Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
siven circuit The maximum values are determined as, Vps = 25 V Ves = 25 V Vas-oF = -4 V Ipss = 30 mA Small Signal JFET Amplifier Self-biased Circuit VDD 20V Design Specifications: Av = 9 (gain ratio) Id = 8mA ŽRD ŞRD A minimum value of transconductance will be C5 Vout 8. = 10000 umhos Drain Q2 2.2µF C4 A. Gate J309 Vin HE 2.2uF Source %3D RG The equation for drain current is obtained as, ERS Von - IosRp- Vns - losRs -0 - Sa V-V Ratk Veo-25 Ins %3D Ips- nfo that may be useful Applying KVL in drain to source circuit -: n the right VDp - IpRp - Vps - IsRs = 0------ 1) Putting value in equation -1) with the help of table (Rp + Rs) - Vn- V 20 – 10 - 1.25 K2--- 2) Applying KVL in gate to source circuit -: -Vas - IsRs = 0 Vas Rs - 3) Ip = Rs - - 4) With the help of the table putting minimum value in equation -4) - 250 2-s) Rs - - 2 Putting equation -5) in equation -1) Rp - 1.25 KO - 250 a

siven circuit The maximum values are determined as, Vps = 25 V Ves = 25 V Vas-oF = -4 V Ipss = 30 mA Small Signal JFET Amplifier Self-biased Circuit VDD 20V Design Specifications: Av = 9 (gain ratio) Id = 8mA ŽRD ŞRD A minimum value of transconductance will be C5 Vout 8. = 10000 umhos Drain Q2 2.2µF C4 A. Gate J309 Vin HE 2.2uF Source %3D RG The equation for drain current is obtained as, ERS Von - IosRp- Vns - losRs -0 - Sa V-V Ratk Veo-25 Ins %3D Ips- nfo that may be useful Applying KVL in drain to source circuit -: n the right VDp - IpRp - Vps - IsRs = 0------ 1) Putting value in equation -1) with the help of table (Rp + Rs) - Vn- V 20 – 10 - 1.25 K2--- 2) Applying KVL in gate to source circuit -: -Vas - IsRs = 0 Vas Rs - 3) Ip = Rs - - 4) With the help of the table putting minimum value in equation -4) - 250 2-s) Rs - - 2 Putting equation -5) in equation -1) Rp - 1.25 KO - 250 a

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

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
G. For the given circuit What could be done to increase Ay for the circuit? and determine the new value of Ay.
Transcribed Image Text:G. For the given circuit What could be done to increase Ay for the circuit? and determine the new value of Ay.
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.
Similar questions
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