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- What is CEMF?Design a rectifier fed from 220-V line through a step-down transformer. The load requires an average voltage of 9 V and average power of 2.5 W. Use each of the three circuits: half-wave, center-tapped full-wave and bridge type full-wave. pls write legibly and show complete process of solutionFAIRCHILD 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…
- Q4) Determine and sketch the output voltage across the load resistor (RL) for the circuit shown below (assume Si diodes) V_DC V DC 0,75 (1+ 0.25 V_SIN V SIN RL -1 V SOR V_SQR 0.75 -0.75 V TRI 1 V_TRI -1Z01 @ 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 6/12For self bias configuration. Solve for Ib and Ic. ... 4 RC 15002 RB VCC 5 16V 100A/A 80 kQ Re 10002 Ib=Blank 1 HA Ic=Blank 2mA
- 3) Draw V-I Charters tics of p-n Junction and explain forward bias & Reverse Bias In a linear power supply, where will you connect the capacitor on order to reduce the ripple output of a rectifier circuit? O in parallel with the transformer O in series with the transformer O in parallel with the load resistor O in series with the load resistorQ4. For the three-phase fully-controlled bridge rectifier circuit shown with purely resistive load: a = 90° a. Sketch the output voltage waveform. a b. Determine the ripple factor. VA= 240 V RL 752 CB A-B A-C B-C B-A CA C-B a Vc 27
- Q-1/ connect the circuit of single phase half-wave uncontrolled rectifier with Ac supply 220V, 50HZ, R=10, L=20m, Vdc=12v and B= 4.04. Find (1)the i/p and o/p waveforms of voltage and current and PIVPower 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 =0:5A RL VL-DC =20V 220V omsb O 001 F O 0.02 F O 0.0167F O None of the above ActivateGiven the following circuit with VDD = 5 V, R = 2.6 k, then the current Iis: Use the CVD model for the diode, with VD= 0.65 V. V DD I a. 0.001673 A O b. 0 A O c. 1.673077 A O d. 1.923077 A e. 0.001923 A R + VD -