Solve for the required parameters : 1.rB = 2.5 QVin = 10 VR= 100 QRequired: V and I using 1st, 2nº, and 3rd approximations (assume a Si diode)2.SiGeKIKHVin = 12 VR= 5.6 kQRequired: output voltage and diode current3. Find the current value, V1,V2, and Vo.R1 = 4.7 k.o VoVin = 10 VR2= 2.2 kQ- 5 V 4. Find Vo , I, Io1, and Io2. Use practical diode model.R1 = 0.33 kQVin = 10 VV.5. Using 2nd approximation, find current I.D1, SiR = 2.2 k.E2 = 4 vE1 = 20 vDz, Si6. Find V. using practical diode model.D1, SiV.E1 = 12 vR1 = 2.2 k.D2, Ge

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Answered: 1. rB = 2.5 Q Vin = 10 V R= 100 Q…

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1. rB = 2.5 Q Vin = 10 V R= 100 Q Required: V and I using 1s, 2n0, and 3rd approximations (assume a Si diode)

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter12: Electronic Control Devices

Section: Chapter Questions

Problem 3RQ: Diodes and rectifiers allow current to _________. flow in one direction only flow in both...

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Concept explainers

Current Division Method

The generalized current divider formula is given by the following relation below,

Question

Solve for the required parameters :

1.
rB = 2.5 Q
Vin = 10 V
R= 100 Q
Required: V and I using 1st, 2nº, and 3rd approximations (assume a Si diode)
2.
Si
Ge
KIKH
Vin = 12 V
R= 5.6 kQ
Required: output voltage and diode current
3. Find the current value, V1,V2, and Vo.
R1 = 4.7 k.
o Vo
Vin = 10 V
R2= 2.2 kQ
- 5 V

4. Find Vo , I, Io1, and Io2. Use practical diode model.
R1 = 0.33 kQ
Vin = 10 V
V.
5. Using 2nd approximation, find current I.
D1, Si
R = 2.2 k.
E2 = 4 v
E1 = 20 v
Dz, Si
6. Find V. using practical diode model.
D1, Si
V.
E1 = 12 v
R1 = 2.2 k.
D2, Ge

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