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F. Use a suitable value for RG Rp 1 ΚΩ 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,? Small Signal JFET Amplifier Self-biased Circuit VDD 20V Design Specifications: ŞRD

F. Use a suitable value for RG Rp 1 ΚΩ 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,? Small Signal JFET Amplifier Self-biased Circuit VDD 20V Design Specifications: ŞRD

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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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 the V-out What would the traces of each look like What is the measured value of A,? Small Signal JFET Amplifier Self-biased Circuit VDD 20V Design Specifications: Av = 9 (gain ratio) Id = 8mA C5 Vout Drain Q2 2.2µF C4 Gate J309 Vin 2.2µF Source RG RS
Transcribed Image Text: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 the V-out What would the traces of each look like What is the measured value of A,? Small Signal JFET Amplifier Self-biased Circuit VDD 20V Design Specifications: Av = 9 (gain ratio) Id = 8mA C5 Vout Drain Q2 2.2µF C4 Gate J309 Vin 2.2µF Source RG RS
The maximum values are determined as, Vps = 25 V VGs = 25 V VGs-OFF = -4 V Ipss = 30 mA A minimum value of transconductance will be, 8. = 10000 umhos A, V %3D %3D Ves+Rslus %3D The equation for drain current is obtained as, 20 VDp - IpsRp – Vps - IpsRs = 0 25 %3D Ips Voo-Vr Ro+ky 0.8-8„RD Vro-25 %3D Ips Applying KVL in drain to source circuit -: VDD IDRD - VDs IşRs = 0----- 1) | Putting value in equation -1) with the help of table (Rp + Rs) = Van - Vs 20 – 10 - 1.25 K2 ---- 2) %3D 8x10 Applying KVL in gate to source circuit -: -VGs IsRs = 0 Vas Rs - 3) Ip Rs = - 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 – 250 2 Rn =1 KQ
Transcribed Image Text:The maximum values are determined as, Vps = 25 V VGs = 25 V VGs-OFF = -4 V Ipss = 30 mA A minimum value of transconductance will be, 8. = 10000 umhos A, V %3D %3D Ves+Rslus %3D The equation for drain current is obtained as, 20 VDp - IpsRp – Vps - IpsRs = 0 25 %3D Ips Voo-Vr Ro+ky 0.8-8„RD Vro-25 %3D Ips Applying KVL in drain to source circuit -: VDD IDRD - VDs IşRs = 0----- 1) | Putting value in equation -1) with the help of table (Rp + Rs) = Van - Vs 20 – 10 - 1.25 K2 ---- 2) %3D 8x10 Applying KVL in gate to source circuit -: -VGs IsRs = 0 Vas Rs - 3) Ip Rs = - 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 – 250 2 Rn =1 KQ
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