The document discusses operational amplifiers and differential amplifiers. It provides details on: 1. Differential amplifiers use two identical input stages to amplify the difference between two input voltages while rejecting signals common to both inputs, reducing noise and drift. 2. Differential amplifiers have very high voltage gain, above 1000, and can amplify very small signals in the microvolt range. They have high input and output impedances. 3. The voltage gain of a differential amplifier is given by Vo = A(V1 - V2) where A is the differential gain, V1 is the non-inverting input, and V2 is the inverting input.

1. Department of Physics
B.SC.III
Sem-V.Paper-XII
DSE-E4- Digital and Analog Circuits and Instrumentation
Topic- Operational Amplifire

2. • Introduction
• The direct coupled amplifier using two
balanced stages of amplifiers to amplify the
difference between two-voltage levels and reject
the voltage common to both is called 'Differential
Amplifier’.
• Transistors, diodes, and resistors are the only
practical components in a monolithic IC.
Capacitors have been fabricated on a chip, but
these are usually less than 50 pF. Therefore, IC
designers cannot use coupling and bypass
capacitors as a discrete-circuit designer can.
Instead, the stages of a monolithic IC have to be
direct-coupled. One of the best direct-coupled
stages is the differential amplifier (diff amp). This
amplifier is widely used as the input stage of an
op- amp. In this section, we focus on the diff. amp
because it determines the input characteristics of
the typical op amp.
• The gain of a single stage amplifier is generally
insufficient for practical use. Hence, to obtain
the desired high gain, two or more amplifier
stages are coupled together.
• Two methods of coupling are RC coupling and
direct coupling. The frequency response of
these amplifiers is fairly wide. But the gain of
RC coupled amplifiers falls at lower as well as
higher frequencies. In direct coupled
amplifiers, the output changes with the age
and change in supply voltage. This is called
as drift.
• These drawbacks of cascaded amplifiers are
removed by using two identical stages of
direct coupled amplifiers to form a differential
amplifier. Since it amplifies the difference
between two input signals, the drift can be
minimized,

3. • Above circuit diagram shows basic differential
amplifier using two transistors with their
respective collector resistances RC1 & RC2 are
joined to form an bridge. If the two transistors
Q1 and Q2 are identical (i.e. their resistive and
transitive characteristics are equivalent or
identical ) then bridge is perfect balanced and
Vout = 0, if the bridge is unbalanced then we get
some Vout.
• In above to compensate current and the
voltage in collector of both transistor RC1 & RC2
are used & across the bridge output is taken is
Vout.
• Consider if we apply differential mode ed . As
ed is on ec (common mode input ) is off. Due to
ed input voltage applied to base of the both the
transistor Q1 and Q2 is 1800 out of phase, hence
the one transistor in differential pair Q1 and Q2 is
become ON at then same time other become
OFF. If the Q1 is ON and Q2 is OFF, then
collector voltage in Q1 is increased and collector
voltage in Q2 is decreased and bridge is
imbalanced and Vout is appeared with shown
polarity.
• If the Q2 is ON and Q1 is OFF, then collector voltage in
Q2 is increased and collector voltage in Q1 is decreased
and bridge is imbalanced and Vout is appeared with
opposite polarity.
• If we apply common mode i.e. ed is OFF and ec is On
the voltage or signal applied to the base of both the
transistor Q1 and Q2 is in same phase , due to that both
the transistor either go to cut-off or conduct state. Due to
this the collector current in the both transistor Q1 and Q2
is same in magnitude but opposite in direction hence
bridge become balanced.

4. • Differential amplifier is a direct coupled amplifier which uses two
balanced stages of amplifiers. Fig. 4.1(a) shows a diiff. Amplifier.
• Differential Amplifier- Double-ended input and output mode
• Transistors are identical Rci = Rc2 . RE is common emitter
resistance of high value. It provides high impedance at
collectors.RE acts as a constant current source. Because of direct
coupling, the input signals can be dc also. ac signals at very high
frequencies can be applied as input signal. Because there are no
coupling or bypass capacitors, there is no lower cut off
frequency.
• The output voltage is zero, when two input voltages are
equal. When V1 is greater than V2, an output voltage with the
polarity is shown in Fig. 4.1(b). When V2 is greater than VI, the
output voltage is inverted and has the opposite polarity.
• The diff. amplifier has two separate inputs. Input
V1 is called the non inverting input because V out
is in phase with V1, V2 is called inverting input
because V out is 1800 out of phase with V2.Output
voltage is given by Vo = A (VI — V2)
• where Vo - Voltage between collectors.
• A - gain =
𝑅𝐶
𝑟′𝑒
where r'eis ac emitter resistance.
• V1 - non-inverting input voltage
• V2 -inverting input voltage
• When both the non-inverting and inverting input
voltages are present, the total input is called a
differential input because the output voltage equals
the voltage gain times the difference of the two
input voltage. There are four types or modes of
operation of differential amplifier.

5. 1. Double-ended input and output.
2. Single-ended input and double-ended output.
3. Double-ended input and single-ended output.
4. Single-ended input and output.
1.Double-ended input and output mode : As
explained in section (4.2), in this mode of
operation, the input is double-ended i.e. inputs V1
and V2 are applied to the bases of two transistors
at the same time. The output is also double-
ended, i.e. Vo is taken across collectors of two
transistors.
• Vo = A (V1 - V2)
2.Single-ended input and double-ended output
mode :
• In this type, the input is single-ended i.e. only one
input is applied and other is grounded as shown in
Fig. 4.3. The output is double-ended i.e. between
two collectors. A double-ended output requires a
floating load. One end of the load is connected to
ground. Therefore, double-ended output is
inconvenient.
• 3.Double-ended input and single-ended output : In this mode of
operation, both inputs V1 and V, are applied simultaneously.
Therefore, input is double-ended. But the output is single-ended i.e.
Vo is taken between the collector of a transistor and ground. Fig. 4.4
shows double-ended input and single-ended output mode. The output
voltage Vo is given by
• Vo = A (V1 — V2)
• where A - gain =
𝑅𝐶
2𝑟′𝑒
where r' e is ac emitter resistance.
• V1 - non inverting input voltage andV2 -inverting input voltage
• The voltage gain A in this type is half of that obtained in double.-
ended input.This mode of operation is widely used, because it can
drive single-ended loads like CE amplifiers, emitter follower or
input stage of an op amp.

6. • . Single-ended input and output
mode : In this mode, input is applied
to the base of one transistor only. The
other input is grounded. Therefore,
input is single-ended. The output is
taken at a collector and ground.
Therefore, output is also single ended.
• Since V2 = 0 , Output voltage
• Vo = A . V1

7. Amplifier and Differential Amplifier
Normal Amplifier Differential Amplifier
1. Normal amplifier uses only one input voltage
2.Single +Vcc power supply is used.
3.It amplifies only one input
4.Output is taken at collector w.r.t. ground
5.Input is applied in base and ground.
6. Gain is limited upto 100.
7. Noise is also amplified.
8. Frequency response is limited in kHz range only.
9. It cannot be used to amplify very small signals in
uvolt or less.
10. It has lower i/p impedance and higher o/p
impedance.
1.Differential amplifier uses two input voltages.
2.Two power supplys are +VCC & -VEE are used.
3.It amplifies the difference between two inputs.
4.outputs is taken in two collectors or one collectors w.r.t.
ground.
5. Input is applied in between two
bases.
6. Very high gain above 1000 can
be easily achieved.
7. Noise signals are completely
eliminated.
8. Frequency response starts from
0 Hz to few μHz
9.Very small signals if μvolt or
less can also be amplified.
10. It has very high o/p impedance
and low i/p impedance.

8. • Fig.4.6. shows the differential amplifier with a non-
inverting inputs 𝑣1 𝑎𝑛𝑑 inverting inputs 𝑣2 one way to
derive the voltage gain is to apply the superposition
theorem. This means working out the voltage gain for each
input separately then combining the two results to get the
total gain.4.6. (a) Diff amp. (b) CE stage drives CB stage. (c) Ac
equivalent circuit. (d) Two ac emitter resistances are in series.
- Let us start by applying 𝑣1 𝑎𝑠
shown in 4.6. b) the circuit has been redrawn to emphasize
the following the input signal drives Q1 which acts like an
emitter follower.The output of the emmiter follower then
drives Q2 which is common- base amplifier.
• Because there is no phase inversion the final output is in
phase with 𝑣1 .This is why 𝑣1 is called the noninverting
input. Fig. 4.6 c) shows the shows the ac equivalent circuit.
Notice that the upper 𝑟
𝑒
′
is the ac emitter resistance of
Q1, and the lower 𝑟
𝑒
′
is the ac input resistance of the
CBamplifier. In any practical circuit,
𝑅
𝐸 is much greater
than𝑟
𝑒
′
, so that the circuitsimplifies to Fig. 4.6.d. This
means that approximately half of the input voltagereaches
the input of the CB amplifier. Stated another way, the ac
emitter current is

9. -
• 𝑖𝑐 =
𝑣1
2𝑟𝑒
′
• Since the ac collector current approximately equalsie,
the ac output voltage is
• Vout =𝑖𝑐𝑅𝐶 =
𝑣1
2𝑟𝑒
′ 𝑅𝐶
•
V
𝑜𝑢𝑡
𝑣1
=
𝑅𝐶
2𝑟𝑒
′
-----------------------(1)
• This is the voltage gain for the noninverting input.
• 𝑖𝑐 =
𝑣1
2𝑟𝑒
′
• Since the ac collector current approximately equalsie, the
ac output voltage is
• Vout =𝑖𝑐𝑅𝐶 =
𝑣1
2𝑟𝑒
′ 𝑅𝐶
•
V
𝑜𝑢𝑡
𝑣1
=
𝑅𝐶
2𝑟𝑒
′
-----------------------(1)
• This is the voltage gain for the noninverting input.
Next, let us find the voltage gain for the inverting input.
This means that we canground vl and redraw the circuit as
shown in Fig. 4.7a. NowQ2drivesQ1whichhas an input
resistance of𝑟𝑒
′
. Figure 4.7.b is the ac equivalent circuit.
Again,REis always much greater than 𝑟𝑒
′
so the circuit
simplifies to Fig. 4.7c. The ac emitter current is-
Equivalents circuits for the inverting inputs.

10. • 𝑖𝑐 =
𝑣2
2𝑟𝑒
′
Since the ac collector current approximately equalsie, the ac
output voltage is
Vout =−𝑖𝑐𝑅𝐶 = −
𝑣2
2𝑟𝑒
′ 𝑅𝐶
V
𝑜𝑢𝑡
𝑣2
= −
𝑅𝐶
2𝑟𝑒
′-------(2)
This is the voltage gain for the inverting input. The minus
sign indicates phaseinversion.
c) Differential gain
• Compare Eqs. (1) and (2). As you can see, the
magnitude of the voltage gainis the same; only the
phase differs. Here is how to find the total voltage gain
with bothinputs active at the same time. Using Eq we
can write
• V𝑜𝑢𝑡(1) =
𝑅𝐶
2𝑟𝑒
′ 𝑣1
• And V𝑜𝑢𝑡(2) = −
𝑅𝐶
2𝑟𝑒
′ 𝑣2
The superposition theorem says that we can add these
individual outputs to get the
total output with both signals present. After factoring
and rearranging, we get
V𝑜𝑢𝑡
= V𝑜𝑢𝑡(1) + V𝑜𝑢𝑡(2) =
𝑅𝐶
2𝑟𝑒
′
𝑣1 + (−
𝑅𝐶
2𝑟𝑒
′
𝑣2)
And V𝑜𝑢𝑡 =
𝑅𝐶
2𝑟𝑒
′ (𝑣1 − 𝑣2)
This can be written as
Vout = A(v1 — v2)--------------(3)
Where A=
𝑅𝐶
2𝑟𝑒
′, The quantityA is called
thedifferential voltage gain because it tells us
how much thedifference (vl—v2 )is
amplified.

11. Any signal that drives both the inputs of a differential amplifier
equally, is called the common-mode signal. Interference, static
and other kinds of undesirable pick ups are common- mode signals.
Fig. 4.8 (a) shows a common-mode signal Vin (cm) which drives both
the inputs of a common mode signal. Differential amplifier does not
amplify common mode signal. If transistors are identical, equal
emitter currents are obtained. Therefore, we can split RE as shown in
Fig. 4.8(b)
• Since equal voltages Vin (cm) drive both inputs simultaneously,
there is almost no current through the wire between the emitters.
Fig. 4.8(c) shows ac equivalent circuit. When a common-mode signal
drives a differential amplifier, a large unbypassed emitter resistance
appears in the ac equivalent circuit.
• A common-mode signal is one that drives both inputs of a diff
amp equally. Most interference, static, and other kinds of
undesirable pickup are common-mode. What happens is this. The
connecting wires on the input bases act like small antennas. If the
cliff amp is operating in an environment with a lot of
electromagnetic interference. each base picks up an unwanted
interference voltage.
• One of the reasons the dill amp is so popular is
because it discriminates against common-mode
signals. In other words, a diff amp refuses to
amplify common-mode signals. Because of this. you
don't get a lot of unwanted interference at the output.
• Let us now find out why the dill amp does not
amplify common-mode signals. Figure 4.8a shows a
common-mode signal driving a diff amp. As you can
see, an equal voltage vin(cm) drives both inputs
simultaneously. Assuming that the transistors are
identical, the equal inputs imply equal emitter
currents. Therefore, we can split RE as shown in Fig.
4.8b. This equivalent circuit has exactly the same
emitter currents as the original circuit.
• Figure 4.8c. shows the ac equivalent circuit. Can
you see what this means? When a common-mode
signal drives a cliff amp, a large unbypassed emitter
resistance appears in the ac equivalent circuit.
Therefore. the voltage gain for a common-

12. •
𝑉𝑜𝑢𝑡
𝑉𝑜𝑢𝑡 𝐶𝑀
=
−𝑅𝐶
𝑟𝑒
′+ 2𝑅𝐸
• Since RE is already much greater than 𝑟𝑒
′
𝑤𝑒 can
approximatethe common mode voltage gain as
−𝑅𝐶
2𝑅𝐸
in
symbols
• A𝐶𝑀 =
−𝑅𝐶
2𝑅𝐸
------------------(4)
Where A𝐶𝑀 = 𝐶𝑜𝑚𝑚𝑜𝑛 𝑚𝑜𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛
RC= collector resistance
RE= emitter resistance, for instance if RC=10 KΩ and
RE= 10 KΩ then,
• A𝐶𝑀 =
−𝑅𝐶
2𝑅𝐸
=
−10KΩ
20KΩ
= −0.5
• Fig. 4.8.a)Common mode signal drives diff. amplifier,
(b)Circuit splitting the emitter resister (c) AC equivalent
circuit.

13. .Common-mode rejection ratio(CMRR)
• Data sheets list thecommon-mode rejection ratio (CMRR).
It is defined as the ratio ofdifferential voltage gain to
common-mode voltage gain. In symbols,
• 𝐶𝑀𝑅𝑅 =
𝐴
−𝐴𝐶𝑀
• where the minus sign is included to get a positive ratio.
For instance, if A= 200 and ACM = —0.5, then
• 𝐶𝑀𝑅𝑅 =
200
−0.5
= 400
• Data sheets almost always specify CMRR in
decibels, using the following formulafor the
conversion:
• CMRR=20 log CMRR, If CMRR
= 400, then
• CMRR = 20 log 400 = 52 dB
• 4.5 OP-AMP Symbol
4.5.1 : Introduction : An operational amplifier is
basically a high gain direct coupled amplifier. It has
two inputs and one output, but generally it is
operated in single ended input-singal ended output
mode. The differential amplifier forms the first stage of
an operational amplifier.
Fig. 4.8 : Schematic symbol of an operational amplifier