Vin D₁ K+ O + C₁7 C₁ RL Vout The rectifier above drives a 640-2 load with a peak voltage of 3.75V. For a 300-µF smoothing capacitor, calculate the ripple amplitude if the frequency is 50 Hz. (Assume VD,on = 0.75 V) ◆
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- What is CEMF?A single-phase half wave uncontrolled rectifier circuit is operating with a purely inductive load and a source voltage of 100 sin (277t) V. An ideal PMMC ammeter is connected in series with the inductive load reads 20 A. Calculate the distortion factor of the rectifier. Answer Choices: a. 0.866 b. 0.707 c. 0.577 d. 0.816Question 15 In half wave rectifier with filter, if the capacitance is increased, the ripple factor increases. True False Moving to another question will save this response. ip
- A full wave rectifier with a capacitor filter provides a dc output voltage of 35 V to a 3.3 kohms load. Determine the maximum value of filter capacitor if the maximum peak to peak ripple voltage is to be 0.5 V. Assume fin = 50 Hz.ier quesuon Will save this response. Quèstion 6 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 =05A RL VLDC =20V 220V omsb 0.01 F 0.02 F 0.0167 F Hows b None of the above DEV ChpA Capacitor filter is used at the output across the load in order to minimize the ripple factor of the output voltage. The expressions of the average output voltage Vpc and the output AC component VAc are given below. If R = 1k2 and the single phase bridge rectifier supplied from a 120V - 50 Hz sinusoidal source, then the capacitor value that will maintain the amount of the output ripples to less than 4% would be equal to: VDC = Vm(4FRC – 1)/4ƒRC) VAC = Vm/(4/2ƒRC) Select one: O a. 63.39uF O b. 33.2uF O c. 93.39UF O d. 126.2uF TOSHIBA
- Question 3 Consider the following cirucit: in + V1 R1 5 PULSE (0 2 1 1 1 1) (Note again the unrealistic diode characteristics.) The input is a ramp: 2.0V- 1.6V- 1.2V- 0.8V- 0.4V- 0.0V- 0.0s 0.2s 0.4s 0.6s 0.8s out .tran 2 .model diode D(Ron=5 Vfwd=1 Roff=1G) V(in) D1 diode 1.0s 1.2s 1.4s 1.6s 1.8s 2.0s The output voltage is similar to the input voltage, except that it breaks at a certain voltage and reaches a different final value. Find the break voltage and the final value of the output.Z01 @ C Asiacell l. positive clipper negative cli 03_diode_clipper_clamper_multiplie... pi 5/12 7. Questions: A. The positive peak voltage of a positive clipper is: 1- 0 V 2- 0.6 V 3- Equal to the input peak voltage 4. 1.2 V B. Why is the positive peak voltage in the negative clipper not cut? 1- The diode is forward biased 2- The diode is reversed biased C. In a positive polarized clipper we found the voltage source in series to the diode equal to be +5V. Which is the cut level of the positive voltage? 1- 0.6 2- Equal to the input peak voltage 3- 5 V 4- 5.6 V 6/12Repeat example 35 for FWD and firing angle Fa) 60°. 215 l65 D1985 Example 35: A full wave rectifier used 220V, with firing angle 10° ,total resistance load 5 KQ, inductance 2.34 H,and frequency 60HZ, Draw and calculate: (a) VD.c and Ipc (b) VD.C(Max) (c) Vn (d) Vorms-
- During the operation of a half-wave rectifier with a capacitor- filter, the ripple factor can be lowered by the value of the filter capacitor or. the load resistors. O increasing, increasing O decreasing, decreasing O increasing, decreasing O decreasing, increasingQ2) The expected average and RMS output voltage of the following PLECS circuit is equal to Single-phase diode rectifier L: 0.05 01 V_R (V i_N Scope V: 120 w: 2*pi*60 v_N R: 2.50 C: 500e-6 Oa) 40V, 84.85 V Ob) 38 V, 84.85V c) 38V, 80.2V O d) 40V, 38V Probe ProbeFAIRCHILD Discrete POWER & Signal Technologies SEMICONDUCTOR ru 1N4001 - 1N4007 Features • Low torward voltage drop. 10 a14 * High aurge eurrent cepablity. 0.160 4.06) DO 41 COLOR BAND DGNOTEs CAT-Cos 1.0 Ampere General Purpose Rectifiers Absolute Maximum Ratings T-26*Cuness atnerwioe rated Symbol Parameter Value Units Average Recttied Current 1.0 375" lead length a TA - 75°C Tsargei Peak Forward Surge Current 8.3 ms single halr-sine-wave Superimposed on rated load JEDEC method) 30 A Pa Total Device Dissipetion 2.5 20 Derste above 25°C Ra Tag Thermal Resistence, Junction to Amblent 5D Storage Temperature Range 55 to +175 -55 to +150 Operating Junetion Temperature PC "These rarings are imithg valuee above whien the serviceatity or any semiconductor device may te impaired. Electrical Characteristics T-20'Cunieas ofherwise roted Parameter Device Units 4001 4002 4003 4004 4005 4006 4007 Peak Repetitive Reverse Vellage Maximum RME votage DC Reverse Voltage Maximum Reverse Current @ rated VR…