Assuming Vin = Vp sin wt, plot the output waveform of the circuit shown below for an initial condition of +0.5 V across C1. Assume Vp = 5 V. Use Ideal diode model for the diode. D1 Vin + Vout + C,=0.5 V
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- How is a solid-state diode tested? Explain.For the circuit shown in the figure below, consider the diode as an ideal diode & R.M.S value of source voltage as Vs. The output voltage waveform at R is most likely to have zero value in the positive half cycle and a peak value of 1.414Vs in the negative half суcle sine-wave nature with a peak value 1.414VS zero value in the negative half cycle and a O peak value of 1.414Vs in the positive half сycle O sine-wave nature with a peak value Vsu= 6sinot V. V. For the circuit of Figure, we have known E=3V, Use the ideal diode model to analyze the circuit and draw the waveform of the output voltage. 6V cot Uj E
- For a given diode network in the figure below, find output the voltage Vo. V, = lova DI D2 lo VB. For the given waveform illustrated below, design a suitable circuit to obtain an output waveform from a sinusoidal Input of peak voltage 20 V. Hint. Select the types of diode as per your design 9.6 V Vot - 5. 2 VIn the clamping circuit below, assume the diode has Von-0.7V and V₁ is a sinusoidal wave with a peak-to-peak amplitude of 10V. How much is the maximum value of the output voltage signal (Vmax)? And how much will it be if we set V₂=2V? Input V1 C1 0.1e-6 9.3 V and 7.3 V 9.3 V and 11.3 V 11.3 V and 7.3 V O 7.3 V and 12.0 V O 12.0 V and 7.3 V Output D1 1N4148 .tran 30-3 V2 0 Vmax V(output) 0.0ms putput MAN Input 0.6ms 1.2ms V(input) 1.8ms 2.4ms 3.0ms
- Design a clipping circuit that will limit the output voltage to 5V when applying an input sinusoidal waveform with a peak value of 10V. Assume available diodes with voltage drop of 0.5V. Sketch the output waveform of the circuit.consider the figure below that shows an approximated reverse recovery turn-off characteristics for a power diode. Show that the following relation can express the total reverse recovery charge, Qrr = 1/2(trr*ts1) di1/dt =1/2(trr*ts21) di2/dt * ip Isl 1s2! -IrFor the given circuit assume that the diode is ideal and the discharging time constant is much greater than the time period of the input signal. The output waveform will vary between. a. -2V to -12V b. -2V to -22V c. 2V to 12V d. 8V to -12V (SHOW COMPLETE SOLUTION AND EXPLANATION)
- 3) Demonstrate clipping of negative waves with an inverted diode in the output. A circuit which removes the peak of a waveform is known as a clipper. Show a negative clipper. Produce a schematic diagram with LTspice schematic capture program. During the positive half cycle of the 5 V peak input, the diode is reversed biased. The diode does not conduct. It is as if the diode were not there. Show the positive half cycle is unchanged at the output V(2). Show the output positive peaks overlays the input sinewave V(1). Use in the LTSPICE display module, the command "plot v(1)+1)" accomplishes this. Voltage values: Resistor valuse: 1 ΚΩ DC offset: 0 V Amplitude: 5 V Freq: 1 KHzDraw the corresponding output waveform of the double diode partial limiter below using simplified diode model. Indicate all the necessary voltage levels and use Silicon diodes. 20 VP-P 1KQ M D1 9V D2 4.5V VOUTA symmetrical Square wave of 2kHz whose voltage varies between +10V and -10V is applied to the given circuit. Find the maximum and minimum value of the voltage for the output waveform. Assume that the forward voltage and the forward resistance of the diode is 0 and reverse resistance is 2MOhm