Manual solution only, no Al used 20. For the self-bias configuration of Fig. 7.92, determine:a. Ipo and VGSQb. Vps and VD.*21. For the network of Fig. 7.93, determine:a. Ipo and VGSOb. Vps and Vs.x14 V1.2 ΚΩ꼭100Dss = 6 mAVp = -4 VVoso10.43 ΚΩ11 ΜΩ18 VJE2.2 ΚΩIDSS = 8 mAVp = -8 V+VGSQ10.39 ΚΩFIG. 7.92Problem 20.8-4 VFIG. 7.93Problem 21.

Step 1: Step 2: Step 3: Step 4:

Answered: 20. For the self-bias configuration of…

Introductory Circuit Analysis (13th Edition)

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Author:Robert L. Boylestad

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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
2. For the self-bias configuration of Fig. 7.80 a. Sketch the transfer curve for the device. b. Superimpose the network equation on the same graph. c. Determine ID and VGS. d. Calculate VDS, VD, VG, and Vs. 18 V 1.5 kQ Dss = 10 mA V, =-4V Vase 1 MQ 750 n
(TRY These) Self Bias PDF Solve these problems on p-channel devices. NOTE:- For p-channel FETS the same calculations and graphs are used, except that the voltage polarities and current directions are reversed. The graphs are mirror images of the n-channel graphs. 32. For the network of Fig. 7.101, determine: a. Ip, and VGso b. VDs- с. Vр- -18 V 2.2 k2 IDss= = 8 mA Vp = 4 V VGSQ 1 ΜΩ 0.51 kS2 FIG. 7.101 Problem 32.
DC EXAMPLE 7.1 Determine the following for the network of Fig. 7.6: VGSQ a. VDD b. IDg c. VDS. d. Vp. ´e. VG. f. Vs. + 1 ΜΩ 2 V G + VGS Ter 16. V 2kQ D -S 1pss = 10 mA DSS V₂ = -8 V FIG. 7.6 Figure ? ? Example of a Fixed-Bias Circuit. 11 VOD - ID Rp 16 - 1
1. For the fixed-bias configuration of Fig. 7.75 a. Sketch the transfer characteristics of the device. b. Superimpose the network equation on the same graph. c. Determine ID, and VDS. d. Using Shockley's equation, solve for ID and then find VDS. Compare with the solutions of part (c). 14 V 1.8 kN Ipss = 12 mA VpsQ V, =-4V 1 MQ 1.5 V-
Fixed bias configuration1. Plot the transfer characteristics of the device and superimpose the network equation on the same graph. 2. Determine IDQ and VDS.3. Using Shockley's equation, solve for IDQ and determine VDSQ. Compare with the solutions in part (c).
7.2 Fixed-Bias Configuration 1. For the fixed-bias configuration of Fig. 7.75: a. Sketch the transfer characteristics of the device. b. Superimpose the network equation on the same graph. c. Determine Ip, and Vpsg d. Using Shockley's equation, solve for Ip, and then find Vpsg. Compare with the solutions of part (c). 14 V 1.8 kn IDQ Ipss = 12 mA Vpso Vp = -4 V 1 M2 1.5 V FIG. 7.75
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1. For the fixed-bias configuration of Fig. 7.75: a. Sketch the transfer characteristics of the device. b. Superimpose the network equation on the same graph. c. Determine In, and Vos d. Using Shockley's equation, solve for Ip, and then find Vps- Compare with the solutions of part (c). 1 MQ 1.5 V 14 V 1.8 ΚΩ IDQ + Vpso FIG. 7.75 pss= 12 mA Vp = -4 V

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