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1. Explain the significance of current mirror in an Op-Amp?
Ans: Current mirrors are basic building blocks of analog design. Fig. 1 shows the basic npn
current mirror. For its analysis, we assume identical transistors and neglect the Early effect, i.e.
we assume VA → ∞. This makes the saturation current IS and current gain β independent of the
collector base voltage VCE . The input current to the mirror is labeled IREF . This current might
come from a resistor connected to the positive rail or a current source realized with a transistor or
another current mirror. The emitters of the two transistors are shown connected to ground. These
can be connected to a dc voltage, e.g. the negative supply rail.
The simplest way to solve for the output current is to sum the currents at the node where IREF
enters the mirror. Because the two transistors have their base-emitter junctions in parallel, it
follows that both must have the same currents. Thus, we can write the equation
Solution for IO yields
2. Discuss in-detail about Op-Amp, Compensating networks
Ans:
An RC filter introduces a phase shift between 0 and π/2. If one cascades these filters, the phase
shifts can accumulate, producing at some frequency ωπ the possibility of a phase shift of ±π. This
is dangerous for op-amp circuits employing negative feedback, as a phase shift of π converts
negative feedback to positive feedback. This in turn tends to compound circuit instabilities and
can lead to oscillating circuits (as we do on purpose for the RC relaxation oscillator).
So it is perhaps easy to simply not include such phase shifts in the feedback loop.
However, at high frequencies (f ~ 1 MHz or more), unintended stray capacitances can become
significant. In fact, within the op-amp circuits themselves, this is almost impossible to eliminate.
Most manufacturers of op-amps confront this issue by intentionally reducing the open-loop gain
at high frequency. This is called compensation. It is carried out by bypassing one of the internal
amplifier stages with a high-pass filter.
Integrated Circuits offer a wide range of applications and could be broadly classified as:
Digital ICs
Linear ICs
Based upon the requirements, several distinctly differ IC technology namely, Monolithic
technology, thin & thick technology and hybrid technology have been developed.
In Monolithic integrated circuits, all circuit components, both active and passive elements
and their interconnections are manufactured into or on top of a single chip of silicon. The
monolithic circuit is ideal for applications where integrated circuits are requires in very large
quantities and hence provides lowest per- unit cost and highest order of reliability. In Hybrid
circuits, separate component parts are attached to a ceramic substrate and interconnected by
means of either metallization pattern or wire bonds. This technology is more adaptable to small
quantity custom circuits. Based upon the active devices used, ICs can be classified as bipolar and
unipolar Bipolar and Unipolar ICs may further be classified depending upon the isolation
technique or type of FET used.
Integrated Circuits
Monolithic ICs
Unipolar
Bipolar
P-N junction
isolation
Hybrid ICs
Di-electric
isolation
MOSFET
JFET
3. Write about classification of ICs on the basis of application & Chip Complexity
Invention of transistor
1947
Development of Silicon Transistor
1955 – 1959
Silicon Planar Technology
Junction transistor diode
1959
First ICs, Small Scale Integration
(SSI)
3-30 gates / chip or 100 transistors / chip (
Logic gates, Flip-Flops)
1960-65
30 to 300 gates / chip or 100 to 1000
transistors / chip
Medium Scale Integration (MSI)
(Counters, Multiplexers, Adders)
1965 – 1970
Large Scale Integration (LSI)
300 to 3000 gates / chip or 1000 to 20000
transistors / chip (8 Bit Microprocessor,
RAM, ROM)
1970 – 1980
Very Large Scale Integration
(VLSI)
More than 3000 gates / chip or 20, 000 to 1,
00, 000 transistors /chip (16 and 32 bit
microprocessors)
1980 – 1990
Ultra Large Scale Integration
(ULSI)
106 – 107 transistors / chip (Special
Processors, virtual reality machines, smart
sensors)
1990 – 2000
Giant Scale Integration
> 107 transistors / chip
4. Differentiate between Monolithic ICs & Hybrid ICs with suitable examples
Ans:
MONOLITHIC IC
It is that the components such as semiconductor materials, passive components (resistor,
capacitor, inductor and transformer) etc ,which is used in ic is fabricated within the single
semiconductor(say silicon) wafer and there is no such shortages between the components
and this have separate pn junctions. All the functions of the components are within the
small silicon wafer and the technology used here to manufacture this ic is PLANAR
TECHNOLOGY
HYBRID IC
It is that the components are connected individualy through the wires and the surrounded
by the special polymer called EPOXY ,this is the technology used while comparing with
the monolithic ic it is big in size.it contain the components all such used in the monolithic
type
5. Write about the temperature ranges and power supply requirements of ICs.
Ans:
6. With a neat diagram explain about construction of differential amplifier with three opamps.
7. Explain how variable gain can be achieved with differential amplifier
8. Discuss about stability of an op-amp.
Ans:
Op-amps are rarely used in open loop configuration because of its high gain. Let us now
consider the effect of feedback on op-amp frequency response. Consider an op-amp amplifier of
fig. it uses resistor feedback network and may be used as an inverting amplifier for v2 = 0 and as
non-inverting amplifier for v1 = 0.
From the negative feedback concepts, we may write the closed loop transfer function as
ACL = A/ (1+A.β)
Where A is the open loop voltage gain and β is the feedback ratio. If characteristic equation
(1+Aβ) = 0, the circuit will become just unstable, that is leads into sustained oscillation.
Rewriting the characteristic equation as 1-(-Aβ) = 0 leads to loop gain, -Aβ = 1
Since Aβ is a complex quantity, the magnitude condition become
|Aβ| = 1
And phase condition is
-Aβ = 0 ( or multiple of 2π)
Or
Aβ = π (or odd multiple of π)
In the given circuit, feedback is a resistive network, so it does not provide any phase shift. Since
op-amp is used in the inverting mode, it provides a phase shift of 180o at low frequencies.
However at high frequencies, due to each corner frequency, an additional phase shift of
maximum -90o can take place in open loop gain A. So for two corner frequencies, a maximum
phase shift that can be associated with gain A is -180o. Thus at high frequencies, it is quite
possible that for some value of β, the magnitude of Aβ becomes unity when A has an additional
phase shift of 180o which makes the total phase shift equal to zero. In this case, there is every
possibility that the amplifier may begin to oscillate as both the magnitude and phase conditions.
This may be noted that oscillation is just the starting point of instability, or, to be more precise, it
is just at the verge instability. This instability means unbounded output;
(1+Aβ) < 1 or
(1+Aβ) < 0 i.e. negative
And then ACL > A, i.e., the coled loop gain increases and leads to instability. The phase
contribution by the resistive feedback network is zero. At low frequency, the additional phase
contribution of A is zero, so Aβ > 0 and obviously ACL < A and the system is stable. But at high
frequencies, the system A having three corner frequencies or three RC pole pair, there is a
chance of open loop gain A to contribute a maximum of -270o phase shift and for which Aβ may
become negative and instability occurs at high frequencies.
9. Draw high frequency model of an op-amp and explain its working
Ideally an op-amp should have an infinite bandwidth. This means that, if its open-loop gain is
90 dB with dc signal its gain should remain the same 90dB through audio and on to high radio
frequencies. The practical op-amp gain, however, decreases at higher frequencies.
When the gain of the op-amp to roll-off after a certain frequency is reached, there must be a
capacitive component in the equivalent circuit of the op-amp. This capacitance is due to the
physical characteristics of the device used and the internal construction of op-amp. For an opamp with only one break (corner) frequency , all the capacitor effects can be represented by a
single capacitor C. The figure represents the high frequency model of an op-amp with a single
corner frequency. There is one pole due to RoC and obviously one -20dB/Decade roll-off comes
into effect.
The open loop voltage gain of an op-amp with one carner frequency is obtained as
or
Where
Is the corner frequency or the upper 3-dB frequency of the op-amp. The magnitude and the phase
angle of the open loop voltage gain are function of frequency and can be written as
The magnitude and phase characteristics from above equations are shown below. It can be seen
that
i. For frequency f<<f1, the magnitude of the gain is 20 log AOL in dB.
ii. At frequency f = f1, the gain is 3gB down from the dc value of AOL in dB. This
frequency f1 is called corner frequency.
iii. For f >> f1, the gain rolls-off at the rate of -20 dB/decade or -6dB/ocatave.
It can further be seen from the phase characteristics that the phase angle is zero at frequency f=0.
At corner frequency f1 the phase angle is -45o (lagging) and at infinite frequency the phase angle
is -90o. This shows that a maximum of 90o phse change can occur in an op-amp with a single
capacitor. the voltage transfer function is s-domain can be written as
10. Explain the role of negative feedback in operational amplifiers.
Ans:
If we connect the output of an op-amp to its inverting input and apply a voltage signal to
the non-inverting input, we find that the output voltage of the op-amp closely follows that input
voltage (I've neglected to draw in the power supply, +V/-V wires, and ground symbol for
simplicity):
As Vin increases, Vout will increase in accordance with the differential gain. However, as Vout
increases, that output voltage is fed back to the inverting input, thereby acting to decrease the
voltage differential between inputs, which acts to bring the output down. What will happen for
any given voltage input is that the op-amp will output a voltage very nearly equal to Vin, but just
low enough so that there's enough voltage difference left between Vin and the (-) input to be
amplified to generate the output voltage.
The circuit will quickly reach a point of stability (known as equilibrium in physics),
where the output voltage is just the right amount to maintain the right amount of differential,
which in turn produces the right amount of output voltage. Taking the op-amp's output voltage
and coupling it to the inverting input is a technique known as negative feedback, and it is the key
to having a self-stabilizing system (this is true not only of op-amps, but of any dynamic system
in general). This stability gives the op-amp the capacity to work in its linear (active) mode, as
opposed to merely being saturated fully "on" or "off" as it was when used as a comparator, with
no feedback at all.
Because the op-amp's gain is so high, the voltage on the inverting input can be maintained almost
equal to Vin. Let's say that our op-amp has a differential voltage gain of 200,000. If Vin equals 6
volts, the output voltage will be 5.999970000149999 volts. This creates just enough differential
voltage (6 volts - 5.999970000149999 volts = 29.99985 µV) to cause 5.999970000149999 volts
to be manifested at the output terminal, and the system holds there in balance. As you can see,
29.99985 µV is not a lot of differential, so for practical calculations, we can assume that the
differential voltage between the two input wires is held by negative feedback exactly at 0 volts.
One great advantage to using an op-amp with negative feedback is that the actual voltage gain of
the op-amp doesn't matter, so long as its very large. If the op-amp's differential gain were
250,000 instead of 200,000, all it would mean is that the output voltage would hold just a little
closer to Vin (less differential voltage needed between inputs to generate the required output).
11. How does negative feedback affect, the performance of an inverting amplifiers
Ans: Effects of Negative Feedback on Op-Amp Impedances
– Negative feedback affects the input and output impedances of an op-amp.
– Let’s examine the effect on both the noninverting and the inverting amplifiers.
i) Impedances of a noninverting amplifier.
a) Input impedance
– Assume there is a small differential voltage, Vd, between the two inputs.
– Thus we are assuming the input impedance is not infinite (or the input current to be
zero).
– The input voltage can be expressed as
Vin = Vd + Vf
– Let B = Ri/(Ri + Rf). Then BVout = Vf. Thus,
Vin = Vd + BVout
– But Vout @ AolVd (i.e. the open loop gain times the differential voltage). So:
Vin = Vd + B AolVd = (1 + AolB)Vd
– Now substitute IinZin for Vd to get
Vin = (1 + AolB) IinZin
– Where Zin is the open-loop input impedance of the op-amp (i.e. without feedback
connections).
– Thus we get that the overall input impedance of a closed-loop noninverting amplifier
configuration is:
Zin(NI) = Vin/Iin = (1 + AolB) Zin
– Therefore, the input impedance of the noninverting amplifier with negative feedback is
much greater than the internal input impedance of the op-amp itself (i.e. without feedback).
a) Output impedance
– Apply Kirchhoff’s law to the output circuit to get:
Vout = AolVd - ZoutIout
– The differential input is Vd = Vin – Vf.
– Assume AolVd >> ZoutIout. Then,
Vout @ Aol(Vin- – Vf)
– Substitute BVout- for Vf to get
Vout @ Aol(Vin – BVout)
= AolVin – AolBVout
AolVin = Vout + AolBVout
@ (1 + AolB)Vout
– The output impedance of the noninverting amplifier is Zout(NI) = Vout/Iout.
– Substitute IoutZout(NI) for Vout:
AolVin = (1 + AolB) IoutZout(NI)
– Or
AolVin/Iout = (1 + AolB) Zout(NI)
– The term on the left is the internal output impedance of the op-amp (Zout) because,
without feedback, AolVin = Vout. Thus,
Zout = (1 + AolB) Zout(NI)
Zout(NI) = Zout/(1 + AolB)
– Thus the output impedance of the noninverting amplifier is much less than the internal
output impedance, Zout, of the op-amp itself.
12. What are the three operating temperature ranges of the IC
Operating temperature Range
Military (µA741A, µA741)
Commercial (µA741E, µA741C)
-55oC to 125 oC
0oC to 70 oC
13. Explain the various techniques used to compensate for thermal Drift in op-amps.
Ans: Thermal Drift
Bias Current, offset current and offset voltage change with temperature. A circuit
carefully nulled at 25oC may not remain so when the temperature rises to 35oC. this is called
drift. Often, offset current drift is expressed in nA/oC and offset voltage drift in mV/oC. these
indicate the change in offset for each degree Celsius change in temperature.
There are very few circuit techniques that can be used to minimize the effect of drift.
Careful printed circuit board layout must be used to keep op-amps away from source of heat.
Forced air cooling may be used to stabilize the ambient temperature.