Differential Amplifier Circuit
The differential amplifier circuit is an extremely popular connection used in IC units. The
basic differential amplifier circuit is shown in figure. The circuit has two separate inputs and
two separate outputs, and the emitters are connected together. Most differential amplifier
circuits use two separate voltage supplies. The circuit can also operate using a single supply.
3 input signal combinations are possible :
If an input signal is applied to either input with the other input connected to ground, the
operation is referred to as “single-ended”.
If two opposite polarity input signals are applied, the operation is referred to as “doubleended”.
If the same input is applied to both inputs, the operation is called “common-mode”.
In single-ended operation, a single input signal is applied. Due to the common emitter
connection, the input signal operates both transistors, resulting in output from both collectors.
In double-ended operation, two input signals are applied. Since the two input signals are
different, the outputs at the two collectors are different and the resultant output is the
difference of these two collector outputs.
In common-mode operation, the two input signals are same. So the two collector outputs are
same and the resultant output, which is the difference of two collector outputs, is zero.
Practically the two collector outputs do not completely cancel and a small signal results.
The main feature of the differential amplifier is the very large gain when opposite signals are
applied to the inputs as compared to the very small gain resulting from common inputs.
1
Current Mirror
A current mirror circuit provides a constant current and is one of the most important building
blocks of integrated circuits. A current mirror exactly reflects or mirrors a current existing on
one side of the circuit. Current mirror is used in integrated circuits as a biasing element or as
a load to amplifier. As a load, current mirror provides high dynamic resistance. Current
mirror is also called current source or current sink. If npn transistor is used it sinks current
and if pnp transistor is used it sources current.
Figure shows a simple current mirror consisting of a resistor and two transistors.
Here the transistor Q2 is connected as a diode. The current through the diode
𝐼𝑅 =
𝑉𝐶𝐶 − 𝑉𝐵𝐸
𝑅
With VBE = 0.7 in the active mode, the current IR is constant and is dependent only on the
supply voltage VCC and R.
Since both Q1 and Q2 are identical, IC1 and IC2 can be expressed as
𝐼𝐶1 = 𝛼𝐼𝑠 𝑒 𝑉𝐵𝐸1 ⁄𝑉𝑇
𝐼𝐶2 = 𝛼𝐼𝑠 𝑒 𝑉𝐵𝐸2 ⁄𝑉𝑇
Since the bases and emitters of two transistors Q1 and Q2 are connected together VBE1 = VBE2.
From the above equations we get IC1 = IC2 = I0. Applying KCL at node X
𝐼𝑅 = 𝐼𝐶2 + 𝐼𝐵2 + 𝐼𝐵1
Since VBE1 = VBE2, IB1 = IB2 = IB. Since both Q1 and Q2 are identical, 𝛽1 = 𝛽2 = 𝛽.
∴ 𝐼𝑅 = 𝐼0 + 2𝐼𝐵 = 𝐼0 +
2𝐼0
2
= 𝐼0 (1 + )
𝛽
𝛽
Since β >> 1, IR = I0. i.e. current through diode is exactly reproduced in transistor Q1. This is
the basis of the current mirror.
2
Consider ‘n’ transistors biased by a transistor diode as shown.
If ‘n’ transistors are connected, then the input current has to supply total base current (n+1)IB
and collector current of transistor Q.
𝑖. 𝑒. 𝐼𝑅 = 𝐼0 + (𝑛 + 1)𝐼𝐵 = 𝐼0 +
(𝑛 + 1)𝐼0
𝑛+1
= 𝐼0 (1 +
)
𝛽
𝛽
where I0 = IC1 = IC2 = ICn. If (n+1) << β then all transistors will exactly reproduce the input
current IR. The value of ‘n’ for which the collector current remains constant is approximately
20. This means using a single resistor and voltage source we can bias 20 transistors.
Wilson Current Mirror
Wilson current mirror uses a negative feedback to make the ratio I0/IR less sensitive to β and
to increase the output impedance.
Here Q3 senses the output current I0 and is fed back using Q1. Wilson current mirror provides
base currents cancellation making I0 very nearly equal to IR. This can be proved as follows.
3
From figure
𝐼𝑅 = 𝐼𝐶1 + 𝐼𝐵
𝐼𝐸2 = 𝐼𝐶3 + 𝐼𝐵1 + 𝐼𝐵3
Since all transistors are identical and VBE1 = VBE3, IB1 = IB3 = IB and IC1 = IC3.
∴ 𝐼𝐸2 = 𝐼𝐶1 + 2𝐼𝐵
𝐴𝑙𝑠𝑜, 𝐼𝐸2 = 𝐼0 + 𝐼𝐵
From the above two equations
𝐼0 = 𝐼𝐶1 + 2𝐼𝐵 − 𝐼𝐵 = 𝐼𝐶1 + 𝐼𝐵 = 𝐼𝑅
In practice, the base current cancellation will not be exact due to mismatches in the
transistors, but the resulting difference between I0 and IR will be extremely small.
From figure the emitter current of Q2 is equal to the collector current of Q3 plus the base
current of Q1 and Q3, i.e.
2
𝐼𝐸2 = 𝐼𝐶3 + 2𝐼𝐵 = 𝐼𝐶3 (1 + )
𝛽
𝐼𝐶2 = 𝛼𝐼𝐸2 =
𝛽
2
𝛽
𝐼𝐸2 = 𝐼𝐶3 (1 + ) (
)
1+𝛽
𝛽 1+𝛽
1
1
∴ 𝐼𝐶3 = 𝐼𝐶2 [
] = 𝐼0 [
]
2
𝛽
2
𝛽
(1 + ) (
)
(1 + ) (
)
𝛽 1+𝛽
𝛽 1+𝛽
Applying KCL at the base of Q2,
𝐼𝐶1 = 𝐼𝑅 − 𝐼𝐵2 = 𝐼𝑅 −
𝐼𝐶2
𝐼0
= 𝐼𝑅 −
𝛽
𝛽
Since IC1 = IC3, using equations of IC3 and IC1, we have
𝐼0 = 𝐼𝑅 (1 −
𝛽2
2
)
+ 2𝛽 + 2
Above equation shows that the output current I0 and the reference current IR differ by only a
factor of the order of 2/β2 as against 2/β of basic current mirror.
Widlar Current Mirror
In operational amplifier low input current is required. Therefore, the input stage of such
amplifier is biased at very low current. For such cases, Widlar current source shown in figure
is used.
4
In the circuit both the transistors are assumed to be identical. The function of R E causes VBE1
and VBE2 to differ. Here VBE1 < VBE2, and hence, IC1 < IC2. The asymmetric nature of the
base-emitter loop causes the circuit to act as a ‘lens’ rather than a ‘mirror’. In effect, Q2, VCC
and R establish the reference current IR, and RE determines the degree to which IC1 is less
than IR. Applying KVL for base-emitter loop in figure
𝑉𝐵𝐸2 − 𝑉𝐵𝐸1 = (𝐼𝐵1 + 𝐼𝐶1 )𝑅𝐸
𝑜𝑟 ∆𝑉𝐵𝐸 = (𝐼𝐵1 + 𝐼𝐶1 )𝑅𝐸
For identical transistors,
𝐼𝐶2 ⁄𝐼𝐶1 = 𝑒 (𝑉𝐵𝐸2 −𝑉𝐵𝐸1 )⁄𝑉𝑇
𝑜𝑟 ∆𝑉𝐵𝐸 = 𝑉𝑇 ln
𝐼𝐶2
𝐼𝐶1
Equating the two equations for ∆VBE we get
𝑅𝐸 =
𝑉𝑇
𝐼𝐶2
ln
𝐼𝐶1 (1 + 1⁄𝛽 ) 𝐼𝐶1
The reference current IR can be obtained as follows
Applying KCL
1
𝐼𝐶1
𝐼𝑅 = 𝐼𝐶2 + 𝐼𝐵2 + 𝐼𝐵1 = 𝐼𝐶2 (1 + ) +
𝛽
𝛽
Since β >> 1 and IC1 is smaller than IC2 the above equation becomes
𝐼𝑅 ≅ 𝐼𝐶2 = 𝐼0
5
The Operational Amplifier
An operational amplifier is a direct-coupled high-gain amplifier usually consisting of one or
more differential amplifiers and usually followed by a level translator and an output stage. An
operational amplifier is available as a single integrated circuit package. The operational
amplifier is a versatile device that can be used to amplify dc as well as ac input signals and
was originally designed for performing mathematical operations such as addition, subtraction,
multiplication and integration. Thud the name operational amplifier stems from its original
use for these mathematical operations and is abbreviated to op-amp. With the addition of
suitable external feedback components, the modern day op-amp can be used for a variety of
applications, such as ac and dc signal amplification, active filters, oscillators, comparators,
regulators and others.
Block Diagram Of A Typical Op-Amp
An op-amp is a multistage amplifier. Its block diagram is shown in figure. The input stage is
the dual-input, balanced-output differential amplifier. This stage generally provides most of
the voltage gain of the amplifier and also establishes the input resistance of the op-amp. The
intermediate stage is usually another differential amplifier, which is driven by the output of
the first stage. In most amplifiers the intermediate stage is dual input, unbalanced (single
ended) output. Because direct coupling is used, the dc voltage at the output of the
intermediate stage is well above ground potential. Therefore, generally, the level translator
(shifting) circuit is used after the intermediate stage to shift the dc level at the output of the
intermediate stage downward to zero volts with respect to ground. The final stage is usually a
push-pull complementary amplifier output stage. The output stage increases the output
voltage swing and raises the current supplying capability of the op-amp. A well-designed
output stage also provides low output resistance.
Schematic Symbol
Figure shows the most widely used symbol of an op-amp. For simplicity, power supply and
other pin connections are omitted. The ‘+’ input is the non inverting input. An ac signal or dc
voltage applied to this input produces an in-phase (or same polarity) signal at the output. On
the other hand, the ‘-‘ input is the inverting input because an ac signal (or dc voltage) applied
to this input produces an 1800 out of phase (or opposite polarity) signal at the output.
6
v1 = voltage at the non inverting input
v2 = voltage at the inverting input
v0 = output voltage
All these voltages are measured with respect to ground.
A = large-signal voltage gain, which is specified on the data sheet for an op-amp.
Characteristics Of An deal Op-Amp
1. Infinite voltage gain A.
2. Infinite input resistance Ri so that almost any signal source can drive it and there is no
loading of the preceding stage.
3. Zero output resistance R0 so that the output can drive an infinite number of other devices.
4. Zero output voltage when input voltage is zero.
5. Infinite bandwidth so that any frequency signal from 0 to ∞ Hz can be amplified without
attenuation.
6. Infinite common-mode rejection ratio so that the output common-mode noise voltage is
zero.
7. Infinite slew rate so that output voltage changes occur simultaneously with input voltage
changes.
Electrical Parameters Of An Op-Amp
Common Mode Rejection Ratio(CMRR) : is the ratio of the differential voltage gain Ad to
the common-mode voltage gain Acm.
𝐶𝑀𝑅𝑅 =
𝐴𝑑
𝐴𝑐𝑚
Being a large value, CMRR is most often expressed in decibels (dB). For the 741C, CMRR is
90 dB. For the 741C precision op-amp, CMRR = 120 dB. This means that this op-amp has a
better ability to reject common-mode voltages, such as electrical noise, than the 741 C.
Slew Rate(SR) : is defined as the maximum rate of change of output voltage per unit of time
and is expressed in volts per microseconds.
7
𝑑𝑉0
|
𝑉 ⁄𝜇𝑠
𝑑𝑡 𝑚𝑎𝑥𝑖𝑚𝑢𝑚
𝑆𝑅 =
Slew rate indicates how rapidly the output of an op-amp can change in response to changes in
the input frequency. The slew rate of an op-amp is fixed; therefore, if the slope requirements
of the output signal are greater than the slew rate, then distortion occurs.
Input Offset Voltage : is the voltage that must be applied between the two input terminals of
an op-amp to null the output. in figure Vdc1 and Vdc2 are dc voltages and Rs represents the
source resistance. The input offset voltage Vio could be positive or negative. The smaller the
value of Vio, the better the input terminals are matched.
Differential Input Resistance(Ri) : is the equivalent resistance that can be measured at
either the inverting or noninverting input terminal with the other terminal connected to
ground.
Input Voltage Range : is the range of common-mode voltage applied to both input terminals
without disturbing proper functioning of the op-amp.
Supply Voltage Rejection Ratio (SVRR) : The change in an op-amp’s input offset voltage,
Vio, caused by variations in supply voltages is called the supply voltage rejection ratio.
𝑆𝑉𝑅𝑅 =
∆𝑉𝑖𝑜
∆𝑉
Large Signal Voltage Gain : Since the op-amp amplifies difference voltage between two
input terminals, the voltage gain of the amplifier is defined as
𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑔𝑎𝑖𝑛 =
𝑜𝑢𝑡𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
That is
𝐴=
𝑉0
𝑉𝑖𝑑
Because output signal amplitude is much larger than the input signal, the voltage gain is
commonly called large-signal voltage gain.
8
Output Voltage Swing : The output voltage swing indicates the values of positive and
negative saturation voltages of the op-amp. The output voltage never exceeds these limits for
given supply voltages +VCC and –VEE.
Output Resistance(R0) : is the equivalent resistance that can be measured between the
output terminal of the op-amp and the ground.
Open Loop Op-Amp Configurations
In case of amplifiers the term open loop indicates that no connection exists between the
output and input terminals. That is, the output signal is not fed back in any form as part of the
input signal, and the loop that would have been formed with feedback is open. When
connected in open-loop configuration, the op-amp simply functions as a high-gain amplifier.
There are three open-loop op-amp configurations:
1. Differential amplifier
2. Inverting amplifier
3. Noninverting amplifier
The Differential Amplifier : Figure shows the open-loop differential amplifier in which
input signals vin1 and vin2 are applied to the positive and negative input terminals. vin1 and vin2
could be either ac or dc. Since the op-amp amplifies the difference between the two input
signals, this configuration is called the differential amplifier. The output voltage
𝑣𝑜 = 𝐴(𝑣𝑖𝑛1 − 𝑣𝑖𝑛2 )
The polarity of output voltage depends on the polarity of the input difference voltage
(vin1 – vin2). A is called open-loop gain. The output of the differential amplifier for two sine
wave inputs is shown below.
9
The Inverting Amplifier : In the inverting amplifier only one input is applied and that is to
the inverting input terminal. The noninverting input terminal is grounded. Since v1 = 0V, and
v2 = vin
𝑣0 = −𝐴𝑣𝑖𝑛
The negative sign indicates that the output voltage is out of phase with respect to input by
1800 or is of opposite polarity. Thus in the inverting amplifier the input signal is amplified by
gain A and is also inverted at the output.
The Noninverting Amplifier : In this configuration the input is applied to the noninverting
input terminal, and the inverting terminal is connected to ground. Since v1 = vin and v2 = 0
𝑣0 = 𝐴𝑣𝑖𝑛
This means that the output voltage is larger than the input voltage by gain A and is in phase
with the input signal.
10
Disadvantages of Open Loop Configurations
In all three open loop configurations any input signal (differential or single) that is only
slightly greater than zero drives the output to saturation level. This results from the very high
gain (A) of the op-amp. Thus, when operated open-loop, the output of the op-amp is either
negative or positive saturation or switches between positive and negative saturation levels.
Besides being large, the open loop voltage gain of the op-amp is not a constant. The voltage
gain varies with changes in temperature and power supply as well as mass production
techniques. The variations in voltage gain are relatively large in open loop op-amps in
particular, which makes the open loop op-amp unsuitable for many linear applications. In
addition, the bandwidth of most open loop op-amps is negligibly small – almost zero. For this
reason the open loop op-amp is impractical in ac applications.
Closed Loop Op-Amp Configurations
We can select as well as control the gain of op-amp by feeding back the output signal to the
input either directly or via another network. If the signal fed back is of opposite polarity or
out of phase by 1800 with respect to the input signal, the feedback is called negative
feedback. An amplifier with negative feedback has a self correcting ability against any
change in output voltage caused by changes in environmental conditions. Negative feedback
is also called degenerative feedback because it degenerates (reduces) the output voltage
amplitude and in turn reduces the voltage gain. If the signal feedback is of same polarity or in
phase with the input signal, the feedback is called positive feedback. In positive feedback the
feedback signal aids the input signal. Hence it is called regenerative feedback. Positive
feedback is necessary in oscillatory circuits.
When used in amplifiers, negative feedback stabilizes the gain, increases the bandwidth,
changes the input and output resistances, decreases the harmonics or nonlinear distortion and
reduces the effect of input offset voltage on output. Negative feedback also reduces the effect
of variations in temperature and supply voltage on output voltage of op-amp.
11
An op-amp that uses feedback is called a feedback amplifier or closed loop amplifier. A
closed loop amplifier can be represented by using two blocks, one for an op-amp and another
for a feedback circuit. There are four ways to connect these two blocks. These connections
are classified according to whether the voltage or current is fed back to the input in series or
in parallel as follows:
1.
2.
3.
4.
voltage series feedback
voltage shunt feedback
current series feedback
current shunt feedback
Offset Compensation
Due to the mismatch of transistors used, a differential input voltage exists between two input
terminals of an op-amp without any external inputs applied. This differential input voltage is
called input offset voltage Vio. This voltage is always dc and can be positive or negative. This
voltage varies with op-amps. This input offset voltage causes an output offset voltage Voo. In
order to reduce the output offset voltage to zero we have to apply an input voltage Vio to any
one of the input terminals which depends on the polarity of the offset voltage. For example
consider an op-amp having positive output offset voltage. This means that the noninverting
input is greater than inverting input. In order to make the output voltage zero we have to
apply either a positive voltage to inverting input or a negative voltage to noninverting input.
If the output offset voltage is negative we have to apply a positive voltage to noninverting
input or a negative voltage to inverting input. It is impossible to predict the magnitude and
polarity of input offset voltage. The polarity of input offset voltage can be obtained from
output offset voltage as explained above. But we don’t know the exact magnitude of voltage
12
to be applied. In datasheets only the maximum offset voltage is specified. So to compensate
the offset voltage we apply an input to any one terminal, which depends on the op-amp
configuration, through a potentiometer so that we can vary the magnitude and polarity of
input voltage till output voltage is zero. The op-amp with such an arrangement is shown in
figure. The offset voltage compensating network consists of potentiometer Ra and resistors Rb
and Rc. Here the compensating network is connected to the noninverting terminal. This
means that we can use the op-amp as inverting amplifier. If we want to use the op-amp as
noninverting amplifier then we have to connect the compensating network to inverting
terminal. By adjusting the wiper on Ra, the differential input voltage and hence the output
offset voltage Voo can be made zero. For example, assume that Voo is positive, which implies
that V1 > V2. This means that V2 should be increased until it equals V1, by moving the wiper
towards VCC. on the other hand, if Voo is negative, that is , V2 > V1, the wiper can be moved
towards –VEE until Voo is reduced to zero.
The maximum thevenin’s equivalent resistance and voltage of potentiometer can be obtained
as follows.
The maximum thevenin’s equivalent resistance Rmax occurs when the wiper is at the center of
the potentiometer as shown in figure.
𝑅𝑚𝑎𝑥 =
𝑅𝑎 𝑅𝑎 𝑅𝑎
∥
=
2
2
4
From the above figure
13
𝑉𝑡ℎ =
𝑉𝐶𝐶 + 𝑉𝐸𝐸
2
Supply voltages +VCC and –VEE are equal in magnitude. i.e.|+𝑉𝐶𝐶 | = |−𝑉𝐸𝐸 | = 𝑉. Thus
Vth = V. The equivalent compensating network is as shown below.
Applying voltage divider rule to the above circuit we get
𝑉2 =
𝑅𝑐
𝑉
𝑅𝑚𝑎𝑥 + 𝑅𝑏 + 𝑅𝑐 𝑡ℎ
But the maximum value of V2 can be equal to Vio, since |𝑉1 − 𝑉2 | = Vio. Thus
𝑉𝑖𝑜 =
𝑅𝑐
𝑉
𝑅𝑚𝑎𝑥 + 𝑅𝑏 + 𝑅𝑐 𝑡ℎ
To simplify the above equation let us make the assumption that Rb > Rmax > Rc. Therefore
𝑅𝑚𝑎𝑥 + 𝑅𝑏 + 𝑅𝑐 ≅ 𝑅𝑏
𝑉𝑖𝑜 =
𝑅𝑐 𝑉𝑡ℎ 𝑅𝑐 𝑉
=
𝑅𝑏
𝑅𝑏
The above equation is used to design the compensating network. The value of Vio is obtained
from the datasheet of given op-amp. The value of V is fixed according to the supply voltages
chosen. Then the ratio of Rc and Rb is found from above equation. Then select a value for Rc
and find the value of Rb. Then Ra is found by using Rb > Rmax⇒ Rb = 10 Rmax, where Rmax =
Ra/4. The value of Rc is selected to be less than 100 Ω so that the Rb and Ra values will not be
too large.
Voltage Series Feedback Amplifier (Noninverting Amplifier)
The schematic diagram of voltage series feedback amplifier is sown in figure. The circuit is
commonly known as noninverting amplifier with feedback or closed loop noninverting
amplifier because it uses feedback, and the input signal is applied to the noninverting input
terminal of the op-amp.
Closed Loop Voltage Gain
The closed loop voltage gain is
14
𝐴𝐹 =
𝑣0
𝑣𝑖𝑛
𝑣𝑜 = 𝐴𝑣𝑖𝑑 = 𝐴(𝑣1 − 𝑣2 )
From figure
𝑣1 = 𝑣𝑖𝑛
𝑣2 = 𝑣𝑓 =
𝑅1 𝑣0
𝑅1 + 𝑅𝐹
𝑠𝑖𝑛𝑐𝑒 𝑅𝑖 ≫ 𝑅1
Therefore,
𝑣0 = 𝐴 (𝑣𝑖𝑛 −
𝑅1 𝑣0
𝐴(𝑅1 + 𝑅𝐹 )𝑣𝑖𝑛
)=
𝑅1 + 𝑅𝐹
𝑅1 + 𝑅𝐹 + 𝐴𝑅1
Thus,
𝐴𝐹 =
𝑣0
𝐴(𝑅1 + 𝑅𝐹 )
=
𝑣𝑖𝑛 𝑅1 + 𝑅𝐹 + 𝐴𝑅1
(1)
Generally A is very large. Therefore,
𝐴𝑅1 ≫ (𝑅1 + 𝑅𝐹 ) 𝑎𝑛𝑑 (𝑅1 + 𝑅𝐹 + 𝐴𝑅1 ) ≅ 𝐴𝑅1
Thus
𝐴𝐹 =
𝑣0
𝑅𝐹
= 1+
𝑣𝑖𝑛
𝑅1
The above equation shows that the gain of the voltage series feedback amplifier is determined
by the ratio of two resistors, R1 and RF. The gain of the feedback circuit is
𝐵=
𝑣𝑓
𝑅1
1
=
=
𝑣0 𝑅1 + 𝑅𝐹 𝐴𝐹
15
This means that the gain of the feedback circuit is the reciprocal of the closed loop voltage
gain and is fixed for given R1 and RF. Also from (1)
𝐴𝐹 =
𝑅 +𝑅
𝐴 (𝑅1 + 𝑅𝐹 )
1
𝐹
𝑅1 + 𝑅𝐹
𝐴𝑅1
𝑅1 + 𝑅𝐹 + 𝑅1 + 𝑅𝐹
=
𝐴
𝑣0
=
1 + 𝐴𝐵 𝑣𝑖𝑛
where AF = closed loop voltage gain
A = open loop voltage gain
B = gain of the feedback circuit
AB = loop gain
Input Resistance with Feedback
Figure shows a noninverting amplifier with the op-amp equivalent circuit. In this circuit Ri is
the input resistance (open loop) of the op-amp, and RiF is the input resistance of the amplifier
with feedback. The input resistance with feedback is defined as
16
𝑅𝑖𝐹 =
𝑣𝑖𝑛
𝑣𝑖𝑛
=
𝑖𝑖𝑛 𝑣𝑖𝑑 ⁄𝑅𝑖
But,
𝑣𝑖𝑑 =
𝑣0
𝐴
𝑎𝑛𝑑 𝑣0 =
𝑣
𝐴
1 + 𝐴𝐵 𝑖𝑛
Therefore,
𝑅𝑖𝐹 = 𝑅𝑖
𝑣𝑖𝑛
𝑣𝑖𝑛
= 𝐴𝑅𝑖
= 𝑅𝑖 (1 + 𝐴𝐵)
𝑣0 ⁄𝐴
𝐴𝑣𝑖𝑛 ⁄(1 + 𝐴𝐵)
This means that the input resistance of the op-amp with feedback is (1+AB) times that
without feedback.
Output Resistance with Feedback
Output resistance is the resistance determined looking back into the feedback amplifier from
the output terminal as shown in figure. To find the output resistance with feedback R oF,
reduce independent source vin to zero, apply an external voltage vo, and then calculate the
resulting current io. Output resistance with feedback is given by
𝑅𝑜𝐹 =
𝑣𝑜
𝑖𝑜
Applying KCL at node N
𝑖𝑜 = 𝑖𝑎 + 𝑖𝑏
𝑆𝑖𝑛𝑐𝑒 [(𝑅𝐹 + 𝑅1 ) ∥ 𝑅𝑖 ] ≫ 𝑅𝑜 𝑎𝑛𝑑 𝑖𝑎 ≫ 𝑖𝑏 ,
𝑖𝑜 ≅ 𝑖𝑎
The current io can be found by writing KVL equation for the output loop:
17
𝑣𝑜 − 𝑅𝑜 𝑖𝑜 − 𝐴𝑣𝑖𝑑 = 0 ⟹ 𝑖𝑜 =
𝑣𝑜 − 𝐴𝑣𝑖𝑑
𝑅𝑜
But
𝑣𝑖𝑑 = 𝑣1 − 𝑣2 = 0 − 𝑣𝑓 = −
∴ 𝑖𝑜 =
∴ 𝑅𝑜𝐹 =
𝑅1 𝑣𝑜
= −𝐵𝑣𝑜
𝑅1 + 𝑅𝐹
𝑣𝑜 + 𝐴𝐵𝑣𝑜
𝑅𝑜
𝑣𝑜
𝑅𝑜
=
(𝑣𝑜 + 𝐴𝐵𝑣𝑜 )⁄𝑅𝑜 1 + 𝐴𝐵
This result shows that the output resistance of the noninverting amplifier with feedback is
1/(1+AB) times the output resistance Ro of the op-amp. That is, the output resistance of the
op-amp with feedback is much smaller than the output resistance without feedback.
Bandwidth with Feedback
The bandwidth of an amplifier is defined as the band (range) of frequencies for which the
gain remains constant. Manufactures generally specify either the gain-bandwidth product or
supply open-loop gain versus frequency curve for the op-amp. Figure shows the open-loop
gain versus frequency curve of the 741C op-amp.
From this curve for a gain of 2x105, the bandwidth is approximately 5 Hz and the gainbandwidth product is 2x105x5 Hz = 1 MHz. When the gain is 1, the bandwidth is
approximately 1 MHz and the gain-bandwidth product is 1x1 MHz = 1 MHz. Thus the gainbandwidth product is constant. For the 741, 5 Hz is the break frequency (f0), the frequency at
which the gain A is 3 dB down from its value at 0 Hz. The frequency at which the gain equals
1 is known as the unity gain bandwidth (UGB). Since the gain-bandwidth product is constant
18
𝐴 × 𝑓0 = 𝐴𝐹 × 𝑓𝐹 ⟹ 𝑓𝐹 =
𝐴 × 𝑓0
𝐴𝐹
For the non inverting amplifier with feedback
𝐴𝐹 =
𝐴
1 + 𝐴𝐵
Therefore bandwidth with feedback
𝑓𝐹 =
𝐴 × 𝑓0
= 𝑓0 (1 + 𝐴𝐵)
𝐴⁄(1 + 𝐴𝐵)
Above equation indicates that the bandwidth of the non inverting amplifier with feedback f F
is increased by (1+AB) times bandwidth without feedback, f0.
Total Output Offset Voltage with Feedback
In an op-amp when the input is zero, the output is also expected to be zero. However, because
of the effect of input offset voltage and current, the output is significantly larger. Let this
output voltage be total output offset voltage V00T. in an open loop op-amp the total output
offset voltage is equal to either the positive or negative saturation voltage. Since with
feedback the gain of the non inverting amplifier changes from A to A/(1+AB), the total
output offset voltage with feedback must also be 1/(1+AB) times the voltage without
feedback. That is,
𝑇𝑜𝑡𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓𝑓𝑠𝑒𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑤𝑖𝑡ℎ 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
𝑡𝑜𝑡𝑎𝑙 𝑜𝑢𝑡𝑝𝑢𝑡 𝑜𝑓𝑓𝑠𝑒𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘
=
1 + 𝐴𝐵
𝑜𝑟
𝑉𝑜𝑜𝑇 =
±𝑉𝑠𝑎𝑡
1 + 𝐴𝐵
So negative feedback significantly reduces the output offset voltage. Negative feedback can
also be used to reduce significantly the effect of noise, variations in supply voltages, and
changes in temperature on the output voltage of a non inverting amplifier.
Voltage Follower
19
The lowest possible gain that can be obtained from a non inverting amplifier with feedback is
1. When the non inverting amplifier is configured for unity gain, it is called a voltage
follower because the output voltage is equal to and in phase with the input. In other words, in
the voltage follower the output follows the input.
All the output voltage is fed back into the inverting terminal of the op-amp. Therefore the
gain of the feedback circuit B=1.
The voltage follower is a special case of the non inverting amplifier with R1 infinity and RF
zero. So the formulas of voltage follower can be obtained from the formulas for non inverting
amplifier.
AF = 1
B=1
V0 = AF * Vin = Vin
1+AB = 1+A =A, since A is very large.
RiF = Ri (1+AB) = ARi
RoF = Ro/(1+AB) = Ro/A
fF = fo (1+AB) = Afo
VooT = ±Vsat/(1+AB) = ±Vsat/A
The voltage follower is also called a non inverting buffer because, when placed between two
networks, it removes the loading on the first network.
Voltage Shunt Feedback Amplifier (Inverting Amplifier)
Figure shows an inverting amplifier with feedback. The input voltage is connected to the
inverting terminal. The output voltage is fed back to the inverting terminal through R F. The
gain of the op-amp depends on RF and R1.
20
Closed Loop Voltage Gain
Applying KCL at the input node v2
𝑖𝑖𝑛 = 𝑖𝐹 + 𝐼𝐵
Since input resistance of op-amp is very large, the input bias current IB is negligibly small.
∴ 𝑖𝑖𝑛 ≅ 𝑖𝐹
𝑣𝑖𝑛 − 𝑣2 𝑣2 − 𝑣0
=
𝑅1
𝑅𝐹
𝑣0 = 𝐴𝑣𝑖𝑑 = 𝐴(𝑣1 − 𝑣2 )
Since 𝑣1 = 0
𝑣0 = −𝐴𝑣2 ⇒ 𝑣2 = −
𝑣0
𝐴
Substituting the value of 𝑣1 and 𝑣2 we get
𝑣𝑖𝑛 + 𝑣0 ⁄𝐴 − 𝑣0 ⁄𝐴 − 𝑣0
=
𝑅1
𝑅𝐹
𝐴𝐹 =
𝑣0
𝐴𝑅𝐹
=−
𝑣𝑖𝑛
𝑅1 + 𝑅𝐹 + 𝐴𝑅1
(𝑒𝑥𝑎𝑐𝑡)
The negative sign indicates that the input and output signals are out of phase by 1800.
Since the open loop gain A is very large (ideally infinity), 𝐴𝑅1 ≫ 𝑅1 + 𝑅𝐹 . Therefore the
above equation can be written as
𝐴𝐹 =
𝑣0
𝑅𝐹
=−
𝑣𝑖𝑛
𝑅1
(𝑖𝑑𝑒𝑎𝑙)
This equation shows that the gain of the inverting amplifier can be set to any value by
choosing RF and R1. Compared to non inverting amplifier the gain of inverting amplifier can
be set below unity.
To write the equation in feedback form, divide the numerator and denominator of exact
equation by (R1+RF)
𝐴𝐹 = −
𝑤ℎ𝑒𝑟𝑒 𝐾 =
𝐴 𝑅𝐹 ⁄𝑅1 + 𝑅𝐹
𝐴𝐾
=−
𝐴𝑅
1 + 𝐴𝐵
1 + 𝑅 + 1𝑅
1
𝐹
𝑅𝐹
, 𝑎 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑡𝑡𝑒𝑛𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
𝑅1 + 𝑅𝐹
21
𝐵=
𝑅1
, 𝑔𝑎𝑖𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑐𝑖𝑟𝑐𝑢𝑖𝑡
𝑅1 + 𝑅𝐹
Block diagram of inverting amplifier with feedback using a voltage summing junction as a
model for current summing
Input Resistance with Feedback
The easiest method of finding the input resistance is to Millerize the feedback resistor R F into
two Miller components, as shown in figure. Input resistance with feedback RiF is
𝑅𝑖𝐹 = 𝑅1 +
𝑅𝐹
∥ 𝑅𝑖
1+𝐴
Since Ri and A are very large
𝑅𝐹
∥ 𝑅𝑖 ≅ 0
1+𝐴
Hence
𝑅𝑖𝐹 = 𝑅1
Output Resistance with Feedback
The output resistance of inverting amplifier can be found by using Thevenin’s theorem.
Thevenin’s equivalent circuit for R0F inverting amplifier is shown in figure. This equivalent
circuit is exactly the same as that for the non inverting amplifier. Thus R0F of both amplifiers
are same.
22
𝑅𝑜𝐹 =
𝑅0
1 + 𝐴𝐵
Offset Minimizing Resistor (Effect of input bias current)
Consider an op-amp with negative feedback as shown. There is no input voltage. But due to
the supply voltages op-amp takes biasing currents IB1 and IB2. Let input bias current IB = IB1 =
IB2. The value of input bias current IB is very small. For 741 IB is 500 nA maximum at supply
voltages = ±15V dc.
Figure 1
Let the input offset voltage Vio = 0. Therefore there is no output offset voltage due to the
input offset voltage Vio. Let VoIB be the output offset voltage due to input bias current IB. The
non inverting terminal is connected to ground; therefore the voltage V1 = 0. The controlled
voltage source AVio = 0V since Vio = 0V is assumed. With output resistance R0 negligibly
small, the right end of RF is essentially at ground potential; that is resistor R1 and RF are in
parallel and the bias current IB2 flows through them. Therefore the voltage at the inverting
terminal is
23
𝑉2 = (𝑅1 ∥ 𝑅𝐹 )𝐼𝐵2 =
𝑅1 𝑅𝐹
𝐼 … … . (1)
𝑅1 + 𝑅𝐹 𝐵2
Applying KCL at node V2 in figure 1
𝐼1 + 𝐼2 = 𝐼𝐵2
0 − 𝑉2 𝑉0𝐼𝐵 − 𝑉2 𝑉2
=
=
𝑅1
𝑅𝐹
𝑅𝑖
𝑉0𝐼𝐵
1
1
1
= 𝑉2 ( +
+ )
𝑅𝐹
𝑅1 𝑅𝐹 𝑅𝑖
where 𝑉0𝐼𝐵 = output offset voltage due to input bias current
Since Ri is extremely high (ideally ∞), 1/Ri ≈ 0. Therefore
𝑉0𝐼𝐵
1
1
𝑅1 + 𝑅𝐹
𝑅1
= 𝑉2 ( + ) = 𝑉2 (
) ⇒ 𝑉2 =
𝑉 … . . (2)
𝑅𝐹
𝑅1 𝑅𝐹
𝑅1 𝑅𝐹
𝑅1 + 𝑅𝐹 0𝐼𝐵
From (1) and (2) we get
𝑉0𝐼𝐵 = 𝑅𝐹 𝐼𝐵
According to the above equation, the amount of output offset voltage depends on the
feedback resistor RF. The amount of 𝑉0𝐼𝐵 can be reduced by using small feedback resistors.
To reduce or eliminate the output offset voltage due to input bias current IB we have to make
V1 = V2. From (1)
24
𝑉2 =
𝑅1 𝑅𝐹
𝐼
𝑅1 + 𝑅𝐹 𝐵2
If we connect a resistor ROM in the inverting terminal, the voltage V1 will be
𝑉1 = 𝑅𝑂𝑀 𝐼𝐵1
To have V1 = V2
𝑅1 𝑅𝐹
𝐼 = 𝑅𝑂𝑀 𝐼𝐵1
𝑅1 + 𝑅𝐹 𝐵2
Now if the currents IB1 = IB2 = IB then
𝑅𝑂𝑀 =
𝑅1 𝑅𝐹
𝑅1 + 𝑅𝐹
Thus the proper value of ROM required in the inverting terminal is the parallel combination of
resistors R1 and RF. However, the use of ROM may not completely eliminate the output offset
voltage 𝑉0𝐼𝐵 because the currents IB1 and IB2 are not exactly equal. But it minimizes the
amount of output offset voltage𝑉0𝐼𝐵 . Therefore ROM is referred to as the offset minimizing
resistor.
Scaling Amplifier
This is used to scale an input. Figure shows an inverting configuration with three inputs Va,
Vb and Vc. The output voltage equation can be derived by applying KCL at node V2
𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 𝐼𝐵 + 𝐼𝐹
25
Since Ri of the op-amp is ideally infinity, IB = 0.
∴ 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 𝐼𝐹
𝑉𝑎 − 𝑉2 𝑉𝑏 − 𝑉2 𝑉𝑐 − 𝑉2 𝑉2 − 𝑉0
+
+
=
𝑅𝑎
𝑅𝑏
𝑅𝑐
𝑅𝐹
Since V1 is grounded, V1 = 0V and V2 is at virtual ground. Therefore V2 ≅ 0V.
∴
𝑉𝑎 𝑉𝑏 𝑉𝑐
𝑉0
+
+
=−
𝑅𝑎 𝑅𝑏 𝑅𝑐
𝑅𝐹
𝑅𝐹
𝑅𝐹
𝑅𝐹
𝑉0 = − ( 𝑉𝑎 +
𝑉𝑏 +
𝑉)
𝑅𝑎
𝑅𝑏
𝑅𝑐 𝑐
This means that output voltage is the weighted or scaled sum of inputs.
Summing Amplifier
It is used to add the input voltages.
If Ra = Rb = Rc = R in the above circuit then
𝑉0 = −
𝑅𝐹
(𝑉 + 𝑉𝑏 + 𝑉𝑐 )
𝑅 𝑎
If RF = R then
𝑉0 = −(𝑉𝑎 + 𝑉𝑏 + 𝑉𝑐 )
This means that the output voltage is the negative sum of all input voltages. If we use Ra = Rb
= Rc = RF = R in the above circuit we will get an inverting summing amplifier.
Averaging Circuit
If we have n number of input voltages and use same series resistance (i.e, R1 = R2 =…= Rn =
R) then
𝑉0 = −
𝑅𝐹
(𝑉 + 𝑉2 + … . +𝑉𝑛 )
𝑅 1
If R = nRF then
1
𝑉0 = − (𝑉𝑎 + 𝑉𝑏 + 𝑉𝑐 )
𝑛
That is the output voltage is the average of input voltages.
26
Integrator
A circuit in which the output voltage is the integral of the input voltage is the integrator. A
basic integrator is obtained by replacing the feedback resistor in the in the inverting amplifier
configuration by a capacitor as shown.
Applying KCL at v2
𝑖1 = 𝐼𝐵 + 𝑖𝐹
Since IB is negligibly small,
𝑖1 ≅ 𝑖𝐹
𝑣𝑖𝑛 − 𝑣2
𝑑(𝑣2 − 𝑣0 )
= 𝐶𝐹
𝑅1
𝑑𝑡
v1 = v2 ≈ 0 because A is very large. Therefore
𝑣𝑖𝑛
𝑑
= 𝐶𝐹 (−𝑣0 )
𝑅1
𝑑𝑡
The output voltage can be obtained by integrating both sides with respect to time
𝑡
∫
0
𝑡
𝑣𝑖𝑛
𝑑
𝑑𝑡 = ∫ 𝐶𝐹 (−𝑣0 ) 𝑑𝑡 = 𝐶𝐹 (−𝑣0 ) + 𝑣0 |𝑡=0
𝑅1
𝑑𝑡
0
Therefore,
𝑡
1
𝑣𝑖𝑛
𝑣0 = −
∫
𝑑𝑡 + 𝐶
𝑅1 𝐶𝐹 0 𝑅1
where C is the integration constant and is proportional to the value of the output voltage at
t=0.
27
Above equation indicates that the output voltage is directly proportional to the negative
integral of the input voltage and inversely proportional to the time constant R 1CF. if the input
is a sine wave, the output will be a cosine wave or if the input is a square wave, the output
will be a triangular wave. The input and output waveforms of an integrator is shown below.
When vin = 0, the integrator works as an open loop amplifier. This is because the capacitor
acts as an open circuit to the input dc offset voltage Vio. In other words, the input offset
voltage Vio and the part of the input current charging capacitor CF produce the error voltage at
the output of the integrator. To reduce the error a resistor RF is connected across the
feedback capacitor. RF limits the low-frequency gain and hence minimizes the variation in the
output voltage. The practical integrator circuit is shown below.
28
The frequency response of integrator is shown in figure.
The frequency at which gain is 0 dB is given by
𝑓𝑏 =
1
2𝜋𝑅1 𝐶𝐹
In the frequency response, for frequencies below fa the gain is constant and the circuit acts as
an inverting amplifier. Between fa and fb the circuit acts as an integrator. The gain limiting
frequency fa is given by
𝑓𝑎 =
1
2𝜋𝑅𝐹 𝐶𝐹
The input signal will be integrated properly if the time period T of the signal is larger than or
equal to RFCF. That is,
T ≥ RFCF
The integrator is most commonly used in analog computers and analog to digital and signal
wave shaping circuits.
Differentiator
The differentiator circuit performs the mathematical operation of differentiation, i.e, the
output waveform is the derivative of the input waveform. The differentiator may be
constructed from a basic inverting amplifier if an input resistor R1 is replaced by a capacitor
C1 .
Applying KCL at node v2
𝑖𝐶 = 𝐼𝐵 + 𝑖𝐹
29
Since 𝐼𝐵 ≅ 𝑜
𝑖𝐶 = 𝑖𝐹
𝐶1
𝑑
𝑣2 − 𝑣0
(𝑣𝑖𝑛 − 𝑣2 ) =
𝑑𝑡
𝑅𝐹
But 𝑣1 = 𝑣2 ≅ 0, because A is very large. Therefore
𝐶1
𝑑𝑣𝑖𝑛
𝑣0
=−
𝑑𝑡
𝑅𝐹
𝑣0 = −𝑅𝐹 𝐶1
𝑑𝑣𝑖𝑛
𝑑𝑡
Thus the output voltage vo is equal to RFC1 times the negative instantaneous rate of change of
the input voltage vin with time.
Since the differentiator performs the reversal of the integrator’s function, a cosine wave input
will produce a sine wave output, or a triangular input will produce a square wave output.
However the differentiator shown above will not do this because it has some practical
problems. The gain of the circuit RF/XC1 increases with increase in frequency at a rate of 20
dB/decade. This makes the circuit unstable. Also, the input impedance XC1 decreases with
increase in frequency, which makes the circuit very susceptible to high frequency noise.
When amplified, the noise can completely override the differentiated output signal. Both the
stability and high frequency noise problems can be corrected by the addition of two
components: R1 and CF as shown below. This circuit is a practical differentiator. Its
frequency response is also shown below. From frequencies below fb, gain increases at 20
dB/decade. After fb the gain decreases at 20 dB/decade. This 40 dB/decade change in gain is
caused by the R1C1 and RFCF combinations. For frequencies below fb the circuit acts as a
differentiator. The gain limiting frequency fb is given by
𝑓𝑏 =
1
2𝜋𝑅1 𝐶1
30
where R1C1 = RFCF. Thus R1C1 and RFCF help to reduce significantly the effect of high
frequency input, amplifier noise, and offsets. Above all, R1C1 and RFCF make the circuit more
stable by preventing the increase in gain with frequency. Generally, the value of fb and in turn
R1C1 and RFCF values should be selected such that
𝑓𝑎 < 𝑓𝑏 < 𝑓𝑐
where
𝑓𝑎 =
1
is the frequency at which gain is 0 dB.
2𝜋𝑅𝐹 𝐶1
𝑓𝑏 =
1
1
=
2𝜋𝑅1 𝐶1 2𝜋𝑅𝐹 𝐶𝐹
𝑓𝑐 = unity gain − bandwidth.
The input signal will be differentiated properly if the time period T of the input signal is
larger than or equal to RFC1. That is,
T ≥ RFC1
A practical differentiator can be designed by the following steps
1. Select fa equal to the highest frequency of the input signal to be differentiated. Then
assuming a value of C1 ˂ 1 µF, calculate the value of RF.
2. Choose fb = 20 fa and calculate the values of R1 and CF so that R1C1 = RFCF.
The differentiator is most commonly used in wave shaping circuits to detect high frequency
components in an input signal and also as a rate of change detector in FM modulators.
Practical Differentiator
31
Frequency Response
Voltage (V) to Current (I) Converter with Floating Load
Figure shows a V to I converter in which load resistor R L is floating (not connected to
ground). The input voltage is applied to the non inverting terminal, and the feedback voltage
across R1 drives the inverting input. This circuit is also called a current series negative
feedback amplifier because the feedback voltage across R1 (applied to the inverting terminal)
depends on the output current io and is in series with the input difference voltage vid.
Applying KVL for the input loop
32
𝑣𝑖𝑛 = 𝑣𝑖𝑑 + 𝑣𝑓
But 𝑣𝑖𝑑 ≅ 0, since A is very large. Therefore
𝑣𝑖𝑛 = 𝑣𝑓 = 𝑅1 𝑖0
𝑖0 =
𝑣𝑖𝑛
𝑅1
This means that an input voltage vin is converted into an output current of vin/R1. The V to I
converter can be used in such applications as low voltage dc and ac voltmeters, diode match
finders, light emitting diodes and zener diode testers.
V to I Converter with Grounded Load
In this circuit one terminal of the load is grounded, and load current is controlled by an input
voltage. Applying KCL at V1
𝐼1 + 𝐼2 = 𝐼𝐿
𝑉𝑖𝑛 − 𝑉1 𝑉0 − 𝑉1
+
= 𝐼𝐿
𝑅
𝑅
𝑉𝑖𝑛 + 𝑉0 − 2𝑉1 = 𝐼𝐿 𝑅 ⇒ 2𝑉1 = 𝑉𝑖𝑛 + 𝑉0 − 𝐼𝐿 𝑅
Since the op-amp is connected in the non inverting mode, the gain of the circuit is 1+R/R = 2.
Then the output voltage is
𝑉0 = 2𝑉1 = 𝑉𝑖𝑛 + 𝑉0 − 𝐼𝐿 𝑅 ⇒ 𝑉𝑖𝑛 = 𝐼𝐿 𝑅
𝑜𝑟 𝐼𝐿 =
𝑉𝑖𝑛
𝑅
33
This means that the load current depends on the input voltage Vin and resistor R.
34