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- The voltage signal coming out of the capacitor is with small ripples compare to the voltage signal out of the diode. True FalseA 10 V peak-to-peak sinusoidal voltage is applied across a silicon diode and series resistor. The positive peak voltage across the diode is а. 9.3 V b. 4.3 V c.0.7 V d. 10 VThe given figure shows a step-down dc chopper. A freewheeling diode is placed across the load. The TON would be TOFF load current varies between the limits of 10 A. The time ratio 100V KH 40 mH vvvvv io 5Q
- A clipper circuit based on diodes are simple way to modify waveform in mechatronics. Assume that the two diodes shown in the circuit below are ideal diodes. If the input voltage in the circuit is a 1 kHz sinusoid with peak amplitude of 8V, sketch the Vaue (t). 10 k. 8V 10 kN. D2 RL Vourlt) Vin= Vin(t) Ims D1 6V -8V 4V Page | 1Step-down chopper circuit connected to 100V d.c source supplies with inductive of 40mH the load resistive of 5Q, and a freewheeling diode is placed to the load . the load current varies between 12A and 16A, determine time ratio of the chopper.A 20-V-rms 60-Hz ac source is in series with an ideal diode and a 100-Ω resistance. Determine the peak current and PIV for the diode.
- What voltage is boosted at the output of a multiplier.Choices: average, rms, peak, or instantaneous Which assumption is not used in the analysis of multipliers?Choices: a capacitor will not discharge, start with the cycle of the source that will forward bias the nearest diode, the output will always be multiplied, or a single diode is forward biased at a time A voltage multiplierChoices: multiplies the output voltage, has very high voltage and low current ,can employ inductors, or may not have equal numbers of diodes and capacitors A voltage regulator maintains constant voltage at the load regardless of variations atChoices: the input and output, the load, source, or in the environment Between a lower percent regulation and a higher one, which is better?Choices: lower or higherUsing a positive Clamper Circuit, with an AC voltage of Vi = 100v with a frequency of 30hz, a value of 1000u for the Capacitor, a silicon diode and a resistor with the value of 100 ohms. Find the voltage value on the resistor at a simulation of Oms. Vm 2Vm Mr. Vm Vo -Vm Input Waveform Output Waveform Positive Clamper circuitwrite the procedure to design phase lead compensator
- Full-wave rectified sine wave circuit is used to measure the RMS value of a half square wave with the help of PMMC meter. The meter was actually calibrated for sine wave. The circuit uses a meter movement with a full scale deflection current of 200uA and internal meter resistance of 5kΩ. Assuming Non-ideal diodes having resistance 1kΩ, Analyze the circuit to determine the value of series multiplier resister and the corrected RMS voltage, if meter is to read 225V RMS full-scale.Using a positive Clamper Circuit, with an AC voltage of Vi = 100v with a frequency of 30hz, a value of 1000u for the Capacitor, a silicon diode and a resistor with the value of 100 ohms. Find the voltage value on the resistor at a simulation of Oms. Vm 2Vm Vo -Vm Input Waveform Output Waveform Positive Clamper circuith=6.626 1034 Js e=1.6 1019 C c=3 10° ms1 NA=6.022 103 atomsmol N(Si)=5 10² atomscm³ n(Si)=1 1010 cm3 He(Si)=1350 cm?Vis Ha(Si)=450 cm?vs