Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
Given circuit The maximum values are determined as, Vos = 25 V Ves = 25 V Ves-OFF = -4 V Small Signal JFET Amplifier Self-biased Circuit VDD 20V Ipss = 30 mA Design Specifications: Av = 9 (gain ratio) Id = 8mA SRD A minimum value of transconductance will be C5 Vout R = 10000 umhos Drain Q2 JJ309 2.2uF C4 A, = Gate Vin HE 2.2µF Source RG The equation for drain current is obtained a ZRS Vnp - IpsRp - Vns - InsRs -0 V-Vn Kk Ins = - TR Ips- 23 V-25 Applying KVL in drain to source circuit -: info that may be useful VDp - IpRp - Vps - IşRs = 0------ 1) on the right Putting value in equation -1) with the help of table (Rp + Rs) - Y V - Vn - Vn. 20- 10 - 1. 25 K2 --- 2) to Applying KVL in gate to source circuit : -Vas s - Is Rs = 0 Va. 3) Ip = Rs - 4) With the help of the table putting minimum value in equation -4) Rs - 250 s2 --s) Putting equation -5) in equation -1) Rp=1.2S KQ - 250n

Given circuit The maximum values are determined as, Vos = 25 V Ves = 25 V Ves-OFF = -4 V Small Signal JFET Amplifier Self-biased Circuit VDD 20V Ipss = 30 mA Design Specifications: Av = 9 (gain ratio) Id = 8mA SRD A minimum value of transconductance will be C5 Vout R = 10000 umhos Drain Q2 JJ309 2.2uF C4 A, = Gate Vin HE 2.2µF Source RG The equation for drain current is obtained a ZRS Vnp - IpsRp - Vns - InsRs -0 V-Vn Kk Ins = - TR Ips- 23 V-25 Applying KVL in drain to source circuit -: info that may be useful VDp - IpRp - Vps - IşRs = 0------ 1) on the right Putting value in equation -1) with the help of table (Rp + Rs) - Y V - Vn - Vn. 20- 10 - 1. 25 K2 --- 2) to Applying KVL in gate to source circuit : -Vas s - Is Rs = 0 Va. 3) Ip = Rs - 4) With the help of the table putting minimum value in equation -4) Rs - 250 s2 --s) Putting equation -5) in equation -1) Rp=1.2S KQ - 250n

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
See similar textbooks
icon
Related questions
Question
100%

Even if the question is to complex can you at least point me in a Direction 

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
E. Using a suitable value for RG within the given circuit Apply a 10mVpPK 10kHz sinewave to the input.
Transcribed Image Text:E. Using a suitable value for RG within the given circuit Apply a 10mVpPK 10kHz sinewave to the input.
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Logic Gate and Its Application
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS