Assignment 11 1) Using the LM741 op-amp IC a circuit is designed as shown, then find the output waveform for an input of 5kHz Vo = 1 x R1Cf 0 Vin t dt, voltage output for the op amp integrator 0.1 m 1 Vo = 10 k x 20 n x 2 0 x dt = 5 x 10 3 x 2 x[ 0 0.1 m] = 1 V Meaning a triangular waveform with voltage + 0.5 V and -0.5 V for the same time period t = 0 to 0.1 m, only figure (a) waveform meets with the conditions. Ans: Figure (a) 2) To obtain an optimum differentiation choose the values of Rf and C for a 7 khz input signal. Optimum differentiation condition is T Rf C1 Given: f = 7 khz, T = 1/f = 1.428 x 10-4 Hz ----(1) Rf C1 = 0.1 µ x 1.3 k = 1.3 x 10-4 -----------------(2) (1) (2), satisfies the conditions Ans: Rf = 1.3 kω, C = 0.1 µf
3) How passive filters are problematic Inductors become large at low frequencies Design of inductors are expensive at low frequency applications Series resistance of inductors degrade its performance All of the above At audio frequencies inductors are problematic due to the its size and bulkiness and hence these are heavy and expensive. Moreover, low frequency applications requires more number of turns of wire which in turn adds to the series resistance degrading inductor s performance. Ans: All of the above 4) Select the correct option for the circuit shown below to get an output Vo as + Vcc The circuit given is a comparator. To obtain an output as +Vcc, the non-inverting terminal voltage input Vref must be greater than the inverting terminal voltage input Vin. Vref > Vin Ans: Vref > Vin 5) Select an op-amp for filter applications to improve the performance µa741 LM318 TL082 LM324 LM318 is a high-speed op-amp that improves the filter s performance through increased slew rate and higher unity gain-bandwidth Ans: LM318 6) What is the 0 db frequency for the circuit given? 0db frequency, f = 1 2π RfCf 1 f = 2π RfCf = 1 = 10.26 Hz 2π x 33 kω x 0.47 µf Ans: 10.26 Hz
7) One terminal of a comparator circuit is connected to ground and a cosine waveform of 10 khz frequency is given as input to the other terminal. The output waveform of the comparator will be a) b) c) d) If we apply reference voltage (Vref) to the inverting terminal and made it ground so that Vref =0V. Now in this case the output of the comparator will be - V max volts as long as the input voltage is below 0V. When input voltage exceeds 0V the output of the comparator switches to +V max volts. Such special case op-amp comparator is called as zero-level detector. This zero-level detector circuit can be used to obtain square waveform from a cosine waveform. Ans: 8) Active filters using op-amps are designed with All passive elements (Resistors, Capacitors and Inductors) Only Resistors and Inductors Only Capacitors and Resistors Only Resistors The active filters use capacitor in the feedback loop and thus the use of inductors can be avoided. Hence, an inductor less RC filters can be achieved with op-amps. Ans: Only Capacitors and Resistors 9) Compute the output voltage for the given circuit. The given circuit is a summing amplifier. Vo = - (V1+V2), output voltage of a summing amplifier Vo = - (5sinωt 3 sinωt) = - 2 sinωt Ans: - 2 sinωt
10) Find out the correct statement about active and passive filters Gain and frequency adjustment is difficult in active filters Passive filters are expensive Active filters cause loading of the source Passive filters are easy to tune or adjust Active filters are more economical than passive filters. This is because of the variety of cheaper op-amp and the absence of inductor s Ans: Passive filters are expensive 11) Compute the duty cycle of the output for the circuit shown below. If input is 8 Vpp sinusoidal signal Note: Consider an ideal comparator Vo = A (V+ - V-) Output of comparator, If V+ > V-, then Vo = +Vcc If V+ < V-, then Vo = -Vee Ton Duty cycle = Ton + Toff = Ton 2 V T From the figure, ωt = 360 0 ωton = 180 2x ωt We know that, Vref = 2 V, Vin = 4 sin ωt and let ωt = x At Vref = 2 V, Vin = 2 = 4 sin x Sin x = ½, x = 30 0 ωton = 180 2 x 30 = 120 0 D = ωton ωt = 120 360 = 0.33 Ans: 0.33 12) Consider a differentiator circuit with RF =10 kω, C1=26 nf and a cosine wave of 4 Vpp at 500 Hz is applied to it. Then find the appropriate output waveform.
Vin = V cos (2π 500 t) Vo = Rf x C1 d Vin, voltage output of a opamp differentiator dt Vo = 10 k x 26 x10 9 x d (4 cos 2π 500t) = 3.26 sin 2 π 500t dt The output is a sine wave with no phase shift. Ans: 13) Compute the output voltage for the given circuit if R=1 kω The circuit is a comparator circuit output of which is given as follows, If V+ > V-, then Vo = +Vcc If V+ < V-, then Vo = -Vee In the circuit given, + 1 V > 0 V, implies that Vo = +Vcc = +15 V Ans: 15 V 14) Select the correct option to make the green LED glow? The green LED will glow when it is forward biased FB, that is when voltage at point A > point B shown in the circuit. Consider comparator A2, When Vi > 3 V, only then the output will be positive Consider comparator A1, When Vi < 6 V, only then the output will be negative Therefore, condition for Vi is, 3 V < Vi < 6 V for the green LED to glow Ans: 3 V< Vi < 6 V
15) Assume the op-amp given in the circuit is ideal and a triangular wave is given as input to it. Then which among the following will be the output waveform. Vo = Rf x C1 d Vin, voltage output of a opamp differentiator dt Assume amplitude is peak to peak and for a triangular wave it is ± At So dv/dt = ± A, therefore it gives a square wave output Ans: Square wave