ANALOG CIRCUITS (20A04302T) UNIT-I

UNIT – I
Multistage and Differential Amplifiers
Introduction – Recap of Small Signal Amplifiers, Multistage Amplifiers, Cascode amplifier, Darlington pair,
the MOS Differential Pair, Small-Signal Operation of the MOS Differential Pair, The BJT Differential Pair,
and other Non-ideal Characteristics of the Differential Amplifier.
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INTRODUCTION
An Amplifier circuit is one which strengthens the signal. Amplifiers are classified according to many
considerations.

Based on number of stages

Depending upon the number of stages of Amplification, there are Single-stage amplifiers and Multi-stage
amplifiers.
 Single-stage Amplifiers − This has only one transistor circuit, which is a single stage amplification.
 Multi-stage Amplifiers − This has multiple transistor circuit, which provides multi-stage
amplification.
Based on its output
Depending upon the parameter that is amplified at the output, there are voltage and power amplifiers.
 Voltage Amplifiers − The amplifier circuit that increases the voltage level of the input signal, is
called as Voltage amplifier.
 Power Amplifiers − The amplifier circuit that increases the power level of the input signal, is called
as Power amplifier.
Based on the input signals
Depending upon the magnitude of the input signal applied, they can be categorized as Small signal and
large signal amplifiers.
 Small signal Amplifiers − When the input signal is so weak so as to produce small fluctuations in
the collector current compared to its quiescent value, the amplifier is known as Small signal
amplifier.
 Large signal amplifiers − When the fluctuations in collector current are large i.e. beyond the linear
portion of the characteristics, the amplifier is known as large signal amplifier.
Based on the frequency range
Depending upon the frequency range of the signals being used, there are audio and radio amplifiers.
 Audio Amplifiers − The amplifier circuit that amplifies the signals that lie in the audio frequency
range i.e. from 20Hz to 20 KHz frequency range, is called as audio amplifier.
 Power Amplifiers − The amplifier circuit that amplifies the signals that lie in a very high frequency
range, is called as Power amplifier.

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Based on Biasing Conditions
Depending upon their mode of operation, there are class A, class B and class C amplifiers.
 Class A amplifier − The biasing conditions in class A power amplifier are such that the collector
current flows for the entire AC signal applied.
 Class B amplifier − The biasing conditions in class B power amplifier are such that the collector
current flows for half-cycle of input AC signal applied.
 Class C amplifier − The biasing conditions in class C power amplifier are such that the collector
current flows for less than half cycle of input AC signal applied.
 Class AB amplifier − The class AB power amplifier is one which is created by combining both
class A and class B in order to have all the advantages of both the classes and to minimize the
problems they have.
Based on the Coupling method
Depending upon the method of coupling one stage to the other, there are RC coupled, Transformer coupled
and direct coupled amplifier.
 RC Coupled amplifier − A Multi-stage amplifier circuit that is coupled to the next stage using
resistor and capacitor (RC) combination can be called as a RC coupled amplifier.
 Transformer Coupled amplifier − A Multi-stage amplifier circuit that is coupled to the next stage,
with the help of a transformer, can be called as a Transformer coupled amplifier.
 Direct Coupled amplifier − A Multi-stage amplifier circuit that is coupled to the next stage
directly, can be called as a direct coupled amplifier.
Based on the Transistor Configuration
Depending upon the type of transistor configuration, there are CE CB and CC amplifiers.
 CE amplifier − The amplifier circuit that is formed using a CE configured transistor combination is
called as CE amplifier.
 CB amplifier − The amplifier circuit that is formed using a CB configured transistor combination is
called as CB amplifier.
 CC amplifier − The amplifier circuit that is formed using a CC configured transistor combination is
called as CC amplifier.
1.1 SMALL SIGNAL AMPLIFIERS
Small Signal Amplifiers
 Small Signal Amplifiers are also known as Voltage Amplifiers.
 Voltage Amplifiers have 3main properties, Input Resistance, Output Resistance and Gain.
 The Gain of a small signal amplifier is the amount by which the amplifiers “Amplifies” the input
signal.

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 Gain is a ratio of output divided by input;therefore, it has no units but is given the symbol (A) with
the most common types of transistor gain being, Voltage Gain (Av), Current Gain (Ai) and Power
Gain (Ap)
Benefits of h-parameters:
1. Real numbers at audio frequencies
2. Easy to measure
3. Can be obtained from the transistor static characteristic curve
4. Convenient to use in circuit analysis and design
5. Most of the transistor manufacturers specify the h-parameters

The relationship between voltages and current in h parameters can be represented as:
V1=h11I1+h12V2------1
I2=h21I1+h22V2-----------2
Equation (i) can be represented as a circuit based on Kirchhoff Voltage Law:

Equation (ii) can be represented as a circuit based on Kirchhoff Current Law:

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Combining these two parts of the network we get:

This can be represented in matrix form as:

Calculation of h Parameter:
Let us consider a two port network. Let V1, I1, V2 and I2 are the input voltage, input current, output voltage
and output current respectively.

Let us short circuit the output port of a two port network as shown below,

Now, ratio of input voltage to input current, at short circuited output port is:

This is referred to as the short circuit input impedance. Now, the ratio of the output current to input current
at the short-circuited output port is:

This is called short-circuit current gain of the network. Now, let us open circuit the port 1. At that condition,
there will be no input current (I1=0) but open circuit voltage V1 appears across the port 1, as shown below:

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Now:

This is referred as reverse voltage gain because, this is the ratio of input voltage to the output voltage of the
network, but voltage gain is defined as the ratio of output voltage to the input voltage of a network.
Now:

It is referred as open circuit output admittance.

Analysis of a transistor amplifier using h-parameters
To form a transistor amplifier it is only necessary to connect an external load and signal source as indicated
in fig. and to bias the transistor properly.

 Consider the two-port network of CE amplifier. RS is the source resistance and ZL is the load
impedence h-parameters are assumed to be constant over the operating range.
 The ac equivalent circuit is shown in fig1.2 (Phasor notations are used assuming sinusoidal voltage
input).
 The quantities of interest are the current gain, input impedence, voltage gain, and output impedence.

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Let us analyze the hybrid model to find current gain, input resistance, voltage gain and output resistance.

Current gain (Ai):

It is defined as the ratio of output to input current. It is given by,

Here IL and I2 are equal in magnitude but opposite in sign. IL = -I2

From above circuit

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Current gain (AIS):
It is given by,

From above figure, using current divider rule,

Input Impedance (Zi):

Ri is the input resistance looking into the amplifier input terminals ( 1, 1’). It is given by,

From the input circuit,

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Substituting V2 = -I2RL = Ai I1 RL in the above equation,

Dividing numerator and denominator by RL we get,

From this equation, note that the input impedance is a function offload impedance.
Voltage gain (Av):
It is the ratio of output voltage to input voltage. It is given by,

By substituting V2 = -I2RL = Ai I1 RL

Voltage gain (Avs):

It is voltage gain including the source. It is given by,

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From above figure, applying potential divider rule, then we get,

Substituting the value of V1/Vs in the equation of

Output Admittance (Yo):

It is the ratio of output current to output voltage. It is given by,

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Dividing above equation by V2, We get,

From transistor amplifier in h-parameter model circuit, with Vs = 0,

RsI1 + hiI1 +hrV2 = 0
(Rs + hi) I 1 = -hr V2

Substituting the value of I1/V2 from above equation in the equation of Yo. We obtain,

From this equation, note that the output admittance is a function of source resistance.
Power gain (Ap):
It is the ratio of average power delivered to the load to the input power. Output power is given as,

Since the input power is P1 = V1I1

The operating power gain Ap of the transistor is given as,

Relation between Avs and AIS

From equation,

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Taking ratio of above two equations we get,

1.2 MULTISTAGE AMPLIFIERS

In practical applications, the output of a single state amplifier is usually insufficient, though it is a voltage or
power amplifier. Hence, they are replaced by multi-stage transistor amplifiers.In Multi-stage amplifiers,
the output of first stage is coupled to the input of next stage using a coupling device. These coupling devices
can usually be a capacitor or a transformer. This process of joining two amplifier stages using a coupling
device can be called as Cascading.
The following figure shows a two-stage amplifier connected in cascade.

The overall gain is the product of voltage gain of individual stages.

Where AV = Overall gain, AV1 = Voltage gain of 1st stage, and AV2 = Voltage gain of 2nd stage.
If there are n number of stages, the product of voltage gains of those n stages will be the overall gain of that
multistage amplifier circuit.

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Purpose of coupling device
The basic purposes of a coupling device are
 To transfer the AC from the output of one stage to the input of next stage.
 To block the DC to pass from the output of one stage to the input of next stage, which means to
isolate the DC conditions.
Types of Coupling
Joining one amplifier stage with the other in cascade, using coupling devices form a Multi-stage amplifier
circuit. There are four basic methods of coupling, using these coupling devices such as resistors, capacitors,
transformers etc. Let us have an idea about them.
Resistance-Capacitance Coupling
This is the mostly used method of coupling, formed using simple resistor-capacitor combination. The
capacitor which allows AC and blocks DC is the main coupling element used here.
The coupling capacitor passes the AC from the output of one stage to the input of its next stage. While
blocking the DC components from DC bias voltages to effect the next stage. Let us get into the details of
this method of coupling in the coming chapters.
Impedance Coupling
The coupling network that uses inductance and capacitance as coupling elements can be called as
Impedance coupling network.
In this impedance coupling method, the impedance of coupling coil depends on its inductance and signal
frequency which is jwL. This method is not so popular and is seldom employed.
Transformer Coupling
The coupling method that uses a transformer as the coupling device can be called as Transformer
coupling. There is no capacitor used in this method of coupling because the transformer itself conveys the
AC component directly to the base of second stage.
The secondary winding of the transformer provides a base return path and hence there is no need of base
resistance. This coupling is popular for its efficiency and its impedance matching and hence it is mostly
used.
Direct Coupling
If the previous amplifier stage is connected to the next amplifier stage directly, it is called as direct
coupling. The individual amplifier stage bias conditions are so designed that the stages can be directly
connected without DC isolation.
The direct coupling method is mostly used when the load is connected in series, with the output terminal of
the active circuit element. For example, head-phones, loud speakers etc.
Role of Capacitors in Amplifiers
Other than the coupling purpose, there are other purposes for which few capacitors are especially employed
in amplifiers. To understand this, let us know about the role of capacitors in Amplifiers.

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The Input Capacitor Cin
The input capacitor Cin present at the initial stage of the amplifier, couples AC signal to the base of the
transistor. This capacitor Cin if not present, the signal source will be in parallel to resistor R 2 and the bias
voltage of the transistor base will be changed.
Hence Cin allows, the AC signal from source to flow into input circuit, without affecting the bias conditions.
The Emitter By-pass Capacitor Ce
The emitter by-pass capacitor Ce is connected in parallel to the emitter resistor. It offers a low reactance path
to the amplified AC signal.
In the absence of this capacitor, the voltage developed across RE will feedback to the input side thereby
reducing the output voltage. Thus in the presence of C e the amplified AC will pass through this.
Coupling Capacitor CC
The capacitor CC is the coupling capacitor that connects two stages and prevents DC interference between
the stages and controls the operating point from shifting. This is also called as blocking capacitor because it
does not allow the DC voltage to pass through it.
In the absence of this capacitor, RC will come in parallel with the resistance R1 of the biasing network of the
next stage and thereby changing the biasing conditions of the next stage.
Amplifier Consideration
For an amplifier circuit, the overall gain of the amplifier is an important consideration. To achieve maximum
voltage gain, let us find the most suitable transistor configuration for cascading.
CC Amplifier
 Its voltage gain is less than unity.
 It is not suitable for intermediate stages.
CB Amplifier
 Its voltage gain is less than unity.
 Hence not suitable for cascading.
CE Amplifier
 Its voltage gain is greater than unity.
 Voltage gain is further increased by cascading.
The characteristics of CE amplifier are such that, this configuration is very suitable for cascading in
amplifier circuits. Hence most of the amplifier circuits use CE configuration.
In the subsequent chapters of this tutorial, we will explain the types of coupling amplifiers.
1.2.1 Two-stage RC Coupled Amplifier
The constructional details of a two-stage RC coupled transistor amplifier circuit are as follows. The two
stage amplifier circuit has two transistors, connected in CE configuration and a common power
supply VCC is used. The potential divider network R1 and R2 and the resistor Re form the biasing and
stabilization network. The emitter by-pass capacitor Ce offers a low reactance path to the signal.

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The resistor RL is used as a load impedance. The input capacitor Cin present at the initial stage of the
amplifier couples AC signal to the base of the transistor. The capacitor C C is the coupling capacitor that
connects two stages and prevents DC interference between the stages and controls the shift of operating
point. The figure below shows the circuit diagram of RC coupled amplifier.

Operation of RC Coupled Amplifier

When an AC input signal is applied to the base of first transistor, it gets amplified and appears at the
collector load RL which is then passed through the coupling capacitor CC to the next stage. This becomes the
input of the next stage, whose amplified output again appears across its collector load. Thus the signal is
amplified in stage by stage action.
The important point that has to be noted here is that the total gain is less than the product of the gains of
individual stages. This is because when a second stage is made to follow the first stage, the effective load
resistance of the first stage is reduced due to the shunting effect of the input resistance of the second stage.
Hence, in a multistage amplifier, only the gain of the last stage remains unchanged.
As we consider a two stage amplifier here, the output phase is same as input. Because the phase reversal is
done two times by the two stage CE configured amplifier circuit.
Frequency Response of RC Coupled Amplifier
Frequency response curve is a graph that indicates the relationship between voltage gain and function of
frequency. The frequency response of a RC coupled amplifier is as shown in the following graph.

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From the above graph, it is understood that the frequency rolls off or decreases for the frequencies below
50Hz and for the frequencies above 20 KHz. whereas the voltage gain for the range of frequencies between
50Hz and 20 KHz is constant.
We know that,

It means that the capacitive reactance is inversely proportional to the frequency.

At Low frequencies (i.e. below 50 Hz)
The capacitive reactance is inversely proportional to the frequency. At low frequencies, the reactance is
quite high. The reactance of input capacitor Cin and the coupling capacitor CC are so high that only small
part of the input signal is allowed. The reactance of the emitter by pass capacitor C E is also very high during
low frequencies. Hence it cannot shunt the emitter resistance effectively. With all these factors, the voltage
gain rolls off at low frequencies.
At High frequencies (i.e. above 20 KHz)
Again considering the same point, we know that the capacitive reactance is low at high frequencies. So, a
capacitor behaves as a short circuit, at high frequencies. As a result of this, the loading effect of the next
stage increases, which reduces the voltage gain. Along with this, as the capacitance of emitter diode
decreases, it increases the base current of the transistor due to which the current gain (β) reduces. Hence the
voltage gain rolls off at high frequencies.
At Mid-frequencies (i.e. 50 Hz to 20 KHz)
The voltage gain of the capacitors is maintained constant in this range of frequencies, as shown in figure. If
the frequency increases, the reactance of the capacitor CC decreases which tends to increase the gain. But
this lower capacitance reactive increases the loading effect of the next stage by which there is a reduction in
gain.
Due to these two factors, the gain is maintained constant.
Advantages of RC Coupled Amplifier
The following are the advantages of RC coupled amplifier.
 The frequency response of RC amplifier provides constant gain over a wide frequency range, hence
most suitable for audio applications.
 The circuit is simple and has lower cost because it employs resistors and capacitors which are cheap.
 It becomes more compact with the upgrading technology.
Disadvantages of RC Coupled Amplifier
The following are the disadvantages of RC coupled amplifier.
 The voltage and power gain are low because of the effective load resistance.
 They become noisy with age.

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 Due to poor impedance matching, power transfer will be low.
Applications of RC Coupled Amplifier
The following are the applications of RC coupled amplifier.
 They have excellent audio fidelity over a wide range of frequency.
 Widely used as Voltage amplifiers
 Due to poor impedance matching, RC coupling is rarely used in the final stages.
1.2.2 Transformer Coupled Amplifier
The amplifier circuit in which, the previous stage is connected to the next stage using a coupling
transformer, is called as Transformer coupled amplifier.
The coupling transformer T1 is used to feed the output of 1st stage to the input of 2nd stage. The collector
load is replaced by the primary winding of the transformer. The secondary winding is connected between the
potential divider and the base of 2nd stage, which provides the input to the 2nd stage. Instead of coupling
capacitor like in RC coupled amplifier, a transformer is used for coupling any two stages, in the transformer
coupled amplifier circuit.
The figure below shows the circuit diagram of transformer coupled amplifier.

The potential divider network R1 and R2 and the resistor Re together form the biasing and stabilization
network. The emitter by-pass capacitor Ce offers a low reactance path to the signal. The resistor RL is used
as a load impedance. The input capacitor Cin present at the initial stage of the amplifier couples AC signal to
the base of the transistor. The capacitor CC is the coupling capacitor that connects two stages and prevents
DC interference between the stages and controls the shift of operating point.
Operation of Transformer Coupled Amplifier
When an AC signal is applied to the input of the base of the first transistor then it gets amplified by the
transistor and appears at the collector to which the primary of the transformer is connected. The transformer
which is used as a coupling device in this circuit has the property of impedance changing, which means the
low resistance of a stage (or load) can be reflected as a high load resistance to the previous stage. Hence the

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voltage at the primary is transferred according to the turns ratio of the secondary winding of the transformer.
This transformer coupling provides good impedance matching between the stages of amplifier. The
transformer coupled amplifier is generally used for power amplification.
Frequency Response of Transformer Coupled Amplifier
The figure below shows the frequency response of a transformer coupled amplifier. The gain of the
amplifier is constant only for a small range of frequencies. The output voltage is equal to the collector
current multiplied by the reactance of primary.

At low frequencies, the reactance of primary begins to fall, resulting in decreased gain. At high frequencies,
the capacitance between turns of windings acts as a bypass condenser to reduce the output voltage and hence
gain.So, the amplification of audio signals will not be proportionate and some distortion will also get
introduced, which is called as Frequency distortion.
Advantages of Transformer Coupled Amplifier
The following are the advantages of a transformer coupled amplifier −
 An excellent impedance matching is provided.
 Gain achieved is higher.
 There will be no power loss in collector and base resistors.
 Efficient in operation.
Disadvantages of Transformer Coupled Amplifier
The following are the disadvantages of a transformer coupled amplifier −
 Though the gain is high, it varies considerably with frequency. Hence a poor frequency response.
 Frequency distortion is higher.
 Transformers tend to produce hum noise.
 Transformers are bulky and costly.
Applications
The following are the applications of a transformer coupled amplifier −
 Mostly used for impedance matching purposes.

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 Used for Power amplification.
 Used in applications where maximum power transfer is needed.
1.2.3 Direct Coupled Amplifier
As no coupling devices are used, the coupling of the amplifier stages is done directly and hence called
as Direct coupled amplifier.
Construction
The figure below indicates the three stage direct coupled transistor amplifier. The output of first stage
transistor T1 is connected to the input of second stage transistor T2.

The transistor in the first stage will be an NPN transistor, while the transistor in the next stage will be a PNP
transistor and so on. This is because, the variations in one transistor tend to cancel the variations in the other.
The rise in the collector current and the variation in β of one transistor gets cancelled by the decrease in the
other.
Operation
The input signal when applied at the base of transistor T 1, it gets amplified due to the transistor action and
the amplified output appears at the collector resistor Rc of transistor T1. This output is applied to the base of
transistor T2 which further amplifies the signal. In this way, a signal is amplified in a direct coupled
amplifier circuit.
Advantages
The advantages of direct coupled amplifier are as follows.
 The circuit arrangement is simple because of minimum use of resistors.
 The circuit is of low cost because of the absence of expensive coupling devices.

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Disadvantages
The disadvantages of direct coupled amplifier are as follows.
 It cannot be used for amplifying high frequencies.
 The operating point is shifted due to temperature variations.
Applications
The applications of direct coupled amplifier are as follows.
 Low frequency amplifications.
 Low current amplifications.

1.3 CASCODE AMPLIFIER

The cascode amplifier has:

1. High output resistance
2. Low input resistance
3. Moderate to high voltage and current gains.
The cascode amplifier is the two stage amplifier in which common emitter stage is connected to common
base stage.
The CE-CB cascode connection is as shown in the figure:

The input signal is applied at Q1 i.e at common emitter stage and output is obtained at Q 2. Vcc, R1, R2, R3, Re
are used to bias transistor Q1 and Q2 in active region. RE is used to make Q-point stable against temperature
variation.
AC output voltage is obtained at RC collector coupling capacitor are used to block dc signal pass a signals.
AC signal is applied at base of Q1 which amplifies it with unity gain, and voltage V01 appears across
collector of Q1. V01 acts as input to Q2 which further amplifies the signal and voltage Vo appears across
collector of CB Configuration.
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To perform small signal analysis, we need to draw ac equivalent circuit of the given amplifier. To draw the
ac equivalent circuit all capacitors must be replaced by short circuit and the DC sources connected to
ground.
The AC equivalent circuit is as shown in the figure:

The overall voltage gain is the product of first stage gain to second stage gain.
AVT = AV2 * AV1
AVT= (Vo/V1) * (V1/Vin)
From the above figure we can see that Vo = Io*Rc
The output current is Io = hfb ie2
hence substituting in equation we get Vo = hfb ie2 * Rc
Now to determine Vo1 we need to apply KVL to the input side of common base connection
and we get Vo1 = ie2 * hib
Vo/V1 =(hfb ie2 * Rc)/( ie2 * hib )
Av2 =(hfb * Rc)/(hib)
Now determine the gain for first stage,
Av1 = Vo1/Vin
from the above figure Vo1 = ie2 * hib
Since ie2 and hfe ib1 are opposite in direction,
we can write ie2 = -hfe *ib2
Substituting in above equation
Vo1 = -hfeib2 * hib
To determine Vin we need to apply KVL at the input side.
Vin – hie ib1=0
Vin = hie ib1
AV1= Vo1/Vin = (−hfe∗ib1)/(hib*hie∗ib1)
AV1= −(hfe∗hib)/hie
But WE know
hib=hie/(1+hfe)
AV1= (−hfe/hie) ×( hie/1+hfe)
AV1= −hfe/(1+hfe)

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AV1 ≈ -1
Multiply AV1 and AV2 to obtain AVT
AVT= AV1×AV2
AVT= (−hfb∗Rc)/(hib).
The negative sign indicates the 180 degree’s phase shift provided by CE stage.
Input impedance: The input impedance is parallel combination of resistors at input side
Rin = R2 // R3 // hie
Output Impedance: The output impedance is given by the output resistance of the second stage
Ro = Rc
In broadband amplifiers cascode configuration is used.
1.4 DARLINGTON PAIR AMPLIFIER

In order to achieve some increase in the overall values of current gain and input impedance, two transistors
are connected as shown in the following diagram, which is known as Darlington configuration.

The emitter of the first transistor is connected to the base of the second transistor. The collector terminals of
both the transistors are connected together.
Analysis:

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The overall current gain of amplifier is
𝐼𝐸2 𝐼𝐸2 𝐼𝐸1
𝐴𝐼 = =
𝐼𝐵1 𝐼𝐵2 𝐼𝐵1

𝐴𝐼 = 𝐴𝐼2 𝐴𝐼1
𝐼𝐵1 + 𝐼𝐶1 𝐼𝐵1 + 𝛽𝐼𝐵1 𝐼𝐵1 (1 + 𝛽)
𝐴𝐼1 = = = =1+𝛽
𝐼𝐵1 𝐼𝐵1 𝐼𝐵1
𝐼𝐵2 + 𝐼𝐶2 𝐼𝐵2 + 𝛽𝐼𝐵2 𝐼𝐵2 (1 + 𝛽)
𝐴𝐼2 = = = =1+𝛽
𝐼𝐵2 𝐼𝐵2 𝐼𝐵2
Overall current gain is
𝐴𝐼 = 𝐴𝐼2 𝐴𝐼1 = (1 + 𝛽 )(1 + 𝛽 ) = (1 + 𝛽)2
Overall voltage gain is
𝐴𝑉 = 𝐴𝑉2 𝐴𝑉1 ≅ 1
Input resistance is
𝑉𝑂 𝐼𝑒2 𝑅𝐿 𝑅𝐿
𝐴𝑉 = = = 𝐴𝐼
𝑉𝑆 𝐼𝑏1 𝑅𝑖𝑛 𝑅𝑖𝑛

𝐴𝐼 𝑅𝐿
𝑅𝑖𝑛 = = (1 + 𝛽)2 𝑅𝐿
𝐴𝑉

1.5 MOS DIFFERENTIAL AMPLIFIER PAIR

MOS differential amplifier consists of two matched transistors Q1 and Q2 whose sources are joined together
and biased by a constant current source I.

Figure : The basic MOS differential-pair configuration.

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ANALOG CIRCUITS (20A04302T) UNIT-I
 Two input signals have same amplitude and phase difference is 180 0.
 The input signal is a combination of common mode signal and differential mode signal. The input
signal can be expressed as
𝑉𝐺1 − 𝑉𝐺2 𝑉𝐺1 + 𝑉𝐺2
𝑉𝐺1 = +
2 2
𝑉𝐺2 − 𝑉𝐺1 𝑉𝐺1 + 𝑉𝐺2
𝑉𝐺2 = +
2 2
Operation of Common Mode input voltage

Figure : The MOS differential pair with a common-mode input voltage vCM.

 Consider case when two gate terminals are joined together.

 Connected to a common-mode voltage (VCM).
 vG1 = vG2 = VCM
 Q1 and Q2 are matched.
 Current I will divide equally between the two transistors.
 ID1 = ID2 = I/2, VS = VCM – VGS
 where VGS is the gate-to-source voltage.
The voltage at each drain terminal will be
𝐼
𝑉𝑑1 = 𝑉𝐷𝐷 − 𝑖𝑑1 𝑅𝐷 = 𝑉𝐷𝐷 − 𝑅𝐷
2

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ANALOG CIRCUITS (20A04302T) UNIT-I
𝐼
𝑉𝑑2 = 𝑉𝐷𝐷 − 𝑖𝑑2 𝑅𝐷 = 𝑉𝐷𝐷 − 𝑅𝐷
2
The output voltage will be
𝑉𝑂 = 𝑉𝑑2 − 𝑉𝑑1
𝐼 𝐼
𝑉𝑂 = 𝑉𝐷𝐷 − 𝑅𝐷 − 𝑉𝐷𝐷 + 𝑅𝐷 = 0𝑉
2 2
Thus, the difference in voltage between the two drains will be zero.
In saturation region the drain current is given by
1 𝑊
𝐼𝑑 = 𝐾𝑛 (𝑉𝐺𝑆 − 𝑉𝑡 )2
2 𝐿
2𝐼𝑑 𝐼
𝑉𝐺𝑆 = √ 𝑊 + 𝑉𝑡 = √ 𝑊 + 𝑉𝑡 = 𝑉𝑂𝑉 + 𝑉𝑡
𝐾𝑛 𝐿 𝐾𝑛 𝐿

Where over driven voltage is

𝐼
𝑉𝑂𝑉 = √ 𝑊
𝐾𝑛 𝐿

Lets vary the value of the common mode voltage VCM, Q1 and Q2 remains in the saturation region, the
current I will divide equally between Q1 and Q2 and the voltages at the drains will not change. Thus the
differential pair does not respond to common mode input signals. An important specification of a differential
amplifier is its input common mode range. This is the range of V CM over which the differential pair operates
properly.
At saturation region the drain voltage is
𝑉𝑑 ≥ 𝑉𝐶𝑀 − 𝑉𝑡
But
𝐼
𝑉𝑑1 = 𝑉𝑑2 = 𝑉𝐷𝐷 − 𝑅𝐷
2
𝐼
𝑉𝐷𝐷 − 𝑅𝐷 ≥ 𝑉𝐶𝑀 − 𝑉𝑡
2
𝐼
𝑉𝐷𝐷 − 𝑅𝐷 + 𝑉𝑡 ≥ 𝑉𝐶𝑀
2
𝐼
𝑉𝐶𝑀(max) ≤ 𝑉𝐷𝐷 − 𝑅𝐷 + 𝑉𝑡
2
The lowest value of VCM is determined by the need to allow for a sufficient voltage across current source I
for it to operate properly.
𝑉𝐶𝑀(min) ≤ −𝑉𝑆𝑆 + 𝑉𝑥 + 𝑉𝐺𝑆
𝑉𝐶𝑀(min) ≤ −𝑉𝑆𝑆 + 𝑉𝑥 + 𝑉𝑂𝑉 + 𝑉𝑡
𝐼
−𝑉𝑆𝑆 + 𝑉𝑥 + 𝑉𝑂𝑉 + 𝑉𝑡 ≤ 𝑉𝐶𝑀 ≤ 𝑉𝐷𝐷 − 𝑅𝐷 + 𝑉𝑡
2

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Operation with a Differential Input Voltage

Figure : The MOS differential pair with a differential input signal vid applied.

 If vid is applied to Q1 and Q2 is grounded, following conditions apply:

 vid = vGS1 – vGS2 > 0
 iD1 > iD2
 The opposite applies if Q2 is grounded etc.
 The differential pair responds to a difference-mode or differential input signals.
 If Vid is positive, VG1> VG2 and hence id1 will be greater than id2 and the difference voltage is V d2-Vd1
is positive.
 If Vid is negative, VG1< VG2 and hence id1 will be smaller than id2 and the difference voltage is
Vd2-Vd1 is negative.
 The differential amplifier responds to differential input signals by providing a corresponding output
signal between two drains. At this point, it is useful to inquire about the value of V id that causes the
entire bias current I to flow in one of the two transistors.
𝒊𝒅𝟏 = 𝑰 & 𝒊𝒅𝟐 = 𝟎
𝑉𝑑1 = 𝑉𝐷𝐷 − 𝑖𝑑1 𝑅𝐷 = 𝑉𝐷𝐷 − 𝐼𝑅𝐷
𝑉𝑑2 = 𝑉𝐷𝐷 − 𝑖𝑑2 𝑅𝐷 = 𝑉𝐷𝐷 − 0 = 𝑉𝐷𝐷
The differential output voltage is given by
𝑉𝑂 = 𝑉𝑑2 − 𝑉𝑑1= 𝑉𝐷𝐷 − 𝑉𝐷𝐷 + 𝐼𝑅𝐷 = 𝐼𝑅𝐷
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𝒊𝒅𝟏 = 𝟎 & 𝒊𝒅𝟐 = 𝑰
𝑉𝑑1 = 𝑉𝐷𝐷 − 𝑖𝑑1 𝑅𝐷 = 𝑉𝐷𝐷 − 0 = 𝑉𝐷𝐷
𝑉𝑑2 = 𝑉𝐷𝐷 − 𝑖𝑑2 𝑅𝐷 = 𝑉𝐷𝐷 − 𝐼𝑅𝐷
The differential output voltage is given by
𝑉𝑂 = 𝑉𝑑2 − 𝑉𝑑1= 𝑉𝐷𝐷 − 𝐼𝑅𝐷 − 𝑉𝐷𝐷 = −𝐼𝑅𝐷

In this positive direction, this happens when VGS1 reaches the value that corresponds to id1=I and VGS2 is
reduced to a value equal to the threshold voltage Vt, at which point Vs= -Vt
1 𝑊
𝐼 = 𝐾𝑛 (𝑉𝐺𝑆1 − 𝑉𝑡 )2
2 𝐿
2𝐼
𝑉𝐺𝑆1 = √ 𝑊 + 𝑉𝑡 = √2𝑉𝑂𝑉 + 𝑉𝑡
𝐾𝑛 𝐿

Where VOV is the overdriven voltage corresponding to a drain current of I/2. Thus the value of V id at which
the entire bias current I is steered into Q1 is
𝑉𝑖𝑑(𝑚𝑎𝑥) = 𝑉𝐺𝑆1 + 𝑉𝑡 = √2𝑉𝑂𝑉 + 𝑉𝑡 − 𝑉𝑡 = √2𝑉𝑂𝑉

−√2𝑉𝑂𝑉 ≤ 𝑉𝑖𝑑 ≤ √2𝑉𝑂𝑉

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ANALOG CIRCUITS (20A04302T) UNIT-I
Large Signal Operation of Differential Amplifier

To derive expressions for drain current id1 and id2 in terms of differential input signal Vid = VG1 – VG2.
Step 1: Expression drain currents for Q1 and Q2.
1 𝑊
𝐼𝑑1 = 𝐾 (𝑉 − 𝑉𝑡 )2
2 𝑛 𝐿 𝐺𝑆1
1 𝑊
𝐼𝑑2 = 𝐾𝑛 (𝑉𝐺𝑆2 − 𝑉𝑡 )2
2 𝐿

Step 2: Take the square roots of both sides of above equations

1 𝑊
√𝐼𝑑1 = √ 𝐾𝑛 (𝑉𝐺𝑆1 − 𝑉𝑡 )
2 𝐿

1 𝑊
√𝐼𝑑2 = √ 𝐾𝑛 (𝑉𝐺𝑆2 − 𝑉𝑡 )
2 𝐿

Step 3: Subtract √𝐼𝑑1 from √𝐼𝑑2 and perform appropriate substitution.

𝑉𝐺𝑆1 − 𝑉𝐺𝑆2 = 𝑉𝐺1 − 𝑉𝐺2 = 𝑉𝑖𝑑

1 𝑊
√𝐼𝑑1 − √𝐼𝑑2 = √ 𝐾𝑛 𝑉𝑖𝑑
2 𝐿

Step 4: Note the constant-current bias constraint

𝑖𝑑1 + 𝑖𝑑2 = 𝐼

1 𝑊
√𝐼𝑑1 − √𝐼𝑑2 = √ 𝐾𝑛 𝑉𝑖𝑑
2 𝐿

Taking square on both sides

1 𝑊
𝑖𝑑1 + 𝑖𝑑2 − 2√𝑖𝑑1 𝑖𝑑2 = 𝐾𝑛 𝑉𝑖𝑑 2
2 𝐿
1 𝑊
2√𝑖𝑑1 𝑖𝑑2 = 𝐼 − 𝐾𝑛 𝑉𝑖𝑑 2
2 𝐿

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ANALOG CIRCUITS (20A04302T) UNIT-I
Squaring on the both sides
2
1 𝑊
4𝑖𝑑1 𝑖𝑑2 = [𝐼 − 𝐾𝑛 𝑉𝑖𝑑 2 ]
2 𝐿
1 𝑊 2 1 𝑊 2 4
(𝑖𝑑1 + 𝑖𝑑2 )2 − (𝑖𝑑1 − 𝑖𝑑2 )2 = 𝐼 2 − 2 ∗ 𝐾𝑛 𝑉 ∗ 𝐼 + (𝐾𝑛 ) 𝑉𝑖𝑑
2 𝐿 𝑖𝑑 4 𝐿
𝑊 2 1 𝑊 2 4
(𝑖𝑑1 − 𝑖𝑑2 )2 = 𝐾𝑛 𝑉 ∗ 𝐼 − (𝐾𝑛 ) 𝑉𝑖𝑑
𝐿 𝑖𝑑 4 𝐿

1 𝑊 2 2 4𝐼
= (𝐾𝑛 ) 𝑉𝑖𝑑 [ 𝑊 − 𝑉𝑖𝑑 2 ]
4 𝐿 𝐾𝑛 𝐿

1 𝑊 4𝐼
𝑖𝑑1 − 𝑖𝑑2 = 𝐾𝑛 𝑉𝑖𝑑 √ 𝑊 − 𝑉𝑖𝑑 2
2 𝐿 𝐾𝑛 𝐿

The output voltage VO=Vd2-Vd1=-RD(𝑖𝑑1 − 𝑖𝑑2 )

𝑖𝑑1 + 𝑖𝑑2 = 𝐼

𝑖𝑑1 − 𝑖𝑑2 = 𝐾
𝐼 𝐾
𝑖𝑑1 =
+
2 2
𝐼 𝐾
𝑖𝑑2 = −
2 2
These two equations describe the effect of applying a differential input signal V id on the currents id1 and id2.

Figure : Normalized plots of the currents in a MOSFET differential pair.

Note that VOV is the overdrive voltage at which Q1 and Q2 operate when conducting drain currents equal to
I/2, the equilibrium situation.
For small signal approximation
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ANALOG CIRCUITS (20A04302T) UNIT-I
4𝐼
𝑉𝑖𝑑 2 ≪ 𝑊
𝐾𝑛 𝐿

𝐼
𝑉𝑖𝑑 ≪ 2√ 𝑊
𝐾𝑛
𝐿

𝑉𝑖𝑑 ≪ 2𝑉𝑂𝑉
Small signal operation of Differential Pair
The MOS differential amplifier with input voltages
1
𝑉𝐺1 = 𝑉𝐶𝑀 + 𝑉𝑖𝑑
2
1
𝑉𝐺2 = 𝑉𝐶𝑀 − 𝑉𝑖𝑑
2
The differential input signal Vid is applied in a complementary manner, that is VG1 is increased by Vid/2 and
VG2 is decreased by Vid/2.

Figure: MOS differential amplifier

1 𝑊 4𝐼
𝑖𝑑1 − 𝑖𝑑2 = 𝐾𝑛 𝑉𝑖𝑑 √ 𝑊 − 𝑉𝑖𝑑 2
2 𝐿 𝐾𝑛 𝐿

𝑉𝑖𝑑 ≪ 2𝑉𝑂𝑉 is the condition for small signal approximation.

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ANALOG CIRCUITS (20A04302T) UNIT-I
1 𝑊 4𝐼
𝑖𝑑1 − 𝑖𝑑2 = 𝐾𝑛 𝑉𝑖𝑑 √ 𝑊
2 𝐿 𝐾𝑛 𝐿

𝑊
= 𝑉𝑖𝑑 √𝐾𝑛 ∗𝐼
𝐿

The output voltage VO=Vd2-Vd1=-RD(𝑖𝑑1 − 𝑖𝑑2 )

Figure: Small signal analysis of MOS differential amplifier

Apply KVL at Vs
VG1-VGS1=Vs
VG2-VGS2=Vs
VG1-VGS1= VG2-VGS2
VG1-VG2= VGS1-VGS2
Apply KCL at node Vs
𝑖𝑑1 + 𝑖2 = 0
𝑔𝑚 𝑉𝐺𝑆1 + 𝑔𝑚 𝑉𝐺𝑆2 = 0

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𝑔𝑚 (𝑉𝐺𝑆1 + 𝑉𝐺𝑆2 ) = 0
𝑉𝐺𝑆1 = −𝑉𝐺𝑆2
𝑉𝐺1 = −𝑉𝐺2
VG1+VG1= VGS1+VGS1
2VG1=2VGS1
Vs =VG1-VGS1=0
The symmetry of the circuit as well as because of the balanced manner in which V id is applied, the signal
voltage at the joint source connection must be zero, acting as a sort of virtual ground.

Figure: Small signal equivalent model for MOS differential amplifier

𝑉𝑑1 = −𝑖𝑑1 𝑅𝐷 = −𝑔𝑚 𝑉𝐺𝑆1 𝑅𝐷
𝑉𝑑2 = −𝑖𝑑2 𝑅𝐷 = −𝑔𝑚 𝑉𝐺𝑆2 𝑅𝐷
The output voltage VO=Vd2-Vd1=−𝑔𝑚 𝑅𝐷 (𝑉𝐺𝑆2 − 𝑉𝐺𝑆1 ) = 𝑔𝑚 𝑅𝐷 𝑉𝑖𝑑
𝑉𝑂
𝐴𝑑 = = 𝑔𝑚 𝑅𝐷
𝑉𝑖𝑑
1.5 BJT DIFFERENTIAL AMPLIFIER
The Differential amplifier is a basic building block of the op-amp. The function of the differential amplifier
is to amplify difference in between two input signals. Let us consider 2 emitter-biased circuits as shown in
the figure.

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ANALOG CIRCUITS (20A04302T) UNIT-I
The 2 transistors Q1 and Q2 have the same characteristics. The resistances of circuits are equal, that is R E1 =
R E2, RC1 = RC2 and magnitude of +VCC is equal to magnitude of -VEE. These voltages are measured with
respect to the ground.
To make a differential amplifier, two circuits are connected as shown in the figure. The two +VCC and -
VEE supply terminals are made common as they are same. The two emitters are also connected and parallel
combination of RE1 and RE2 is replaced by the resistance RE. The two input signals v1 & v2 are applied at
base of Q1 and at base of Q2. The output voltage is taken between the two collectors. The collector
resistances are equal and hence denoted by RC = RC1 = RC2.
Ideally, output voltage is zero when the two inputs are equal. When v1 is greater than v2 output voltage with
the polarity shown appears. When V1 is less than V2, output voltage has the opposite polarity.
Differential Amplifier
The differential amplifier amplifies the difference between two input voltage signals. Hence it is also called
difference amplifier. The need for differential amplifier in many physical measurements arises where
response from d.c to many megahertz is required. It is also the basic input stage of an integrated amplifier.

The output signal in a differential amplifier is proportional to the difference between the two input signals.
Vo α (V1 – V2)
Where,
V1 & V2 – Two input signals
Vo – Single ended output Each signal is measured with respect to the ground.

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Differential Gain (Ad):
The output voltage
V0=Ad(V1-V2)
Where,
Ad is the constant of proportionality.
Ad is the gain with which differential amplifier amplifies the difference between two input signals.
Hence it is known as ‘differential gain of the differential amplifier’.
𝑉0
𝐴𝑑 =
𝑉𝑑
V1-V2= Difference of two voltage
Generally, differential gain is expressed in decibel (dB) as
Ad= 20 log10 (Ad) in dB
Common Mode Gain (Ac)
If we apply two input voltages which are equal in all the respect to the differential amplifier i.e V 1=V2 then
ideally the output voltage V0= (V1-V2) Ad, must be zero.
But the output voltage of the practical differential amplifier not only depends on the difference voltage but
also depends on the average common level of the two inputs. Such an average level of the two input signals
is called common mode signal denoted as Vcm.
𝑉1 + 𝑉2
𝑉𝑐𝑚 =
2
Practically, the differential amplifier produces the output voltage proportional to such common mode signal.
The gain with which it amplifies the common mode signal to produce the output is called common mode
gain of the differential amplifier denoted as Ac.
Vo = Ac Vc Where Ac is the common mode gain.
Therefore, there exists some finite output for V1 = V2 due to common mode gain Ac.
Hence the total output of any differential amplifier can be given as,
Vo = Ad Vd+ Ac Vc
For an ideal differential amplifier, the differential gain Ad must be infinite while the common mode gain
must be zero. This ensures zero output for V1=V2.
But due to mismatch in the internal circuitry, there is some output available for V 1=V2 and gain Ac is not
practically zero. The value of such common mode gain Ac is very small while the value of the differential
gain Ad is always very large.
Common Mode Rejection Ratio (CMRR):
The ability of a differential amplifier to reject a common mode signal is defined by a ratio called ‘Common
Mode Rejection Ratio’ denoted as CMRR.

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ANALOG CIRCUITS (20A04302T) UNIT-I
CMRR is defined as the ratio of the differential voltage gain Ad to common mode gain Ac and is expresses
𝐴
in dB. 𝐶𝑀𝑅𝑅 = 20 𝑙𝑜𝑔 | 𝐴𝑑 | 𝑑𝐵
𝑐

Ideally the common mode voltage gain is zero, hence the ideal value of CMRR is infinite. For a practical
differential amplifier Ad is large and Ac is small hence the value of CMRR is also very large.
Modes of operation of Differential Amplifier (DA)
• There are two modes of operations of DA
– Differential mode
– Common mode
Differential mode:
• Two input signals are of same magnitude but opposite polarity are used (1800 out of phase)
Common mode
• Two input signals are of equal in magnitude and same phase are used
Differential mode operation
Assume sine wave on base of Q1 is +ve going signal while on the base of Q2 –ve going signal
• An amplified –ve going signal will appear at collector of Q1
• An amplified +ve going signal will appear at collector of Q2
• Due to +ve going signal of base of Q1, current increases in RE & hence a +ve going wave is
developed across RE
• Due to -ve going signal of base of Q2, -ve going wave is developed across RE because of emitter
follower action of Q2
• So, signal voltages across RE, due to effect of Q1 &Q2 are equal in magnitude &180 0 out of phase-
due to matched transistors
• Hence the two signals cancel each other & there is no signal across RE
• No AC signal flows through it
• Vo= +10-(-10)= 20
• Vo is difference voltage in two signals

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Common Mode Operation

• Two input signals are of equal in magnitude and same phase are used
• In phase signal develops in phase signal voltages across RE
• Hence RE carries a signal current & provides -ve feedback
• This –ve f/b decreases AC In signal voltages of equal magnitude will appear across two
collectors of Q1 &Q2
• Vo= 10-10=0 Negligibly small
• Ideally it should be zero

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DIFFERENTIAL AMPLIFIER CONFIGURATIONS
The differential amplifiers are of the various configurations. The 4 differential amplifier configurations are
given as follows:
1. The Dual input, balanced output differential amplifier.
2. The Dual input, unbalanced output differential amplifier.
3. The Single input balanced output differential amplifier.
4. The Single input unbalanced output differential amplifier.
Here V1& V2 are input voltages and the difference between them is called “Differential Input Voltage”. It
is very essential that the transistors Q1& Q2 match perfectly for proper functioning of the differential
amplifier. The differential input voltage gets amplified and appears as the output voltage V 0 across the
collector terminals C1& C2.
If the output is measured between C1& C2 it is termed as Balanced Output. If the output is measured
between either of the collector terminals and ground, it is termed as Unbalanced Output.
On observing phase relationship between output (V0) and inputs (V1& V2), V0& V1 are in phase and V0& V2
are 180° out of phase. Hence B1 at which input signal V1 is applied is called “Non-Inverting Input Terminal”
and is at which input signal V2 applied is called “Inverting Input Terminal”.
∴V0 = A (V1 - V2)
Where A - Gain of the amplifier.
If B2 is grounded i.e., V2 = 0, V0 = AV1
If B1 is grounded i.e., V1 = 0, V0 = -AV2.

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ANALOG CIRCUITS (20A04302T) UNIT-I

Figure : Differential Amplifier Configurations

Dual Input Balanced Output Differential Amplifier Configuration

The two input signals V1 and V2 are applied to the bases B1 and B2 of transistors Q1 and Q2. The output
V0 is measured between the two collectors C1 and C2, which are at the same DC potential. Because of the
equal dc potential at the two collectors with respect to ground, the output is referred to as a balanced output.
DC Analysis (Determine VCEQ, ICQ)
To determine the operating point values (VCEQ, ICQ) for the differential amplifier, a dc equivalent circuit is to
be obtained. The dc equivalent circuit can be obtained by reducing the input signals V1 and V2 to zero. Here
the internal resistances of the signals are denoted as RS1=RS2=RS and the circuit is symmetrical.
Hence, in order to determine the Q-point (ICQ, VCEQ) it is only necessary to consider any one half of the
figure

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ANALOG CIRCUITS (20A04302T) UNIT-I

To Find ICQ:
Consider LHS half of the circuit
By applying KVL to the Base - Emitter loop
-RSIB - VBE - RE (2IE) + VEE = 0
VEE - RSIB - VBE - 2IERE = 0
IC
We know IB = ; where β = hfe is the current amplification factor and also, since the transistor is in CE
β

mode, IC = IE.
I
∴VEE -VBE - RS βE - 2REIE = 0
RS
VEE -VBE = IE (2RE + β )
VEE − VBE
IE = RS
(2RE + )
β

RS
Since 2RE>>
β
VEE − VBE
IE = 2RE

VEE − VBE (∵IE = IC = ICQ)

∴ ICQ =
2RE
By selecting a proper value of R E, we can obtain a desired value of IE for a known value of -VEE. Note that
the IE in transistors Q1 and Q2 is independent of collector resistance Rc.
Next is to determine the collector to emitter voltage VCE. The voltage at the emitter of the transistor Q1 is
approximately equal to –VBE. If we assume the voltage drop across Rs to be negligibly small. Knowing the
value of IE , we can obtain the voltage at the collector Vc as follows:
To Find VCEQ
The voltage at the emitter of transistor Q1 is approximately equal to -VBE. If we assume the voltage drop
across Rs to be negligible small. We can obtain the voltage at the collector V C as follows:

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ANALOG CIRCUITS (20A04302T) UNIT-I
From the figure
VCE = VC - VE
But VC = VCC - ICRC
VE ≈ - VBE, neglecting the small voltage across source resistance RS.

∴VCEQ = VCC + VBE - ICRC

Hence, for both transistors we can determine the I CQ and VCEQ by using equations above, because at the
operating point IE=ICQ and VCE=VCEQ.
NOTE: The DC analysis is applicable for all four configurations as long as we use the same biasing
arrangements for each of them.
AC Analysis:
To perform AC analysis to derive the expression for the voltage gain A d and the input resistance Ri of the
differential amplifier. To obtain the small signal AC equivalent circuit to perform AC analysis, the DC
voltages VCC and VEE are set equal to zero, and both transistors are replaced by its equivalent r e model.The
26𝑚𝑣
AC resistances of both transistors are equal and it is 𝑟𝑒 = 𝐼𝐸

Since both transistors are in CE mode both collector terminals C1 and C2 are negative with respect to
ground. But assume that C1 is more negative and hence the polarity of output voltage V 0 is as shown in the
figure.
The voltage across each collector resistor is shown out of phase by 180 0 with respect to the input voltage
V1& V2. This polarity assignment is in accordance with the common emitter configuration.

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ANALOG CIRCUITS (20A04302T) UNIT-I

Differential Voltage Gain (Ad)

It is the ratio of output voltage to the differential input voltage.
V0 V0
Ad= Vid = V1−V2 [∵V1 = Vi1& V2 = Vi2]

and can be derived as follows

On applying KVL to the loops ①&②
Vin1– Rin1𝑖𝑏1 - re ie1 - RE (ie1 + ie2) = 0
Vin2– Rin2𝑖𝑏2 - re ie2 - RE (ie1 + ie2) = 0
𝐼𝐸
We know that 𝐼𝐵 = 𝛽
𝑖𝑒1 𝑖𝑒2
Substituting 𝑖𝑏1 = and 𝑖𝑏2 =
𝛽 𝛽
ie1
Vin1– Rin1 𝛽 - re ie1 - RE (ie1 + ie2) = 0
ie2
Vin2– Rin2 𝛽 - re ie2 - RE (ie1 + ie2) = 0
𝑅𝑖𝑛
In practice, is very small and hence we may neglect the terms containing them
𝛽

∴Vin1 = ie1 (re + RE) + ie2 RE

Vin2 = ie1 RE + ie2 (re + RE)
The above equations are solved have to solve for ie1 and ie2 using Cramer’s Rule.
𝑉𝑖𝑛1 𝑟 + 𝑅𝐸 𝑅𝐸 𝑖
[ ]=[𝑒 ] [ 𝑒1 ]
𝑉𝑖𝑛2 𝑅𝐸 𝑟𝑒 + 𝑅𝐸 𝑖𝑒2
𝑟 + 𝑅𝐸 𝑅𝐸
∆= |𝑒 |
𝑅𝐸 𝑟𝑒 + 𝑅𝐸
= (re + RE) 2 - R2E
∆ = re² + 2 re RE
𝑉 RE
| 𝑖𝑛1 |
𝑉𝑖𝑛2 re +RE
ie1 = ∆
𝑉𝑖𝑛1 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛2 𝑅𝐸
= 𝑟𝑒 2+ 2 𝑟𝑒 𝑅𝐸

Similarly
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ANALOG CIRCUITS (20A04302T) UNIT-I
𝑉𝑖𝑛2 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛1 𝑅𝐸
ie2 = 𝑟𝑒 2 + 2 𝑟𝑒 𝑅𝐸

The output voltage VO = VC2 - VC1

= -iC2 RC - (-iC1 RC)
= -iC2 RC+iC1 RC
= RC (iC1 - iC2)
VO = RC (ie1 - ie2) ∵ ic ≈ iSubstitute ie1 and ie2 values in above equation
𝑉𝑖𝑛1 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛2 𝑅𝐸 𝑉𝑖𝑛2 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛1 𝑅𝐸
VO = RC[ − ]
𝑟𝑒 2 + 2 𝑟𝑒 𝑅𝐸 𝑟𝑒 2 + 2 𝑟𝑒 𝑅𝐸

(𝑟𝑒 +𝑅𝐸 )𝑉𝑖𝑛1 −𝑉𝑖𝑛2 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛2 𝑅𝐸 𝐼+𝑉𝑖𝑛1 𝑅𝐸

= RC[ ]
𝑟𝑒 2+ 2 𝑟𝑒 𝑅𝐸

(𝑟𝑒 +𝑅𝐸 )(𝑉𝑖𝑛1 −𝑉𝑖𝑛2 )+𝑅𝐸 (𝑉𝑖𝑛1 −𝑉𝑖𝑛2 )

𝑉𝑂 = 𝑅𝐶 [ ]
𝑟𝑒 2+ 2 𝑟𝑒 𝑅𝐸

(𝑟𝑒 + 2𝑅𝐸 )(𝑉𝑖𝑛1 − 𝑉𝑖𝑛2 )

𝑉𝑂 = 𝑅𝐶 [ ]
𝑟𝑒 (𝑟𝑒 + 2𝑅𝐸 )
|∵ Vid = Vin1 - Vin2
𝑅𝐶
𝑉𝑂 = 𝑉
𝑟𝑒 𝑖𝑑
RC
Therefore, output is in phase with the differential input voltage and it gets amplified by times.
re
VO RC
∴The DifferentialVoltage Gain , Ad = Vid = =gmRc
re

Thus, a differential amplifier amplifies the difference between two input signals.
Differential Input Resistance (Ri)
It is defined as the equivalent resistance that would be measured at either input terminal with the other
terminal grounded. This means that the input resistance Ri1 seen from the input signal source Vin1 is
determined with the signal source Vin2 set at zero. Similarly, the input signal source Vin1 is set to zero to
determine the input resistance Ri2 seen from the input signal source Vin2.
The differential input resistance at any one input terminal is defined as the equivalent resistance that would
be measured at that terminal while the other terminal is grounded
∴Ri = Ri1 (or) Ri2
Vi1 Vi2
|Vi2 = 0 (or) |Vi1 = 0
𝑖𝑏1 𝑖𝑏2

Vi1 iC ie
|
ie1 Vi2 = 0 |∵ib = =𝛽
𝛽
𝛽

Substitute the value of ie1 in above equation

Vin1𝛽
Ri = Vin1 (re+RE )−Vin2 RE | Vi2 = 0
re²+ 2 re RE

Vin1 𝛽(re²+ 2 re RE)

= ∵ Vi2 = 0
Vin1 (re+RE )

𝛽(re²+ 2 re RE)
Ri = (re+RE )

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ANALOG CIRCUITS (20A04302T) UNIT-I
But in practice RE>> re
𝛽( 2 re RE)
∴Ri = RE

∴
Ri = 2 βre

Output Resistance (Ro)

It is defined as the equivalent resistance that would be measured at either of the collector terminals with
respect to ground.
From the above AC equivalent circuit
The output resistance at C1 or C2 is
Ro = RC1 (or) RC2
| ∵ RC1 = RC2 = RC
Ro = RC

The current gain of the differential amplifier is undefined. Therefore, the current gain equation will not be
defined for any of the four differential amplifier configurations.
Common Mode Voltage Gain (Acm)
Since emitter voltage at emitter E1 and E2 is changing, therefore, the emitter resistance of the half circuit
should be 2RE instead of RE after splitting into two half circuits.
𝑉𝑂𝐶𝑀 = 𝑖𝐶 𝑅𝐶
𝑉𝐶𝑀 = 𝑖𝑒 2𝑅𝐸 = 𝑖𝐶 2𝑅𝐸
Common mode voltage gain
𝑉𝑂𝐶𝑀 𝑖𝐶 𝑅𝐶 𝑅𝐶
𝐴𝐶𝑀 = = =
𝑉𝐶𝑀 𝑖𝐶 2𝑅𝐸 2𝑅𝐸
Common Mode Rejection Ratio (CMRR)
𝑅𝐶⁄
𝑟𝑒 2𝑅𝐸
𝑪𝑴𝑹𝑹 = 𝑅𝐶 = = 2𝑔𝑚 𝑅𝐸
𝑟𝑒
2𝑅𝐸

Dual - Input Unbalanced Output Differential Amplifier

Here two input signals are used; the output is measured at only one of the two collectors with respect to
ground. The output is referred as an unbalanced output because the collector at which the output voltage is
measured is at some finite DC potential with respect to ground.
In other words, there is some DC voltage at the output terminal without any input signal applied. The output
is measured at collector of Q2 with respect to ground.
DC Analysis:
Note: The DC analysis is applicable for all four configurations as long as we use the same biasing
arrangements for each of them.

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ANALOG CIRCUITS (20A04302T) UNIT-I

AC Analysis:

For AC analysis, the DC voltages VCC and VEE are set equal to zero. Rin1 = Rin2 = Rin
Voltage Gain (Ad)
Apply KVL for loop ① and ②

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ANALOG CIRCUITS (20A04302T) UNIT-I
Vin1 - Rin ib1 - reie1 - RE (ie1 + ie2) = 0
Vin2 - Rin ib2 - reie2 - RE (ie1 + ie2) = 0
iC1 ie1 iC2 ie2
ib1 = = and ib2 = =
𝛽 𝛽 𝛽 𝛽
Rin
Vin1 - ie1- reie1 - RE (ie1 + ie2) = 0
𝛽
Rin
Vin2 - ie2- reie2 - RE (ie1 + ie2) = 0
𝛽
Rin
is very small, neglect this term.
𝛽

Vin1 = (re + RE) ie1 + REie2

Vin2 = REie1 +(re + RE) ie2
The above equations are solved have to solve for ie1 and ie2 using Cramer’s Rule.
𝑉𝑖𝑛1 𝑟 + 𝑅𝐸 𝑅𝐸 𝑖
[ ]=[𝑒 ] [ 𝑒1 ]
𝑉𝑖𝑛2 𝑅𝐸 𝑟𝑒 + 𝑅𝐸 𝑖𝑒2
𝑟 + 𝑅𝐸 𝑅𝐸
∆= |𝑒 |
𝑅𝐸 𝑟𝑒 + 𝑅𝐸
= (re + RE) 2 - R2E
∆ = re² + 2 re RE
𝑉 RE
| 𝑖𝑛1 |
𝑉𝑖𝑛2 re +RE
ie1 = ∆
𝑉𝑖𝑛1 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛2 𝑅𝐸
= 𝑟𝑒 2+ 2 𝑟𝑒 𝑅𝐸

Similarly
𝑉𝑖𝑛2 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛1 𝑅𝐸
ie2 = 𝑟𝑒 2 + 2 𝑟𝑒 𝑅𝐸

The output voltage is

VO = VC2 = -RCiC2 Since ic≈ ie
VO = -RC ie2
𝑉𝑖𝑛2 (𝑟𝑒 +𝑅𝐸 )−𝑉𝑖𝑛1 𝑅𝐸
VO = -RC[ ]
𝑟𝑒 2 + 2 𝑟𝑒 𝑅𝐸

Generally, RE>> re, hence (re + RE) ≈ RE

RE (Vin2 −Vin1)
∴VO = -RC[ ]
2reRE
RC (Vin1 −Vin2)
VO = 2re
RC (Vid)
VO = 2re

VO RC
Ad =
Vid = 2re

Thus the voltage gain of the Dual Input Unbalanced Output differential amplifier is half the gain of the Dual
Input Balanced Output differential amplifier.

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ANALOG CIRCUITS (20A04302T) UNIT-I
Differential Input Resistance (Ri)
The differential input resistance at any one input terminal is defined as the equivalent resistance that would
be measured at that terminal while the other terminal is grounded
∴Ri = Ri1 (or) Ri2
Vi1 Vi2
|Vi2 = 0 (or) |Vi1 = 0
𝑖𝑏1 𝑖𝑏2

Vin1 iC ie
ie1 |Vi2 = 0 |∵ib = =𝛽
𝛽
𝛽

Substitute the value of ie1 in above equation

Vi1𝛽
Ri = Vin1 (re+RE )−Vi2 RE | Vin2 = 0
re²+ 2 re RE

Vin1 𝛽(re²+ 2 re RE)

= ∵ Vin2 = 0
Vin1 (re+RE )

𝛽(re²+ 2 re RE)
Ri = (re+RE )

But in practice RE>> re

𝛽( 2 re RE)
∴Ri = RE

Ri = 2 βre

Output Resistance (Ro)

It is defined as the equivalent resistance that would be measured at either of the collector terminals with
respect to ground.
From the above AC equivalent circuit
The output resistance at C1 or C2 is
Ro = RC1 (or) RC2
| ∵ RC1 = RC2 = RC
Ro = RC

Single Input Balanced Output Differential Amplifier

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ANALOG CIRCUITS (20A04302T) UNIT-I
Here the output is applied to the base of transistor Q 1 and the output is measured between the two collectors
which are at the same DC potential. Hence the output is said to be balanced output.

Single Input Unbalanced Output Differential Amplifier

This configuration is not commonly used for the following reasons.

1. Although, identical to the CE amplifier it not only requires more components but also yields less
voltage gain than the former one.
2. There exists a DC output voltage without any input signal applied.

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ANALOG CIRCUITS (20A04302T) UNIT-I

Transfer Characteristics

Collector currents ic1 and ic2 for transistors Q1 and Q2 biased in the forward active mode may be given by
𝑉𝐵𝐸1
⁄𝑉
𝑖𝑐1 = 𝛼𝐹 𝐼𝐸𝑆 𝑒 𝑇

𝑉𝐵𝐸2
⁄𝑉
𝑖𝑐2 = 𝛼𝐹 𝐼𝐸𝑆 𝑒 𝑇

Here, IES is the reverse saturation current of emitter base junction and VT is volts equivalent of
temperature.
𝑖𝑐1 (𝑉𝐵𝐸1 −𝑉𝐵𝐸2 )
⁄𝑉
=𝑒 𝑇
𝑖𝑐2
Write KVL for the loop containing two emitter base junctions as
𝑉1 − 𝑉𝐵𝐸1 + 𝑉𝐵𝐸2 − 𝑉2 = 0
𝑉1 − 𝑉2 = 𝑉𝐵𝐸1 − 𝑉𝐵𝐸2 = 𝑉𝑖𝑑
Where Vid is the difference of two input voltages
𝐼𝑄 = 𝑖𝐸1 + 𝑖𝐸2
𝑖𝐶1 𝑖𝐶2 𝑖𝐶1 𝑖𝐶2
𝐼𝑄 = + = (1 + )
𝛼𝐹 𝛼𝐹 𝛼𝐹 𝑖𝐶1

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ANALOG CIRCUITS (20A04302T) UNIT-I
𝛼𝐹 𝐼𝑄
𝑖𝐶1 = −𝑉𝑖𝑑
⁄𝑉
1+𝑒 𝑇

𝛼𝐹 𝐼𝑄
𝑖𝐶2 = 𝑉𝑖𝑑
⁄𝑉
1+𝑒 𝑇

The normalized transfer characteristics for a differential amplifier as shown in figure

The following important points are observed from the transfer characteristics:
1. For Vid>4 VT, ic1=αFIQ and ic2=0
V01=Vcc- αFIQRc
V02=Vcc
2. For Vid<4 VT, ic2=αFIQ and ic1=0
V02=negligible
V01=Vcc
Thus, for 4VT<Vid<-4VT, a differential amplifier can be made to function as a switch.
3. The differential amplifier functions as a very good limiter for Vid>±4VT
4. Between the values -2VT<Vid<2VT , differential amplifier functions as a linear amplifier.
1.6 OTHER NON IDEAL CHARACTERISTICS OF THE DIFFERENTIAL AMPLIFIER

 Differential amplifier amplifies the difference component, Reject the noise component.
 Ideal characteristics of differential amplifier
o Infinite differential gain
o Infinite input resistance
o Zero output resistance
o Infinite CMRR
o Infinite Bandwidth
o Zero output voltage for zero difference input signal
Due to the mismatches in load resistors, deviates from its ideal characteristics:
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ANALOG CIRCUITS (20A04302T) UNIT-I
1. Input offset voltage of the MOS differential pair
Consider the basic MOS differential amplifier with both inputs grounded.

If the two sides of the differential pair were properly matched, then current I would split equally between Q1
and Q2 and VO would be zero. Practical circuits exhibits mismatches that result in a dc output voltage V O
even with both inputs grounded.
𝑉𝑂
𝐴𝑑 =
𝑉𝑂𝑆
𝑉𝑂
𝑉𝑂𝑆 =
𝐴𝑑
If we apply a voltage –VOS between the input terminals of the differential amplifier, then the output voltage
will be reduced to zero. For the differential amplifier, consider first the case where Q 1 and Q2 are perfectly
matched but RD1 and RD2 show a mismatch ∆RD.

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ANALOG CIRCUITS (20A04302T) UNIT-I
∆𝑅𝐷
𝑅𝐷1 = 𝑅𝐷 +
2
∆𝑅𝐷
𝑅𝐷2 = 𝑅𝐷 −
2
Because Q1 and Q2 are matched, the current I will split equally between them. Because of the mismatch in
load resistances the output voltages Vd1 and Vd2 will be
𝐼 ∆𝑅𝐷
𝑉𝑑1 = 𝑉𝐷𝐷 − (𝑅𝐷 + )
2 2
𝐼 ∆𝑅𝐷
𝑉𝑑2 = 𝑉𝐷𝐷 − (𝑅𝐷 − )
2 2
Thus the differential output voltage VO will be
𝐼
𝑉𝑂 = 𝑉𝑑2 − 𝑉𝑑1 = ∆𝑅
2 𝐷
The corresponding input offset voltage is obtained by dividing V O by the gain Ad.
𝑉𝑂 𝐼∆𝑅𝐷
𝑉𝑂𝑆 = =
𝐴𝑑 2𝑔𝑚 𝑅𝐷
2. Input offset voltage of the Bipolar Differential Amplifier
The output offset results from mismatches in the load resistances Rc1 and Rc2 and from junction area, β and
other mismatches in Q1 and Q2.

If we apply a voltage –VOS between the input terminals of the differential amplifier, then the output voltage
will be reduced to zero. For the differential amplifier, consider first the case where Q 1 and Q2 are perfectly
matched but RC1 and RC2 show a mismatch ∆RC.
∆𝑅𝐶
𝑅𝐶1 = 𝑅𝐶 +
2
∆𝑅𝐶
𝑅𝐶2 = 𝑅𝐶 −
2

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ANALOG CIRCUITS (20A04302T) UNIT-I

Because Q1 and Q2 are matched, the current I will split equally between them. Because of the mismatch in
load resistances the output voltages VC1 and VC2 will be
𝛼𝐼 ∆𝑅𝐶
𝑉𝐶1 = 𝑉𝐶𝐶 − (𝑅𝐶 + )
2 2
𝛼𝐼 ∆𝑅𝐶
𝑉𝐶2 = 𝑉𝐶𝐶 − (𝑅𝐶 − )
2 2
Thus the differential output voltage VO will be
𝛼𝐼
𝑉𝑂 = 𝑉𝐶2 − 𝑉𝐶1 = ∆𝑅𝐶
2
The corresponding input offset voltage is obtained by dividing V O by the gain Ad.
𝑉𝑂 𝛼𝐼∆𝑅𝐶
𝑉𝑂𝑆 = =
𝐴𝑑 2𝑔𝑚 𝑅𝐶
Input Bias current:
In an ideal differential amplifier, we assumed that no current is drawn from the input terminals. However,
practically, input terminals do conduct a small value of dc current to bias the input transistors. The base
currents entering into the inverting and non-inverting terminals are shown as 𝐼𝐵− and 𝐼𝐵+ respectively. Even
though both the transistors are identical, 𝐼𝐵− and 𝐼𝐵+ are not exactly same.
The input bias current IB is the average of the current entering the input terminals of a balanced amplifier
𝐼𝐵− + 𝐼𝐵+
𝐼𝐵 =
2
Input offset current:
Bias current compensation will work if both bias currents 𝐼𝐵+ and 𝐼𝐵− are equal. Since the input transistors
cannot be made identical, there will always be some small difference between 𝐼𝐵+ and𝐼𝐵− . This difference is
called the offset current Iio and can be written as
Iio = | IB1- IB2 |

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ANALOG CIRCUITS (20A04302T) UNIT-I
Two Marks Questions
1. Define CMRR.
2. What is the meaning of balanced and unbalanced output?
3. Define differential amplifier.
4. Draw the hybrid equivalent model of CE amplifier.
5. Mention four h-parameters of CE amplifier.
6. Identify different types of coupling used in amplifiers.
7. Write the advantages of multistage amplifier.
8. Draw the circuit of transformer coupled amplifier.
9. Distinguish between cascade and cascade amplifiers.
10. Explain the effect of bypass capacitor in multistage amplifier.
11. Draw the differential amplifier using BJT.
12. Write the differences between direct coupled and RC coupled amplifiers.
13. Draw the circuit diagram of Darlington pair.
14. List out the ideal characteristics of differential amplifier.
15. Define input bias current and offset current of differential amplifier.

Ten Marks Questions

1. Explain DC and AC analysis of dual input balanced output differential amplifier.
2. What is a differential amplifier? Explain balanced and unbalanced output in differential amplifier.
3. Estimate ICQ, VCEQ, re, voltage gain, input and output resistances for a dual-input, balanced-output
differential amplifier with circuit parameters RC = 2.2 kΩ, RE = 4.7 kΩ, Rin1 = Rin2 = 50 Ω, VCC =
+10 V, VEE = -10 V, 𝛽𝐷𝐶 = 𝛽𝐴𝐶 = 100 and 𝑉𝐵𝐸 = 0.71 V typical.
4. A dual input unbalanced-output differential amplifier has the following specification: |Vcc| = 10 V,
|-VEE| = 10 V, Rc1 = Rc2 = 2.7 kΩ, Rin = 50 Ω and RE = 3.9 kΩ and the transistor is CA3086
with βac = βdc = 100 and VBE = 0.715 V. Calculate: (i) ICQ and VCEQ values. (ii) Voltage gain.
(iii) Input and output resistances.
5. Explain different types of coupling. When two identical stages are cascaded, obtain voltage gain,
current gain and power gain.
6. Draw and explain the circuit of cascade amplifier and mention the advantages.
7. Sketch two stage RC coupled amplifier and derive the expression for current gain.
8. Derive an equation for voltage gain, current gain, input impedance and output impedance of
Darlington emitter follower.
9. Draw the circuit of a differential amplifier using BJT and derive an expression for CMRR.
10. Draw and analyze two stage RC coupled amplifier and derive the expression for input resistance,
output resistance and voltage gain.

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ANALOG CIRCUITS (20A04302T) UNIT-I
11. Derive the expression for differential voltage gain of MOS differential amplifier.
12. Derive the expression for range of voltages of common mode signal in MOS differential amplifier.
13. Explain the large signal analysis of MOS differential amplifier.
14. Explain non ideal characteristics of differential amplifier.

N GUNASEKHAR REDDY, S V ENGINEERING COLLEGE, TIRUPATHI Page 53