Determine V P for the characteristics of Fig. 6.22 using I DSS and I D at some
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Determine VP for the characteristics of Fig. 6.22 using IDSS and ID at some value of VGS. That is, simply substitute into Shockley’s equation and solve for VP. Compare the result to the assumed value of -3 V from the characteristics.
Fig. 6.22
Transcribed Image Text:
Vas = 0 V oss = 9 mA Vertical Sens. 1 mA per div. Vas =-0.5 V Нorizontal Sens. per div. Vas =-1 V oss = 45 mA (Vas = -0.9 V) 500 mV per slep. -Vas =-1.5 V 2 m per div. ImA div Vas =-2 V V-3 V div
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To determine VP for the characteristics of Fig 622 we can use Shockleys equation VP 2IDSSCoxVGS VP W...View the full answer
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Related Book For
Electronic Devices And Circuit Theory
ISBN: 9781292025636
11th Edition
Authors: Robert Boylestad, Louis Nashelsky
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