Frequency Response of Amplifier Circuits › resources › 042223011 › 6.04222301_Fre… · Common-emitter Amplifier - Low-frequency ac equivalent circuit R S R B R C R L R E C

AJAMANGALA UNIVERSITY of TECHNOLOGY SRIVIJAYAMUTSV, ELECTRONIC ENGINEERINGR

การตอบสนองความถีข่องวงจรขยายFrequency Response of Amplifier Circuits

อาจารย์ผู้สอน : อภริักษ์ เสือเดช

Email : [email protected]

Website : www.suadet.yolasite.com

The Decibel (dB)

A logarithmic measurement of the ratio of power or voltage Power gain is expressed in dB by the formula:

where ap is the actual power gain, Pout/Pin

Voltage gain is expressed by:

If av is greater than 1, the dB is +ve, and if av is less than 1, the dB gain is –ve value & usually called attenuation

vdBV aA log20)(

PP aA log10

2A. Suadet Electronic Circuits Analysis : Frequency Response of Amplifiers

Amplifier gain vs frequency

Midband range

Gain falls of due to the effects of CC and CE

Gain falls of due to the effects of stray capacitance and

transistor capacitance effects

LHBW fff


Definition Frequency response of an amplifier is the graph of its gain versus the

frequency. Cutoff frequencies : the frequencies at which the voltage gain equals

0.707 of its maximum value. Midband : the band of frequencies between 10fL and 0.1fH where the

voltage gain is maximum. The region where coupling & bypass capacitors act as short circuits and the stray capacitance and transistor capacitance effects act as open circuits.

Bandwidth : the band between upper and lower cutoff frequencies Outside the midband, the voltage gain can be determined by these

equations:

21 /1 ff

AA mid

22/1 ff

AA mid

Below midband Above midband


Lower & Upper Critical frequency

Critical frequency a.k.a the cutoff frequency The frequency at which output power drops by 3 dB.

[in real number, 0.5 of it’s midrange value. An output voltage drop of 3dB represents about a

0.707 drop from the midrange value in real number. Power is often measured in units of dBm. This is

decibels with reference to 1mW of power. [0 dBm = 1mW], where;

.dBm0mW1mW1log10


Gain & frequencies

Gain-bandwidth product : constant value of the product of the voltage gain and the bandwidth.

Unity-gain frequency : the frequency at which the amplifier’s gain is 1

BWAf midT


LOW FREQUENCY

At low frequency range, the gain falloff due to coupling capacitors and bypass capacitors.

As signal frequency , the reactance of the coupling capacitor, XC - no longer behave as short circuits.


Short-circuit time-constant method (SCTC)

To determine the lower-cutoff frequency having ncoupling and bypass capacitors:

n

i iiSL CR1

1

RiS = resistance at the terminals of the ith capacitor Ci with all the other capacitors replaced by short circuits.


Common-emitter Amplifier

30 k

VCC = 12V

10 k

RS C1

2 F

C2

C310 F

0.1 F

1 k

1.3 k

4.3 k

R1 RC

RER2vS

vO

RL

100 k

Given :

Q-point values : 1.73 mA, 2.32 V

= 100, VA = 75 V

Therefore,

r = 1.45 k,

ro =44.7 k


Common-emitter Amplifier- Low-frequency ac equivalent circuit

RS

RB

RC RL

RE

C1

C2

C3

vs

vo

In the above circuit, there are 3 capacitors (coupling plus bypass capacitors). Hence we need to find 3 resistances at the terminals of the 3 capacitors in order to find the lower cut-off frequency of the amplifier circuit.


Circuit for finding R1S

22201450750010001 rRRRRRR BSinCEBSS

7500 where 21 RRRB

sradFkCR S

/22500.222.2

11

11

RS

RB

R1S

RC RL

RinCE

Replacing C2and C3 by

short circuits



kkkkrRRRRRR oCLoutCECLS 1047.443.41002

sradFkCR S

/1.96100.0104

11

22

RLRS

RB

R2S

RC

RoutCE


short circuits

kr 7.44 where 0



7.22

10188214501300

13 TH

EoutCCESRrRRRR

sradFCR S

/4410107.22

11

33


short circuitsRS RB

R3SRE

RC||RL

RoutCC

RTH

882BSTH RRR


Estimation of L

sradCRi iiS

L /473044101.9622513

1

Hzf LL 753

2


Common-base Amplifier

Given :

Q-point values : 0.1 mA, 5 V

= 100, VA = 70 V

Therefore,

gm = 3.85 mS, ro = 700 k

r = 26

RE

RS

RCRL

C1 C2

vO

vS 22 k

100

43 k 75 k

1 F4.7 F

+VCC-VEE


Common-base Amplifier

RE

RS

RCRL

C1 C2

vo

vs

16

- Low-frequency ac equivalent circuit

A. Suadet Electronic Circuits Analysis : Frequency Response of Amplifiers


Replacing C2by short circuit

10026.0430010011

rRRRRRR ESinCBESS

sradFCR S

/1013.27.4100

11 3

11

RC || RL

R1S

RE

RS

RinCB



kkkRRRRRR CLoutCBCLS 9722752

sradFkCR S

/309.10197

11

22

Replacing C1by short circuit RS || RE

R2S

RC RL

RoutCB


Estimation of L

sradCRi iiS

L /309.10309.101013.21 32

1

Hzf LL 64.1

2


Common-collector Amplifier

RB

RS

RERL

C1

C2

vSvO

-VEE

+VCC

100 k

1 k

3 k47 k

100 F

0.1 F

Given :

Q-point values : 1 mA, 5 V

= 100, VA = 70 VTherefore,

r = 2.6 k, ro =70 k


Common-collector Amplifier - Low-frequency ac equivalent circuit

RB

RS

RE RL

C2

C1

vovs




LEoBSinCCBSS RRrrRRRRRR 11

sradFkCR S

/18.1361.043.74

11

11

k43.74

RB

RS

RE || RL

R1S

RinCC



oTH

ELoutCCELS rrRRRRRRR12

sradFkCR S

/213.0100038.47

11

22


k038.47

RE RL

RTH = RS || RB

R2S

RoutCC


Estimation of L

sradCRi iiS

L /393.136213.018.13612

1

Hzf LL 7.21

2


Example

vS

62 k

VCC = 10V

22 k

RS C1

0.1 F

C2

C310 F

0.1 F

600

1.0 k

2.2 k

R1 RC

RER2

vO

RL

10 k

Given :

Q-point values : 1.6 mA, 4.86 V

= 100, VA = 70 V

Therefore,

r = 1.62 k, ro = 43.75 k, gm = 61.54 mS

Determine the total low-frequency response of the amplifier.


Low frequency due to C1 and C2 C3

kkkrRRR BSS 07.262.124.166001

kRRRB 24.1621

HzHzFkCR

fS

C 76986.7681.007.22

12

1

111

kkkkrRRR oCLS 09.1275.432.2102

HzHzFkCR

fS

C 13264.1311.009.122

12

1

222

Low frequency due to C1



32.21

10158.062.11

13kkkRrRR TH

ES

HzHzFCR

fS

C 7475.7461032.212

12

1

333

kRRR BSTH 58.0




HIGH FREQUENCY

The gain falls off at high frequency end due to the internal capacitances of the transistor.

Transistors exhibit charge-storage phenomena that limit the speed and frequency of their operation.

Small capacitances exist between the base and collector and between the base and emitter. These effect the frequency characteristics of the circuit.

C = Cbe ------ 2 pF ~ 50 pF

C = Cbc ------ 0.1 pF ~ 5 pF

reverse-biased junction

capacitance

forward-biased junction

capacitance


Cob = CbcCib = Cbe

Output capacitance Input capacitance

Basic data sheet for the 2N2222 bipolar transistor


Miller’s Theorem

This theorem simplifies the analysis of feedback amplifiers.

The theorem states that if an impedance is connected between the input side and the output side of a voltage amplifier, this impedance can be replaced by two equivalent impedances, i.e. one connected across the input and the other connected across the output terminals.


Miller Equivalent Circuit

I2I1

V1 V2

AZV

ZAVI

VAV

ZVVI

1

)1( 111

12

211

A

Z

VI

ZA

VI

VAV

ZVVI

11

11

22

2

2

12

122

Impedance Z is connected between the input side and the output side of a voltage amplifier..


V1 V2

A

ZZ

A

ZIV

A

Z

VI

M 11

11

11

2

2

2

22

AZZ

AZ

IV

AZVI

M 1

1

1

1

1

1

11

Miller Equivalent Circuit (cont)

.. The impedance Z is being replaced by two equivalent

impedances, i.e. one connected across the input (ZM1) and the

other connected across the outputterminals (ZM2)


)1(

)1(11

1

1

1

1

1

1

ACC

ACC

AXX

AZZ

M

M

CCM

M

)11(

)11(

11

11

11

2

2

2

2

ACC

ACC

A

XX

A

ZZ

M

M

CCM

M

Miller Capacitance Effect

I2I1

V1 V2

C

CM = Miller capacitanceCM = Miller capacitance

Miller effectMultiplication effect of CµMiller effectMultiplication effect of Cµ


r ro

C

V gmVC

-

+

C = Cbe C = Cbc

High-frequency hybrid- model


ACACC bcMi 11

AC

ACC bcMo

1111

Miin CCC Moout CC

High-frequency hybrid- model with Miller effect

r roCMi gmVC CMo

A : midband gain


High-frequency in Common-emitter Amplifier

vO22 k

VCC = 10V

4.7 k

RS C1

10 F

C2

C310 F

10 F

600

470

2.2 k

R1 RC

RER2vS

RL2.2 k

Given :

= 125, Cbe = 20 pF, Cbc = 2.4 pF, VA= 70V, VBE(on) = 0.7V

Determine :

1. Upper cutoff frequencies

2. Dominant upper cutoff frequency

Calculation Example


pFpACC bcMi 66.13736.574.21

36.5621

21

LCo

Sm RRr

rRRRrRR

gA

pFpA

CC bcMo 44.2018.14.211

R1||R2

RS

RC||RLvs

vo

r roC CMi CMogmV

midband gain

Miller’s equivalent capacitor at the input

Miller’s equivalent capacitor at the output

High-frequency hybrid- model with Miller effect for CE amplifier


pFppCCC Mibein 66.15766.13720

pFCC Moout 44.2

47.38955.17.42260021 kkkrRRRR Si Thevenin’s equivalent

resistance at the input

kkkkrRRR oLCo 08.162.472.22.2 Thevenin’s equivalent resistance at the output

total input capacitance

total output capacitance

MHzpCR

fini

Hi 59.266.15747.3892

12

1

MHzpkCR

fouto

Ho 39.6044.208.12

12

1

upper cutoff frequency introduced by input capacitance

upper cutoff frequency introduced by output capacitance

Calculation (Cont..)


How to determine the dominant frequency

The lowest of the two values of upper cutoff frequencies is the dominant frequency.

Therefore, the upper cutoff frequency of this amplifier is

MHzf H 59.2


TOTAL AMPLIFIER FREQUENCY RESPONSE

f (Hz)fC3fC1 fC2

fC4 fC5

A (dB)

Amid

fHfL

ideal

actual-3dB


Total Frequency Response of Common-emitter Amplifier

33 k

VCC = 5V

22 k

RS C1

1 F

C2

C310 F

2 F

2 k

4 k

4 k

R1 RC

RER2vS

RL5 k

vO

Given :

= 120, Cbe = 2.2 pF, Cbc = 1 pF, VA = 100V, VBE(on) = 0.7V

Determine :

1. Midband gain

2. Lower and upper cutoff frequencies

Calculation Example


Step 1 - Q-point Values

ARR

onVVIEB

BEBBB

615.2

1)(

VVRR

RV CCBB 221

2

kRRRRRRRB 2.13||

21

2121

mAII BCQ 314.0


Step 2 - Transistor parameters value

kIVrCQ

T 94.9

kIVr

CQ

Ao 47.318

mSVI

gT

CQm 08.12


Step 3 - Midband gain

LCo

BS

Bmmid RRr

rRRrR

gA

kkkRRr LCo 18.222.247.318

kkkRr B 67.52.1394.9

kkkkRrR BS 67.72.1394.92

47.1918.2

67.767.508.12 k

kkmAmid


Step 4 - Lower cutoff frequency (fL)

sradCR S

/38.1301

111 krRRR BSS 67.71

sradCR S

/87.551

222 krRRR oCLS 95.82

sradCR S

/9.10601

333

26.9413

BSES

RRrRR

sradi

L /15.124732

3

11

Hzf LL 49.198

2

Due to C1

Due to C2

Due to C3

SCTCmethod

Lower cutoff frequency


Step 5 - Upper cutoff frequency (fH)

pFpACC bcMi 47.2047.2011

pFpA

CC bcMo 05.1051.1111

pFCCC Mibein 67.22

pFCC Moout 05.1

krRRRR Si 48.121

krRRR oLCo 18.2

Miller capacitance

Input & output resistances


MHzpkCR

fini

Hi 74.467.2248.12

12

1

MHzpkCR

fouto

Ho 53.6905.118.22

12

1

MHzfH 74.4

Input side

Output side

Upper cutoff frequency(the smallest value)

Step 5 - Upper cutoff frequency (fH)


Exercise

Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007

Exercise 7.11



49

Frequency Response of FET Amplifiers



LOW-FREQUENCY AMPLIFIER RESPONSE

Input RC Circuit Output RC Circuit Bypass RC Circuit


51

Amplifier Gain Versus Frequency


FET Amplifier

RL

RS

RD

RG

C1

C2

Vi

+VDD

C3

RSi

Vo

Common-source FET amplifier


Input RC circuit Output RC

circuit

Bypass RC circuit

Low-frequency Equivalent Circuit


Input RC circuit

The cutoff frequencies defined by the input , output and bypass circuits can be obtained by the following formulas:

1121

CRf

Cc where

RC1=RSi+RGInput RC circuit


Output RC circuit

2221

CRf

Cc where

RC2=RD+RL

Output RC circuit


Bypass RC circuit

3321

CRf

Cc where

RC3=RS||1/gm

Bypass RC circuit


Low cut-off frequency

Hence,

fC = the largest of the three low cut-off frequency


Example

RL

RS

RD

RG

C1

C2

Vi

+VDD

C3

RSi

Vo

1K

20V

10K

4.7K

2.2K1M

0.5F0.01F

2F

Determine the lower cutoff frequency for the FET amplifier. Given K = 0.4mA/V2, VTN= 1V, = 0

58

mgm 2


HzMKCR

fC

c 8.15)01.0)(110(2

12

1

11

HzmKCR

fC

c 73.238)2)(211(2

12

1

33

HzKKCR

fC

c 13.46)5.0)(2.27.4(2

12

1

22

Input RC circuit

Output RC circuit

Bypass RC circuit

Solution


Since fc in bypass RC circuit is the largest of the three cutoff frequencies, it defines the low cutoff frequency for the amplifier:

fc = 238.73Hz



HIGH-FREQUENCY AMPLIFIER RESPONSE

Input RC Circuit Output RC Circuit


HIGH-FREQUENCY

Small capacitances exist between the gate and drain and between the gate and source. These affect the frequency characteristics of the circuit.

ro

Cgd

Vgs gmVgsCgs

-

+

hi-frequency hybrid- model


Basic data sheet for the BS 107 n-MOSFET

Cgs = Ciss - Crss

Cgd = Crss


Unity-Gain Bandwidth

Unity gain frequency / bandwidth, fT is defined as a frequency at which the magnitude of the short-circuit current gain goes to 1

It is a parameter of FET & is independent of circuit

)(2 gdgs

mT CC

gf

Page 521


FET Amplifier

In high-frequency analysis, coupling and bypass capacitorsare assumed to have negligible reactances and are considered to be shorts.

vo

RL

RS

RDR1

C1

C2

vi

+VDD

C3

RSi

R2


R1||R2 CMoVi Cgs

RSi

RD||RLCMi gmVgs

Vo

ACC gdMi 1

ACC gdMo

11

RTH1 RTH2

High-frequency hybrid- model with Miller effect

Migsin CCC Moout CC

A : midband gain


The cutoff frequencies defined by the input and output circuits can be obtained by first finding the Thevenin equivalent circuits for each section as shown below:

RTH1

Cin

vi

(a) Input circuit

inTHc CR

f12

1

where RTH1 = RSi||R1||R2 and

Cin = Cgs + CMi

RTH2

Cout

vi

(b) Output circuit

where RTH2 = RD||RL and

Cout = CMo

outTHc CR

f22

1


ExampleFind the cutoff frequency of the input and output RC circuit for the FET amplifier in figure below. Given that Cgd=0.1pF, Cgs=1pF, K =0.5mA/V2 and VTN=2V, =0.

vo

RL

RS

RDR1

C1

C2

vi

+VDD

C3

RSi

R2

4 k234 k

10 k

166 k

0.5 k

20 k

10 V


Solution

VVRR

RV DDG 15.421

2

GSGSS

SD VVV

RVI and

S

GSGTNGS R

VVVVK 2

GSGSGS VVVk 15.4445.05.0 2

VVGS 55.3

mSVVKg TNGSm 55.12

DC Analysis


68.421

21

SiLDm RRR

RRRRgA

pFpACC gdMi 579.079.51.01

pFCCC Migsin 58.1579.01

MHz

pkCRf

inTHinc

85.1058.128.92

12

1

1)(

Input RC circuit

kRRRR SiTH 28.9211

Midband gain

Thevenin’s equivalent resistance at the input

total input capacitance

upper cutoff frequency introduced by input capacitance


pFpA

CC gdMo 5.051.011

pFCC Moout 5.0

MHz

pkCRf

outTHoutc

49.955.033.32

12

1

2)(

Output RC circuit

kRRR LDTH 33.32

total output capacitance

Thevenin’s equivalent resistance at the output

upper cutoff frequency introduced by output capacitance


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Frequency Response of Amplifier Circuits › resources › 042223011 › 6.04222301_Fre… · Common-emitter Amplifier - Low-frequency ac equivalent circuit R S R B R C R L R E C