5. Given the information appearing in Figure 5, determine: +22 V a. IB b. lc c. Vc d. VCE 470 ΚΩ B =90 19.1 ΚΩ le VCE 19.1 ΚΩ Figure 5
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- For 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 2mAHw@: : A555 IC is Connected 95 slhown in fingune below; fingune Detarmine the range of oscillation finaguerney? Ans: VCC =1Ov 1ok trag. range (621H2 -4.7 kllz 4 7. loke 555 IC 2. CPotentiometa10oke D.olHFFAIRCHILD 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…
- ide cal diode effechive value Frequency equal to 60 He. the voltage ripple at the terminals of R is equeal the source is sinu soidal, with equal to120V and a to 6.06 % of the arerage value of he voltage of R. at the terminals value of Hhe vollage the cueraye value of the volkage R is ĕ guual to: across O lo5 V l10 v (3 165 V 160 VPower 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 euerRepeat 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-
- 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 6/12As 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,)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 -1
- After replacing R5 with diode IN4007 (switch K3 to diode side) .What conclusion can you draw from analysis of this step ?Q4) Determine and sketch the output voltage across the load resistor (RL) for the circuit shown below. (assume Si diodes) V_DC 0.75 0.25 V_SIN Y SIN as RL -1 v SOR V_SQR 0.75 -0.75 V_TRI V TRI asier 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 Chp