Bipolar Differential Amplifiers: Qualitative Analysis

Different modes of operation of the BJT differential pair: (a) The differential pair with a common-mode input signal vCM. (b) The differential pair with a “large” differential input signal. (c) The differential pair with a large differential input signal of polarity opposite to that in (b). (d) The differential pair with a small differential input signal vi. Note that we have assumed the bias current source I to be ideal (i.e., it has an infinite output resistance) and thus I remains constant with the change in vCM.

Common Mode Differential Mode

Bipolar Differential Amplifiers: Large Signal Analysis

1 2

1 2

( ) / ( ) /1 2

( ) /1

2

The exponential relationship applied to each of the two transistors may be written as:

and

These two equations can be combined to obta

in

w

hic

B E T B E T

B B T

v v V v v VS SE E

v v VE

E

I Ii e i e

i ei

α α− −

−

= =

=

2 1 1 2

1

1 2( ) / ( ) /

1 2 1 2

1 2

1 (

2

1 11 1

1

h can be manipulated to yield

and

From the circuit we have

Which may be used to obtain the following expressions for and

B B T B B T

E Ev v V v v V

E E E E

E E

E v

E E

i ii i e i i e

i i I

ie

iI

i

− −= =+ + + +

+ =

=+ 2 1 1 22) / ( ) /

1 2

1 2

1and

and may be obtained by multiplying and by which is almost unity and

plotted as shown in the fi

gure

B B T B B TEv V v

C C

E E

v V

Iie

i ii i α

− −=+

Bipolar Differential Amplifiers: LinearizationFor Your Information

The transfer characteristics of the BJT differential pair (a) can be linearized (b) (i.e., the linear range of operation can be extended) by including resistances in the emitters.

What is the role of the degeneration resistance (Re)?

Bipolar Differential Amplifiers: DC Analysis (Example 1)

Problem: Find the Q-points of transistors in the shown differential amplifier.

Given data: VCC=VEE=15 V, REE=RC=75kΩ, β =100Analysis:

A3.95)3102(75

V7.015μ=

Ω×

−=

−=

⎟⎠⎞⎜

⎝⎛

EE2RBEVEEV

EI

100 94.4 A101

I I IE EC α μ= = =

94.4 A 0.944 A100

ICIBμ μ

β= = =

15 7.92V

7.92V-(-0.7V) 8.62V

V I RC C C

V V VECE C

= − =

= − = =

Due to symmetry, both transistors are biased at Q-point (94.4 μA, 8.62V)

Bipolar Common-mode Input Voltage Range

For symmetrical power supplies (VEE=VCC) , VEE >> VBE, and RC = REE,

0

2

1

1

V V I R VCB CC C C ICV V VIC BE EEIC REE

V VR EE BEC2R VEE CCV VIC CC RC

2REE

α

α

α

⎛ ⎞⎜ ⎟⎝ ⎠

= − − ≥

− +=

−−

∴ ≤+

53

VVCCV IC =≤

We want to find max. VIC while the C-B junction is reverse biased.

Problem: Find the max. VIC before saturation in the shown differential amplifier.

Given data: VCC=VEE=15 V, REE=RC=75kΩ, β =100Analysis:

Bipolar Differential Amplifiers: DC Analysis (Example 2)

Problem: Find vE ,vc1 , and vc2 in the shown differential amplifier.Given data: VCC=VEE= 5 V, REE=RC=1kΩ, α ≈, |VBE|=0.7 VAnalysis:

1

2

2 2

2 2

0.75 * 5 0 5

(5 0.7) /1 4.34.3

5 * 5 4.3 0.7

V V

mA mA

V

E

C1 C C

E E

C E

C C C

vv R II II Iv R I

α

+− + − + = −

= = − == =

= − + − + = −

== =

=

We can assume Q1 to be off and Q2 on

Differential-mode Gain and Input Resistance

( 1/ )(v v ) 1/ ve3 4v (1/ 2/ 2 ) 0 v 0e e

g r Rm EER r gEE mπ

π

+ + =

∴ + + = → =

2id

v4v −=

ev2id

v3v −= ev

2id

v4v −−=

Output signal voltages are:

2id

vc1v CRmg−=

2id

vc2v CRmg+=

idvodv CRmg−=∴

2id

v3v =∴

Emitter node in differential amplifier represents virtual ground for differential-mode input signals.

Differential-mode gain for balanced output, is:

If either vc1 or vc2 is used alone as output, output is said to be single-ended.

Differential-mode Gain and Input Resistance (contd.)

CRmgddA −=

=

=

0ic

vidvod

vc2vc1vodv −=

220

icvid

vc1

v1

ddA

CRmg

ddA =−=

=

=22

0ic

vidvc2

v2

ddA

CRmg

ddA −==

=

=

Differential-mode input resistance is small-signal resistance presented to differential-mode input voltage between the two transistor bases.

If vid =0, . For single-ended outputs,

πridR 2b1i/idv ==∴

CRorCRodR 2)(2 ≅= CRodR ≅πr

)2/id

v(b1i =

Common-mode Gain and Input Resistance

Both arms of differential amplifier are symmetrical. So terminal currents and collector voltages are equal. Characteristics of differential pair with common-mode input are similar to those of a C-E amplifier with large emitter resistor.

Output voltages are:

vicib 2( 1)r REEβπ=

+ +

v v i vc1 c2 b ic2( 1)RCRC r REE

ββ

βπ

−= =− =

+ +

v 2( 1)ie b2( 1)

v vic ic2( 1)

REEREE

r REE

βββπ

= +

+= ≅

+ +

Common-mode gain is given by:

For RC=REE, common-mode gain =0.5. Thus, common-mode output voltage and Acc is 0 if REE is infinite. This result is obtained since output resistances of transistors are neglected. A more accurate expression is:

Therefore, common-mode conversion gain is found to be 0.

Common-mode Gain and Input Resistance (contd.)

vocv 2( 1) 2ic v 0id

R RC CAcc r R REE EE

β

βπ= =− ≅−

+ +=

0c2vc1vodv =−=

1 12

A Rcc C r Ro EEβ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

≅ −

v 2( 1)ic ( 1)2i 2 2b

r R rEER REEicβπ π β

+ += = = + +

Common-Mode Rejection ratio (CMRR)

Represents ability of amplifier to amplify desired differential-mode input signal and reject undesired common-mode input signal.For differential output, common-mode gain of balanced amplifier is zero, CMRR is infinite. For single-ended output,

/ 2 1C M R R1 12 2r g

A Adm dd g Rm EEA Acm ccg Ro m m EEβ

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

≅= = =−

Analysis of Differential Amplifiers Using Half-Circuits

Half-circuits are constructed by first drawing the differential amplifier in a fully symmetrical form- power supplies are split into two equal halves in parallel, emitter resistor is separated into two equal resistors in parallel.None of the currents or voltages in the circuit are changed.For differential mode signals, points on the line of symmetry are virtual grounds connected to ground for ac analysisFor common-mode signals, points on line of symmetry are replaced by open circuits.

Bipolar Differential-mode Half-circuits

Applying rules for drawing half-circuits, the two power supply lines and emitter become ac grounds. The half-circuit represents a C-E amplifier stage.

2id

vc1v CRmg−=

2id

vc2v CRmg+=

v v v vc1 c2od idg Rm C= − =−

Direct analysis of the half-circuits yield:

πridR 2b1i/idv == )(2 orCRodR =

CRmgddA −=

=

=

0ic

vidvod

v

220

icvid

vc1

v1

ddA

CRmg

ddA =−=

=

=

Bipolar Common-mode Half-circuits

Applying rules for drawing half-circuits, the points at the line of symmetry are open circuited. The half-circuit represents a C-E amplifier stage with an emitter resistance.

v v v 0c1 c2od = − =

Direct analysis of the half-circuits yield:

v v i vc1 c2 b ic2( 1)RCRC r REE

ββ

βπ

−= =− =

+ +

v 2( 1)ic ( 1)2i 2 2b

r R rEER REEicβπ π β

+ += = = + +

/2CMRR

A Adm dd g Rm EEA Acm cc≅= =

vocv 2( 1) 2ic v 0id

R RC CAcc r R REE EE

β

βπ= =− ≅−

+ +=

Biasing with Electronic Current Sources

Differential amplifiers are biased using electronic current sources to stabilize the operating point and increase effective value of REE to improve CMRRElectronic current source has a Q-point current of ISSand an output resistance of RSS as shown.DC model of the electronic current source is a dc current source, ISS while ac model is a resistance RSS.

SSR

0V

SSIDCI −=