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- What is the difference between a diode and rectifier?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 =20VC4 R1 Vout For the precision rectifier circuit shown in Figure C4 what is the correct operation for the circuit if a Sine wave is 10k D1 R2 LM324 VEE applied to the input signal VIN? Vin D2 10k OUT V3 U1A Figure C4 A. When VIn is negative the circuit operates as unity voltage follower providing an in phase sine wave at VouT. When VIN is positive the circuit conducts but only to one diode drop, B. When VIn is negative the circuit operates as an inverting amplifier providing an inverted sine wave at VouT. When VIN is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, c. When VIn is negative the circuit operates as an inverting amplifier providing an in-phase sine wave at VOUT. When VIn is positive the circuit is driven to very close to zero due to amplifier gain reducing the diode drop voltage, D. When VIn is negative the circuit operates as unity voltage follower providing an inverted sine wave at Vour. When VIn is positive the…
- 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 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 ActivatePower 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 omsh RL VI-DC20V 0.01 F 0.02 F 0.0167 F None of the aboveA three phase full wave rectifier is shown below along with peak phase voltage Vm-169.7V. The load is purely resistive. The rectifier delivers Ipc = 100 A and the source frequency is 60 Hz. The DC output voltage is %3D VDc=280.7V and the output RMS voltage is equal to VRMS=280.93V. The efficiency, FF and RF are respectively equal to: Secondary D D D, R AD. Z D. D, Select one: O a. 99.83%, 100.08%, 4% O b. 99.83%, 55.08%, 4% O C. 99.83%, 100.08%, 16% O d. 87.83%, 100.08%, 4%
- A full-wave rectifier uses 2 diodes. The internal resistance of each diode is 20 Q. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 Q. Mean load current will beA single-phase full-wave bridge rectifier circuit is fed from a 220 V, 50 Hz supply. It consists of four diodes, a load resistance 20 Q and a very large inductance so that the load current is constant. What is the average or dc output voltage?A full wave bridge rectifier is supplied from 230V, 50Hz and uses a transformer of turns ration 15:1. It uses load resistance of 50 ohms. Calculate load voltage and ripple voltage. Assume ideal diode and transformer. Assume ripple factor equals to 0.482. a.Load Voltage= 13.8 V; Ripple Voltage=6.652 V b.Load Voltage= 15.6 V; Ripple Voltage=7.611 V c.Load Voltage= 21.3 V; Ripple Voltage=8.410 V d.Load Voltage= 25.1 V; Ripple Voltage=10.442 V
- 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 aboveTiguie Question 3 Figure 2 shows a center-tapped full wave rectifier circuit when a 100 V peak sine wave is applied to the primary winding. If the peak input in Figure 2 is changed to 220V draw the voltage waveforms across each half of the secondary winding and across Ri for TWO (2) complete input cycle voltage. Also determine the Vout, current through R. and PIV rating for each diode. Assume Dị and D2 are practical diode model. D 2:1 +100 V -- IN4001 V OV RL -100 V 10 k2 IN4001 Figure 2A full-wave rectifier uses 2 diodes. The internal resistance of each diode is 20 02. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 2. Mean load current will be