5. Squaring both sides and substituting for i1 + ip2 = | gives ipi – 2/ipiiD2 + ip2 1 knT Vid 2 inim = I-;K, W 2 U jd %3D 2 6. Substituting for İ22 as ip2 = 1– ip1 and squaring both sides of the resulting equation provides a quadratic equation in i that can be solved to yield

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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How to drive id1 and id2 the steps are written but I did not understand point 5 to the whole and I prefer to explain it for me by steps
5. Squaring both sides and substituting for ip1 + ip2 = | gives
1
W 2
V id
W
ip1 – 2/ipiiD2 + ip2
kn7 Vid
6. Substituting for Ip2 as Ip2 = /- ip1 and squaring both sides of the resulting
equation provides a quadratic equation in i1 that can be solved to yield
W, ("id
Kn L
(v/2)
2.
W
I/ k,
(v/2)
1
(v ja/ 2)*
W
I/ K
W
1
2
W
k'n
%3D
%3D
W
I/ K,
W
2 2L
W
(Vos - V, = K,- Vov
Then, replace K,(W/L) by VVov
Since,
/2 2
ipi
Vov 2
1
Vov
ipz
Vov 2.
1.
Vov
Transcribed Image Text:5. Squaring both sides and substituting for ip1 + ip2 = | gives 1 W 2 V id W ip1 – 2/ipiiD2 + ip2 kn7 Vid 6. Substituting for Ip2 as Ip2 = /- ip1 and squaring both sides of the resulting equation provides a quadratic equation in i1 that can be solved to yield W, ("id Kn L (v/2) 2. W I/ k, (v/2) 1 (v ja/ 2)* W I/ K W 1 2 W k'n %3D %3D W I/ K, W 2 2L W (Vos - V, = K,- Vov Then, replace K,(W/L) by VVov Since, /2 2 ipi Vov 2 1 Vov ipz Vov 2. 1. Vov
+Vom HANential inf
1. Expression drain currents for Q, and Q,.
1
W
İpi
Kn 7 (v GSI - V)
L.
W
(v GS2
-
%3D
2. Take the square roots of both sides of both equations
JiDi =
W
Kn7 (v CsI - V.)
Im =
W
K, (v cs2- V)
GS2
Q2
%D
I GS2
3. Subtract the two equations and perform
UGI
オVL
Vid
results in
22
W
Jim - Jim -
%3D
Transcribed Image Text:+Vom HANential inf 1. Expression drain currents for Q, and Q,. 1 W İpi Kn 7 (v GSI - V) L. W (v GS2 - %3D 2. Take the square roots of both sides of both equations JiDi = W Kn7 (v CsI - V.) Im = W K, (v cs2- V) GS2 Q2 %D I GS2 3. Subtract the two equations and perform UGI オVL Vid results in 22 W Jim - Jim - %3D
Expert Solution
Step 1

Transistors :

A semiconductor device that is used to boost electronic signals and electrical power.

Transistors are fundamental components of modern electronics.

It is generally made of semiconductor.

A transistor has at least three terminals for connecting to an external circuit, which are :

1. Collector 2. Base 3. Emitter


The collector current is controlled by the base current, hence, it is a current-driven device.

 

There are two types of transistors, which are :

PNP transistor : The PNP transistor is a kind of transistor that has one n-type material and two p-type materials.

NPN transistor :The NPN transistor is a kind of transistor that has one p-type material and two n-type materials.

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