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- Q5. For the following center tapped transformer, show the waveform across each half of the secondary winding and across Ri, when a 100 V peak sine wave is applied to the primarywinding. What is the PIV rating must the diode have? (Use constant voltage drop model for the Silicon diode) IN4001 Ry OV 10k) D IN4001Refer to the circuit on the right. VEB = 0.7 V, when the BE junction is conducting. VECsat = 0.2 V B Vref = 2.2 V (Use two decimal places.) = ∞ VREF IL +5 V R LOAD Figure 2 a. What should be the value of R in ohms if IL is to be maintained at 146 mA under changing load conditions? b. In order to maintain this load current, what should be the maximum allowable load resistance in ohms?C. A single phase full converter feeding RL load has the following data: source voltage 230V(RMS), SOHZ.R-2.52 L=very large, firing angle -30"f the load inductrance is large enough to make the load current constant then i Sketch the time variation of sounce voltage source current, load voltage, load current current through one thyristor and voltage across it H. Compute the average value of load voltage and load current, Derive the formula.
- 3 A peak rectifier (peak tollower) Cir cuit is gian below. Vs is a GOH3 Sinuscidal Vollge with peak value Upz50V. 50V. The load resistance R= Sk . Find the Vallke of the capacitance C Such that the peak- to-peak ripple 2 volts. (idenl diede) Voltage Vr VsThree phase Bridge rectifier operated from a 380V, 50Hz source throughout three-phase Y/Y connected transformer with ration 2:1. The load current has continuous character with negligible ripples and average thyristor current of 15A. The circuit inductance per phase is 1.2mH. The rectified voltage due to overlapping angle is 90 % of that voltage when this angle is neglected. 1- How much will the voltage drop be due to circuit inductance * 2- How much will the dc average voltage Vdc(alpha) be. 3- How much will the firing angle be ? 4- The overlapping angle. 5- How much is the percentage change of the output dc power if the thyristors T2, T4 and T6 are replaced by three diodes D2, D4, D6. 6- How much will be the maximum RMS phase current for the mentioned in previous task circuit ( semi-converter). 7- If an additional of 1 mH inductances are added to the existing inductances in the phases , how much will the overlapping angle be. The dc load current is kept constant & the…Please show steps and work
- Q1. a) Figure Q1 is a single phase 2-level voltage source converter (VSC) with a total DC voltage V-1000V. A PWM scheme, by comparing a reference value with a 1kHz triangular waveform, is used to control the switches S₁ and S2, and output 340v at Vo. With the aid of graphs explain how pulse signals for the switches are generated, sketch the output voltage Vo, determine the duty cycle and the modulation index. Va + 2 it Va d + TS₁ V {1+5₂ = Figure Q1 A single phase VSC1) AC power in a load can be controlled by using a. Two SCR's in parallel opposition b. Two SCR's in series c. Three SCR's in series d. Four SCR's in series 2) The advantage of using free - wheeling diode in half controlled bridge converter is that a. There is always a path for the ac current independent of the ac line b. There is always a path for the dc current independent of the ac line c. There is always a path for the dc current dependent of the ac line d. There is always a path for the ac current dependent of the ac line 3) Silicon controlled rectifier can be turned on a. By applying a gate pulse and turned off only when current becomes zero b. And turned off by applying gate pulse c. By applying a gate pulse and turned off by removing the gate pulse d. By making current non zero and turned off by making current zeroA 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.
- Q1: Draw the d.c load line for the circuit shown in figure. NO SIGNAL VBB Vcc= 12.5 V www Re=2.5 k2In the full-wave rectifier circuit of figure , the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 230V (rms), 50 Hz. The load resistor is 50ohms. For this circuit, determine the peak-to-peak ripple in the output voltageIn the full-wave rectifier circuit of figure, the transformer has a turns ratio of 1:2. The transformer primary winding is connected across an AC source of 230V (rms), 50 Hz. The load resistor is 50ohms. For this circuit, determine the peak-to- peak ripple in the output voltage * Di i v2 = VmSin ot RL Vs(N V1 Vo + V3 = VmSin ot D2 Full-wave rectifier- Circuit operation during positive half cycle 460 V 207 V 325.3 V None of the above ell ell