H.W2 The single-phase half wave rectifier has R-L load with R= 50 and L= 6.5mH. The input voltage V, 220V at 50Hz, determine: BU a) The average diode current b) The rms diode current c) The rms output current d) The average output current » V₂ 'CH
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- How is a solid-state diode tested? Explain.A- If V, is a sinusoidal voltage with Vm = 40 V, and V= 15 V. Plot the waveform of the output voltage in each of the following clippers circuits assuming ideal diodes. B- Repeat part (A) if the diodes are silicon diodes. R R R (a) (b) (c) (d)If the positive cycle of the waveform for full wave bridge rectifier , Which of the diodes is working . ..... ...... 10:1 120 V ms Output 60 Hz Puec) RL 220 N Dz 1000µF O D1, D2 D3, D2 D4, D2 D1, D4 not all above D1, D3 ellee
- Draw output waveform for the following circuit. Suppose input is a pure square wave with a positive half at 10 V and negative half at -10 V for ideal diodes. D1 D2 220 R3 R1 1k R2 1kIn 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 3.0ms7) sine wave. Sketch the waveform (1 period is enough) for the resulting Vout- What are its positive and negative peak values? (AVforward,ideal = 0) For each of the ideal-diode circuits below, the input Vin is a 1kHz 5-V peak a) b) Vin Vaut Vin Vaut 1kN 1kN c) c) Vin Vout Vin Veut 1kN 1kN e) f) Vin 1kN Vaut Vin 1kN Vout 太太
- What voltage is boosted at the output of a multiplier. a) average b) rms c) peak d) instantaneous Which assumption is not used in the analysis of multipliers? a) a capacitor will not discharge b) start with the cycle of the source that will forward bias the nearest diode c) the output will always be multiplied d) a single diode is forward biased at a time A voltage multiplier a) multiplies the output voltage b) has very high voltage and low current c) can employ inductors d) may not have equal numbers of diodes and capacitors A voltage regulator maintains constant voltage at the load regardless of variations at a) the input and output b) the load c) source d) in the environment Between a lower percent regulation and a higher one, which is better? a) lower b) higherI ln. l. . :D docs.google.com/forms/d/e/1l Diode clippers are wave-shaping circuits in that they are not used to prevent signal voltage from going above or below certain * .levels TO F like that of the clipper, the shape of the .input signals of a clamper is changed F The clamping circuit clamps or shift the * .input sinusoidal signal up or down FO IIPower supply circuit is delivering 0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to IL-DC =0:5A RL VL-DC =20V 220V omsb O 001 F O 0.02 F O 0.0167F O None of the above Activate
- Consider a junction diode has dc biased current at operating point ID=0.8 mA. A sinusoidal voltage is superimposed on the VD such that the peak-to-peak sinusoidal current id(t) = 0.05ID. Find the value of the applied peak-to-peak sinusoidal voltage vd(t).Make the described voltage multiplier circuit, determine the voltage charge in each capacitor, and the output voltage of the circuit. Plot the output waveform in proper phase alignment with the input voltage. Indicate the peak values with the support of your solutions. Use silicon diodes in your analysis and an input V=10 Vrms. a) Half-wave voltage doubler b.) Full-wave voltage doublerPower supply circuit is delivering 0.5 A and an average voltage 20 V to the load as shown in the circuit below. The ripple voltage of the half wave rectifier is 0.5 V and the diode is represented using constant voltage model. The smoothing capacitor value is equal to 220V ams 5OHZ İL-DC =05A RL VL-DC =20V