Q3VBR₁330523R₂27052Figure 3.18(a) Circuit symbolof the pr junctiondiode; (b) voltage-current character-istic of a silicondiode with I, =10-12 A and n = 1.For negative valuesof up, the diodecurrent is too smallto plot precisely onthe graph.b) is forLDFind a) is for VB = 2 V+UD-2.0c) is and VD ifdiode is reversed.XVDVB = 6 V(a)iDip=-1₁-1.0Reverse-biased region} Use load-line techniqueVB = 6V and theip (mA)201510Given the diodehas the characteristicgiven in the vi plot.Is = 10-1²2 A5Forward-biasedregion1.0Turn-on voltage2.0UD (V)

Answered: Image /qna-images/answer/4a052a08-8739-4319-b2b6-3e34595d9de6.jpg

Q3 VB Find Figure 3.18 (a) Circuit symbol of the pr junction diode; (b) voltage -current character- istic of a silicon R₁ 33052 diode with I, = 10-12 A and n = 1. For negative values of up, the diode current is too small to plot precisely on the graph. 3R₂ 27052 a) iD for VB = 2 V b) is for Up NB = 6 V is and no it -2.0 + diode is reversed. (a) ip=-1₁ VD J -1.0 Reverse-biased region ip (mA) 20 15 10 5 Given the diode has the characteristic given in the vi plot 12 Is = 10-1¹²A Use load-line technique VB = 6V and the I Forward-biased region A 1.0 Turn-on voltage 2.0 UD (V)

Q3 VB Find Figure 3.18 (a) Circuit symbol of the pr junction diode; (b) voltage -current character- istic of a silicon R₁ 33052 diode with I, = 10-12 A and n = 1. For negative values of up, the diode current is too small to plot precisely on the graph. 3R₂ 27052 a) iD for VB = 2 V b) is for Up NB = 6 V is and no it -2.0 + diode is reversed. (a) ip=-1₁ VD J -1.0 Reverse-biased region ip (mA) 20 15 10 5 Given the diode has the characteristic given in the vi plot 12 Is = 10-1¹²A Use load-line technique VB = 6V and the I Forward-biased region A 1.0 Turn-on voltage 2.0 UD (V)

Electric Motor Control

10th Edition

ISBN:9781133702818

Author:Herman

Publisher:Herman

Chapter59: Motor Startup And Troubleshooting Basics

Section: Chapter Questions

Problem 12SQ: How is a solid-state diode tested? Explain.

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