*10. For the network of Fig. 7.83, determine:a. Ip.b. VDS.c. VD.d. Vs.20 V2.2 ΚΩID-4 VIDSS= 4.5 mAVDSVp = -5V0.68 ΚΩ

Answered: Image /qna-images/answer/4867ffa2-001c-4faf-9dc5-97bf6f695cfb.jpg

Answered: *10. For the network of Fig. 7.83,…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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determine:a. IDQ.b. VGSQ and VDSQ.c. VD and VS.d. VDS.
Transistor AC Analysis
14. For the network of Fig. 7.86, Vp = 12 V. Determine: a. Ip. b. Vs and Vps- c. Vg and VGs. d. Vp.
Find a. IDQ.b. VGSQ and VDSQ.c. VD and VS.d. VDS
2. For the fixed-bias configuration of Fig. 7.76, determine: a. Ip, and VGS, using a purely mathematical approach. b. Repeat part (a) using a graphical approach and compare results. c. Find VDs, VD, VG, and Vs using the results of part (a). 16 V -3 V 2.2 k2 1.2 M2 = 10 mA Ipss Vp =-4.5 V Vaso FIG. 7.76
14. For the network of Fig. 7.86, Vp = 12 V. Determine: a. Ip. b. Vs and Vps- c. Vg and VGs- d. Vp. 18 V 2 k2 oV,= 12 V %3D Ve 12 V o WW 680 k2 Vas Loss =8 mA Vas Vs 110 k2 0.68 kC2
Q 2) For the network in Fig. 7.64: a. Find the total resistance R, b. Find the source current I, and currents /, and Iy. c. Find current ls. d. Find voltages V; and Va. Ry R 120 40 R4 14 V V, R 120 Ry
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For the network of Fig. 7.93, determine: a. ID, and VGS₂- 18 V b. Vps and Vs. Shockley's equation, VGS ID= Vp) ID Vos V 1- VIDSS VGSQ VG = R₂VDD R₁ + R2 VGS VG-IDRS VDS VDD-ID(RD + Rs) (a) ID = 9 mA, VGS₁ = 0.5 V (b) VDs = 7.69 V, Vs = -0.5 V • 2.2 ΚΩ Dss = 8 mA Vp=-8V • 0.39 ΚΩ 8-4 V FIG. 7.93
For the network of Fig. 7.86, Vp = 12 V. Determine: a. Ip. b. Vs and Vps- c. Vg and VGs- d. Vp. 18 V 2 k2 O Vp = 12 V + VG 12 VoMw VDS Ipss = 8 mA 680 k2 VGS Vs '110 k2 0.68 k2
3. For the network of Fig. 7.85 , determine: a. VG. b. ID and VGS. c. VD and VS. d. VDS. O 20 V 2.2 kn 910 kQ Dss = 10 mA V, =-3.5 V Vasa 110 kQ 1.1 ka
Q/ For the circuit shown below find I7, Vr, and V.. 1K(ohm) D R1 ITA Ge ideal D3 D4 Si 12v VT Si 2K(ohm) 3K(ohm) 2.5K(ohm)

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*10. For the network of Fig. 7.83, determine: a. Ip. b. VDS. c. VD. d. Vs. 20 V 2.2 ΚΩ ID -4 V IDSS = 4.5 mA VDS Vp = -5V 0.68 ΚΩ