G.For the given circuitWhat could be done to increase Ay for the circuit?and determine the new value of Ay. Given circuitThe maximum values are determined as,Vps = 25 VVGs = 25 VVGS-OFF = -4 VSmall Signal JFET AmplifierSelf-biased CircuitVDD 20VIpss = 30 mADesign Specifications:Av = 9 (gain ratio)Id = 8mA3RDA minimum value of transconductance will beC5Vout8s = 10000 umhosDrainQ2J3092.2µFC4VA,=GateVin2.2µFSourceIns RoSRGThe equation for drain current is obtained as,ERS20VDp - IpsRp – Vps - IpsRs = 0Voo-VrsIps = Rok,0.8-RDIps = V 5Applying KVL in drain to source circuit -:info that may be usefulon the rightVDD - IDRD – Vps - IşRs = 0------ 1)Putting value in equation -1) with the help of table(Rp + Rs) =Vrn - Vos20 – 10 = .25 K2 ---- 2)Applying KVL in gate to source circuit -:-VGs - 1sRs = 0Ip =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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Answered: G. For the given circuit What could be…

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G. For the given circuit What could be done to increase Ay for the circuit? and determine the new value of Av.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Concept explainers

Superposition Theorem

The superposition theorem is applied to linear systems. For such systems, the output from two different inputs equals the sum of the outputs from each input separately. On applying this theorem many times we get the output from any number of inputs equal to the sum of the outputs from the individual inputs separately.

Question

G.
For the given circuit
What could be done to increase Ay for the circuit?
and determine the new value of Ay.

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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