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Common Emitter Amplifier Frequency Response

Objectives

To analyze the CE amplifiers and to calculate different poles and zeros which

determine its frequency response

To calculate the dominant poles and the CE amplifier bandwidth To analyze the effect of the emitter resistor on the CE amplifier gain and

bandwidth

Introduction

In the last lecture the frequency response of the amplifier circuits were examined

!lso" the frequency response of the common source amplifier was calculated and the

dominant poles were determined In this lecture the frequency response of the

common emitter amplifier will be considered using the #CTC and OCTC techniques

introduced in the last lecture The design trade$off using the emitter resistor will be

explored

Short-Circuit Time Constant Method to Determine ω L

% !s mentioned in the last lecture midband gain and upper and lower cutoff

frequencies that define bandwidth of the amplifier are of more interest than

complete transfer function

% In the next example the low cutoff frequency of the CE amplifier will be

determined using the #CTC method

Example &' (erive an expression for the low cutoff frequency for the CE amplifier

circuit in )igure & using the #CTC method

)ig$& in *+ec,-.,/er,-&vsd0Figure Common Emitter Amplifier Circuit

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#olution'

The small signal circuit may be obtained by short circuit the (C supply as shown in

)igure 1

)ig$1 in *+ec,-.,/er,-&vsd0Figure ! CE amplifier circuit used for small signal analysis

2 3 is the equivalent resistance for 2 & and 2 1 given by'

& 144 B R R R=

5sing #CTC method )or C 1 the resistance seen across the capacitor terminals may be

calculated from )igure 6

& * 0 * 0CE

S sig B in sig B R R R R R R r π = + = +

#o the time constant associated with C& equals C&*2 sig72 344r π 0

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)ig$6 in *+ec,-.,/er,-&vsd0Figure " Circuit used to calculate the short-circuit time constant associated #ith C 1

)or C 2 the resistance seen across the capacitor terminals may be calculated from

)igure 8

1 * 0 * 0CE

S L C out L C o

L C

R R R R R R r

R R

= + = +

≅ +#o the time constant associated with C1 equals C1*2 +72 C0

)ig$9 in *+ec,-.,/er,-&vsd0Figure $ Circuit used to calculate the short-circuit time constant associated #ith C 2

)or C E the resistance seen across the capacitor terminals may be calculated from

)igure :

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* 0

&

sig BCC out ES E E

o

r R R R R R R

π

β

+= =

+

#o the time constant associated with CE equals CE2 E#

)ig$. in *+ec,-.,/er,-&vsd0Figure % Circuit used to calculate the short-circuit time constant associated #ith C E

The low cutoff frequency is calculated using the #CTC as'6

&& 1

& & & &

* * 00 * 0 * 0

&

Li

iS i sig B L C sig B

E E

o

R C C R R r C R R r R RC R

π π

ω

β

=≅ = + +∑

+ + + ÷ ÷+

;lease note that due to the finite input resistance of the CE amplifier compared to the

C# amplifier" the lower cutoff frequency will be higher in the CE amplifier compared

to the C# amplifier

Example 1' Calculate the low cutoff frequency for the common emitter amplifier in

)igure & using the #CTC method !ssume' /CC<&1/" 2 sig<&= Ω" 2 &<&-= Ω" 2 1<6-= Ω"

2 C<96= Ω" 2 E<&6= Ω" 2 +<&--= Ω" C&<1µ)" C1< -&µ)" and CE<.-µ)

#olution'

2efer to Example & solution

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& 1

& & &

* * 00 * 0 * 0

&

L

sig B L C sig B

E E

o

C R R r C R R r R RC R

π π

ω

β

≅ + ++ + +

÷ ÷+

To calculate ω+ we need to find r π" this value may be found from the dc operating point of the amplifier

)ig$8 in *+ec,-.,/er,-&vsd0Figure & DC circuit for the CE amplifier in Figure

The (C circuit of the amplifier may be obtained from the CE amplifier circuit in

)igure & by considering all the capacitors are open circuits and the !C sources are

zero

The circuit in )igure 8 may be further simplified using Thevinin>s Theorem as shown

in )igure :

)ig$: in *+ec,-.,/er,-&vsd0Figure ' Simplified circuit for DC analysis

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& 144 :.th B R R R R k = = = Ω

&

& 1

&- &1 6

9-th CC

RV V V

R R= = =

+

!pplying ?/+ in the 3$E loop and assuming that the transistor is in the active mode'

* *& 0 0

6 -:&88

* *& 0 0 :. &-&@&6

th B th BE E E

BE B th o E

th BE B

th o E

V I R V I R

V I R R

V V I A

R R

β

µ β

∴ = + += + + +

− −∴ = = =

+ + +&88C o B E I I I mAβ = ≅ =

!pplying ?/+ at the C$E loop

&1 &88@*&6 960 1:CC C C CE E E

CE

V I R V I R

V V ∴ = + +∴ = − + =#ince /CE A /CEsat " then the transistor is in active region as assumed

B$point < *&88m!" 1:/0

1.&.

&88

T

B

V mV r k

I Aπ

µ = = = Ω

#ubstitute in the expression of ω+

111 8 :.- &-8D sec L rad ω ≅ + + =

Direct Determination of (igh-Frequency )oles and *eros+

#imilar to what we did in the last lecture The high frequency poles and zeros may be

determined by direct analysis In this case all the coupling and bypass capacitors may

be considered short circuit Fowever" the high frequency model of the transistor

should be used since the transistor internal capacitances can>t considered open circuit

at high frequencies The following example will illustrate this method for the common

emitter amplifier

Example 6' (rive an expression for the high frequency response !vF*s0 of the

common emitter amplifier shown in )igure & Fence" determine the midband gain

!mid" high frequency poles and zeros" and the high cutoff frequency

#olution'

The CE amplifier is redrawn for small signal analysis as shown in )igure D by

replacing the transistor in )igure 1 by its high frequency small signal model

!ll the coupling and bypass capacitors are considered short circuit at high frequency

The biasing resistors 2 & and 2 1 are combined into 2 3 & 144 B R R R∴ =

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)ig$D in *+ec,-.,/er,-&vsd0Figure , The CE amplifier in Figure ! redra#n #ith the transistor replaced y its high frequency

small signal model

The small$signal model can be simplified by using Thevinin>s theorem as shown in

)igure The simplified circuit is shown in )igure &-

)ig$ in *+ec,-.,/er,-&vsd0Figure . /sing The0inin1s Theorem to simplify the CE amplifier circuit in Figure ,

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Bth sig

sig B

Rv v

R R=

+ and sig B

th

sig B

R R R

R R=

+

1

thth

th x

v r v

R r r

π

π

=+ + and 1 * 0th th xr r R r π = +

!lso" 44 44 L L C o R R R r ′ =

)ig$&- in *+ec,-.,/er,-&vsd0Figure 2 Simplified circuit to calculate the frequency response of the CE amplifier

Griting the two nodal equations at nodes ! and 3 in the frequency domain"

1

1

* 0 -

* 0 -

be thbe be o

th

om gs o be

L

v vv sC v v sC

R

v g v v v sC

R

π µ

µ

′ −′ ′+ + − =

′+ + − =′

#olving the last two equations by eliminating vH be yields"

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( )

th1o

1

1

1 1

1

1

1

* $ 0*s0*s0

& & &

* $ 0* 0* 0

* 0 * 0* 0

+et us define &

&

mv

th

m

L th L L th

mo BvH

sig th x sig B

LT m L

th

T

L L th

sC g v

R

C s C C s C g

R R R R R

sC g v s R A s

v s R r R R

RC C C g R

R

sC s C C

R R R

µ

π π µ µ

µ

π µ

π µ

=∆

∆ = + + + + + ÷ ÷ ÷′ ′ ′

∴ = =

+ + ∆

′′= + + + ÷

∴∆ = + +′ ′

1

1

* $ 0* 0

&* 0* 0

m BvH

T th x sig B

L L th

sC g R A s

sC R r R R C C s R C C R R C C

µ

π µ

π µ π µ

∴ =+ +

+ +′ ′

!mid may be found from the last expression by assuming s -

1$ $

* 0* 0 * 0* 0

$

* 0* 0

m L B th m L Bmid

th x sig B th x sig B

o L B

th x sig B

g R R R g R R r A

R r R R r R r R R

R R

r R r R R

π

π

π

β

′ ′∴ = =

+ + + + +

′=

+ + +Figh$frequency response is given by 1 poles" one finite zero and one zero at infinity

)inite right$half plane zero" ω Z < 7 g mC µ A ω T can easily be neglected

)or a polynomial s17 sA&7 A- with roots a and b" if one root is much larger than the

other one" we can assume that a < A& and b< A- A&

-&

& 1

1

1

1

&

!ssume that the midband gain is very large * ie & "and 0

& &&

th T

Lm L m L

th

m m

m th m L

A

A R C

R g R g R C C

R

g g

C g R g R C

µ π

π π

ω

ω

∴ = ≅

′′ ′+ ÷

∴ ≅ + + ≅ ÷′

? ?

#mallest root that gives first pole limits frequency response and determines ω H

#econd pole is important in frequency compensation as it can degrade phase margin

of feedbac= amplifiers

p& p1 p&

&

1

!ssuming * is the dominanat pole0

& H

th T R C

ω ω ω

ω ω ∴ ≅ =

=

(ominant pole model at high frequencies for C# amplifier is shown in )igure &&

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)ig$&& in *+ec,-.,/er,-&vsd0Figure Equi0alent circuit assuming dominant pole model

Example 9' )ind the midband gain" high frequency poles and zeros" ωF" and the

bandwidth for the CE amplifier in Example 1 iven the following transistor

parameters ! T <.-- JFz" β o <&--" C µ <-. p)"and r x <1.-Ω, /!<&--/.

#olution'

2eferring to example 1 solution" the B$point is *&88m!" 1:/0

9- 889

&.

&--8-1

&88

C m C

T

Ao

C

I g I mS

V

r k

V r k

I

π

∴ = = =

= Ω

= = = Ω

1-8p)1

m

T

g C C

! π µ

π = − =

44 &-- 96 44 8-1 6D8=K L L C o R R R r ′ = = =

:. & -DD=Kth B sig R R R= = =

1 * 0 -89th th x R r R r k π = + = Ω

1& &.16p)

L

T m Lth

R

C C C g R Rπ µ

′

′= + + + = ÷

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&

1

&&-18J radsec

th T R C ω = =

1

1

& && 66& radsecm

m th m L

g

C g R g Rπ

ω

≅ + + ≅ ÷′

&61D radsec g

m # C

ω

µ

= =

It is clear that ω p& is much lower than ω p1 and ωz which means that ω p& is the dominant

pole

&&-18J radEsec

&861

H

H H $H#

ω ω

ω

π

∴ ≅ =

= =

&86 &:- &86 H L B % ! ! $H# H# $H# = − = − ≅

$ &1. //

* 0* 0

o L B

th x sig B

R R A

mid r R r R Rπ

β ′= = −

+ + +

3ain-4and#idth )roduct 5imitations of the C-E Amplifier

!s we can see from the expressions of the midband gain and the CE amplifier

bandwidth 2 th is appearing in both expressions

If Rth is reduced to zero in order to increase the bandwidth" then Rth2 would not be zero

but would be limited to approximately r x

&3G

1

o Lv H

th x

R A

R r r R C th T π

β ω

′ ÷= ≤ ÷ ÷+ +

If Rth < -" r x LLr π so that r x < Rth2 and * 0T m LC C g R µ ′=

&

I3G xr C µ ∴ ≤

If we used the same values in Example 9

I3G DI radEsec∴ ≤

The !ctual 3G in Example 9 is &669 radsec

6pen-Circuit Time Constant Method to Determine ω H

% !s mentioned in the last lecture the open$circuit time constants associated

with the transistor capacitances may be used to simplify the determination of

the high cutoff frequency of the amplifier

% In the next example the high cutoff frequency of the CE amplifier will bedetermined using the OCTC method

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Example .' (erive an expression for the high cutoff frequency for the circuit in

Example 9 using the OCTC method

#olution'

5sing OCTC method for the circuit in )igure &- )or C π the resistance seen across the

capacitor terminals may be calculated from )igure &1

1& th R Rπ =#o the time constant associated with Cπ equals Cπ 2 th1

)ig$&1 in *+ec,-.,/er,-&vsd0Figure ! Circuit used to calculate the open-circuit time constant associated #ith C

)or C µ the resistance seen across the capacitor terminals may be calculated from

)igure &6

x1

x 1

v*& 0

i

m

Lo th L

th

R R R g R

R

µ

′′= = + +

#o the time constant associated with Cµ equals Cµ 2 th1 *&7gm2H+ 72H+2 th0

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)ig$&6 in *+ec,-.,/er,-&vsd0Figure " Circuit used to calculate the open-circuit time constant associated #ith C

The high cutoff frequency is calculated using the OCTC as'

1

&

& & &

1 H

io ii

R C R C r C R C o o th T

ω

π π µ µ =

≅ = =+∑

Ghich is the same expression found before using direct calculations in Example 6

3ain-4and#idth Trade-off /sing Emitter Resistor

The emitter resistance of the common emitter amplifier may be used to trade off

between the gain and the amplifier bandwidth In this case no bypass capacitor is

used

The small signal circuit is shown in )igure &9 Ghere 2 th and vth are the same as

calculated in Example 6

Bth sig

sig B

Rv v

R R=

+ and sig B

th

sig B

R R R

R R=

+

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)ig$&9 in *+ec,-.,/er,-&vsd0Figure $ Small signal CE Amplifier circuit at high frequency #ith no ypass capacitor

The midband gain may be calculated from )igure &9 by assuming the transistor

capacitors to be open circuit

* * &0 0* 0

o L B Lmid

th x o E sig B E

R R R A

R r r R R R Rπ

β

β

′ ′= − ≅ −

+ + + + +

for th xr R r π >> + " B sig r R>> and &m E g R >>

The midband voltage gain decreases as the emitter resistance increases and the

bandwidth of CE amplifier will correspondingly increase

To find the effect of 2 E on the amplifier bandwidth the high cutoff frequency may becalculated as follows'

5sing OCTC method for the circuit in )igure &9 )or C π the resistance seen across the

capacitor terminals may be calculated as r π 44 2 eq Ghere 2 eq is calculated from )igure

&.

x

x

v

i &

th x E e'

m E

R r R R

g R

+ += =

+

&

th x E & e'

m E

R r R R r R

g Rπ π

+ += ≅

+#o the time constant associated with Cπ equals Cπ 2 πo

)ig$&. in *+ec,-.,/er,-&vsd0Figure % Circuit used to calculate the open-circuit time constant associated #ith C

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)or C µ the resistance seen across the capacitor terminals may be calculated from

)igure &8

)ig$&8 in *+ec,-.,/er,-&vsd0Figure & Circuit used to calculate the open-circuit time constant associated #ith C

To simplify the calculation the test source i x is first split into two equivalent sources

as shown in )igure &: and then superposition is used to find v x

<*vb

$ v(0

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)ig$&: in *+ec,-.,/er,-&vsd0Figure ' Modified circuit used to calculate the open-circuit time constant associated #ith C

!ssuming that β o AA& and ( ) ( )* &0th x o E R r r Rπ β + << + +

x

x

v* 0 &i &

m L Lo th x

m E th x

g R R R R r g R R r

µ

′ ′= = + + + ÷+ +

#o the time constant associated with Cµ equals Cµ 2 µo

The high cutoff frequency is calculated using the OCTC as'

1

&

& &

* 0 & && &

H

E m L Lio ith xi

m E th x m E th x

C R g R R R C R r C g R R r g R R r

π

ω

µ =

≅ = ′ ′∑ + + + + + ÷ ÷ ÷ ÷+ + + +

If we used the same amplifier discussed in Example 9 with C E<-

Ge will obtain'

6D81: //

&6

Lmid

E

R A

R

′ −≅ − = = −

@

&

* 0 & && &

&

1-8 &6 889@6D8 6D8

*-DD -1.0 & -. && 889@&6 -DD -1. & 889 &6 -DD -1.

1&&J radsec

H

E m L Lth x

m E th x m E th x

C R g R R R r C

g R R r g R R r

)*

k )*

π

ω

µ

≅ ′ ′

+ + + + + ÷ ÷ ÷ ÷+ + + +

=

+ Ω + + + + ÷ ÷ ÷+ + + + =

!s we can see we got a higher bandwidth on the expense of lower midband gain

lec 05 wfig ver 01

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