Two Marks Questions with Answers

Two Marks Questions with Answers

 

Q. 1 One op-amp has CMRR of 100 dB and other has CMRR of 40 dB. Which you will prefer and why?

Ans. : The dB value of the CMRR is 20 Log (CMRR). More the value of CMRR, better is the performance of the differential amplifier as it greatly rejects the unwanted common mode signals and produce the output almost proportional to the difference voltage at its input terminals. Thus the differential amplifier with higher CMRR is always preferred so that having CMRR 100 dB is preferred over the other having CMRR of 40 dB.

 

Q. 2 What is an operational amplifier ?

Ans. : The operational amplifier is basically an excellent high gain differential amplifier. It amplifies the difference between its two inputs. Due to its use in performing mathematical operations it has been given a name operational amplifier. 

 

Q. 3 Draw the op-amp symbol and state its important terminals.

Ans. : For the op-amp symbol, refer the Fig. 2.2.1. The important terminals are,

1. Positive supply 2. Negative supply 3. Output 4. Inverting input 5. Noninverting input.

 

Q. 4 State the various blocks of IC op-amp.

Ans. : 1. Internal stage   2. Intermediate stage

3. Level shifting stage 4. Output stage.

 

Q. 5 Which circuits are commonly used for the various stages of an 1C op-amp ?

Ans. :

1. Internal stage : Dual input balanced output differential amplifier.

2. Intermediate stage : Multistage amplifiers without coupling capacitors and single ended

3. Level shifting stage : Various transistor circuits to reduce the d.c. output level.

4. Output stage: Push-pull complementary symmetry amplifier in Class AB operation.

 

Q. 6 List the requirements of the internal stage of IC op-amp.

Ans. : 1. High voltage gain 2. High input impedance 3. Two input terminals

4. Small offset voltages 5. Small offset currents 6. High CMRR.

 

Q. 7 Why level shifting is required in op-amp ?

Ans. : As coupling capacitors are not used in the intermediate stage of the op-amp, the d.c. biasing voltage level propagates through the chain of the amplifiers. Finally it appears as a significant d.c. component present at the output along with the desired a.c. output. Such a d.c. level distorts the output and limits the maximum output voltage swing. Hence it is necessary to reduce this d.c. level before the output stage for which level shifting circuit is used.

 

Q.8 List the requirements of the output stage of IC op-amp.

Ans. : 1. High output voltage swing 2. High output current swing 3. Low output impedance 4. Low quiescent power dissipation 5.  Various protection circuits.

 

Q.9 List the important characteristics of ideal op-amp.

Ans. : 1. Infinite voltage gain 2. Infinite input impedance 3. Zero output impedance

4. Infinite bandwidth 5. Infinite CMRR 6. Infinite slew rate 7. Zero PSRR 

 

Q.10 Draw the equivalent circuit of practical op-amp.

Ans. : 

 

Q. 11 What is input offset voltage ?

Ans. : When both the input terminals of the op-amp are grounded, ideally the output should be zero. But practically op-amp produces the small output voltage. To nullify this voltage, some voltage is required to be applied to either of the two input terminals in the particular direction. This voltage is called input offset voltage.

How much voltage, to which input and with what polarity is specified by the manufacturer in the data sheet of the op-amp.

 

Q.12 What is input bias current and input offset current ?

Ans. : The base currents of the transistors used in the input stage of the differential amplifier are practically finite and the two currents differ in magnitude as perfect matched transistors are not possible in practice. These currents are responsible for the input bias current and input offset current of op-amp.

The input bias current is the average of the two base currents Ib,and Ib2 while the input offset current is the difference between the two base currents Ib,^d Ib2.

 

Q.13 What is output offset voltage ? Which parameters are responsible to produce it ?

Ans. : The output voltage present when both the input terminals of the op-amp are grounded is called output offset voltage. The input offset voltage, input bias current and input offset current are responsible to produce output offset voltage.

 

Q. 14 What is thermal drift ?

Ans. : The dependence of the op-amp parameters on the temperature is indicated by a factor called temperature drift. The change in the op-amp parameter corresponding to the change in the temperature is defined as the thermal drift of that parameter. Thus Δ V_ios / ΔT is called input offset voltage temperature drift.

 

Q. 15 What is PSRR ? What should be its ideal value ?

Ans. : PSRR is power supply rejection ratio. It is defined as the change in the input offset voltage due to the change in one of the two supply voltages when other voltage is maintained constant Its ideal value should be zero.

 

Q. 16 What is slew rate ? State slew rate equation.

Ans. : The maximum rate of change of output voltage with respect to time is called slew rate of the op-amp.

The slew rate equation is, S = 2π fV_m V / sec 

 

Q.17 Draw the typical frequency response of the op-amp.

Ans. : Refer the Fig. 2.15.2 of the section 2.15. The students may draw it without semilogpaper only indicating the nature of the two plots.

 

Q.18 Compare the ideal and practical op-amp characteristics.

Ans. :

 

Q.19 What is frequency response of the op-amp ?

Ans. : The plot showing the variations in magnitude and phase angle of the gain of op-amp due to the changes in the input frequency is called the frequency response of the op-amp.

 

Q.20 What is frequency compensation ?

Ans. : The method of modifying loop gain frequency response of the op-amp so that it behaves like single break frequency response which provides sufficient positive phase margin is called frequency compensation technique.

 

Q.21 List the methods used to provide the external frequency compensation.

Ans. : The methods used to provide the external frequency compensation are,

1. Dominant pole compensation      

2. Pole-Zero compensation

3. Feed-Forward compensation.

 

Q. 22 Define unity gain bandwidth of an Op-amp. What is its value for IC741 ?

Ans. : At a certain frequency, the open loop gain reduces to 0 dB. This indicates that 20 log | A_OL(f) | is 0 dB i.e. | A_OL(f) | = 1. Such a frequency at which magnitude of the open loop gain is unity is called gain cross-over frequency or Unity Gain Bandwidth (UGB). It is also called closed loop bandwidth. For IC 741, UGB is about 1 MHz.

 

Q.23 Find the maximum frequency for an op-amp with sine wave output voltage of 10 V peak and slew rate is 2 V/µs.

Ans : 

 

Q.24 In what way 741S is better than 741 ? Dec.-03

Ans.

S stands for slew rate. Thus 7418 has higher slew rate than 741.

 

Q.25 Draw a circuit using operational amplifier to convert square wave to triangular wave.

Dec.-04

Ans. : This circuit is integrator circuit. Refer section for the circuit of an integrator.

 

Q.26 What is a practical op-amp? Draw the equivalent circuit. (Refer section 2.11)

 

Q.27 Define slew rate and CMRR. (Refer sections 2.13.2 and 2.4.3)

 

Q.28 Draw the circuit diagram of differentiator using op-amp. (Refer section 2.30)

 

Q.29 The two input bias currents of an operational amplifier are 22 μA and 26 μА. What is the value of input offset current? Input bias current ? May-07

Ans. :

 

Q.30 Why operational amplifier open loop configurations are not used in linear applications ? May-07

Ans. : Op-amp has a saturable property in open loop mode. Its output saturates at V_sat which is 0.9 times its supply voltages. Its open loop gain is very very high, of the order of 2 × 10⁵. Hence for very small range of input, of the order of few µV the output varies linearily with the input. This range is not sufficient for the practice linear applications. Hence the open loop op-amp is not used in the linear applications.

 

Q.31 Find V_o for the circuit shown R_f = 10 kΩ, R₁ = 2 kΩ, R₂ = 5kΩ.

Ans. : This is inverting summing amplifier with the output expression as

 

Q. 32 Define slew rate. What is its significance ?

(Refer section 2.13.2)

 

Q. 33 What is the need for compensating network in op-amps ?

Ans. : Due to slight mismatch in the first stage of op-amp, there exists input offset voltage and input bias current. Both are responsible to produce output offset voltage. This voltage must be ideally zero. For making this voltage as small as possible, the compensating network is necessary in op-amps.

 

Q. 34 An op-amp data sheet gives 15 V power bandwidth of 25 kHz. What is the slew rate ?

Ans. :

 

Q. 35 Discuss the cause of output offset voltage in op-amp.

Ans. : The output offset voltage (V_oos) is the difference between d.c. voltage present at the output terminals when both the input terminals are grounded. The two main causes for output offset voltage in op-amp are,

1. Input offset voltage (V_ios) 2. Input bias current (I_b)

The output voltage caused by input offset voltage can be either positive or negative with respect to ground. Similarly, the output voltage caused by input bias current can be either positive or negative with respect to ground. If these voltages are of the same polarity then the output offset voltage is large.

The output offset voltage due to these two parameters are given by,

V_oos = (l + R_f / R₁) V_ios + I_b R_f

where R_f is feedback resistance and Rx resistance in series with the source.

To make this V_oos as small as possible, compensating network is used in the op-amps.

 

Q.36 Draw the circuit diagram of an op-amp integrator. Mention its applications.

(Refer section 2.29)

Q. 37 Compare ideal and practical characteristics of an op-amp.

(Refer section 2.19)

Dec.-08

Q. 38 Design an amplifier with a gain of -10 and input resistance of 10 kΩ.

Ans. : The gain is -10 hence it is inverting amplifier and its input resistance is R₁ = 10 kΩ. 

Gain = R_f / R₁

- 10 = - R_f / 10

R_f = 100 kΩ

The designed circuit is shown in the Fig. 2.2.

 

Q.39 Give the ideal characteristics of op-amp and give its equivalent circuit.

(Refer sections 2.10 and 2.11)

Q.40 Draw the circuit diagram of an integrator and give its output equation.

Ans. : Refer Fig. 2.29.1 of the section 2.29. The output equation of an integrator is,

Q.41 What are the different kinds of packages of 1C 741 ?

Ans. : The three packages in which IC741 op-amp is available are,

1. Metal can package (TO)

2. Dual in line package (DIP) 3. Flat package.

 

Q. 42 Mention two linear and two non-linear applications performed by an operational amplifier.

Ans. : The linear applications are inverting amplifier, non-inverting amplifier, voltage follower, integrator, differentiator, instrumentation amplifier etc.

The non-linear applications are comparator, Schmitt trigger, zero crossing detector, log amplifier, precision rectifier etc.

 

Q. 43 State the two realistic simplifying assumptions used for the op-amp.

Ans. : The two realistic simplifying assumptions used for the op-amp are,

1. The input current of the op-amp is always zero.

2. Concept of virtual ground which says that the two input terminals of the op-amp are always at the same potential. Thus if one is grounded then other can be assumed to be at ground potential, which is called virtual ground. 

 

Q.44 Draw the circuit diagram of inverting amplifier and state gain. expression for its gain

Ans. : Refer section 2.21. Gain = - R_f / R₁  .

 

Q. 45 Draw the circuit diagram of non-inverting amplifier and state expression for its gain.

Ans. :  Refer section 2.22. Gain = 1 + (R_f / R₁       )

Q.46  When inverting amplifier is called phase inverter ?   

Ans. :  When the gain of the inverting amplifier is unity and it is       used to change the phase of the input to produce the output then it is called phase inverter.

 

Q.47 What is sign changer or inverter ?

Ans. : If in the inverting amplifier if Rf is made equal to R|then gain of the amplifier becomes -1. Thus the circuit changes the sign of the input i.e. phase of the input without changing its magnitude at its output. This circuit is called sign changer or phase inverter.

 

Q.48 How an op-amp can be used as a voltage follower ?

(Refer section 2.24)

Q.49 State the advantages of voltage follower.

Ans. :

1) Very large input resistance, of the order of MQ.

2) Low output impedance, almost zero. Hence it can be used to connect high impedance source to a low impedance load, as a buffer.

3) It has large bandwidth.

4) The output follows the input exactly without a phase shift.

 

Q. 50 Draw the circuit diagram of inverting summer.

(Refer Fig. 2.27.1 of section 2.27.1)

Q.51 Draw the circuit diagram of non-inverting summer.

(Refer Fig. 2.27.2 of section 2.27.2)

Q.52 How to obtain the average circuit from the inverting summer ?

Ans. : In an inverting summer if the values of resistances are selected such that R₁ = R₂ = R and R_f = R/2 then the output is the average of the two inputs applied to the circuit.

 

Q.53 Draw the circuit diagram of subtractor.

(Refer Fig. 2.28.1 of the section 2.28) 

 

Q. 54 Draw the input and output waveforms for an integrator for step and sinusoidal type of inputs.

(Refer section 2.29.2)

 

Q. 55 State the errors in an ideal integrator.

Ans. :

1. The output is likely to saturate in absence of input.

2. The output may be distorted due to the error voltage.

3. The bandwidth is very small.

4. It can be used only for small range of frequencies.

 

Q. 56 Draw the circuit diagram of practical integrator circuit.

(Refer Fig. 2.29.8 of the section 2.29.4)

Q. 57 List the applications of integrator circuit.

Ans. : i) In the analog computers, ii) In solving differential equations, iii) In analog to digital converters, iv) Various signal wave shaping circuits, v) In ramp and triangular waveform generators.

 

Q. 58 Draw the circuit diagram of an ideal differentiator. Write expression for its output.

Ans . :

 Its output expression is ,

Vo = - C₁ R_f  (dV_in / dt)

Q. 59 Draw the input and output waveforms for differentiator for step and square type of inputs.

(Refer section 2.30.2)

 

Q. 60 State the errors in an ideal differentiator.

Ans. :

1. The gain increases as the frequency increases hence the circuit becomes unstable at high frequency.

2. The input impedance decreases as the frequency increases hence circuit becomes sensitive to the noise which can completely override the differential output.

 

Q. 61 Draw the circuit diagram of the practical differentiator circuit.

(Refer Fig. 2.30.6 of the section 2.30.4)

Q. 62 List the applications of differentiator circuit.

Ans. : i) In the wave shaping circuits, ii) To detect high frequency components in the input, iii) As a rate of change detector in the FM demodulators.

 

Q. 63 In an inverting amplifier the values of Rf and R1 are 100 kΩ and 10 kΩ respectively. What is the gain of the amplifier?

Ans : Gain of the inverting amplifier is given by – R_f / R₁ = - 100 / 10 = -10.

 

Q. 64 How will you increase the output of a general purpose op-amp ?

Ans. : A simple method of increasing the output current of a general purpose op-amp is to connect a power booster circuit in series with the op-amp. This is shown in the Fig. 2.3

 

Q 65 List any four non ideal dc characteristics of op-amp.

May-12

i) Input bias current       iii) Input offset current

iii) Input offset voltage  iv) Output offset voltage.

 

Q. 66 Calculate the value of the current I flowing through resistance R in the operational amplifier circuit shown in Fig. 2.4.

Ans. : Due to the concept of virtual ground, the non-inverting terminal voltage must be same as inverting terminal. 

V_B = V_A = 5 V.. Virtual ground

As no current can flow into the op-amp, current through 1 kΩ is I only.

 

Q. 67 Define CMRR.

Ans. : The ratio of the differential voltage gain Ad to the common mode voltage gain A_c is called CMRR of an op-amp.

 

Q. 68 What is V to I converter ?

Ans. : The circuit in which the output load current is proportional to the input voltage is called V to I converter. Mathematically it is given by I_L ∝ V_in

 

Q. 69 What is I to V converter ?

Ans. : The circuit in which the output voltage is proportional to the input current is called I to V converter. Mathematically it is given by Vo oc Iin.

 

Q. 70 Draw the circuit of current to voltage converter using op-amp and bring out the relation. State one application.

(Refer section 2.26)

Q. 71 State the applications of voltage to current converter.

Ans. :

1. Low voltage d.c. voltmeter 2. Low voltage a.c. voltmeter

3. Diode tester and match finder 4. Zener diode tester.

 

Q. 72 State the applications of current to voltage converter.

Ans. : 1. Photodiode detector 2. PhotoFET detector.

 

Q. 73 Draw the frequency response characteristic of an A.C. integrator and indicate the part where it behaves as a true integrator.

 Ans. : The frequency response of the practical integrator is shown in the Fig. 2.5.

 

Q. 74 State the causes for slew rarte in an operational amplifier ? How it is indicated?

Ans. : The causes of slew rate are,

1. Limited charging rate of the compensating capacitor.

2. Current limiting of the internal stages of an op-amp. 

3. Saturation of the internal stages of an op-amp.

The slew rate is indicated as, S = dV_o / dt | _maximum

 

Q 76 A 100 pF capacitor has a maximum charging current of 100 microamps. What is the slew rate ?

Ans : S = I_max / C = 100 × 10^-6 / 100 × 10^-12 = 1 × 10⁶ V/sec = 1 V/µ sec

 

Q.77 Explain how the average circuit can be derived from the summer.

(Refer section 2.27)

Q. 78 Why IC 741 is not used for high frequency applications ?

Ans. : Due to predominant capacitance effect, the slew rate of IC 741 is very small. At high frequencies, the output gets distorted due to small slew rate. Hence IC 741 is not used for high frequency applications. 

 

Q.79 What is integrator ?

Ans. : An integrator is a circuit which produces output which is integration of the input applied to it. Then refer Q.40.

Q.80 Draw the circuit diagram of a symmetrical emitter coupled differential amplifier.

(Refer Fig. 2.5.2 of section 2.5)

Q. 81 For the circuit diagram shown below determine the output voltage V0.

Ans. : It is a subtractor with

Q. 82 What is the drawback of IC 741?

Dec-17

Ans. : The main drawback of IC 741 is that its slew rate is small hence it cannot be used for high frequency applications. Compared to modem op-amp ICs, 741 has low input impedance, high offset voltage and current and minimum power supply required is ± 10 V. The drift in parameters with respect to temperature is high.

 

Q. 83 The slew rate of an op-amp is 0.6 V/micro sec. What is the maximum undistorted sine-wave that can be obtained for a 10 V peak and 1 V peak ?

May-18

Ans. : From slew rate equation,