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F. Use a suitable value for RG Rp = 1 K2 Rs = 250 2 . 8x10 For the given circuit What would it look like using an oscilloscope to Measure the V-in and th V-out What would the traces of each look like What is the measured value of Av?

F. Use a suitable value for RG Rp = 1 K2 Rs = 250 2 . 8x10 For the given circuit What would it look like using an oscilloscope to Measure the V-in and th V-out What would the traces of each look like What is the measured value of Av?

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
10th Edition
ISBN:9781337399128
Author:Russell E. Smith
Publisher:Russell E. Smith
Chapter8: Basic Electric Motors
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F. Use a suitable value for RG Rp = 1 KQ Rs = = 250 2. 8x103 For the given circuit What would it look like using an oscilloscope to Measure the V-in and the V-out What would the traces of each look like What is the measured value of A,?
Transcribed Image Text:F. Use a suitable value for RG Rp = 1 KQ Rs = = 250 2. 8x103 For the given circuit What would it look like using an oscilloscope to Measure the V-in and the V-out What would the traces of each look like What is the measured value of A,?
Given circuit The maximum values are determined as, Vps = 25 V VGs = 25 V VGS-OFF = -4 V Small Signal JFET Amplifier Self-biased Circuit VDD 20V Ipss = 30 mA Design Specifications: Av = 9 (gain ratio) Id = 8mA 3RD A minimum value of transconductance will be C5 Vout 8s = 10000 umhos Drain Q2 J309 2.2µF C4 V A, = Gate Vin 2.2µF Source Ins Ro SRG The equation for drain current is obtained as, ERS 20 VDp - IpsRp – Vps - IpsRs = 0 Voo-Vrs Ips = Rok, 0.8-RD Ips = V 5 Applying KVL in drain to source circuit -: info that may be useful on the right VDD - IDRD – Vps - IşRs = 0------ 1) Putting value in equation -1) with the help of table (Rp + Rs) = Vrn - Vos 20 – 10 = .25 K2 ---- 2) Applying KVL in gate to source circuit -: -VGs - 1sRs = 0 Ip = 3) Rs = 4) With the help of the table putting minimum value in equation -4) Rs - - - 250 2----5) Putting equation -5) in equation -1) Rp = 1.25 KN – 2502
Transcribed Image Text:Given circuit The maximum values are determined as, Vps = 25 V VGs = 25 V VGS-OFF = -4 V Small Signal JFET Amplifier Self-biased Circuit VDD 20V Ipss = 30 mA Design Specifications: Av = 9 (gain ratio) Id = 8mA 3RD A minimum value of transconductance will be C5 Vout 8s = 10000 umhos Drain Q2 J309 2.2µF C4 V A, = Gate Vin 2.2µF Source Ins Ro SRG The equation for drain current is obtained as, ERS 20 VDp - IpsRp – Vps - IpsRs = 0 Voo-Vrs Ips = Rok, 0.8-RD Ips = V 5 Applying KVL in drain to source circuit -: info that may be useful on the right VDD - IDRD – Vps - IşRs = 0------ 1) Putting value in equation -1) with the help of table (Rp + Rs) = Vrn - Vos 20 – 10 = .25 K2 ---- 2) Applying KVL in gate to source circuit -: -VGs - 1sRs = 0 Ip = 3) Rs = 4) With the help of the table putting minimum value in equation -4) Rs - - - 250 2----5) Putting equation -5) in equation -1) Rp = 1.25 KN – 2502
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