

2

k RC eqC 4.157

1



Standard Capacitors

Design requires

CC1 = 1.11μF & CC2 =3.06 μF

Standard Capacitor Available commercially are

CC1 = 1.0μF & CC2 =3.3 μF

So selecting Standard capacitor

Dominant Low frequency pole @ f L=9.44 Hz

L f C 1 & C 2



Low frequency response : Cc1 & Cc2



High Frequency Response

fH(s)

Frequency Response : Exact Method



Frequency Response : Exact Method

F R E t M th d



Frequency Response : Exact MethodReference Chapter 6-6.3

Hi h F R



High Frequency Response

Mid band Gain



Mid-band Gain

' L M

in sig

in M R g

R R

R A



Open Circuit Time Constant for f H• If poles and zeros can be determined easily so

calculate f H

• If not simple to find poles & zero then use opencircuit time constant method.

• Consider one capacitor at a time while reducing

all other capacitance to zero – (open circuit) andsignal source to zero



Open Circuit Time Constant for f H

2

2

2

1

2

2

2

1

112

11

1

z z p p

H

•Determine resistance Rieq seen by Ci –

•Repeat process for other capacitors

•Compute b1 by Summing individual time constant – called open

circuit time constant

R for C Open Cct Time Constant



Req for Cgs Open Cct Time Constant

R

1

R

||R R

0V(ii)

circuitopenis (i)as||R sees

gs

p1

gsgs

siggs

sig

sig

gs

gs

in

gd in gs

C

C

R

C RC

To find ωz1,

Find the value of ‘s’ when Vo(s) = 0

when 1/sCgs= 0, Causes Vgs=0 -- gmVgs= 0

so ωz1=∞

To find ωp1,

R for C Open Cct Time Constant



Req for Cgd Open Cct Time Constant

To find ωz1,Find the value of ‘s’ that causes Vo(s) = 0

Vo(s) will reduce to zero, if current equal

to gmVgs flow through Cgd from left to

right, allowing no current thru RL.

m gd

z gs gd gsmC

so

gsm gd so gsC

g C

sV sC V g I

V To have

V g sC V V I

gd

gd

1

1

0

2

)(

)(

gmVgs

R f C



Req for Cgd

circuitopenC(i) gs gd C

There is one dependent current source so apply test current IXFind Vtest, & independent source to be suppressed

To find ωp1,



• @ node G

||where'

sig in

'

X gs R R R R I V

)(

1

)2(

''''

'

''

'

''

''

'

b RC

R R g R R I

V R

R

R R g I

R

V

R

V

R

R I R I g I

R

V V V g I

gd gd gd

Lm L X

X

L

m X

L

X

L

X

L

X L X m X

L

X gs gsm X

gd

@ Node D



||

' R I V

R R R

x gs

sig in

'

gd gd

p gd gd gd

L Lm Lm L

x

x gd

Lm L x L xm L x x

L gsm L x x

RC RC

R R g R R R g R R I

V R

R R g R R I R R I g R R I V

RV g R R I V

1

1'''

''''

'

2

''''

''''

''

gsm s V g I I

Frequency Response



q y p

2

2

1

1

1

1

1

)(

)(

p

z

p

M

sig

o

s

s

s A sV

sV

L sig inm L sig in gd

m

gd

sig in gs

Lm

sig in

in

sig

o

R R R g R R R sC

g C

s

R R sC R g

R R

R

sV

sV

'||'||1

1

||1

1'

)(

)(

m

gd

z

g

C 1

12

''2||||

1

L sig inm L sig in gd

p R R R g R R RC

01 z

in gs RC ||R 1sig

p1



Miller Theorem

Miller effect



Miller effect• The miller effect is the effective multiplication of a impedance

across a negative gain device.

• Miller effect accounts for an increase in the equivalent inputcapacitance of an inverting voltage amplifier due to amplificationof capacitance between the input and output terminals.

• Although Miller effect normally refers to capacitance, anyimpedance connected between the input and another nodeexhibiting high gain can modify the amplifier input impedance viathe Miller effect.

• This increase in input capacitance is given by

• where Av

is the gain of the amplifier and C is the feedbackcapacitance

• The Miller effect was named after John Milton Miller When Miller published his work in 1920.

v M AC C 1

Miller Effect ACC 1

http://en.wikipedia.org/wiki/Capacitance

http://en.wikipedia.org/wiki/Amplifier

http://en.wikipedia.org/wiki/John_Milton_Miller

http://en.wikipedia.org/wiki/John_Milton_Miller

http://en.wikipedia.org/wiki/Amplifier

http://en.wikipedia.org/wiki/Capacitance



Miller Effect

• As most amplifiers are inverting amplifiers (i.e. Av < 0) theeffective capacitance at the input is larger .

• For non-inverting amplifiers, the Miller effect results in anegative capacitor at the input of the amplifier (compareNegative impedance converter ).

• The circuit is also referred as a Capacitance Multiplier.

• The capacitance on the output is often neglected since it sees

C (1 − 1 / A v ) and amplifier outputs are typically low impedance.However if the amplifier has a high impedance output, such as if a gain stage is also the output stage, then this RC can have asignificant impact on the performance of the amplifier. This iswhen pole splitting techniques are used.

Pole splitting causes the pole next in frequency (usually anoutput pole) to move to a higher frequency. This pole movementincreases the stability of the amplifier and improves its stepresponse at the cost of decreased speed

v M AC C 1

http://en.wikipedia.org/wiki/Negative_impedance_converter

http://en.wikipedia.org/wiki/Pole_splitting

http://en.wikipedia.org/wiki/Step_response




http://en.wikipedia.org/wiki/Pole_splitting

http://en.wikipedia.org/wiki/Negative_impedance_converter



The Miller equivalent circuit.



q

K

Z

I

V Z

Z

KV V

Z

KV I

Z

V I

K

Z Z

I

V

Z

K V

Z

KV V I

Z

V I

11

00

1

1

2

2

2

11

2

1

2

2

2

1

1

1

1

11

1

1

1

Miller Theorem



Miller Theorem

• Miller theorem states that bridging impedance Z can be

replaced by two impedances: Z1 connected between node 1 and

ground and Z2 connected between node 2 and ground where

circuit equivalent theobtainto

function gainV

V k where

K

Z

Z K

Z

Z

11

& 1

1

2

21



• Miller equivalent circuit is valid only as long as theconditions that existed in the network when ‘k’ wasdetermined are not-changed

• Miller theorem is very useful in determining the inputresistance and the gain of the amplifier.

• Miller equivalent circuit cannot be used directly todetermine the output resistance of an amplifier. It is dueto the fact that test source is applied to the outputterminal and input signal is eliminated – obviously itchanges the value of ‘k’.

Miller Theorem



Miller Theorem

Miller theorem



Miller theorem

K=-100 V/V, Z = 1 M Ω

Example1 M ohm



Example

k k

K

Z Z 9.9

1001

100

11

V V R Z

Z

V

V

V

V

V

V

sig sig

O

sig

O /4971001

11

1

M

K

Z Z 99.0

11

2

v

sig

o

A

R R

R

R

V

V

1

2

1

2

11



OBSERVATIONS



OBSERVATIONS• The Miller replacement for

a negative feedback resultsin a smaller resistance [bya factor of (1-K)] at theinput.

• The multiplication of afeedback impedance by afactor (1-k) is referred as

Miller Multiplication or Miller Effect



Application of Miller Effect



• If the feedback impedance is a capacitor ,thenthe Miller capacitance reflected across the inputterminal is .

• Therefore, connecting a capacitance from theinput to output is equivalent to connecting acapacitance

• Due to Miller effect, a small feedbackcapacitance appears across the input terminalsas a much larger equivalent capacitance with alarge gain (e.g. ).

• At high frequencies, this large capacitance has alow impedance that tends to short out the inputsignal

f C

Application of Miller Effect

)1(1

,

v f

Miller in AC j

Z

f C

)1( v f AC

80|| v A

Analysis using Miller’s Theorem



Analysis using Miller s Theorem

Since Cgd is small, the current through it will be much smaller

than that of the controlled source gmVgs.

Thus neglect the current through Cgd in determining the output voltage Vo.

‘

‘



Intrinsic/Cut-off / Unity-gain

Frequency

Gain-Bandwidth Product

Frequency

h Bode Plot



hfe Bode Plot

Bode Plot

f



Bode plot for (h fe ) .

84

Cct for hf (s) = I /Ib



85

Cct for h fe ( s) = I c /I b

Unity Gain Frequency ω



86

Unity Gain Frequency ωT

p

z

M H M b

c

s

s

s A s F s A I

I

shfe

1

1

)()()()(

0 B

C M

I

I A

r C C s s

s

s F

p

z H

1

1

1

1)(

r C C

p

z

1

pT 0

)(2

C C

g f

C C

g

mT

mT

The Cutoff Frequency



87

The Cutoff Frequency

sC sC

r

I C C r I V b

b

1||||

V sC g V sC V g I mmC

Common Emitter short circuit gain, as a function of frequency in terms of Hybrid ‘π’

components

@ Output node

@ Input node



Single Stage BJT Amplifier



Single Stage BJT Amplifier Small Signal Output

Model Resistance

Pi Model Include

‘T’ Model Include

‘T’ Model Don't

Include

‘T’ Model Dent

Include

Common Emitter (CE)

Common Collector (CC)

Common Emitter (CE)

with Emitter Resistance

Common Base (CB)



CE with RE



E E

To analyze this configuration, we first set down the complete nodal equations

Using the relationship , the nodal equations can be rewrite in a more homogeneous form:

CE with RE



Eliminating vo from the last two nodal equations we find that

and if we substitute this expression into the first nodal equation we find that

E E

CE with RE



E E

•When but it reduces to

•When this expression reduces to

Finally, substituting these two expressions into the second nodal equation

we find the following expression for the voltage gain:

Lec2 Frequency Response

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