For the last two parts, use the diagram to the right. d. V = -1.14V, R = 62, and the voltage across the diode is equal to 0.68V when it is "on". V= i= e. Vo = -0.16V, R = 52, and the voltage across the diode is equal to 0.64V when it is "on". V = i = Vo R KHI + DI
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- What is CEMF?Question 1: In the circuit shown below, the output (Vo = 10V Max.) Unipolar. The frequency of Primary is 60 Hz. The diodes are Silicon with VD = 0.7V. a. Sketch the output without a Capacitor. b. Determine Voc without a Capacitor. c. Sketch Vs (at the Secondary). d. Determine Voc with a Capacitor of 10 uF across RL. e. Determine the RMS Value of Vp (at the Primary). f. PIV (Peak Inverse Voltage). 10:1 Output C. 22 k1 All diodes are IN4001. | 00000How do you identify the anode of an unmarked diode? When the forward current of a diode increases, its forward resistance The current flowing in a reverse-biased diode circuit is extremely while the resistance of the diode is extremely The Vz of a zener diode will fairly constant even if the power supply voltage The series resistor Rs is used with the zener diode to the zener current Iz to a level.
- 5V 3000 1000 . 2m 1 Si 4000 500 12V The circuit shown in Figure 1 contains a Si diode with a diode resistance on 1000. Use the piecewise linear model to solve the circuit. • How many nodes are in the circuit? • Determine the Thevenin voltage across the diode. Replace the diode with the appropriate model and then solve the circuit to determine the power dissipated by each resistor and the diode.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 c 20:5A L VI-DC20V 220V omsh T 0.01 F 0.02 F 0.0167 F None of the above euerA 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 Va.. (t) 10 kO 8V 10 kO Vin Vin(t) D2 Vout(t) RL Ims D1 6V -8V 4V Page | 1
- Given the following circuit with VDD= 9.2 V, R=2.3 k2, then the current Iis: Use the CVD model for the diode, with VD = 0.65 V. I VDD a. 0.004000 A O b. 3.717391 A OC. 0 A d. 4.000000 A e. 0.003717 A R + VD -A- 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)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 higher
- In the circuit shown below, the ideality factorn of the diode is unity and the voltage drop across it is 0.7 V. The dynamic resistance of the diode at room temperature is approximately 1.7 V 31 k2As shown is a positive parallel clipper circuit which has an input voltage of E=t5V. The negative output voltage is to be -4.5V when lo is 5mA. Assume ideal diode model approximation. Determine the value of R1 in ohms. Note: Write only the numeric value of the answer, round to 4 decimal places. No need to include the unit. R, +E ww Output Input D, -E -(E-1,R,)Exercises: Determine VO and Ifor each circuit in figures. Assume that each of the diode in these circuits has a forward voltage drop of 0.7 V. V₁. D 1.1.2 D V. $5.140 1.V₁=V₁=OV. 2.V₁ =V₁=SV, and 3.V=OV & V, -5V. (d) C C ΣΙΩ T-15V D₂ +201 20 V₂ 14.02 I-15V 1. No pulses at either A or B. 2.A 30 V positive pulse at A or B, and 3. Positive pulses (30 V) at both A and B. (e)