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

R sCR R R

sC R R

R R sC

Low-Pass Filter

sC R R

Low-Pass Filter

R R sC

sC R R

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

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

.......

....... )(

s s s F

.......11

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

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...

.......

....... )(

z z p p L

s s s F

.....11

2.........11

.......11

.......11 )(

z z p p

s s s F

Normalized High frequency Response

Low Frequency Response

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

DC analysis

External & Internal Capacitors open Circuit

DC analysis

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

r R Rr g

o /5.22

)(|| r r R R x Bin

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

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

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

3 Rad RC

R sC R sC

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

s B xe E CE

R Rr r R R

sec/2470

rad RC

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

Low frequency Response

LC om M R R

r R Rr g A

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

s B x E CE

R Rr r R R

LC oC R Rr R ||2

02 zCC

.......

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

z z L H L M

s s s F s F s F A s A

Low Frequency : Dominant Pole

Hz rad

Hz Rad Hz rad

Hz rad

393sec/2470PoleDominanatApprox

394sec/24792470103189PoleDominantExact

5.0sec/4.4

393sec/247015sec/103

30sec/189

Hz f frequencylow L 393 3

Low Frequency Response CE : CE, CC1 & CC2 Included

393 hZ30 hZ

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

L 39.05.40

pF C C

poleother f c

L 30103.5

L L L f Hz f 2100

pF C C poleother f cC

L 5.161066.9

k RC eqC

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

E E BC

BE CC E

084.51

Mid b d G i A

Mid-band Gain AM

Configuration is CC so ‘T’ Model

k R Rr RR

R Rr g

r R R g

L E e Bin

sig in

L E em

sig in

e L E m

/922.0

4.147||1||

Low Frequency Response : C

Low Frequency Response : Cc1

(β+1)IB

k Rr R R R

Le B sig eq

in sig eq

4.157)')(1(||1

1 104.157

ceqc p C RC

Low Frequency Response : C

Low Frequency Response : Cc2

r R R Rsig B

e E Leq 72.5

1072.5

I/β+1

Iβ /β+1

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

antpoledo f c

L 06.31072.5

1min211

F C C poleother f cC

L 11.11042.157

L L Lf Hz f 210

k RC eqC 72.5

k RC eqC 4.157

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

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

in sig

in M R g

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

z z p p

•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

0V(ii)

circuitopenis (i)as||R sees

gd in gs

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.

z gs gd gsmC

gsm gd so gsC

sV sC V g I

V To have

V g sC V V I

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

R R g R R I

R R g I

R I R I g I

V V V g I

gd gd gd

Lm L X

X L X m X

X gs gsm X

@ Node D

' R I V

sig in

p gd gd gd

L Lm Lm L

Lm L x L xm L x x

L gsm L x x

R R g R R R g R R I

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

RV g R R I V

gsm s V g I I

Frequency Response

s A sV

L sig inm L sig in gd

sig in gs

sig in

R R R g R R R sC

R R sC R g

'||'||1

''2||||

L sig inm L sig in gd

p R R R g R R RC

in gs RC ||R 1sig

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

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

The Miller equivalent circuit.

KV V I

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

• 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

K=-100 V/V, Z = 1 M Ω

Example1 M ohm

Example

Z Z 9.9

V V R Z

sig sig

O /4971001

Z Z 99.0

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

Application of Miller Effect

Miller in AC j

)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

Bode plot for (h fe ) .

Cct for hf (s) = I /Ib

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

Unity Gain Frequency ω

Unity Gain Frequency ωT

M H M b

s A s F s A I

)()()()(

r C C s s

The Cutoff Frequency

I C C r I V b

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

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

CE with RE

•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:

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