BJT and FET Frequency Response

BJT and JFET Frequency Response

Effects of Frequency on Operation of Circuits

The frequency of a signal can affect the response of circuits. The reactance of capacitors increases when the signal frequency decreases,

and its reactance decreases when the signal frequency increases. The reactance of inductors and winding of transformers increases when the

signal frequency increases, and its reactance decreases when the signal frequency decreases.

Devices such as BJTs, FETs, resistors, and even copper wires have intrinsic capacitances, whose reactance at high frequencies could change the response of circuits.

The change in the reactance of inductors and capacitors could affect the gainof amplifiers at relatively low and high frequencies.

At low frequencies, capacitors can no longer be treated as short circuits, because their reactance becomes large enough to affect the signal.

At high frequencies, the reactance of intrinsic capacitance of devices becomes low enough, that signals could effectively pass through them, resulting to changes in the response of the circuit.

At low frequencies, reactance of primary of transformers become low, resulting to poor low frequency response. Change in magnetic flux at low frequencies become low.

At high frequencies, the stray capacitance of transformer windings reduces the gain of amplifiers.

Effects of Frequency on Operation of Circuits Increase in the number of stages could also affect the frequency response of a

circuit. In general, the gain of amplifier circuits decreases at low and high

frequencies. The cutoff frequencies are the frequencies when the power delivered to the

load of the circuit becomes half the power delivered to the load at middle frequencies.

Voltage gain

Frequency

0.707 AVmid

AVmid

f1 f2BandwidthAvmid = voltage gain of amplifier at middle frequencies0.707 Avmid = voltage gain of amplifier at lower cutoff frequency and higher cutoff frequency

(when output power is half the output power at middle frequencies)f1 = low cutoff frequency PO(HPF) = output power at higher cutoff frequency Vi = input voltagef2= high cutoff frequency PO(LPF) = output power at lower cutoff frequency Pomid= output power at middle frequencies

Omid

2vmid

2vmid

O(LPF)(HPF) P 0.5Ro

Vi A5.0Ro

Vi 0.707APPo

Bandwidth = f2-f1

Frequency Response of Amplifier Circuits f1 and f2 are called half power, corner, cutoff, band, break, or -3db

frequencies. f1 is the low cutoff frequency and f2 is the high cutoff frequency. When the amplitude of a signal is 0.707 of its original amplitude, its power

becomes half of its original power.

PHP = PMF / 2 = power at half power frequency

where: PHP = Power at half power point (f1 or f2)PMF = Power at middle frequencies

The bandwidth of the signal is equal to f2 f1

B = f2 f1 = bandwidth

Effects of Frequency on Operation of Circuits The 180 degrees phase shift of most amplifiers (Common emitter, common

source) is only true at middle frequencies. At low frequencies, the phase shift is more than 180 degrees. At high frequencies, the phase shift is less than 180 degrees.

Phase shiftbetween Voand Vi

Frequency

18002700

f1 f2

Phase shift between Vo and Vi

900

Frequency Response of Amplifier Circuits The graph of the frequency response of amplifier circuits can be plotted

with a normalized gain. (gain is divided by the gain at middle frequencies.)

Frequency

0.707 AVmid

1 AVmid

f1 f2

Normalized Gainin Ratio

frequency middleat gain voltageA ffrequency at gain voltageA :where

AAGain Normalized

Vmid

V

Vmid

V

Normalized Plot of Voltage Gain Versus Frequency

Frequency Response of Amplifier Circuits A decibel plot of the gain can be made using the following formula:

Voltage gain

Frequency0.707 AVmid

1 AVmid

f1 f2

Normalized Gain in db

frequency middleat gain voltageA ffrequency at gain voltageA :where

dbin gain normalizedAA20log

AA

Vmid

V

Vmid

V

Vmid

V

db

Decibel plot of Normalized Voltage Gain Versus Frequency

0 db-3 db-6 db-9 db

Capacitor Coupled Amplifier Circuit Frequency Response For capacitor coupled (also called RC-coupled) amplifiers:

The drop in gain at low frequencies is due to the increasing reactance of the coupling capacitors (Cc), and bypass capacitors (Cb, CE, and Cs).

The drop in gain at high frequencies is due to the parasitic capacitanceof network and active devices, and frequency dependence of the gain of BJTs, FETs, or vacuum tubes.

Voltage gain

Frequency

0.707 AVmid

AVmid

f1 f2

Drop in gain is due to increase in reactance of coupling and bypass capacitors

Avmid = voltage gain of amplifier at middle frequencies0.707 Avmid = voltage gain of amplifier at lower cutoff frequency and higher cutoff frequency

(when output power is half the output power at middle frequencies)f1 = low cutoff frequencyf2= high cutoff frequency

Bandwidth

Transformer Coupled Amplifier Circuit Frequency Response For transformer coupled amplifier circuits:

The drop in gain at low frequencies is caused by shorting effect of the input terminals (primary) of the transformer at low frequencies. The reactance of the primary of a transformer becomes very low at low frequencies and becomes zero at 0 hertz.

At low frequencies, change in magnetic flux becomes low, resulting to lower output voltage.

The drop in gain at high frequencies is due to the stray capacitance at the primary and secondary of a transformer, and frequency dependence of gain of devices. At high frequencies, the reactance of the stray capacitances becomes low enough) that high frequency signals are also shorted out.

Voltage gain

Frequency

0.707 AVmid

AVmid

f1 f2

Drop in gain is due to shorting effect of primary of transformerat low frequencies.

Drop in gain is due to stray capacitance at primary and secondary of transformer and other components, and frequency dependence of gainof active devices.

Bandwidth

Direct Coupled Amplifier Circuit Frequency Response For direct coupled amplifier circuits:

There are no coupling or bypass capacitors, or transformers to cause a drop in the gain at low frequencies. The gain at low frequencies is typically the same as that at middle frequencies.

The drop in gain at high frequencies is due to stray capacitance of the circuit and the frequency dependence of the gain of active devices.

Voltage gain

Frequency

0.707 AVmid

AVmid

f2

Drop in gain is due to stray capacitance of the circuit, and the frequency dependence of the gain of active devices.

Bandwidth

Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit A capacitor coupled circuit which acts as a high pass filter is shown below. At middle and high frequencies, the capacitor C can be considered a short

circuit because its reactance becomes low enough that the voltage appearing across RL is almost equal to Vi (input voltage of combination of C and RL).

At low frequencies, the coupling capacitor C could no longer be treated as a short circuit because its reactance becomes high enough that the voltage appearing at the load (RL) becomes significantly lower than Vi.

R can represent any resistance or resistance combination in a circuit. At low frequencies, the RC combination of the coupling capacitor (C) and the

resistance (R) determines the frequency response of the amplifier circuit. The reactance of the coupling capacitor C can be computed as:

R

Cc

IRVi =Input voltage to RC network

Capacitor Coupled Circuit Which Acts As A High Pass Filter

Vo = Outputvoltage

(Farad) Cc of ecapacitanc C (Hz) signal offrequency f :where

ffrequency at Cc of reactancefC21Xc

Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit

1ViVo

R load theacross voltageViVo

sfrequenciehigh at Cc of reactance0fC2

1Xc

L

At high and middle frequencies, Xc becomes low enough that it can be assumed to be zero (0) and Cc is assumed to be a short circuit. The voltage across R (Vo) can be assumed to be equal to the input voltage

of the RC network (Vi).

If the frequency is equal to zero (0) such as when the signal is a DC voltage, the reactance of Cc is equal to infinity, and the capacitor Cc can be assumed to be an open circuit. The voltage across R (Vo) is equal to zero (0).

Between the two extremes, the ratio between Vo and Vi will vary between zero and one (1).

0ViVo

R load theacross voltage0Vo

hz 0f when Cc of reactancefC2

1Xc

L


2ViVo when sfrequencie middleat Rat dissipatedpower

sfrequencie middleat Power R

Vi R

Vi21

R1

2Vi

RVoP

XcR when tageoutput vol 0.707ViVo

0.707Vi2

ViRRR

ViRXcR

Vi)(R)I(Vo

Xc,R When below.shown as Xc R when occurs thisand s,frequencie middleat power output theof that half is Rat power output the

(f1),frequency cutoff low the toequal is signal theoffrequency When the

RXcR

Vi)(R)I(Vo

2222

2R

2R

222

2

The magnitude of the output voltage can be computed as:


(Hz)frequency cutofflower RC21f1

f1C21XcR

When the frequency is equal to the low cutoff frequency (f1), R=XC and f1 can be computed as follows:

The normalized voltage gain at lower cutoff frequency (f1) can be computed as:

The normalized voltage gain at middle frequencies (fmid) can be computed as:

3dbA

0.707Alog20dbA

AVmid

Vmid

Vmid

cutoffVlower

db 0AAlog20db

AA

Vmid

Vmid

Vmid

Vmid


(Hz)frequency cutofflower RC21 f1 :where

(unitless) ffrequency at gain voltage

ff1j1

1Av

fCR21j1

1

RjXc1

1jXcR

RjXcR I

R IViVoAv

At frequency f, the voltage gain can be computed as:

In magnitude and phase form, the voltage gain at any frequency can be computed as:

fi/f/Tan

ff11

1Av 12


(db)

ff11

1log20ViVolog20Av

2db

In db (logarithmic form), the voltage gain at frequency f can be computed as:

When f=f1= lower cutoff frequency,

db 3-

f1f11

1log20ViVolog20Av

2db


22/12

2db

ff11log10

ff11-20log

(db)

ff11

1log20ViVolog20Av

The voltage gain at frequency f can be written as:

When f

Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit Using the points in the preceding slide, a bode plot can be made as shown below. A Bode plot is a piecewise linear plot of the asymptotes and associated

breakpoints. A Bode plot for the low frequency region is shown below. One octave is equivalent to a change in frequency by a factor of two (2). One octave results to a 6 db change in the normalized gain. One decade is a change in frequency by a factor of 10. One decade results to a 20 db change in the normalized gain.

Frequency (log scale)0.707 AV

1 AV

f1Normalized Gain in db

Bode Plot for Low Frequency Region

0 db-3 db-6 db-9 db

-12 db-15 db-18 db-21 db

f1/2f1/4f1/10

ff1log20X

(db)

ff11

1log20Av2

db

Actual Response Curve

Asymptote

Asymptote

Frequency Analysis of High Pass Resistor Capacitor (RC) Circuit The plot in the preceding slide shows two asymptotes. One for f> f1 (horizontal line 0 db). The plot of the line corresponding to f


Example: For the RC network shown below, Determine the break frequency (cutoff frequency), sketch the asymptotes and the frequency response curve.

Frequency (log scale)0.707 AV

1 AV

f1 = 99.47Normalized Gain

in db

Bode Plot for Low Frequency Region

0 db-3 db-6 db-9 db

-12 db-15 db-18 db-21 db

f1/2=49.74

f1/4=24.87

f1/10=9.947

RL =8 kohm

Cc = 0.2 microfarad

IRLVi =Input voltage to RC network

Vo = Outputvoltage Hz 99.47

)0x12.0)(000,8(21

CR21f1 6

L

Asymptote

Asymptote

-3db point

Low Frequency Analysis of Capacitor Coupled BJT Amplifier A capacitor coupled (also called RC coupled) BJT amplifier circuit is shown

below. At middle and high frequencies, the capacitors Cc, Cs, and Ce can be

considered short circuits because their reactance become low enough, that there are no significant voltage drops across the capacitors.

At low frequencies, the coupling capacitors Cc, Cs, and Ce could no longer be treated as short circuits because their reactance become high enough that the there are significant voltage drops across the capacitors.

Rc

Q1

Vcc

RB2

Cc

Vo = Outputvoltage

RL

ViVs

Rsig

Cs

RB1

RE Ce

Zi Zo

Vs = Signal sourceRsig = internal resistance of signal sourceCs =coupling capacitor for VsCc= coupling capacitor for RLCe= bypass capacitor for RE

Low Frequency Analysis of Capacitor Coupled BJT Amplifier The frequency analysis of high pass RC network can be used for capacitor

coupled BJT amplifier circuits. The values of R and C are taken from the equivalent resistances and capacitances in the BJT amplifier circuit.

For the portion of the circuit involving the coupling capacitor Cs, the equivalent circuit is shown below. Equivalent circuit assumes that the input impedance of the amplifier (Zi) is

purely resistive and is equal to Ri.

Cs

Zi = RiVi

Equivalent Circuit of Vs, Cs and Zi

IiCs

Zi = Ri

Vi

Equivalent Circuit of Vs, Cs and Zi

Ii

RB1//RB2 hie = rere

Zi = Ri

Vs

Rsig

Vs

Rsig

Low Frequency Analysis of Capacitor Coupled BJT Amplifier

CsjXRsigRiRi VsVi

The value of the input impedance (resistance) of the amplifier can be computed as:

Zi = Ri = RB1 // RB2 // hie= RB1 // RB2 // re

The voltage Vi can be computed using voltage divider rule.

The voltage Vi at middle frequencies (Cs can be considered short circuit) can be computed as:

The lower cutoff frequency can be computed as:RsigRiRi VsVi mid

Cs involving

circuit theofportion for thefrequency offcut lower Ri)Cs(Rsig2

1fLs

Low Frequency Analysis of Capacitor Coupled BJT Amplifier For the portion of the circuit involving the coupling capacitor Cc, the

equivalent circuit is shown below. Equivalent circuit assumes that the output impedance of the transistor is

purely resistive and is equal to ro.Cc

Zo= Ro

VRL

Equivalent Circuit of Circuit Portion Involving Cc

IRL

Rc RLib ro

Cc

Zo= Ro = Rc // ro

VRL

IRLRc // ro

RL

Low Frequency Analysis of Capacitor Coupled BJT Amplifier The value of the output impedance (resistance) of the amplifier can be

computed as:Zo = Ro = RC // ro

The lower cutoff frequency can be computed as:

Cc involving

circuit theofportion for thefrequency offcut lower )CR(Ro2

1fCL

LC

Low Frequency Analysis of Capacitor Coupled BJT Amplifier For the portion of the circuit involving the bypass capacitor Ce, the equivalent

circuit is shown below. The equivalent circuit uses the re model.

The resistance (Re) seen looking into RE from the output side can be computed as:

Ce(Rs/ + re

Equivalent Circuit of Portion of Circuit Involving RE and CE

RE

(Ampere)current quiescent Emitter (Ampere)current DCEmitter I

(ohms) I

10 X 26r

//RRsig//RRs' :Where

(ohms) sideoutput thefrom R into lookingseen impedancerRs'//RRe

E

E

3-

e

B2B1

EeE

eE rRs'//RRe


The lower cut-off frequency of the portion of the circuit involving the bypass capacitor Ce can be computed as:

The voltage gain of the amplifier without considering the effects of the voltage source resistance Rsig can be computed as: At middle frequencies, RE is shorted out because the reactance of Ce is

very low. Voltage gain can be computed as:

At low frequencies, the reactance of Ce becomes high and RE should be considered in the computation of the voltage gain.

side.output thefrom R into looking resistance equivalent Re :where

circuit theofportion for thefrequency offcut lower CeRe2

1f

E

LE

)considerednot (ro sfrequencie lowat amplifier theofgain voltageRr

Rc//RViV

ViVoAv

Ee

LRL

sfrequencie middleat amplifier theofgain voltager

ro//Rc//RViV

ViVoAv

e

LRL

Low Frequency Analysis of Capacitor Coupled BJT Amplifier Overall, the effects of the capacitors Cs, Cc, and Ce must be considered in

determining the lower cutoff frequency of the amplifier. The highest lower cutoff frequency among the three cutoff frequencies will

have the greatest impact on the lower cutoff frequency of the amplifier. If the cutoff frequencies due to the capacitors are relatively far apart, the

highest lower cutoff frequency will essentially determine the lower cutoff frequency of the amplifier.

If the highest lower cutoff frequency is relatively close to another lower cutoff frequency, or if there are more than one lower cutoff frequencies, the lowercutoff frequency of the amplifier will be higher than the highest lower cutoff frequency due to the capacitors.

fLT = overall lower cutoff frequency of amplifierfLT > fLSfLT > fLCFlt > fLE

Low Frequency Analysis of Capacitor Coupled BJT Amplifier Example: A voltage divider BJT amplifier circuit has the parameters listed below.

Determine the low cutoff frequency of the amplifier and sketch the low frequency response.Cs=12 uF Rsig = 2 kohm RL= 2 kohm RB1=50 kohm RB2= 10 kohmCc=2 uF RE = 2 kohm RC= 4 kohm =100 Vcc= 20 voltsCE=20 uF Assume that output resistance of transistor to be infinite.

742.1910 x 1.317

10 x 26re

CurrentQuiescent Emitter

A10 x 1.3172000

7.0333.3R

VVRVI

ground torelative baseat voltageDC

volts3.33350,00010,000

0)(20)(10,00R R

R V V

done. becan ionsapproximat following theand Rohms 20,0000)(100)(2,00 R

3

3

3

E

BEB

E

REE

B1B2

B2CCB

B2 E


0.4442,0001,596.08

1,596.08RsigRi

RiVsVi

amplifier of impedanceinput ohms 1,596.08 2.974,1

1000,101

000,501

1r////RRRiZi

Rs) gconsiderin(not sfrequencie midat gain voltage538.67 742.19

000,2000,4)000,2)(000,4(

rRc//R

ViVoAv

:sfrequencie middleAt

2.974,1)742.19(100)(r

eB2B1

e

Lmid

e


Cc involvingcircuit offrequency cutofflower

Hz 13.26310 X 2 )000,2(4,0002

1)CcR(Rc2

1f

:Cc of effects thegConsiderin

Cs involvingcircuit theofportion for thefrequency offcut lower Hz 3.688 10 X 21,596.08)1(2,0002

1Ri)Cs(Rsig2

1f

amplifier of impedanceinput ohms 1,596.08 2.974,1

1000,101

000,501

1r////RRRiZi

:Cs of effects thegConsiderin

Vs) of resistance (internal Rs gconsideringain voltage9.9872

0.444)(-67.538)(VsVi

ViVo

VsVoAvs

6-L

LC

6-LS

eB2B1

mid


circuit.amplifier whole theoffrequency cutoff affect thetly predominan willfrequency cutofflower its Cc, and Cs todue

frequecies cutoff the tocomparedhigh relatively isC involvingcircuit theofportion theoffrequency cutofflower theBecause

Hz 225.82210 X 20 5.239)3(2

1CRe2

1f

ohms 5.2393742.91100

1612.903//2000re

'R//RRe

ohms 1612.903

000,101

000,501

20001

1//RRsig//RRs'

:C of effects thegConsiderin

E

6-E

LE

SE

B2B1

E

Low Frequency Response of JFET Common Source Amplifier

CcID = Drain current(ac)

VGSVDSIG (Gate

Current) = 0

Drain (D)

Source

Gate (G)

VO= Output voltage

RG1

Vi = Input

voltage

CG

RD

VDD

Zi Zo

CsRs

The analysis of low frequency response of FET amplifiers is similar to that of BJT amplifiers.

At middle and high frequencies, the capacitors Cc, Cs, and CG can be considered short circuits because their reactance become low enough, that there are no significant voltage drops across the capacitors.

At low frequencies, the coupling capacitors Cc, Cs, and CG could no longer be treated as short circuits because their reactance become high enough that the there are significant voltage drops across the capacitors.

Vs = Source voltage

Rsig

Ii

RL

RG2

Low Frequency Response of JFET Common Source Amplifier The frequency analysis of high pass RC network can be used for capacitor

coupled FET amplifier circuits. The values of R and C are taken from the equivalent resistances and capacitances in the FET amplifier circuit.

For the portion of the circuit involving the coupling capacitor CG, the equivalent circuit is shown below. Equivalent circuit assumes that the input impedance of the amplifier (Zi) is

purely resistive and is equal to Ri.

CG

Zi = Ri= RG1 // RG2Vi

Equivalent Circuit of Vs, CG and Zi

IiCG

Vi

Equivalent Circuit of Vs, CG and Zi

Ii

RG1 // RG2Vs

Rsig

Vs

Rsig

Zi = Ri= RG1 // RG2

Zi = Ri= RG1 // RG2


CGjXRsigRiRi VsVi

The value of the input impedance (resistance) of the amplifier can be computed as:

Zi = Ri = RG1 // RG2 Zi = RG2 if RG1 is not present (RG1 = infinity)

The voltage Vi can be computed using voltage divider rule.

The voltage Vi at middle frequencies (when CG can be considered as short circuit) can be computed as:

The lower cutoff frequency (half power frequency) can be computed as:

RsigRiRi VsVi mid

G

GLG

C involving

circuit theofportion for thefrequency offcut lower Ri)C(Rsig2

1f

Low Frequency Response of JFET Common Source Amplifier For the portion of the circuit involving the coupling capacitor Cc, the

equivalent circuit is shown below. Equivalent circuit assumes that the output impedance of the transistor is

purely resistive and is equal to Ro.

rd

Drain (D) ID

RD

Zo = Ro

RLgmVgs

+

- --

-

+ +

gmVgs

Ird IRD IRL

Cc

Low Frequency Response of JFET Common Source Amplifier The value of the output impedance (resistance) of the amplifier can be

computed as:

Zo = Ro = RD // rd Zo = Ro = RD if rd is equal to infinity

The lower cutoff frequency can be computed as:

Cc involving

circuit theofportion for thefrequency offcut lower )CR(Ro2

1fCL

LC

Low Frequency Response of JFET Common Source Amplifier For the portion of the circuit involving the bypass capacitor Cs, the equivalent

circuit is shown below. The resistance (Req) seen looking into Rs from the output side can be

computed as:

Cs

Equivalent Circuit of Portion of Circuit Involving RS and CS

RS

gm1Rs//Req

be willaboveequation theinfinity,rwhen

(ohms) sideoutput thefrom R into lookingseen impedance

//RRrr gm1Rs1

RsReq

d

S

LDd

d

Req

System


The low cut-off frequency of the portion of the circuit involving the bypass capacitor Cs can be computed as:

side.output thefrom R into looking resistance equivalent Req :where

circuit theofportion for thefrequency offcut lower Cs Req 2

1f

S

Ls

Overall, the effects of the capacitors CG, Cc, and CS must be considered in determining the low cutoff frequency of the amplifier.

The highest lower cutoff frequency among the three cutoff frequencies will have the greatest impact on the low cutoff frequency of the amplifier.

If the cutoff frequencies due to the capacitors are relatively far apart, the highest low cutoff frequency will essentially determine the low cutoff frequency of the amplifier.

If the highest lower cutoff frequency is relatively close to another lower cutoff frequency, or if there are more than one lower cutoff frequency, the low cutoff frequency of the amplifier will be higher than the highest lower cutoff frequency due to the capacitors.

involving Cs

Low Frequency Response of JFET Common Source Amplifier Example: Given a common source FET amplifier with the following

parameters, determine the lower cutoff frequency of the amplifier.CG =0.02F Cc = 0.6 F Cs = 2 F Rsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 KIDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 volts Since RG1 is not present, configuration is self bias FET.

negative) more goes V when reachedfirst is value(ThisA 10 x 2.9806I Choose

1016.49x I A 10 x 2.9806I 2(183)

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I183I565.310 x 90

I183I565.210 x 920,408II285110 x 97-

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(Ampere)current Drain V

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GS3

DQ

3DQ

3DQ

32

DQ D

DD3

DD3

DD3

2 D3

D

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P

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P

GSDSSDQD

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3volts volts2.9806)1000)(10 x -(2.9806))(R-(IV 3S DGSQ

ctancetransconduSiemens x1033.19-3-1

9) x10(9 2

VV1

VI 2gm 3-

-3

P

GS

P

DSS

Hz 7.86 10 x )0.02000,000,1(12,0002

1)CR(Rsig2

1Ri)C(Rsig2

1f 6-GGG

LG

Hz 89.730.6x10 )000,2(5,0002

1)CR(R2

1)CR(Ro2

1f 6-CLDCL

LC

amplifier theoffrequency cutoff low on theimpact highest thehasit s,frequencie cutofflower three theoflargest theis f Since

Hz 185)x10(429.18)(22

1Cs Req 2

1f

LS

6-Ls

ohms 18.29410 x 33.1/11000

)10 x 33.1/1)(1000(gm1Rs//Req infinity,rd Since 3-

-3


sfrequencie midlleat gain voltage

9.12,0005,000

000)(5,000)(2,)10 x (1.33)//Rgm(RViVo Avmid 3-LD

Av / Avmid (db)Normalized Voltage gain


1 AVmidfLSfLC


Low Frequency Response (Normalized Voltage Gain Versus Frequency

0 db-3 db-5 db

-25 db

-20 db

-15 db

-10 db

1 10 100 1K 10K 100K 1M 10M 100MfLG fHi fHof

- 20 db / decade

High Frequency Response of Low Pass RC Network At the high frequency end, the frequency response of a low pass RC network

shown below is determined by the decrease in the reactance of the capacitor as frequency of operation increases.

Because of the decrease in the capacitance, there is a shorting effect across the terminals of the capacitor at high frequencies, and the voltage drop across the capacitor decreases as frequency increases.

R

IR

Vi =Input voltage to RC network

Vo = Outputvoltage

Frequency (log scale)

0.707 AV

1 AV

f2Normalized Gain in db

Bode Plot for High Frequency Region

0 db

-3 db-6 db / octave

CAv = Vo / Vi

High Frequency Response of Low Pass RC Network The voltage gain of the low pass RC network can be computed as:

sfrequencie midat gain voltage the times0.707 isgain gewhen voltafrequency

(Hz)frequency cutoffhigh RC21 f2 :where

(unitless) ffrequency at gain voltage

f2fj1

1Av

ff21j1

1

fRC211j1

1

R1R1

fC21Rj1

1

fC21Rj1

1

jXcR1

1jXcR

jXc-jXcR I

(-jXc) IViVoAv

The above equation results to plot that drops off at 6db per octave with increasing frequency.

Miller Effect Capacitance When the frequencies being processed by an amplifier are high, the frequency response

of the amplifier is affected by: Interelectrode (between terminals) capacitance internal to the active device Wiring capacitance between leads of the network

The coupling and bypass capacitors are considered short circuits at mid and high frequencies because their reactance levels are very low.

The diagram below shows the existence of a feedback capacitance whose reactance becomes significantly low at high frequencies, that it affects the performance of an amplifier.

The input and output capacitance are increased by a capacitance level sensitive to the interelectrode (between terminals) capacitance (Cf) between the input and output terminals of the device and the gain of the amplifier.

Because of Cf, an equivalent capacitance, called Miller capacitance, is produced at the input and output.

Vo

+

-Vi

+

-

Zi

Av =Vo / Vi

CfI2

I1

Ii

Ri

Miller Effect Capacitance The value of the Miller effect input capacitance can be computed as:

CMi

fMi

CMff

21

21

X1

Ri1

Zi1

ecapacitancinput effect Miller C Av) (1C

ecapacitancinput effect Miller of ReactanceXfC2 Av) (1

1C Av) (1

1Av) (1

XcfAv) (1

Xcf1

Ri1

Zi1

XcfAv) (1

Ri1

Zi1

XcfAv) Vi(1

RiVi

ZiVi

IIIiXcf

Av) Vi(1Xcf

Av ViViXcf

VoViI RiViI

ZiViIi

Miller Effect Capacitance The equivalent circuit due to the Miller Effect Capacitance is shown below.

Above results show that for any inverting amplifier (negative AV), the input capacitance will be increased by a Miller effect capacitance, which is a function of the gain of the amplifier and the interelectrode (parasitic) capacitance between the input and output terminals of the active device.

If the voltage gain is negative (with phase reversal), Miller Effect capacitance (CM) is positive and higher than the interelectrode capacitance.

If the voltage gain is positive (no phase reversal) and greater than 1, Miller Effect capacitance (CM) is negative.

+

-

Vi

Zi

IiCM i= (1-AV)Cf Ri

Miller Effect Capacitance At high frequencies, the voltage gain Av is a function of the Miller effect

capacitance (CM). There is difficulty in solving the value of the Miller effect capacitance (CM)

since it is a function of the gain AV which in turn is a function of the Miller effect capacitance. In general, the midband value of the voltage gain is used for AV, to get

the worst case scenario for the Miller effect capacitance, since the highest value of Av is the midband value.

The Miller effect also increases the level of the output capacitance, and it must also be considered in determining the high cutoff frequency.

The diagram below shows the feedback capacitor as seen in the output sideof the amplifier.

Vo

+

-Vi

+

-Zo

Av =Vo / Vi

Cf I2

I1

Io

Ro

Miller Effect Capacitance The Miller effect output capacitance can be determined as follows:

ecapacitancoutput effect Miller of ReactanceC 1

C Av11

1

Av11

XcfIoVo

XcfAv11

VoIo

XcfAv11Vo

XcfAvVoVo

XcfViVoIo

:by edapproximat becan Io

largely sufficientusually is Ro because smallry usually ve is RoVobut

XcfViVo

RoVoIo

XcfViVoI

RoVoI

ZoVoIo

IIIo

Mof

21

21

Miller Effect Capacitance

ecapacitancoutput effect Miller CfC

:by edapproximat becan ecapacitancoutput effect Miller the1,an greater thmuch is AvWhen

ecapacitancoutput effect Miller Cf Av11C

Mo

Mo

BJT High Frequency Response At the high frequency end, the high cutoff frequency (-3 db) of BJT circuits is

affected by: Network capacitance (parasitic and induced) Frequency dependence of the current gain hfe

At high frequencies, the high cutoff frequency of a BJT circuit is affected by: the interelectrode capacitance between the base and emitter, base and

collector, and collector and emitter. Wiring capacitance at the input and output of the BJT.

At high frequencies, the reactance of the interelectrode and wiring capacitance become significantly low, resulting to a shorting effect across the capacitances.

The shorting effect at the input and output of an amplifier causes a reduction in the gain of the amplifier.

For common emitter BJT circuits, Miller effect capacitance will affect the high frequency response of the circuit, since it is an inverting amplifier.

BJT High Frequency Response The figure below shows the RC network which affects the frequency response of

BJT circuits at high frequencies.

CcIRC

Cwo

IB C

E

B

VO= Output voltage

REVi =

Input voltage

CS

RC

VCC

RB2

RB1

IRB2

IRB1

Ce

Zi ZoZix Zox

IRL

CwiVs = Source voltage

Rsig

Ii Cbc

CceCbe

Cbe = capacitance between the base and emitter of transistorCce = capacitance between collector and emitter of transistorCbc = capacitance between base and collector of transistorCwi = wiring capacitance at input of amplifierCwo = wiring capacitance at output of amplifier

BJT High Frequency Response The figure below shows the ac equivalent circuit of the BJT amplifier in the

preceding slide. At mid and high frequencies, Cs, Cc, and Ce are assumed to be short circuits

because their impedances are very low. The input capacitance Ci includes the input wiring capacitance (Cwi), the

transistor capacitance Cbe, and the input Miller capacitance CMi. The output capacitance Co includes the output wiring capacitance (Cwo), the

transistor parasitic capacitance Cce, and the output Miller capacitance CMo. Typically, Cbe is the largest of the parasitic capacitances while Cce is the

smallest

E

Ci

IcIb

Vo=Vce

Ri re= re

ro

Zix Zox

Vi Ib

E

RL

BIi

CoRCRiRB1// RB2Vs

Rsig

Ci = Cwi + Cbe + CMi Co = Cwo + Cce + CMo

Thi Tho

BJT High Frequency Response The Thevenin equivalent circuit of the ac equivalent circuit of the BJT amplifier

is shown below. For the input side, the -3db high cutoff frequency can be computed as:

Vo=VceVi

Co

Ri

RThi = Rsig // RB1// RB2 // Ri

VThi

Thi Tho

VTho

CiRTho= Rc // RL// ro

circuit of ecapacitancinput Av)C-(1 C C C C C C

sideinput at resistance equivalentThevenin 1)re( // R// R // Rsig Ri // R// R // RsigR

frequency) (-3db sideinput for thefrequency offcut higher Ci R 2

1f

bcbewiMibewii

B2 B1B2 B1THi

ThiHi

BJT High Frequency Response At the high frequency end, the reactance of capacitance Ci will decrease as

frequency increases, resulting to reduction in the total impedance at the input side. This will result to lower voltage across Ci, resulting to lower base current,

and lower voltage gain. For the output side, the -3db high cutoff frequency can be computed as:

At the high frequency end, the reactance of capacitance Co will decrease as frequency increases, resulting to reduction in the total impedance at the output side. This will result to lower output voltage Vo, resulting to lower voltage and

power gain.

circuit of ecapacitancoutput C Av1-1 C C C C C C

sideoutput at resistance equivalentThevenin //roR // RcR

frequency) (-3db sideoutput for thefrequency offcut higher Co R 2

1f

bccewoMocewoo

LTHo

ThoHo

BJT High Frequency Response The Hybrid or Giacolletohigh frequency equivalent circuit for common

emitter is shown below. The resistance rb includes the base contact resistance (due to actual connection

to the base) , base bulk resistance (resistance from external base terminal to the active region of transistor), and base spreading resistance (actual resistance within the active region of transistor).

The resistances r, ro, and ru are the resistances between the indicated terminalswhen the BJT is in the active region.

Cbe and Cbc are the capacitances between the indicated terminals.

E

Cu = Cbc

IcIb

ro = 1 / hoe

Zix ZoxE

B

C = Cber re

rbC

ru

Ib =hfe Ib

Hybrid High Frequency Equivalent Circuit (Common Emitter)

BJT High Frequency Response At the high frequency end, hfe of a BJT will be reduced as frequency increases. The variation of hfe (or ) with frequency can approximately be computed as:

rtransistoofcurrent emitter DC I

I26mVr

))(r(hfe r r

CuC r 2

11 f

CuC r 21

hfe1

CuC r 21 f f

sheet) specsat given usually one (thefrequency middleat hfe hfe:where

ffrequency at hfe

ffj1

hfehfe

E

Ee

e midemid

mid

midhfe

mid mid

mid

e

e

BJT High Frequency Response Since re is a function of the DC emitter current IE, and f is a function of re, f is a

function of the bias condition of the circuit. hfe will drop off from its midband value with a 6 db / octave slope.

For the common base configuration: It has improved frequency response compared to common emitter configuration. Miller effect capacitance is not present because of its non-inverting

characteristics. f is higher than f.


f

Normalized hfe in db

Bode Plot for hfe () in the High Frequency Region

0 db

-3 db-6 db / octave (for f)

hfe / hfe mid

BJT High Frequency Response The relationship of f (-3db high cutoff frequency for ) and f db high

cutoff frequency for ) is shown below.

The upper cutoff frequency of the entire system (upper limit for the bandwidth) is lower than the lowest upper cutoff frequency (lowest among fHi, fHo, and f)

The lowest upper cutoff frequency has the greatest impact on the bandwidth of the system. It defines a limit for the bandwidth of the system.

The lower is the upper cut off frequency, the greater is its effect on the bandwidth of the entire system.

for frequency offcut db 3)1(ff


f

Normalized hfbin db

Bode Plot for hfb () in the High Frequency Region (Common Base)

0 db-3 db

-6 db / octave (for f)

hfb / hfb mid f

BJT High Frequency Response The gain-bandwidth product of a transistor is defined by the following condition:

productbandwidth gain CuC r 2

1f

CuC r 211)( ))(f( f

bandwidth ff

bandwidthf andgain since product,bandwidth gain ))(f( ))(fhfe(f

1

ff

hfe

ff1

hfe:as computed is )f(f when hfe of magnitude theand

fby denoted is 0db toequal ishfeat which frequency The

db 01log20

ffj1

hfelog20hfe and 1

ffj1

hfe hfe

T

midmidmidT

mid

T

midmidmidT

T

mid

2

T

mid

T

T, db

middb

mid

e

e

BJT High Frequency Response Example: Given a common emitter BJT amplifier with the following parameters,

determine the following:a. High cutoff frequency for the input of the circuit (fHi)b. High cutoff frequency for the output of the circuit (fHo)c. High cutoff frequency for fd. Gain bandwidth product (fT)

e. Sketch the frequency response for the low and high frequency range

Specs similar to example on BJT low frequency response:Cs=12 uF Rsig = 2 kohm RL= 2 kohm RB1=50 kohm RB2= 10 kohmCc=2 uF RE = 2 kohm RC= 4 kohm Vcc= 20 voltsCE=20 uF = hfemid = 100 ro = infinite

Additional specs:C = Cbe= 35 pF Cu = Cbc= 3 pF Cce = 1 pFCwi = 5 pF Cwo = 6 pF

BJT High Frequency Response

response)frequency low(for amplifier of impedanceinput ohms 1,596.08 RiZiRs) gconsiderin(not sfrequencie midat gain voltage538.67 Av

742.19rresponsefrequency high for Ri2.974,1)742.19(100)(r

response,frequency lowon example previous theFrom

mid

e

e

Ce involvingcircuit offrequency cutofflower Hz 225.822fCc involvingcircuit offrequency cutofflower Hz 263.13 f

Cs involvingcircuit theofportion for thefrequency offcut lower Hz 3.688 fVs) of resistance (internal Rs gconsideringain voltage9.9872 Avs

LE

LC

LS

mid

ohms 678.887

2.974,11

000,101

000,501

000,21

1

Ri // R // R // RsigRpF 614.452pF 3 (-67.538))-(1 pF 35 pF 5

Cbc Av)-(1 Cbe Cwi Ci:responsefrequency high For the

B2B1Thi

BJT High Frequency Response

Hz 359,884,11)10 x )(10.04433.333,1(2

1)(Co)R(2

1f

pF 044.01pF 3538.67

11pF 1pF 6

CbcAvmid

11 Cce CwoC Cce Cwo Co

ohms 33.333,12,0004,000

000)(4,000)(2, R // Rc R

Hz 981,729)10 x 614.45)(2678.887(2

1)(Ci)R(2

1f

12-Tho

Ho

Mo

LTho

12-Thi

Hi

1hfewhen frequency Hz 800,092,215) 892,150,2)(001() )(fhfe(f

Hz 892,150,2 x103 x1035 19.472)( 2

1(100)

1

CuC r 21

hfe1

CuC r 21 f f

midT

1212

midhfe

e

BJT High Frequency Response In the low frequency region, the lower cutoff frequency due to the emitter capacitor

(fLE) has the highest value. Consequently, it has the greatest impact on the bandwidth of the system, among the three lower cutoff frequencies.

In the high frequency region, the high cutoff frequency due to the input capacitors and resistors (fHi) has the lowest value. Consequently, it has the greatest impact on the bandwidth of the system, among the three high cutoff frequencies.



1 AVmidfLEfLC


Full Frequency Response (Normalized Voltage Gain Versus Frequency)

0 db-3 db-5 db

-25 db

-20 db

-15 db

-10 db

1 10 100 1K 10K 100K 1M 10M 100MfLS fHi fHof

Bandwidth

- 20 db / decade(-6 db / octave)

+20 db / decade(6 db/octave)

FET High Frequency Response The high frequency response analysis for FET is similar to that of BJT. At the high frequency end, the high cutoff frequency (-3 db) of FET circuits is

affected by the network capacitance (parasitic and induced). The capacitances that affect the high frequency response of the circuit are

composed of: the interelectrode capacitance between the gate and source, gate and drain,

and drain and source. Wiring capacitance at the input and output of the circuit.

At high frequencies, the reactance of the interelectrode and wiring capacitance become significantly low, resulting to a shorting effect across the capacitances.

The shorting effect at the input and output of an amplifier causes a reduction in the gain of the amplifier.

For common source FET circuits, the Miller effect will be present, since it is an inverting amplifier.

FET High Frequency Response The figure below shows the RC network which affects the frequency response of

FET circuits at high frequencies.

CcIRD

Cwo

IG D

S

G

VO= Output voltage

RSVi =

Input voltage

CG

RD

VCC

RG2

RG1

IRG2

IRG1

CS

Zi ZoZix Zox

IRL

CwiVs = Source voltage

Rsig

Ii Cgd

CdsCgs

Cgs = capacitance between the gate and source of transistorCds = capacitance between drain and source of transistorCgd = capacitance between gate and drain of transistorCwi = wiring capacitance at input of amplifierCwo = wiring capacitance at output of amplifier

Common Source FET Amplifier Circuit

FET High Frequency Response The figure below shows the ac equivalent circuit of the FET amplifier. At mid and high frequencies, CG, CS, and Cc are assumed to be short circuits because

their impedances are very low. The input capacitance Ci includes the input wiring capacitance (Cwi), the transistor

capacitance Cgs, and the input Miller capacitance CMi. The output capacitance Co includes the output wiring capacitance (Cwo), the

transistor parasitic capacitance Cds, and the output Miller capacitance CMo. Typically, Cgs and Cgd are higher than Cds. At high frequencies, Ci will approach a short-circuit and Vgs will drop, resulting to

reduction in voltage gain. At high frequencies, Co will approach a short-circuit and Vo will drop, resulting to

reduction in voltage gain.

S

Ci

Id

Vo=Vdsrd

Zix Zox

Vi = Vgsgm Vgs

S

RL

GIi

CoRDRG1// RG2

Vs

Rsig

Ci = Cwi + Cgs + CMi Co = Cwo + Cds + CMo

Thi Tho

D

IRL

FET High Frequency Response The Thevenin equivalent circuit of the ac equivalent circuit of the FET amplifier

is shown below. For the input side, the -3db high cutoff frequency can be computed as:

Vo=VdsVi

CoRThi = Rsig // RG1// RG2

VThi

Thi Tho

VTho

CiRTho= RD // RL// rd

scenario case worst get the toAvfor used is Avmid where

sideinput at ecapacitanceffect Miller Cgd Av1C

circuit of ecapacitancinput C Cgs C C

sideinput at resistance equivalentThevenin R// R // RsigR

frequency) (-3db sideinput for thefrequency offcut high Ci R 2

1f

Mi

Miwii

G2 G1THi

ThiHi

FET High Frequency Response For the output side, the -3db high cutoff frequency can be computed as:

scenario case worst get the toAvfor used is Avmid

sideoutput at the ecapacitanceffect Miller Cgd Av11C

circuit of ecapacitancoutput C Cds C C

sideoutput at resistance equivalentThevenin //rdR // RR

frequency) (-3db sideoutput for thefrequency offcut high Co R 2

1f

Mo

Mowoo

LDTHo

ThoHo

FET High Frequency Response Example: Given a common source FET amplifier with the following parameters,

determine the following:a. High cutoff frequency for the input of the circuit (fHi)b. High cutoff frequency for the output of the circuit (fHo)c. Sketch the frequency response for the low and high frequency range

Specs similar to example on FET low frequency response:CG =0.02mF Cc = 0.6 mF Cs = 2 mF Rsig = 12 K RG= 1 Mohm RD= 5 K Rs= 1 K RL = 2 KIDSS= 9 mA Vp= -7 volts rd= infinity VDD = 20 volts Since RG1 is not present, configuration is self bias FET.

Additional specs:Cgd= 3 pF Cgs = 5 pF Cds = 1 pFCwi = 5 pF Cwo = 6 pF

FET High Frequency Response

Cs involvingcircuit toduefrequency cutofflower Hz 851fCc involvingcircuit toduefrequency cutofflower Hz 37.89f

C involvingcircuit toduefrequency cutofflower 7.86HzfRs) gconsiderin(not sfrequencie midat gain voltage9.1 Av

response, frequenc lowon example previous theFrom

LS

LC

GLG

mid

Hz 800,717)10 x 8.71)(857,11(2

1)(Ci)R(2

1f

ohms 857,11000,000,1000,12

)000,000,1)(000,12(

R // RsigRpF 7.18pF 3 (-1.9))-(1 pF 5 pF 5

Cgd Av)-(1 Cgs Cwi Ci:responsefrequency high For the

12-Thi

Hi

GThi

FET High Frequency Response

Hz 605,621,9)10 x 579.1)(1 57.428,1(2

1)(Co)R(2

1f

pF 579.11pF 39.1

11pF 1pF 6

CgdAvmid

11 Cds CwoC Cds Cwo Co

ohms 57.428,12,0005,000

000)(5,000)(2, R // R R

12-Tho

Ho

Mo

LDTho

FET High Frequency Response In the low frequency region, the lower cutoff frequency due to the source capacitor (fLS)

has the highest value. Consequently, it has the greatest impact on the bandwidth of the system, among the three lower cutoff frequencies.

In the high frequency region, the high cutoff frequency due to the input capacitors and resistors (fHi) has the lowest value. Consequently, it has the greatest impact on the bandwidth of the system, among the two high cutoff frequencies.



1 AVmidfLSfLC


Full Frequency Response (Normalized Voltage Gain Versus Frequency)

0 db-3 db-5 db

-25 db

-20 db

-15 db

-10 db

1 10 100 1K 10K 100K 1M 10M 100MfLG fHi fHo

Bandwidth

-20 db / decade(-6 db / octave)

+20 db / decade(6 db / octave)

Frequency Response of Multistage (Cascaded) Amplifiers If there are several stages in a cascaded amplifier system, the overall bandwidth

of the system will be lower than the individual bandwidth of each stage. In the high frequency region:

The output capacitance Co must now include the wiring capacitance (Cwi), parasitic capacitance (Cbe or Cgs), and input Miller capacitance (CMi) of the next stage.

The input capacitance Ci must now include the wiring capacitance (Cwo), parasitic capacitance (Cce or Cds), and input Miller capacitance (CMO) of the preceding stage.

The lower cutoff frequency of the entire system will be determined primarily by the stage having the highest lower cutoff frequency.

The upper cutoff frequency of the entire system will be determined primarily by the stage having the lowest higher cutoff frequency.

For n stages having the same voltage gain and lower cutoff frequency (f1), the overall lower cutoff frequency (f1) can be computed as:

stages ofnumber n stageeach offrequency cutofflower f1 :where

amplifier entire theoffrequency cutofflower overall12

f1f1'1/n

Frequency Response of Multistage (Cascaded) Amplifiers

For n stages having the same voltage gain and higher cutoff frequency (f2),the overall higher cutoff frequency (f2) can be computed as:

stages ofnumber n stageeach offrequency cutoffhigher f2 :where

amplifier entire theoffrequency cutoffhigher overall12 f2f2' 1/n

Square Wave Testing A square wave signal can be used to test the frequency response of single

stage or multistage amplifier. If an amplifier has poor low frequency response or poor high frequency

response, the output of the amplifier having a square wave input will bedistorted (not exactly a square wave at the output).

A square wave is composed of a fundamental frequency and harmonics which are all sine waves.

If an amplifier has poor low or high frequency response, some low or high frequencies will not be amplified effectively and the output waveform will be distorted.

Square Wave Testing The figures below show the effect of poor frequency response of an

amplifier using a square wave input.

V

t

V

t

tt

VV

No distortion (Good Frequency Response)

Poor High FrequencyResponse

Poor Low FrequencyResponsePoor High and Low

Frequency Response

tilt

long rise time

tilt

V

t

t

V

Very Poor High FrequencyResponse

Very Poor Low FrequencyResponse

tilt

very long rise time

Square Wave Testing The high cutoff frequency can be determined from the output waveform by

measuring the rise time of the waveform. Rise time is between the point when the amplitude of the waveform is 10 %

of its highest value up to the point when the amplitude is 90 % of its highest value.

The high cutoff frequency can be computed as:

The lower cutoff frequency can be determined from the output waveform by measuring the tilt of the waveform.

(seconds) timerise tr :where

)f toequalely approximat is (bandwidthamplifier ofBandwidth tr

0.35fBW

(Hz)amplifier offrequency cutoffupper tr

0.35f

HH

H

(Hz) wavesquare offrequency fs

(unitless) tilt V

V'-V P :where

(Hz)amplifier offrequency cutofflower fsPfLO

t

V tilt

VV

Rise time (tr)

Square Wave Testing Example: The output waveform of an amplifier with a 4 Khz square wave

input has the following characteristics:

Rise time = 15 microseconds Maximum amplitude (V) = 40 millivoltsMinimum voltage of tilt (V) = 30 millivolts

Determine: high cutoff frequency, bandwidth, low cutoff frequency.

Hz 23,333.3315x10

0.35fBW

Hz 23,333.3315x10

0.35tr

0.35f

6-H

6-H

frequency cutofflower Hz 318(4,000)0.25fsPf

(unitless) tilt 0.2510 x 40

10 x 30-10 x 40 V

V'-V P

LO

3-

-3-3

Documents

BJT and FET Frequency Response