A) For a single-phase half-wave uncontrolled rectifier with (RL) load, find the rms value of the output voltage (Vr) and form factor (FF%) with supply voltage: v,(t) = 10 sin(wt). Draw the circuit diagram and sketch the voltages and current waveforms. The conduction angle B 220°. %3D %3D
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- How is a solid-state diode tested? Explain.What is CEMF?A R-load half-wave rectifier used sinusoidal voltage source with a peak value (35.353kV). The firing angle is (30). The total resistance is 50k2 and frequency is 50HZ. Calculate (in details): load direct voltage, maximum direct voltage, normalized voltage and effective voltage, power delivered to the load, power factor, form factor, ripple factor?
- 3 Given the bridge-type full-wave rectifier circuit shown 220V 60Hz 220V:Es Es D1 D2 D3 D4 The load resistor is 220 and the transformer average current is 300mA. Determine: (a) Average value, RMS value, and frequency of load voltage VRL. (b) Average current and PIV of each diode. RL. A half-wave rectifier is needed to supply 15-V dc to a load that draws an average current of 300 mA. The peak-to-peak ripple is required to be 0.2 V or less. What is the minimum value allowed for the smoothing capacitance?H.W:- For a single phase half wave uncontrolled rectifier if the input supply voltage is (v= 310 sin314t), at frequency (f= 50 Hz). Then determine the followings:- Vm. Vmean- i. Draw the rectifier circuit. un %3D ii. Draw the input and the output voltages waveforms. iii. Determine the mean and rms values of the output voltage. s Ec iv. If the load is (R=10 S), then determine the mean and rms values of the current. 2019
- The Figure 2 shows an electronic circuit designed for supplying power to a load (R1). The supply voltage 235V (RMS, AC) at frequency of 50HZ. The required DC voltage and power for the load are 24V and 3.6 W respectively. The Electrical Components of this AC to DC converter are: A full-wave rectifier to convert AC voltage to DC voltage. A regulator with transistor and Zener diode to ensure a constant voltage and power for the load. D1 Iide 91 Vaut VLoad D2 D3 Vde VI sine R1 RL Regulate Reetifier Figure 2. Complete Circuit Assume that the diodes are real diodes (NOT ideal diodes). The following information is available: The collector to base resistor of the regulator R1 = 5.0 k The transistor Q1 with B value of 24 is used for the regulator circuit. Determine the following quantities for this electronic device and fill the table below: Question Answer The voltage of Zener Diode (Vz) The current in R1 The DC current into the regulator | (Idc) Base current of transistor (IB) Collector…For the circuit shown below, compute the maximum current through each resistor when terminal A is a) positive and, b) negative. Assume the diodes are ideal. (Will we convert to the RMS value to it's actual value?)Consider a peak rectifier ted by a 50 Hz sinusoid having a peak value Vp-100 V. Let the load resistance R 10 kohm. The value of the capacitance C that will result in a peak to peak ripple of 3 V is:
- A 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.A half-wave rectifier with ohmic load given in the figure below feeds a 50 Q load and the peak value of the source voltage is 311V and the frequency is 50 Hz. Accordingly, find the following desired circuit parameters: a) Effective (rms) voltage of the source, b) the peak current of the load, c) Average stress of the load, d) Average current of the load, e) The effective (rms) voltage of the load, f) Effective (rms) current of the load, g) Power of the load + vD RPower 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 above