2008. Lecture 12 _361-1-3661_

7/28/2019 2008. Lecture 12 _361-1-3661_

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Lecture 12: Introduction to electronic analog circuits 361-1-3661 1

Our aim is to develop the transistor small-signal model forhigh frequencies and to find the high-frequency response of elementary transistor circuits. We will do this for the BJ Ttransistors only; the analysis of JFET and MOSFET transistorcircuits is very similar.

8.1. Transistor small-signal model for high frequencies

At high-frequencies, the impedances of the parasiticcapacitances related to the transistor junctions (see Fig. 1)become comparable to the values of the corresponding h-parameters of the transistor model, which was obtained for dc

and relatively low frequencies. Thus to adjust the transistormodel for high frequencies, we include in it the parasiticcapacitances of the transistor emitter and collector junctions,

C π andC μ , and also the ohmic resistance of the base, r b. Wealso rename some of the transistor small-signal parameters toallow the hie parameter represent the input impedance of thenew model:

π

ϖ ϖ

ϖ r r

ji

jvh b

b

beie +=≡

=0)(

)(, (1)

where

e fe r hr )1( +=π . (2)

Unity-gain frequency (GBP)

For a given transistor, the ohmic resistance r b of the base

and the capacitance C μ of the collector junction can be found

analytically. The capacitance C π of the emitter junction isusually measured experimentally.

The experimental setup is shown in Fig. 2. The aim is to

measure the frequency ω t ("omega test"), at which the CEamplifier with the collector, which is short-circuited for small

signals, has the current gain |Ai( jω t )|=1. The unity-gainfrequencyω t is then translated into C π .

Note from Fig. 2 that in the small-signal analysis, r b can beneglected because it is connected in series with the currentsource i s and, hence, does not affect the base current. Thetransistor output resistance, r o, can also be neglected because

it is short-circuited. The capacitor C μ can be connected in

parallel with C π , provided that the current flowing through C μ is much smaller than the current generated by the dependentsource:

vπ

E

p+n++

C

p Eo

pCo

iC

n

i B

V CE

i B i

R

i E

n Bo

B

r B

B' C π j

C π d

C μ

C π

dQπ d

T be BE V vv

Boen/)( +

T BE V V

Boen/

π

π

π v

QC

j

j d

d

≡

π

π

π v

QC d

d d

d

≡

μ

μ

μ v

QC

j

d

d

≡

π ≡ E B' μ ≡C B'

C π

C μ

r B

B

C

E

B'

ib

h fe

iπ

g mvπ

r o

vce

C

B B' r B

C π

ib

h fe

iπ

g mvπ

r o

vce

C

vπ

r π

B

iπ

B' r B

C π

C μ

(1+h fe

)iπ

iπ

r e

C μ

Fig. 1. Parasitic capacitances of the BJ T transistor and its high-frequencysmall-signal models: T and hybrid-π .

8. High-Frequency Response© Eugene Paperno, 2008

7/28/2019 2008. Lecture 12 _361-1-3661_



μ

π

μ

π ϖ

ϖ

C

g v g

C

v mm <<⇒<<

1. (3)

As a result, the circuit current gain

c

fe fe

fec

i s

ci

j

h

r C C j

h

h

r C C j

C C ji

ji

ji j A

s

ϖ ϖ ϖ

ϖ

ϖ

ϖ

ϖ ϖ

π μ π

π

μ π

μ π

/1)(1

)(

1

)(

1

)(

)()(

1

+=

++=

++

+==≡

=

, (4)

whereω c =1/[( C π +C μ )r π ] is the circuit corner frequency.Assumingh fe>>1,

c fet

fect

fect

ct

fet i

h

h

h j

j

h j A

ϖ ϖ

ϖ ϖ

ϖ ϖ

ϖ ϖ ϖ

⋅=⇒

=⇒

>>=+⇒

=+

=

/

1/1

1/1

|)(|

. (5)

According to (5), ω t has an additional meaning: "gain-bandwidth product", whereh fe is the maximum, dc gain of the

circuit, andω c is the circuit bandwidth.

Since in the vicinity of ω t the circuit is not lamped and the

current gain does not meet (5) (see Fig. 2), ω t is measured

indirectly, namely, a current gain |Ai'( jω ' )| is measured at a

frequencyω t >>ω ' >>ω c

'|)('|

'|)('|

/'

/'1|)('|

'

ϖ ϖ ϖ

ϖ ϖ ϖ ϖ

ϖ ϖ

ϖ ϖ ϖ

ϖ ϖ

t it

t c fet i

c

fe

c

fet i

j A

h j A

h

j

h j A

c

=⇒

==⇒

=

+=

>>

. (6)

i s

h fe

iπ

g mvπ

r o

C

vπ

r π

B

iπ

r B

C π

C μ

V CE

I Bi

s

i s

ic

ic

icμ

icμ

<< ic

i s

h fe

iπ

g mvπ

C

vπ

r π

B

iπ

C π +C

μ

ic

i s

Ai= f (ω ) ?

|Ai( jω )| (dB)

ω 0

ω c

ω t

ω '

|Ai'|

ω '<< g

m

C μ

Fig. 2. Experimental setup for measuringω t andC π .

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Havingω t , we can eventually findC π as follows

μ π

μ μ π μ π μ π

μ π π μ π

ϖ

α

ϖ ϖ

C V

I C

C

g

C C

g

V C C

I

V C C

I

hr C C

h

r C C

hh

t T

C

mm

T

C

T

E f

fee

fe fec fet

−=⇒

<+

=+

=+

=

++=

+=⋅=

1

)()(

)1()()(

. (7)

This experimental value of C π is given in the data sheets of thetransistor for a particular I C , which was set in the experimentalsetup.

For a different I C , the value of C π should be adjusted. Since

the diffusion capacitance of the emitter junction is usuallymuch greater than the junction capacitance, and the diffusioncapacitance is directly proportional to the emitter and collector

currents, the C π value, which given in the date sheets, cansimply be multiplied by the ratio of the collector current that isset in the developed circuit and that is given in the data sheets.

8.2. Miller's theorem for voltages

The Miller theorem for voltages allows splitting (see Fig. 2)the impedance connected to an arbitrary port of a circuit into

two impedances connected to ground, provided the ratio μ (Miller gain) is known of the port terminal voltages relative to

ground. As the Miller theorem for current, this new theoremwill help us to divide a circuit into two separate parts and tosolve them independently, thus, simplifying the analysis.

s s

s s

Z μ 1

Z μ 2

Electronic circuit

Electronic circuit

Z μ

iμ

v1

v2

in out

in out

iμ

iμ

v1

v2

μ v2v1

vμ =v

1−v

2

iμ

v1−v

2v

1(1−μ )

μ v2

v1

− v2(1−1/μ )

Z μ

Z μ

Z μ

iμ

v1

v1(1−μ )

Z μ

Z μ

1−μ

iμ

− v2

Z μ o

Z μ

1−1/μ

1)

2) Z μ o Z

μ

Z μ in

Z μ in

− v2(1−1/μ )

Fig. 3. Miller's theorem for voltages.

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Example circuit: CE amplifier at high frequencies

Our aim is to find the upper limit, ω H , of the amplifiersmall-signal frequency band. To simplify the analysis, we split

C μ into C μ in and C μ o (see Fig. 4) by applying the Miller

theorem. This allows us to separate the circuit input andoutput loops. To find C μ in, we suppose thatω H depends on thecorner frequency of the circuit input part only, and that the

C μ o capacitor impedance is high enough to be neglected in the

circuit the output part. Thus, we can find C μ in=C μ (1−μ ), where

μ =μ dc does not depend on C μ o. To find C μ o, we suppose that

|μ |>>1, which gives usC μ o=C μ (1−1/μ ) ≈ C μ .

The corner frequenciesω in andω o of the separate input andoutput parts of the circuit can very easily be found (see Fig.

5). If ω o>4ω in (see Fig. 4), thenω H ≈ω in, and the impedance of

theC μ o capacitor is indeed high enough, and its effect on the

Miller gain μ can be neglected. If ω o<4ω in, then our above

assumption is not correct, and we cannot solve this circuit byhand.

Note that for negative |μ |>>1, theC μ in=C μ (1−μ ) capacitancein the input part of the CE amplifier can be much greater than

C π . Such an increase of the equivalent input capacitance andthe corresponding reduction of the circuit frequency band iscalled Miller effect.

vπ

in out

h fe

iπ

g mvπ

r o

r π

iπ

r B

C π

C μ

RB2

RC

R B1

R E

C C

C B

vO

+V CC

v s

v s

RC

vo

h fe

iπ

g mvπ

r o

vπ

r π

iπ

C π v s' R

C

vo

h fe

iπ

g mvπ

r o

vπ

r π

iπ v

s R

C

vo

C μ oC μ in

μ dc

= − g m( R

C ||r

o) for r

o>> R

C >>1/ g

m, |μ

dc| >>1

= C μ (1−1/μ ) =C

μ C μ o

= C μ (1−μ

dc)C

μ in

|Av/ A

v max|, (dB)

0

log (ω )−3

ω H

ω in

ω in

ω in− ω

H <6% for

ω in

>4ω

o

ω in

=(C

μ in+C

π )(r

b||r

π )

1ω ο

=C μ o

( RC ||r

o)

1

|Z μ o

| 04ω

oω

o/4

in out

r B

r B

ω o>4ω

in

R B

Fig. 4. Example circuit: CE amplifier at high frequencies..

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Example circuit: CB amplifier at high frequencies

The aim is to find the upper limit, ω H , of the amplifier

small-signal frequency band. To findω H easily, we neglect r o and also the voltage drop across r b (because both ib and r b are

small). The latter allows us to ground r b and the controlledsource (see Fig. 5) and to separate the circuit input and outputparts.

The corner frequenciesω in andω o of the separate input andoutput parts can very easily be found (see Fig. 5). When oneof these frequencies is greater than the other by at least of a

factor of 4, then ω H is nearly equal to the lower corner

frequency. When ω in=ω o, then ω H =(20.5−1)0.5

ω in/o ( prove

this!). In all the other cases, ω H cannot by found in a simpleway.

Note that the CB amplifier does not suffer from the Millereffect, and, thus, has a much wider frequency band than the

CE amplifier.

REFERENCES

[1] S. Sedra, K. C. Smith, Microelectronic Circuits, 4th ed. New York:Oxford University Press, 1998.

h fe

iπ

g mvπ

r o

vπ

r B

C π

RB2

RC

C B

vO

+V CC

v s

ω in

=C π (r

s||r

e)

1ω ο

=C μ ( R

C ||r

o)

1

R B1

v s

RC

vo

C μ

iπ

r e

ib

iπ

v s

r e

r s

r s

C π

vπ

h fe

iπ

g mvπ

RC

vo

C μ

∞=

|Av/ A

v max|, (dB)

0

log (ω )−3

ω o>4ω

inω

H ω

in

|Av/ A

v max|, (dB)

0

log (ω )−3

ω o

ω H

ω in=

ω H

=(20.5−1)0.5ω in/o

Fig. 5. Example circuit: CB amplifier at high frequencies.

2008. Lecture 12 _361-1-3661_

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