EC3451 / Linear Integrated Circuits UNIT-1

UNIT I
BASICS OF OPERATIONAL AMPLIFIERS
Current mirror and current sources, Current sources as active loads, Voltage sources, Voltage References, BJT
Differential amplifier with active loads, Basic information about op-amps – Ideal Operational Amplifier -
General operational amplifier stages -and internal circuit diagrams of IC 741, DC and AC performance
characteristics, slew rate, Open and closed loop configurations – JFET Operational Amplifiers – LF155 and
TL082
Constant current source (Current
Mirror):
Draw and explain the circuit diagram of a basic current mirror.
Explain the working of current source with a circuit diagram.
Draw the circuit of simple bipolar transistor current source and show that its output current is
dependent on the β of the transistor.
A current mirror is a circuit designed to copy a current through one active device by controlling the
current in another active device of a circuit, keeping the output current constant regardless of loading.
A constant current source use the transistor in the active mode of operation, the collector
current is independent of the collector voltage.

Fig. Current mirror circuit Current source output current characteristics

 Transistors Q1&Q2 are matched as the circuit is fabricated using IC technology.
 Base and emitter of Q1& Q2 are tied together and have the same VBE.
 Transistor Q1 is connected as a diode by shorting its collector to base.
 The input current Iref flows through the diode connected transistor Q1 and thus establishes a
voltage across Q1.
 This voltage appears between the base and emitter of Q2.
 Since Q2 is identical to Q1, the emitter current of Q2 will be equal to emitter current of Q1 which
is approximately equal to Iref.

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 As long as Q2 is maintained in the active region ,its collector current IC2=Iout will be
approximately equal to Iref.
 Since the output current Io is a reflection or mirror of the reference current Iref, the circuit is
often referred to as a current mirror.
Analysis:
 If the effect of finite β is considered, this may lead to an output current not equal to the input
reference current, since

where the expression for IREF is found by using KCL at the

collector of Q1.
The current gain of the current mirror
is

which approaches unity for β very large.

 Another deviation of Iout from Iref has to do with the Early effect. Since the VBE of Q2 is
constant as determined by IREF, the output resistance of Q2 determines the dependence of Iout.
 This may be perceived as a disadvantage of this configuration – the output resistance of the
current mirror is limited by the ro of Q2, or

Widlar Current Source

Draw the circuit of widlar current source and derive an expression for its output current.

 The Widlar current source differs from the basic

current mirror in a resistor (R2) is added to the emitter
circuit of transistor Q2.
 Since multistage amplifier systems often have high
gain, bias currents must be small. Instead of the large
resistors required to create small currents, the Widlar
current source generates small constant currents using
relatively small resistors.

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EC3451 / Linear Integrated Circuits

 This allows considerable savings in chip real estate – which, as was mentioned before and will
be mentioned often, is considered one of the ultimate goals in IC design.

Where, Io is the reverse saturation current , VT is the thermal voltage (kT/q≈26mV at room
temperature)
Solving Equation for VBE, we get

Using this information, and neglecting base currents, we can express VBE1 and VBE2 in the
circuit above as follows:

Assuming we have matched devices, VT and IO are the same for Q1 and Q2. Subtracting
VBE2 from VBE1, and using the appropriate property of logarithms (i.e., lnA-lnB=ln(A/B)), we get

Writing a KVL around the base loop of the two transistors,

Assuming that IC2=IE2=Iout, and using our expression for VBE1-VBE2,

Since IREF (=IC1) is usually defined in design can be solved for the required value of R2.
Wilson Current Source
With neat diagram explain Wilson current source. (Nov 2009)

 Another current source configuration that possesses increased

output resistance is the Wilson current source.
 The increased ro of the Wilson current source is due to the negative
feedback provided by Q3.
 It provides an output current Io which is very nearly equal to Vref
and also exhibits a very high output resistance.

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Writing a KCL equation at the emitter of Q2

If all three transistors are matched VBE1=VBE2=VBE3, β1=β2=β3, IB1=IB3, and IC1=IC3.
With this information and the relationship between base and collector currents (IB=IC/β), we can
rewrite the expression for IE2 as

Using IC2=αIE2=βIE2/(β+1) and simplifying,

Now, if we sum the currents at the base of Q2,

Solving for IC2, we get an expression for the output current (IC2) in terms of the transistor
parameter β and the input current, IREF:

For reasonable values of β, the second term will be negligible and IC2=Iout=IREF.
Therefore, in addition to the increased output resistance, the Wilson configuration provides an
output that is almost independent of the internal transistor characteristics.
Current sources as Active loads
 The current source can be used as an active load in both analog and digital IC‘s.
 The active load realized using current source in place of the passive load (i.e. a resistor) in the
collector arm of differential amplifier.
 The active load used to achieve high voltage gain without requiring large power supply voltage.
Voltage Sources
Explain the voltage sources with neat circuit diagram.
 A voltage source is a circuit that produces an output voltage V0, which is independent of the load.
 A number of IC applications require a voltage reference point with very low ac impedance and a
stable dc voltage.
 There are two methods used to produce a voltage source, namely,
1. Using the impedance transforming properties of the transistor,
2. Using an amplifier with negative feedback.

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Voltage source circuit using Impedance transformation:

 The voltage source circuit using the impedance

transforming property of the transistor is shown
in.
 The source voltage Vs drives the base of the
transistor through a series resistance RS and the
output is taken across the emitter.
 From the circuit, the output ac resistance
looking figure into emitter is given by

 It is to be noted that, equation is applicable only for small changes in the output current.
 The load regulation parameter indicates the changes in Vo resulting from large changes in output
current Io.
 Reduction in Vo occurs as Io goes from no-load current to full-load current and this factor determines
the output impedance of the voltage sources.
Emitter– follower or Common Collector Type Voltage source:

 The figure shows an emitter follower or common

collector type voltage source.
 This voltage source is suitable for the differential gain
stage used in op-amps. This circuit has the advantages of
1. Producing low ac impedance and
2. Resulting in effective decoupling of adjacent
gain stages.
 The low output impedance of the common-collector stage simulates a low impedance voltage
source with an output voltage level of Vo represented by

 The diode D1 is used for offsetting the effect of dc value VBE , across the E-B junction of the
transistor, and for compensating the temperature dependence of VBE drop of Q1.
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 The load ZL shown in dotted line represents the circuit biased by the current through Q1.
 The impedance R0 looking into the emitter of Q1 derived from the hybrid π model is given by

Voltage Source Using Temperature compensated Avalanche Diode

 The voltage source using common

collector stage has the limitations of its
weakness for changes in bias voltage
VN and the output voltage Vo with
respect to changes in supply voltage
Vcc.
 This is overcome in the voltage source
circuit using the breakdown voltage of
the base- emitter junction shown below.
 The emitter – follower stage of common – collector is eliminated in this circuit, since the
impedance seen looking into the bias terminal N is very low.
 The current source I1 is normally simulated by a resistor connected between Vcc and node n. Then, the
output voltage level V0 at node N is given by V0 = VB +VBE Where VB is the breakdown voltage
of diode DB and VBE is the diode drop across D1.
 The breakdown diode DB is normally realized using the base-emitter junction of the transistor.
The diode D1 provides partial compensation for the positive temperature coefficient effect of VB.
 In a monolithic IC structure, DB and D1 can be conveniently realized as a single transistor with two
individual emitters as shown in figure.
Voltage source using breakdown voltage of the base- emitter junction
 The structure consists of composite connection of two transistors which are diode connected back-
to back.
 Since the transistors have their base to collector terminals common, they can be designed as a
single transistor with two emitters.
 The output resistance R0 looking into the output terminal in figure is given by Ro=RB+VT/ I1
 where RB and VT / I1 are the ac resistances of the base–emitter resistance of diode DB and D1
respectively.
 Typically RB is in the range of 40Ω to 100Ω, and V0 in the range of 6.5V to 9V.

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Voltage Source using VBE as a reference:

 The output stage of op-amp requires stabilized bias voltage source, which can be obtained using
a forward-biased diode connected transistor.
 The forward voltage drop for such a connection is approximately 0.7V, and it changes slightly
with current.
 When a voltage level greater than 0.7V, is needed, several diodes can be connected in series,
which can offer integral multiples of 0.7V.
 Alternatively, the figure shows a multiplier circuit, which can offer voltage levels that need not
be integral multiplied of 0.7V.
 The drop across R2 equals VBE drop of Q1. Considering negligible base current for Q1,
current through R2 is the same as that flowing through R1.
 Therefore, the output voltage V0 can be expressed as

 Hence, the voltage V0 can be any multiple of VBE by

properly selecting the resistors R1 and R2.
 Due to the shunt feedback provided by R1, the
transistor current I1 automatically adjusts itself,
towards maintaining I2 and V0 relatively independent
of the changes in supply voltage.
 The ac output resistance of the circuit R0 is given by,

Voltage References
Explain the voltage references with neat circuit diagram.
 The circuit that is primarily designed for providing a constant voltage independent of changes
in temperature is called a voltage reference.
 The most important characteristic of a voltage reference is the temperature coefficient of the
output reference voltage TCR , and it is expressed as

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 The desirable properties of a voltage reference are:

1. Reference voltage must be independent of any temperature change.
2. Reference voltage must have good power supply rejection which is as independent of the
supply voltage as possible and
3. Output voltage must be as independent of the loading of output current as possible, or in
other words, the circuit should have low output impedance.
 The voltage reference circuit is used to bias the voltage source circuit, and the combination can
be called as the voltage regulator.
 The basic design strategy is producing a zero TCR at a given temperature, and thereby achieving
good thermal ability.
0
 Temperature stability of the order of 100ppm/ C is typically expected.
Voltage Reference circuit using temperature compensation scheme
The voltage reference circuit using basic temperature
compensation scheme is shown below.
 This design utilizes the close thermal coupling achievable among
the monolithic components and this technique compensates the known
thermal drifts by introducing an opposing and compensating drift
source of equal magnitude.
 A constant current I is supplied to the avalanche diode DB and it
provides a bias voltage of VB to the base of Q1.
 The temperature dependence of the VBE drop across Q1 and
those across D1 and D2 results in respective temperature coefficients.
 Hence, with the use of resistors R1 and R2 with tapping across them at point N compensates for
the temperature drifts in the base- emitter loop of Q1. This results in generating a voltage reference
VR with normally zero temperature coefficient.
Voltage Reference circuit using Avalanche Diode Reference:
 A voltage reference can be implemented using the breakdown phenomenon condition of a
heavily doped PN junction.
 The Zener breakdown is the main mechanism for junctions, which breakdown at a voltage of 5V
or less.

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 For integrated transistors, the base-emitter breakdown voltage falls in the range of 6 to 8V.
Therefore, the breakdown in the junctions of the integrated transistor is primarily due to avalanche
multiplication.
 The avalanche breakdown voltage VB of a transistor incurs a positive temperature coefficient,
0
typically in the range of 2mV/0 C to 5mV/ C.
 Figure depicts a current reference circuit using avalanche diode reference.
 The base bias for transistor Q1 is provided through register R1 and it also provides the dc
current needed to bias DB, D1 and D2.
 The voltage at the base of Q1 is equal to the Zener voltage VB added with two diode drops due to
D1 and D2.
 The voltage across R2 is equal to the voltage at the base of Q1 less the sum of the base –
emitter voltages of Q1 and Q2.

 Hence, the voltage across R2 is approximately equal to that across DB = VB. Since Q2 and Q3 act as
a current mirror circuit, current I0 equals the current through R2.

 It shows that, the output current Io has low temperature coefficient, if the temperature coefficient of
R2 is low, such as that produced by a diffused resistor in IC fabrication.
 The zero temperature coefficients for output current can be achieved, if diodes are added in series
with R2, so that they can compensate for the temperature variation of R2 and VB.
 The temperature compensated avalanche diode reference source circuit is shown in figure.
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 The transistor Q4 and Q5 form an active load current mirror circuit. The base voltage of Q1 is
the voltage VB across Zener DB.
 Then, VB = (VBE * n) +VBE across Q1 + VBE across Q2 + drop across R2. Here, n is the number
of diodes.
It can be expressed as

Differentiating for VB, I0, R2 and VBE partially, with respect to temperature T, we get

Dividing throughout by I0 R2, we get

Therefore, zero temperature coefficient of I0 can be obtained, if the above condition is satisfied.
Differential amplifier
Explain the operation of basic differential amplifier.
 The function of a differential amplifier is to amplify the difference between two signals.
 This forms the basic input stage of an integrated amplifier.
 The basic differential amplifier has the following important properties of
o Excellent stability
o High versatility and
o High immunity to interference signals
 The differential amplifier as a building block of the op-amp has the advantages of
o Lower cost, Easier fabrication as IC component and closely matched components.

 The above figure shows the basic block diagram of a differential amplifier, with two input
terminals and one output terminal.
 The output signal of the differential amplifier is proportional to the difference between the two
input signals. V0 = Adm (V1 – V2 )

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 If V1 = V2, then the output voltage is zero. A non-zero output voltage V0 is obtained when V1 and
V2 are not equal.
 The difference mode input voltage is defined as Vm = V1 – V2 and the common mode input voltage
is defined as

 These equation show that if V1 = V2, then the differential mode input signal is zero and common
mode input signal is Vcm = V1 =V2.
Differential Amplifier with Active load:
Draw the circuit of a differential amplifier with current mirror load. Drive an expression for its gain.

Differential amplifier is designed with active loads to increase the differential mode voltage gain.

 The open circuit voltage gain of an op-amp is needed to be as large as possible.

 This is got by cascading the gain stages which increase the phase shift.
 The gain can be increased by using large values of collector resistance. For such a circuit, the
voltage gain is given by Adm = gm Rc
 To increase the gain the Ic Rc product must be made very large.
 However, there are limitations in IC fabrication such as,
1. A large value of resistance needs a large chip area.
2. For large RC, the quiescent drop across the resistor increase and a large power supply will
be required to maintain a given operating current.
3. Large monolithic resistor introduces large parasitic capacitances which limits the frequency
response of the amplifier.
4. for linear operation of the differential pair, the devices should not be allowed to enter into saturation.
 This limits the max input voltage that can be applied to the bases of transistors Q1 and Q2 the
base- collector junction must be allowed to become forward-biased by more than 0.5V.
 The large value of load resistance produces a large dc voltage drop (IEE / 2) Rc, so that the
collector voltage will be VC=Vcc - (IEE/2) RC and it will be substantially less than the supply voltage
Vcc.
 This will reduce the input voltage range of the differential amplifier.
 Due to the reasons cited above, an active load is preferred in the differential amplifier configurations.
BJT Differential Amplifier using active loads:
 A simple active load circuit for a differential amplifier is the current mirror active load as shown
in figure.
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 The active load comprises of transistors Q3 and Q4 with the transistor Q3 connected as a Diode with
its base and collector shorted. The circuit is shown to drive a load RL.
 When an ac input voltage is applied to the differential amplifier, the various currents of the circuit
are given by IC4 = IC3 = IC1 = gmVid/2, where IC4 = IC3 due to current mirror action. IC2 = -
gmVid/2 .
 We know that the load current IL entering the next stage
is IL= IC2-IC4 = - gmVid/2 - gmVid/2 = - gmVid
 Then, the output voltage from the differential amplifier is given by V0= - ILRL = gm RLVid.
 The ac voltage gain of the circuit is given by

 The amplifier can amplify the differential input signals and it provides single-ended output with
a ground reference since the load RL is connected to only one output terminal.
 This is made possible by the use of the current mirror active load.
 The output resistance Ro of the circuit is that offered by the parallel combination of transistors
Q2 (NPN) and Q4 (PNP). It is given by Rr = ro2 || ro4
Analysis of BJT differential amplifier with active load:
 The collector currents of all the transistors are equal. IC1 = IC2 = IC3 =IC4 = IEE/2 .
 The Collector -emitter voltages of Q1 and Q2 are given by
VCE1-VCE2 =VC-VE=V CC - VEB-(-VEB)=
VCC
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 Eqation shows that, the offset is higher than that of a resistive loaded differential amplifier A.

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 This can be reduced by the use of emitter resistors for Q 3 and Q 4 , and a transistor Q5 in the
current mirror load.
Common Mode Rejection Ratio (CMRR)
The common mode rejection ratio is a measure of the differential amplifier’s ability to
reject the common mode signal and amplify the differential mode signal.

Thus, the larger the input transconductance or REE, the larger the common mode rejection ratio.
Input common-mode range (ICMR)
The input common-mode range is the range of common-mode voltages over which
the differential amplifier continues to sense and amplify the difference signal with the same gain.
Typically, the ICMR is defined by the common-mode voltage range over which all
MOSFETs remain in the saturation region and all BJTs remain in the active region.
Output offset voltage ( VOS(out))
The output offset voltage is the voltage which appears at the output of the differential
amplifier when the input terminals are connected together.
Input offset voltage ( VOS(in) = VOS)
The input offset voltage is equal to the output offset voltage divided by the differential voltage
gain.

----------------------------------------------------------------------------------------------------------------------------------
Basic information about operational amplifiers
Explain about Ideal Op-amp in detail.
An operational amplifier is a direct coupled high gain amplifier consisting of one or
more differential amplifiers, followed by a level translator and an output stage.
It is a versatile device that can be used to amplify ac as well as dc input signals & designed
for computing mathematical functions such as addition, subtraction ,multiplication, integration &
differentiation.

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Ideal characteristics of OPAMP

1. Open loop gain infinite
2. Input impedance infinite
3. Output impedance zero
4. Bandwidth infinite
5. Zero offset, ie, Vo=0 when V1=V2=0
Ideal Vs Practical Op-Amp
Parameter Ideal Practical
5
Open Loop gain A µ 10
Bandwidth BW µ 10-100Hz
nput Impedance Zin µ >1MW
Output Impedance Zout 0W 10-100 W
Output Voltage Vout Depends only on Depends slightly on
Vd = (V+-V-) Differential average input Vc = (V+
mode signal +V-)/2 Common- Mode
signal
CMRR µ 10-100Db

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General operational amplifier stages:

Explain the general stages of op-amp with block diagram.

Input stage:
 The input stage, or differential amplifier, provides the common-mode rejectionl to op-amp operation.
 It also supplies the large input impedance that allows coupling to a high impedance source without
loss of signal level.
 The differential amplifier may also provide voltage gain if the output of the differential amplifier
is connected to an emitter follower with a large emitter resistor.
 The Emitter Follower amplifier will provide a high impedance load to the differential amplifier
to obtain a high voltage gain.
Intermediate stage:
 The intermediate stages are shown as the voltage gain stage and the level shifter in the figure above.
 The voltage gain stage usually consists of one or more CE amplifiers to provide the bulk of the
overall voltage gain.
 Linear operational amplifiers are direct coupled to eliminate the need for coupling capacitors that
are too large to be placed on an IC chip.
 The level shifter stage may be one or more level shifters that are included to ensure that there is no
dc offset in the output signal.
 In addition, these intermediate stages may be used to convert the signal from differential
(double- ended) mode to single-ended mode.
Output stage:
 The output stage, or power output stage, also serves a dual purpose.
 It must supply the current required by the load without dissipating too much power in the
output transistors.
 Also, the output stage should provide a low output impedance to allow coupling to a low
impedance load without loss of gain.

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Internal circuit diagrams of IC 741

AC Characteristics:
Explain the AC characteristics of op-amp.
For small signal sinusoidal (AC) application one has to know the ac characteristics such
as frequency response and slew-rate.
1.9.1 Frequency Response:
 The variation in operating frequency will cause variations in gain magnitude and its phase angle.
 The manner in which the gain of the op-amp responds to different frequencies is called the
frequency response.
 Op-amp should have an infinite bandwidth BW =∞ (i.e.) if its open loop gain in 90dB with
dc signal its gain should remain the same 90 dB through audio and onto high radio frequency.
 The op-amp gain decreases (roll-off) at higher frequency.
 There is a capacitive component in the equivalent circuit of the op-amp, decrease gain after a
certain frequency reached.
 For an op-amp with only one break (corner) frequency all the capacitors effects can be
represented by a single capacitor C.
 Below fig is a modified variation of the low frequency model with capacitor C at the output.

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Equivalent circuit of practical circuit

 There is one pole due to RoC and one -20dB/decade. The open loop voltage gain of an op-amp with
only one corner frequency is obtained from above fig.
 f1 is the corner frequency or the upper 3 dB frequency of the op-amp.
 The magnitude and phase angle of the open loop volt gain are f1 of frequency can be written as,
The magnitude and phase angle characteristics:
1. For frequency f<< f1 the magnitude of the gain is 20 log AOL in db.
2. At frequency f = f1 the gain in 3 dB down from the dc value of AOL in db. This frequency f1 is
called corner frequency.
3. For f>> f1 the fain roll-off at the rate off -20dB/decade or -6dB/decade.

Frequency response of op amp

1.9.2 Circuit
Stability:
Explain various stability criteria of op-amp.
 A circuit or a group of circuit connected together as a system is said to be stable, if its output
reaches a fixed value in a finite time.
 A system is said to be unstable, if its output increases with time instead of achieving a fixed value.
 Bode plots are compared of magnitude Vs Frequency and phase angle Vs frequency. Any system
whose stability is to be determined can represented by the block diagram.

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Fig. 1.21 Feedback loop system

 The block between the output and input is referred to as forward block and the block between the
output signal and feedback signal is referred to as feedback block.
 The content of each block is referred as transfer frequency. From fig. we represented it by AOL (f)
which is given by
AOL (f) = V0 /Vin if Vf = 0 ----- (1)
where AOL (f) = open loop volt gain.
 The closed loop gain Af is given by AF = V0 /Vin = AOL / (1+(AOL ) (B) ----(2)
 B = gain of feedback circuit. B is a constant if the feedback circuit uses only resistive components.
 Once the magnitude Vs frequency and phase angle Vs frequency plots are drawn, system stability
may be determined as follows
1. Method 1:
Determine the phase angle when the magnitude of (AOL) (B) is 0dB (or) 1.
If phase angle is >-180, the system is stable.
However, the some systems the magnitude may never be 0, in that cases method 2, must be used.
2. Method 2:
Determine the phase angle when the magnitude of (AOL) (B) is 0dB (or) 1.
If phase angle is > - 180 , If the magnitude is –ve decibels then the system is stable.
However, the some systems the phase angle of a system may reach -1800, under such
conditions method 1 must be used to determine the system stability.
Frequency Compensation:
Need:
• Frequency compensation is needed when large bandwidth and lower closed loop gain is desired.
• Compensating networks are used to control the phase shift and hence to improve the stability.
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Frequency compensation methods:

1. External compensation
-Dominant- pole compensation
-Pole- zero compensation
2. Internal compensation
DC Characteristics of op-amp:
Explain the DC characteristics of op-amp.
DC characteristics of opamp are,
 Input bias current
 Input offset current
 Input offset voltage
 Thermal drift
Input bias current:
Input bias current IB as the average value of the base currents entering into terminal of an op-
+ -
amp IB=(IB + IB ) / 2

Input offset current

The difference between the bias currents at the input terminals of the op- amp is called as input
offset current. The input terminals conduct a small value of dc current to bias the input transistors. Since
the input transistors cannot be made identical, there exists a difference in bias currents
+ -
IOS= IB - IB
Input offset voltage
A small voltage applied to the input terminals to make the output voltage as zero when the two input
terminals are grounded is called input offset voltage

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THERMAL DRIFT
Bias current, offset current and offset voltage change with temperature. A circuit carefully nulled at
o o
25 c may not remain so when the temperature rises to 35 c. This is called drift.
Slew Rate
Define slew rate and describe the methods to improve slew rate.
• The slew rate is defined as the maximum rate of change of output voltage caused by a step input
voltage.
• An ideal slew rate is infinite which means that op-amp’s output voltage should change instantaneously
in response to input step voltage
• Slew rate is the maximum rate of change of output voltage with respect to time. Specified in V/μs.
Reason for Slew rate:
• There is usually a capacitor within 0, outside an op-amp oscillation. It is this capacitor
• which prevents the o/p voltage from fast changing input. The rate at which the volt across
the capacitor increases is given by
dVc/dt = I/C --------(1)
• I -> Maximum amount furnished by the op-amp to capacitor C. Op-amp should have the
either a higher current or small compensating capacitors.
Open – loop op-amp Configuration:
Explain the open loop configuration of op-amp with example circuits. (Nov 2016)
The term open-loop indicates that no feedback in any form is fed to the input from the output.
When connected in open – loop the op-amp functions as a very high gain amplifier. There are three
open – loop configurations of op-amp namely,
1. Differential amplifier
2. Inverting amplifier
3. Non-inverting amplifier
The above classification is made based on the number of inputs used and the terminal to which
the input is applied. The op-amp amplifies both ac and dc input signals. Thus, the input signals can be
either ac or dc voltage.
Open Loop Differential Amplifier:
In this configuration, the inputs are applied to both the inverting and the non- inverting
input terminals of the op-amp and it amplifies the difference between the two input voltages.

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Figure shows the open-loop differential amplifier

configuration. The input voltages are represented by
Vi1 and Vi2.
The source resistance Ri1 and Ri2 are
negligibly small in comparison with the very high
input resistance offered by the op-amp, and thus the
voltage drop across these source resistances is
assumed to be zero.

The output voltage V0 is given by

V0 = A (Vi1 – Vi2)
where A is the large signal voltage
gain.
Thus the output voltage is equal to the voltage gain A times the difference between the two
input voltages. This is the reason why this configuration is called a differential amplifier. In open
– loop configurations, the large signal voltage gain A is also called open-loop gain A.
Inverting amplifier:

 In this configuration the input signal is applied to the

inverting input terminal of the op - amp and the non-inverting
input terminal is connected to the ground.
 Figure shows the circuit of an open loop inverting amplifier.
 The output voltage is 180 out of phase with respect to the
input and hence, the output voltage V0 is given by,
V0 = -AVi.
 Thus, in an inverting amplifier, the input signalis amplified
0
by the open-loop gain A and in phase shifted by 180 .
Non-inverting Amplifier:

 Figure shows the open – loop non- inverting

amplifier.
 The input signal is applied to the non-
inverting input terminal of the op-amp and

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the inverting input terminal is connected to the ground.

 The input signal is amplified by the open – loop gain A and the output is in-phase with input signal.
V0 = AVi
Limitations of Open – loop Op – amp configuration:
 In the open – loop configurations, clipping of the output waveform can occur when the output
voltage exceeds the saturation level of op-amp. This is due to the very high open – loop gain of the
op-amp.

 The open – loop gain of the op – amp is not a constant and it varies with changing temperature and
variations in power supply. Also, the bandwidth of most of the open- loop op amps is negligibly
small. This makes the open – loop configuration of op-amp unsuitable for ac applications.
Closed – loop op-amp configuration:
Present the INV and Non INV amplifier of closed loop configurations. (Dec 2018)
 The op-amp can be effectively utilized in linear applications by providing a feedback from the
output to the input, either directly or through another network.
 If the signal feedback is out - of- phase by 1800 with respect to the input, then the feedback is
referred to as negative feedback or degenerative feedback.
 Conversely, if the feedback signal is in phase with that at the input, then the feedback is referred
to as positive feedback or regenerative feedback.
 An op – amp that uses feedback is called a closed – loop amplifier. The most commonly
used closed – loop amplifier configurations are 1. Inverting amplifier (Voltage shunt amplifier)
2. Non- Inverting amplifier (Voltage – series Amplifier)

The inverting close‐loop configuration

 External components R1 and R2 form a close loop
 Output is fed back to the inverting input terminal
 Input signal is applied from the inverting terminal

Inverting configuration using ideal op amp

 The required conditions to apply virtual short for op-amp circuit:
o Negative feedback configuration
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EC3451 / Linear Integrated Circuits
o Infinite open‐ loop gain
 Closed loop gain: G ≡ VO /VI = -R2 /R1
 Infinite differential gain: V2 - V1 = VO /A = 0
 Infinite input impedance: i2 = i1 = 0
 Zero output impedance: VO = V1 - i1R2 = - vI R2 /R1
 Voltage gain is negative: Input and output signals are out of phase
 Closed loop gain depends entirely on external passive components (independent of op-amp gain)
 Close loop amplifier trades gain (high open loop gain) for accuracy (finite but accurate closed
loop gain)

Equivalent circuit model for the inverting configuration

 Input impedance: Ri ≡vI /iI = vI / (vI /R1) = R1

 For high input closed loop impedance, R1
should be large, but is limited to provide sufficient
G
 In general, the inverting configuration suffers
 Output impedance: Ro = 0 from a low input impedance
 Voltage gain: Avo = - R2/R1
Non inverting closed loop
Configuration:

 External components R1 and R2 form a close loop

 Output is fed back to the inverting input terminal
Input signal is applied from the noninverting
terminal
Non inverting configuration using ideal op amp
 The required conditions to apply virtual short for op‐ amp circuit:
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o Negative feedback configuration

o Infinite open loop gain
 Closed loop gain: G ≡ VO /VI = 1 + R2 /R1
 Infinite differential gain: V+ - V- = VO /A = 0
 Infinite input impedance: i2 = i1 = V- /R1
 Zero output impedance: VO = V- + i1R2 = VI (1 + R2 /R1)
 Closed loop gain depends entirely on external passive components (independent of op amp gain)
 Close loop amplifier trades gain (high open loop gain) for accuracy (finite but accurate closed
loop gain)

 Equivalent circuit model for the noninverting configuration

Input impedance: Ri = Infinite

 Output impedance: Ro = 0
 Voltage gain: Avo = 1 + R2 /R1

Virtual Ground:
Virtual ground (or virtual earth) is a node of a circuit that is maintained at a steady
reference potential, without being connected directly to the reference potential.
A voltage divider, using two resistors, can be used to create a virtual ground node. If two
voltage sources are connected in series with two resistors, it can be shown that the midpoint becomes
a virtual ground if

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Fig: Op-amp inverting amplifier

 An active virtual ground circuit is sometimes called a rail splitter.

 Such a circuit uses an op-amp or some other circuit element that has gain.
 Since an operational amplifier has very high open-loop gain, the potential difference between
its inputs tend to zero when a feedback network is implemented.
 To achieve a reasonable voltage at the output (and thus equilibrium in the system), the output supplies
the inverting input (via the feedback network) with enough voltage to reduce the potential
difference between the inputs to microvolts.
 The non-inverting (+) input of the operational amplifier is grounded; therefore, its inverting (-)
input, although not connected to ground, will assume a similar potential, becoming a virtual
ground if the opamp is working in its linear region (i.e., outputs have not saturated).

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 EC3451 / Linear Integrated Circuits

UNIT I-BASICS OF OPERATIONAL AMPLIFIERS

1. Define an integrated circuit. (or) what is an integrated circuit?

An integrated circuit (IC) is a miniature, low cost electronic circuit consisting of active
and passive components fabricated together on a single crystal of silicon. The active components are
transistors and diodes and passive components are resistors and capacitors.
2. Mention the advantages of integrated circuits.
The advantages of integrated circuits are,
 Miniaturization and hence increased equipment density.
 Cost reduction due to batch processing.
 Increased system reliability due to the elimination of soldered joints.
 Improved functional performance.
 Matched devices.
 Increased operating speeds. Reduction in power consumption
3. Define Current mirror circuit & List out its advantages. (or) What is current
mirror?

A circuit in which the output current is equal to input current is called current mirror
circuit. In current mirror circuit, the output current is the mirror image of the input current.
Advantages:
a. High CMRR
b. Easy to design
4. Draw the current mirror circuit.

5. Give the basic concept of current source.

A constant current source makes use of the fact that for a transistor in the active mode of
operation, the collector current is relatively independent of collector voltage.

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6. What are the applications of current sources?

The applications of current sources are:
i. The emitter resistance in differential amplifier to increase CMRR.
ii. As an active load to provide high a.c.resistance without disturbing the d.c conditions.
7. Give the limitation of current mirror circuit.
The limitation of current mirror circuit is whenever we need low value of current, the value of
resistance is high and it cannot be fabricated economically in IC circuits
8. Justify the reasons for using current sources in integrated circuits.
1. Superior insensitivity of circuit performance to power supply variations and temperature.
2. More economical than resistors in terms of die area required providing bias currents of
small value.
3. When used as load element, the high incremental resistance of current source results in high
voltage gains at low supply voltages.
9. What is the advantage of widlar current source over constant current source?
Using constant current source, the output current of small magnitude (microamp range)
is not attainable due to the limitations in chip area. Widlar current source is useful for obtaining
small output currents. Sensitivity of widlar current source is less compared to constant current source.
10. Mention the advantages of Wilson current source.
The advantages of Wilson current sources are:
 Provides high output resistance.
 Offers low sensitivity to transistor base currents.
11. What is voltage source?
A voltage source is a circuit that produces an output voltage (V0), which is independent of
the load driven by the voltage source or the output current supplied to the load.

12. Give advantages of emitter follower voltage source.

The advantages of emitter follower are:
a. It produces low ac impedance
b. Gives effective decoupling of adjacent gain stages.
13. Write the limitation of emitter follower or common collector voltage source.
Emitter follower voltage source is weak and without protection for changes in bias voltage and
the output voltage with respect to changes in supply voltage.

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14. What is voltage reference?

The circuit that is designed for providing a constant voltage independent of changes in
temperature is called a voltage references.
15. Define temperature coefficient.
Temperature coefficient is the measure of the ability of the circuit to maintain the
standard output voltage under varying temperature conditions.
o
TC(Vo) = dVo/ dT in mV/ C
16. What are the properties of voltage reference?
The Properties of voltage reference are:
i. Reference voltage must be independent any temperature change.
ii. It must have good power supply rejection which is independent of the supply voltage.
iii. The circuit should have low output impedance.
17. What are the parameters of voltage reference circuits?
The Parameters of voltage reference are:
i. Line regulation
ii. Load regulation
iii. Long term stability
iv. Ripple rejection ratio.
18. Define line regulation.
Line regulation is defined as the ratio of change in output voltage to the change in input voltage.
Line regulation = ∆Vo/∆Vi
Where ∆Vo – changes in the output line voltage
∆Vi - changes in the input line voltage
It is expressed in mV/V
19. Define load regulation.
Load regulation is defined as the ratio of change in output voltage to the change in load
current. Load regulation = ∆Vo/∆IL
Where ∆Vo – changes in the output line voltage
∆IL - changes in the load current
It is expressed in mV/mA
20. What is long term stability?
The ability of the circuit to maintain the output voltage constant with respect to time is given
by the parameter long term stability.It is measured in ppm/1000 hours.
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21. Define Ripple Rejection Ratio (RRR).

The ability of the circuit to reject input ripples and an indication of how much ripples are
present at the output due to input is given by the factor RRR.It is defined as RRR = 20 log10[Vri/Vro]
Where Vri - Input ripple and Vro –Output ripple
22. What are the limitations in a temperature compensated zener-reference source?
Limitations of zero reference source:
A power supply voltage of at least 7 to 10 V is required to place the diode in the
breakdown region and that substantial noise is introduced in the circuit by the avalanching diode.
23. What is differential amplifier?
A circuit which amplifies the difference between two input voltage signals. Hence it is
also called as difference amplifier.

24. Draw the block diagram for differential amplifier.

25. What are the properties of differential amplifier?

The Properties of differential amplifier are:

i. Excellent stability
ii. High versatility
iii. High immunity to interference signals.
26. Give the advantages of differential amplifier.
The advantages of differential amplifier are:

1. Lower cost
2. IC fabrication is easy
3. Closely matched components
27. Define differential gain.
Differential gain is defined as the ratio of the output voltage to the difference voltage.

Ad = Vo /Vd , In decibel 20 log10 Ad = 20 log10Vo /Vd

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28. Why are active loads preferred than passive loads in the input stage of an operational amplifier?

Differential amplifier designed with active load to increase the CMRR. The gain increased
by using large value of collector resistances. If the collector resistance is large then limitation in IC
fabrications are large chip area and large bias voltage need.
29. Define CMRR.
The common Mode Rejection ratio (CMRR) is defined as the ration of difference mode

gain to the common mode gain

CMRR = |ADM/ACM|
It is expressed in decibel (dB)

30. What are the draw backs of using large RC in differential amplifier?
The main draw backs are: It requires large chip area. For larger RC quiescent drop, a large
power supply will be required to maintain a given quiescent collector current
31. What is active load? Where it is used and why?
The requirement to increase the gain is same that the collector resistance (RC) should not disturb d.c
conditions while it must provide large resistance for a.c purposes. The current mirror which has very low
d.c resistance (dV/dI) and higher a.c resistance (dv/di) can be used as a collector load instead of RC. Such
a load is called as active load.
32. What is an operational amplifier?
An operational amplifier is a direct-coupled, high gain amplifier consisting of one or
more differential amplifier. By properly selecting the external components, it can be used to perform a
variety of mathematical operations.

33. What are the popular IC packages available?

The IC packages available are:
 Metal can package.
 Dual-in-line package.
 Ceramic flat package.
34. List out the ideal characteristics, and draw the equivalent diagram of an OP-AMP Mention
any four important characteristics of ideal operational amplifier.
The Ideal characteristics of op-amp are:
 Open loop voltage gain is infinity.

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 Input impedance is infinity.

 Output impedance is zero.
 Bandwidth is infinity.
 Zero offset
 Infinite CMRR
 Infinite slew rate
 PSRR = 0
35. Draw the circuit symbol for op-amp.

36. Define Virtual ground property of an OP-AMP.

A virtual ground is a ground which acts like a ground. It may not have physical
connection to ground. This property of an ideal op-amp indicates that the inverting and non-
inverting terminals of op-amp are at the same potentials. The non-inverting input is grounded for
the inverting amplifier circuit. This means that the inverting input of the op-amp is also at ground
potential.
37. Draw the equivalent diagram of an OP-AMP.

38. What are the stages in op-amp?

The stages of op-amp are:
a. Input stage
b. Gain stage
c. Output stage
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39. What are the requirements of the input stage of an op-amp?

The differential amplifier eliminates the need for an emitter bye-pass capacitor. So, differential
amplifier is used as an input stage in op-amp ICs
40. In what way 741S is better than 741?
741S is a military grade op-amp with higher slew rate
41. Why is IC741 op-amp not used for high frequency applications?
Op-amp IC741 has very low slew rate and therefore cannot be used for high frequency
applications
42. Define Input bias current.
Input bias current IB is the average of the currents that flow into the inverting and non-
inverting input terminals of the op-amp.
i.e. IB = (IB1+IB2)/2
43. Define Input offset current.
The algebraic difference between the current into the inverting and non-inverting terminals
is referred to as input offset current Iio. Mathematically it is represented as Iios= |IB+ - IB- |
Where I B+ is the current into the non-inverting input terminals.
IB- is the current into the inverting input terminals.
44. What is Input offset voltage?
Input offset voltage is the voltage required to be amplified at the input for making output
voltage to zero volts.
45. Determine the slew rate of the op-amp. Or Define slew rate.
Slew rate can be defined as the maximum rate of change of output voltage of op-amp
with respect to time. It is expressed as S = (dVo / dt) max in V/Sec.Where slew rate S = 2П f Vm in
V/Sec.
46. What is a compensating network?
The networks formed by components such as resistors and capacitors for modifying the rate
of change of gain and the phase shift is called as compensating network
47. When are the internally compensating systems used?
In applications where the op-0amp is required to amplify relatively slow changing signals and does not
require good high frequency response, internally compensating systems are used.
48. In response to a square wave input, the output of an op-amp changed from -3V to +3V over a
time interval of 0.25μs, find slew rate.
Slew rate= ΔVo/Δt = 6/0.25=24 V/µs.
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49. List the types of frequency compensation.

a. External frequency compensation
1. Dominant pole freq. compensation
2. Pole zero freq. compensation
b. Internally frequency compensation.
50. List the DC characteristics of opamp.
DC characteristics of op-amp are:
1. Input bias current
2. Input offset current
3. Input offset voltage
4. Thermal drift.
51. List the AC characteristics of opamp. (Dec 2018)
AC characteristics of op-amp are:
1. Frequency response
2. Stability of an op-amp,
3. Frequency compensation
4. Slew rate.
52. What is the need for frequency compensation in practical op-amps?
Frequency compensation is needed when large bandwidth and lower closed loop gain
is desired. Compensating networks are used to control the phase shift and hence to improve the stability.
53. What are the merits of Dominant-pole compensation?
The merits of Dominant-pole compensation are:
1. Noise immunity of the system is improved.
2. Open-loop bandwidth is reduced.
54. What is the maximum undistorted amplitude, that a sine wave input of 10 KHz, can produce, at
the output of an op-amp whose slew rate is 0.5V/µs?
Given F=10KHZ and slew rate (S) =
0.5V/µs Solution: Slew rate (S) = 2πfVm V/
µs. Maximum amplitude (Vm) = S/2πf
=7.92V
55. The op-amp has a gain of 12 million. Express the gain in dB.
Gain in dB=20log(gain)
6
20 log(20*10 )=
141.6dB
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56. What is the need for frequency compensation in practical op-amps?

Frequency compensation is needed when large bandwidth and lower closed loop gain is desired.
Compensating networks are used to control the phase shift and hence to improve the stability.
57. Why open loop op-amp configurations are not used in linear applications?

a.The open loop gain of the op-amp is very high. Therefore only the smaller signals having low
frequency may be amplified accurately without distortion.
b.Open loop Voltage gain of the op-amp is not a constant voltage gain varies with changes in
temperature and power supply as well as mass production techniques. This makes op-amp
unsuitable for many linear applications
c.Bandwidth of most open loop op-amps is negligibly small or almost zero therefore op-amp is
impractical in ac applications.
58. In practical op-amps, what is the effect of high frequency on its performance?
The open-loop gain of op-amp decreases at higher frequencies due to the presence of
parasitic capacitance. The closed-loop gain increases at higher frequencies and leads to instability.
59. What are the advantages of negative feedback?
Advantages of negative feed back are:
1. It reduces the gain and makes it controllable
2. It reduces the possibility of distortion
3. It increases the bandwidth
4. It increases the input resistance of the op-amp
5. It decreases the output resistance of the op-amp
6. It reduces the effect of temperature, power supply on the gain of the circuit.

60. Define Differential Mode gain.

Gain of an amplifier is defined as VOUT/VIN. For the special case of a differential amplifier, the input VIN
is the difference between its two input terminals, which is equal to (V1-V2) as shown in the following
diagram.

34
61. What are the two methods can be used to produce voltage sources?
There are two methods used to produce a voltage source, namely,
3. Using the impedance transforming properties of the transistor,
4. Using an amplifier with negative feedback.