23. For the voltage-divider configuration of Fig. 7.95, determine:a. Ipo and VGSb. VD and Vs.9 24 V2.2 ΚΩLIDO10 ΜΩ6.8 ΜΩwww* VGSQVGS(Th) = 3 VID(on) = 5 mAVGS(on) = 6 V0.75 ΚΩFIG. 7.95Problem 23.

We need to find out current and voltage for given circuit

23. For the voltage-divider configuration of Fig. 7.95, determine a. Ipo and VGS b. VD and Vs. 9 24 V 2.2 ΚΩ De 10 ΜΩ 6.8 ΜΩ ww VGSQ VGS(Th) = 3 V ID(on) = 5 mA VGS(on) = 6 V 0.75 ΚΩ FIG. 7.95 Problem 23.

23. For the voltage-divider configuration of Fig. 7.95, determine a. Ipo and VGS b. VD and Vs. 9 24 V 2.2 ΚΩ De 10 ΜΩ 6.8 ΜΩ ww VGSQ VGS(Th) = 3 V ID(on) = 5 mA VGS(on) = 6 V 0.75 ΚΩ FIG. 7.95 Problem 23.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Concept explainers

Current Division Method

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

Question

23. For the voltage-divider configuration of Fig. 7.95, determine:
a. Ipo and VGS
b. VD and Vs.
9 24 V
2.2 ΚΩ
LIDO
10 ΜΩ
6.8 ΜΩ
www
* VGSQ
VGS(Th) = 3 V
ID(on) = 5 mA
VGS(on) = 6 V
0.75 ΚΩ
FIG. 7.95
Problem 23.

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