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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.