1.) IS THE POSSIBLE SWING MA XIMI ZED? a-) YES. THE SELECTED VALUE OF Vour IS EXACTLY HAF of THE SUPPLY RANGE. B.) VET. THE VorsAt Of EAlH TRANSISTOR ARE EQUAL of VALUE MIOWNG MAXIMUM OWING AT THE GIVEN VAUE OF Vour· c) NO. THE TRANSISTORS HAVE DIFFERENT PROPERTIES TO BEGIN WITH 8o BOTH CAN'T BE BIASEO n STURATION REGION AT THE SAME TIN'E.

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...
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Q UESTIO NS:
1.) IS THE POSSIBLE SWING MA XIMI 2ED?
a.) YES. THE SELECTED VALUE OF Vour I5 EXACTLY HME OF
THE SUPPLY RANGE.
B.) VET. THE Vorsar Of EACH TRANSISTOR ARE EQUAM of
VALUE ALLOWING MAXIMUM owING AT TH GIVEN VAUE
OF Vour
G) NO. THE TRANSISTORS HAVE DIFFERENT PROPERTIES TO
BEGIN WITH 8o BOTH CAN'T BE BIASEO SATURATION
REBION AT THE
SAME TIN'E .
Transcribed Image Text:Q UESTIO NS: 1.) IS THE POSSIBLE SWING MA XIMI 2ED? a.) YES. THE SELECTED VALUE OF Vour I5 EXACTLY HME OF THE SUPPLY RANGE. B.) VET. THE Vorsar Of EACH TRANSISTOR ARE EQUAM of VALUE ALLOWING MAXIMUM owING AT TH GIVEN VAUE OF Vour G) NO. THE TRANSISTORS HAVE DIFFERENT PROPERTIES TO BEGIN WITH 8o BOTH CAN'T BE BIASEO SATURATION REBION AT THE SAME TIN'E .
5V
THE CURCUET
ONTHE
LEFT IS BASED SUCH
THAT Vour 2.5 V. GIVON THE FOLLOWING
VBIAS
M2
TRANSISTOR PROPERTIO:
mA
kp: 10. 25
an pl
Kn :
Vout
0.7 V
: 0.5V
dp= 0.04 +
: 0. 01
Vin
COMMON SOURCE WITH ACTIVE LOND
Transcribed Image Text:5V THE CURCUET ONTHE LEFT IS BASED SUCH THAT Vour 2.5 V. GIVON THE FOLLOWING VBIAS M2 TRANSISTOR PROPERTIO: mA kp: 10. 25 an pl Kn : Vout 0.7 V : 0.5V dp= 0.04 + : 0. 01 Vin COMMON SOURCE WITH ACTIVE LOND
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