Expt. No. 3

DIFFERENTIAL AMPLIFIER USING BJT

Objective: 1)The aim of the experiment is to study the operation of differential amplifiers using BJT and obtain CMRR and input resistance.

Components and equipments required:

1. Oscilloscope (Scope/CRO).2. Function Generators (FG).3. DC power supply.4. Project Breadboard. 5. Resistors.6. Capacitors7. BC 107 or SL100 BJTs8. Connection Wires. 9. Oscilloscope Probes.

Circuit diagram:

a) Differential Amplifier using BJT:i) Differential Mode, Balanced Output (for differential gain)

Fig. 1

ii) Common mode, balanced output (for common mode gain)

Fig. 2

Different Modes:

a) Dual input balanced output

The circuit is same as shown in Fig. 1

b) Dual input unbalanced output

The input is same as above, but output is taking across any of the transistor’s collector and ground.

c) Single input balanced output.

One of the inputs set to ground and output is same as first mode.

d) Single input unbalanced output.

One of the inputs set to ground and output is same as second mode.

Power Supply Arrangement for Dual Voltage Source:

Fig. 3

Note: For unbalanced output use coupling capacitor at the output

Theory:

The differential amplifier, or differential pair, is an essential building block in all integrated amplifiers. In general, the input stage of any analog integrated circuit with more than

one input consists of a differential pair or differential amplifier. The basic differential pair circuit consists of two-matched transistors Q1and Q2 , whose emitters are joined together and biased by the constant current source circuit. Q1, D1, D2, R2 and RE acts as current source, which introduce constant current IEE to the differential amplifier with high ac source resistance, since ac equivalent of this dc current source is ideally open circuit.

The important characteristics of the differential amplifier are: the common-mode rejection ratio CMRR, the input differential resistance R id, and the differential-mode gain Ad.

Differential gain: Ad=V od

V id, where Vid is differential input voltage,

V id=V 1−V 2 and Vod is differential output. For a perfectly matched pair transistor differential gain should be infinite.

CMRR: Common Mode Rejection Ratio,CMRR=Ad

Ac , where Ac is common mode gain

and it is Ac=V oc

V cm, where Vcm is common mode input voltage, V cm=

V 1+V 2

2 and Voc is common

mode output. For perfectly matched pair transistor common mode gain should be zero, and then CMRR should be infinite.

Differential input resistance: The resistance between the differential input and ideally it should be infinite.

DC Analysis:

I CQ=V EE−V BE

2 REE

V CEQ=V CC+V BE−RC I CQ

I E1=I E2=I E3

2

For constant current source circuit:

I E3=V D

RE

I 2=V EE−2V D

R2

AC Analysis: Balanced output Mode:

Differential gain Ad=−gm RC

Common mode gain Acm=−RC

2REE

CMRR ¿2 gm REE

Differential mode input resistance Rid=2r π

Output resistance Ro=RC

Unbalanced output Mode:

Differential gain Ad=−gm RC

2

Common mode gain Acm=−RC

2REE

CMRR ¿ gm REE

Differential mode input resistance Rid=2r π

Output resistance Ro=RC

Resistance of current source circuit:REE=¿¿

Design:

Given that |V CC|=|V EE|=15 V , I CQ=2 mA , Ad=50

From data sheet of BC 107, hoe=60 μS,β=150

RE=?

R2=?

gm=?

RC=?

I B=?

r π=?

REE=?

From the above design: |Ac|=?, CMRR=?=? (¿ dB)

Pre-lab Assignments:

1. Define CMRR.

2. Draw the basic circuit of differential amplifier using BJT. Give the significance of RC

and REE

3. Design differential amplifier with current source circuit using BJT using the following specifications

|V CC|=|V EE|=10 V , I CQ=10 mA , hoe=80 μS , β=200 , Acm=0.001 ,CMRR=10000

4. Give the use of current mirror circuit in differential amplifier.5. Give the advantages of MOSFET differential amplifier.6. Design differential amplifier with current source circuit using MOSFET using the

following specifications

|V DD|=|V SS|=5V , I D=1 mA , Acm=0.001 ,CMRR=80 dB

7. Draw the differential amplifier circuit using BJT with active load.8.

Model Waveforms:

Differential Input and Balanced Output

Differential Input and Unbalanced Output

Common Mode Input Unbalanced Output

Simulated Results:

Differential Gain:

For balanced output:

v¿=0.5 mV −0V , vo=25.8 mV

¿ Ad∨¿51.6

For unbalanced output:

v¿=0.5 mV −0V , vo 1=−12.9 mV , vo 2=12.9 mV

¿ Ad∨¿51.6

Common Mode Gain:

v¿=1 V , vo 1=964 μV , vo2=964 μV

¿ Acm∨¿0.00964

CMRR=74.57 dB

For input resistance measurement use the following setup:

Measure vin and iin using DMM and then

R¿=v¿

i¿

Viva Questions:

1) Define differential amplifier and give its application2) Draw a current mirror circuit. Differentiate it with constant current source.3) In a differential amplifier |VCC| ≠ |VEE|, list out the problems4) Differentiate different configurations of differential amplifiers. Which one is

commonly used?5) Give the main advantage of constant current bias over emitter bias.6)