l2-bjtce

Imperial College London EEE 1L2 Autumn 2009 E2.2 Analogue Electronics

The Miller Theorem

and the

Frequency Response of the Common Emitter Amplifier


Consider a shunt admittance connected between the input and output of an inverting voltage amplifier of gain G.

Assuming that Y does not affect the gain, KCL at the input says that:

This is the same current as we would get if we connected Y(1+G) in parallel to the input and Y(1+1/G) in parallel to the output (on the right)

Note that this is an approximation; Y always changes the gain,unless Z out // Z L = 0. Nonetheless, very often it is a good approximation.

A useful approximation: The Miller Theorem

( ) ( )( ) ( )

1 1 1

2 2 2

1

1 1/in F in out in

out F in out out

i i i i v v Y i G Y v

i i i i v v Y i G Y v

= + = + = + + = = = + +

VIn VOut

-G

Y

Iin

IF

I1 IoutI2

- IFIn Out

-G(1+G)Y (1+1/G)Y


An application of the Miller theorem

Input impedance of the inverting amplifier (if op-amp has zero Yin):

We can even use the Miller theorem to calculate the gain of this circuit for finite op-amp gain G:

This is the correct answer for the closed loop gain as we shall later see!

R2

R1

Gvin

Vout ( )21 1inRZ RG

= + +

( )( )2 2 21 2 1 1 2 1

/ 1) , lim

/ 1out

v vGin

R Gv GR RA G Av R R G RG R R R

+ = = = =+ + + +

SF1.6p34


The small signal model the of BJT The model below is useful to frequencies of up to 100s of MHz The model is strictly valid for B-E voltage variations smaller than VT=25mV. To analyse a circuit:

Each transistor in a circuit is replaced by this model The base spreading resistance RBB is lumped with the thevenin

impedance of the signal source. It is not shown explicitly in these notes.

What does RBB do: RBB is responsible for most of the noise of transistor amplifiers RBB is responsible for the power gain rolling-off at high frequencies RBB is often ignored at low (f < MHz) frequencies


BJT model: impedances and transconductance

The large signal BJT equation gives the transconductance:

However, since IE=(1+1/)IC ,

Same equation gives output conductance:

( )/ /0 01 126 mV at room temperature

be th be thV V V VCEC

A

Bth

C Cm

be th

VI I e I eV

k TVeI IgV V

= + ==

/ /BE T C mB

VR V I gI

= = =

1/ C CCE CECE A

I IG RV V= = =


BJT model Capacitances BC junction is reverse biased:

depletion capacitance overlap capacitance

BE junction is forward biased: depletion capacitance storage (diffusion capacitance) overlap capacitance

CE capacitance is small overlap capacitance

Fringing/Overlap capacitances:

Junction capacitance: Reverse biased diode M=1/2-1/3, from doping approx 0.8V M, usually determined from two

bias points

Diffusion capacitance from transit time

0BC BC jC C C= +

0BE BE j DC C C C= + +

0CE CEC C0 0 0 , , BC BE CEC C C

0

1j M

BE

CCV

=

, 1/ 2 D m T T TC g f = =


An important figure of merit:the BJT Current gain and the transit frequency fT

fT is usually specified by the transistor manufacturer as a figure of merit.

fT is a rough estimate of the highest frequency at which the transistorcan be used as an amplifier. Typically fT / 10 is this highest frequency.

The transistor is connected as a current amplifier: Driven by a current source Driving a current meterfT is the frequency at which the current gain drops to unity.


An important figure of merit:the BJT Current gain and the transit frequency fT

( )( )

( ) ( )( ) ( )( )

0

0

0

0 0

0

/0

1/

" "1/ 1 1 / 2

where , since: 2 2

the current gain is li 1

m/ T

B BE BE BC

C m BE

C m mfe

B BE BC BE BC T

m mT

BE BC B

f f

E BC BE m

feT

i v s C C R

i g vi g g Rhi s C C R sR C C s f

g gf RR C C C C C g

h fj f f

h

>

= + + = = = = =+ + + + +

= = =2 + += +

1 with 2

Tfe T

T

f ff = =

fT allows the easy calculation of CBE + CBC since gm only depends on the collector bias current! Note that CBE>>CBC so knowing fT defines CBE


The Common Emitter amplifier

( ) 11 1

1

/

/ 25mV at 290K=17C

C Cm

BE T

C CB

BE BE BE m

C ACE

CE C

T

I IgV V

I IIRV V V g

I VRV I

V kT q

= = = = = = = = =

Model parameters:

Load

Source Amplifier


1

1

1

/ /1

SS S

S S in

S SS S

S S in

R VV VR Z Z Y

R Z ZR R ZR Z Z Y

= =+ += = =+ +

The Common Emitter amplifier (2)

Combine into a Thevenin circuit:, ,S SV Z R

2Consider and together: / / 1and note that

CE L LCE L CE L

CE L L out

S S BB

R Z ZR Z R R ZR Z Z Y

Z R R

= = =+ += +


The Common Emitter amplifier (3)

Gain:

( ) ( )12 21

1 1out s m L

v ms s s s in L out

v V g ZVA g Rv V V R Y Z Y

= = = + +If Yin= 0 and Yout = 0 this is the familiar:

outv m L

s

vA g Zv

= Note that lowercase letters represent small signal variations, i.e. differentials


1

1

BE Bin

B BE

C Cout CE

out CE

V IZ RI V

V IZ RI V

= = = = = =

Input Impedance:

The Common Emitter amplifier (4):Input and output impedance

Output Impedance:


Frequency response of the Common Emitter amplifier

If we momentarily neglect CBC this would be the same as before but with:

1 / /1

1 / /1

in BEin BE

CEout CE CE

out CE CE

RZ R CY j C R

RZ R CY j C R

= = = += = = +

Input Impedance:

Output Impedance:

We still have: ( ) ( )1

1 1out m L

Vs s in L out

v g ZAv R Y Z Y

= = + + and ( )1out m LB L outv g ZGv Z Y

= = +


Dealing with CBC

In the absence of CBC we know that: ( )1out m LB L outv g ZGv Z Y

= = +the current through CBC is:

( ) ( ) ( )1 1/ 1CBC BC B C BC B BC CI j C V V j C G V j C G V = = =

ICBC

We can account for CBC by introducing an additional input Miller admittance:( ), 1M in BCY j C G= and an additional output Miller admittance:

( ), 1 1/M out BCY j C G=


Frequency response of the CE amplifier

Use the Miller Theorem to get:

1 2 2The low requency gain from to is: m m LV V G g R g Z= ( ) ( ) ( )2 3with 1 1 and 1 1/BE BC BC CE BC BCC C G C G C C C G C C= + + + = + +

VS1 and RS1 is the hevenin equivalent of RS VS andR


Frequency response of the CE amplifier (2)

( )( )2 21 1 2 2 3 1 2 2 3input pole output pole

the gain of this amplifier is:

1 1 1 1m

S S S

gdV R GdV sR C sR C sR C sR C

= =+ + + +

If the gain is large the input pole can be at a much lower frequency than the output pole:

22

21 2 2

1. lim . 1 1 2 2 (1 ) 22 2

This is independent of G! The tiny and the s can define the gain-bandwidth product oource impedance f the CE a

m m Lm

GS S BC S BC

S S

BC

G g R g Zg R GGBW GBW

BW R C R C G R CR C R C

C

= == +

1 2 1 2 2

mp.This suggests that feedback is at work!Note that the input pole will be at a lower frequency than the output pole if

/ 1S m S m S SG R R g R R R g R R R> > = >

( ) ( )( ) ( ) ( )( )( )( )

1 2

2 3

/ / 1 1 since

1 1/ / / (since )IN S S BE BC S BC S

OUT BC CE CE L BC L BC CE

R C R R C G C R G C R R

R C C G C R Z C Z C C

= = + + += = + +

The gain-bandwidth product of this amplifier is:if in out >


Importance of the gain-bandwidth product

( )( )2 21 1 2 2 3 1 2 2 3input pole output pole

the gain of the CE amplifier is:

1 1 1 1m

S S S

gdV R GdV sR C sR C sR C sR C

= =+ + + +

One of the two pole frequencies is usually much lower than the other.We then say that we have a dominant pole amplifier whose frequency response is:

( ) 0 01 1vG GA ss j = =+ + for a DC gain G0 and a bandwidth 1/B =

0 0 /B BG G G = =We have already seen that in the limit of large G0 the Gain-Bandwidth product:

is independent of the amplifier gain.

We will see later in the course that this is a very general result which applies to all one pole feedback amplifiers regardless of how they are constructed.

l2-bjtce

Documents