7) Figure shows a chopper operating from a 100 V dc input. The duty ratio of the main switch S is 0.8. The load is sufficiently inductive so that the load current is ripple free. The average current through the diode D under steady state is (a) 1.6 A (b) 6.4 A (c) 8.0 A 10.0A 100V La [2004] 100
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- How is a solid-state diode tested? Explain.In the circuit shown below. Determine the following a. Vs (at the secondary) b. Vout (across RL) c. Vrip (ripple votage) d. VDC e. PIV (Peak Inverse Voltage) 10:1 Output 115 V mis 60 Hz RL 2.2 k) D, 50μF- All diodes are IN4001. Tund AlMannai EENG261 Page 5/110:- Consider the circuit in Figure a) What type of circuit is this? b) Find and Sketch the voltage waveform across RL, assume the diodes are practical. c) If 100uf capacitor parallel with the resistor, calculate the ripple is connected factor I O o D
- 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! -IrIn the circuit of the following figure, the input voltage Vs is 15 volts rms with a frequency of 60 Hz, R equals 150 Ohms and C equals 100,000 Pico Farads. The diodes are Germanium (Vd = 0.2 volts) and the Zener diode is 12 volts. a) The magnitude of the ripple voltage at Cb) The Magnitude of the Peak Inverse Voltage (PIV) for D1 and D2.Power 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
- Power 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 rmsh soHz} VL-DC =20V 0.01 F 0.02 F 0.0167 F None of the aboveThe reverse recovery time for fast recovery diodes is a. 20 micro seconds b. 10 micro seconds c. 5 micro seconds d. 15 micro secondsA single phase – half wave controlled rectifier with freewheeling diode is supplying a load consistingseries connected a resistor and an inductance from a 70.7V (RMS), 50Hz sinusoidal AC source.The firing delay of the thyristor is 90° and the load values are R=10Ω, L=0.1 H. Define the loadcurrent expression and draw the load current by calculating for first two periods. And calculate theaverage values of the load voltage and current.
- 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 1kA- 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)Consider the circuit in the figure below 4:1 120 V rms ooooo gll reelee D₁ L a. What type of circuit is this? b. What is the total peak secondary voltage? D₂ c. Find the peak voltage across each half of the secondary. d. Sketch the voltage waveform across RL. e. What is the peak current through each diode? f. What is the PIV for each diode? RL 1.0 ΚΩ