Amplifiers
Differential Differential Amplifier Circuit
Differential amplifier circuits have 2 inputs and 2 outputs.
It can be operated with a dual power supply: VCC to VEE;
or with a single supply: VCC to GND
Features of Differential Amplifiers
It amplifies the difference between the 2 inputs
It is a high gain, low noise amplifier
3 Modes of Operation
1.
2.
3.
Single-ended
an input signal is applied to one of the inputs and the other input is grounded.
Double-ended
two opposite polarity input signals are applied to its inputs.
Common-mode
the same input signal is applied to both inputs.
DC Bias
VE 0V VBE 0.7V
Both inputs are grounded:
IE
VE ( VEE ) VEE 0.7
RE
RE
assuming both transistors are well matched:
I C I C1 I C 2
IE
2
VC VC1 VC 2 VCC I C R C
VCE VCEQ1 VCEQ 2 VCC I C R C 2R E VEE
AC Operation
R E 1 h fe i b1
v i1 h ie R E 1 h fe
h ie R E 1 h fe i b 2
v i 2 R E 1 h fe
R E 1 h fe v i1
i b1
h ie R E 1 h fe
1
i h 2R 1 h h R 1 h
h ie R E 1 h fe v i 2
b2
E
fe
ie
E
fe
ie
v o1 R C i C1 h fe R C i b1
v o 2 R C i C 2 h fe R C i b 2
Separate signals are applied to the inputs: Vi1 and Vi2
Separate signals are available at the outputs: Vo1 and Vo2
i b1
h ie R E 1 h fe v i1 R E 1 h fe v i 2
h ie 2R E 1 h fe h ie
i b2
h ie R E 1 h fe v i 2 R E 1 h fe v i1
h ie 2R E 1 h fe h ie
v o1 R C i C1 h fe R C i b1
v o 2 R C i C 2 h fe R C i b 2
Separate signals are applied to the inputs: Vi1 and Vi2
Separate signals are available at the outputs: Vo1 and Vo2
Single-Ended Mode AC Voltage Gain
In this mode a signal is connected to one input and the other is grounded, i.e. vi2 = 0.
Assuming the transistor circuits are perfectly matched:
Av
Vo1
hR
fe C
Vi1
2h ie
Common-Mode AC Gain
Common-mode applies the same signal to both inputs, vi1 = vi2 = vi. Because the amplifier
amplifies the difference between the inputs. The common-mode gain should be quite
small.
Ac
Vo1
h fe R C
h ie 2(h fe 1)R E
Vi
where Ac = common mode gain
Differential-Mode (Double-ended) Voltage Gain
Double-ended mode connects two opposite polarity signals, i.e. vi2 = -vi1 to its inputs.
Here the differential input is vd = vi1-vi2 (the difference between the inputs)
Let us define
v i1
vd
2
and
vi2
vd
2
So the differential voltage gain, Ad is given by
Ad
v o1
v
hR
o 2 fe C
vd
vd
2h ie
Common-Mode Rejection = Noise Rejection
In common-mode, the signal common to both inputs will have a low gain (Ac).
In differential-mode (single- or double-ended), any signal that is common to both inputs
will have a low gain. Any signal, in differential-mode that is common to both inputs is
noise.
The ability of the amplifier to have a low common-mode gain, i.e. not amplify signals that
are common to both inputs, is called Common-Mode Rejection.
CMRR
Ad
Ac
h ie 21 h fe R E
2h Common-Mode ie
Improving Rejection
To improve common-mode rejection:
Ad must increase
Ac must decrease
One method is to increase the value of RE by replacing it with a constant-current source
circuit.
Linear Operation
v d v i1 v i 2
v avg
v i1 v i 2
2
vd
v i1 v avg
2
v i 2 v avg
vd
2
Thus
v o1 Ac v avg Ad v d
v o 2 Ac v avg Ad v d
v o v o1 v o 2
v o 2 Ad v d
Constant-Current Source Circuits
This increases the AC impedance for RE.
Constant-current sources can be built using FETs, BJTs and a
combination of these devices.
BJT Constant Current Source
IE IC
Transistor/Zener Constant Current Source
Replacing resistor R2 with a Zener improves the constant current source.
I IE
VZ VBE
RE
Current Mirror Sources
Current Mirror circuits are used to provide constant current in integrated circuits.
Differential Amplifier with a Current Source
Diode equation
I D IS e
VD / VT
I E1 I ES e
I E 2 I ES e
I 0 I E1 I E 2
1 I S e VD / VT
VBE1
VBE 2 VBE 1
I0
1 e
I E1
v i 2 v i1
kT
q
VT 26mV
VBE 2
I E1
I
I 0 I E1 1 E 2 I 0 I E 2 1
I E2
I E1
I0
1 e
I E1
where VT
I0
1 e
I E2
v i1 v i 2
I E1
I0
I E2
I0
1
1 e
v i 2 v i1
VT
1
1 e
v i1 v i 2
VT
If
Vi1 - Vi2 30mV
We are in linear region
and
v o1 Ac v avg Ad v d
v o 2 Ac v avg Ad v d
holds
Example
1.
For the circuit below find vo for
vi1 = 0V and vi2 = 58.5mV
VCC = 15V
RC = 1k
Io = 10mA
Differential Amplifier Parameters
Input offset voltage : VIO
Input voltage difference (VB1 - VB2)
when vo = 0V.
Due to the VBE difference of the two BJTs
Input offset current : IIO
Input current difference (IB1 - IB2)
when vo = 0V.
Due to the hfe difference of the two BJTs
FET Differential Amplifier
Separate signals are applied to the inputs: Vi1 and Vi2
Separate signals are available at the outputs: Vo1 and Vo2
R S g m v gs1
v i1 1 R S g m
v R g
1 R S g m v gs 2
i2 S m
v gs1
1
v
2
22
gs 2 1 R S g m R S g m
1 R S g m R S g m v i1
R g
1 R Sgm vi 2
Sm
Ac
Vo1
R D gm
Vi
1 2 R S gm
Ad
Vo1
Rg
D m
Vd
2
Ac
Vo1
R D gm
Vi
1 2 R S gm
vo1 Ac vavg Ad v d
v o 2 Ac v avg Ad v d
Ad
Vo1
Rg
D m
Vd
2
v o v o1 v o 2
vo 2 Ad v d
Uses of differential amplifiers
1.
Gain amplifiers in operational amplifiers
2.
Due to high voltage gain
Comparators
e.g. circuit below
Rth: thermistor
(resistance varies
with temperature)
for V1 = V2, Vo = 0
Examples
1.
For the circuit below
a)
b)
Calculate the value of RC for vo = 0V when vi = 0V.
Find vo when vi = 1mV
h = h = 100 , = 1, V
fe
FE
BE
= 0.6V
2.
For the circuit below
a)
b)
Calculate the value of R for vo = 0V when vi = 0V.
Sketch vo when vi = 20mV sin(t)
h = h = 20 , = 1, V
fe
FE
BE
= 0.6V, VT = 25mV
3.
For the circuit below calculate the value of R2/R1 for vo = 0V when vi = 0V.
hfe = hFE = 100 , = 1, VBE = 0.6V