01__Field effect transistor - Teamwork

heartlustElectronics - Devices

Nov 2, 2013 (3 years and 7 months ago)

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W. Bergholz GEE 2 (Spring 2012)

1

01_Field Effect transistors

Objectives:


Understand how a
f
ield
e
ffect
t
ransistor works


Know the essential parameters


Know the types of FETs


Understand simple circuits


Acknowledgement: most figures from ITT training material








W. Bergholz GEE 2 (Spring 2012)

2

01_Field Effect transistors

Motivation and introduction:



A bipolar transistor does many useful things



Assembled in integrated circuits, even more



So why would you want to have something else ?







W. Bergholz GEE 2 (Spring 2012)

3

01_Field Effect transistors

Motivation and introduction:









high

W. Bergholz GEE 2 (Spring 2012)

4

01_Field Effect transistors

Motivation and introduction:









W. Bergholz GEE 2 (Spring 2012)

5

01_Field Effect transistors

Motivation and introduction:









W. Bergholz GEE 2 (Spring 2012)

6

01_Field Effect transistors

Motivation and introduction:


Key idea: influence the flow of charge carriers between
source

and
drain

by an electric field


The electric field is set up by the
gate

(insulated from the
channel by a pn
-
junction in reverse bias








W. Bergholz GEE 2 (Spring 2012)

7

01_Field Effect transistors

Motivation and introduction:









W. Bergholz GEE 2 (Spring 2012)

8

01_Field Effect transistors


Before looking at this in detail, transform the 3
-
dimensional view to a
2
-
dimensional

representation








W. Bergholz GEE 2 (Spring 2012)

9

01_Field Effect transistors transistors


3
-
terminal device, like a bipolar transistor:








W. Bergholz GEE 2 (Spring 2012)

10

01_Field Effect transistors

How does it work?


First simplest approach: connect gate to
source


So U
GS

= 0


Explore I
DS

versus U
DS







W. Bergholz GEE 2 (Spring 2012)

11

01_Field Effect transistors

Field effect:


The silicon between drain and source is a
distributed resistive medium


Simple model: a network of resistors


As one moves from the source to the drain,
there is a rise in voltage


In the center, where the gate is, this is at
about 5V


The gate is at 0V !


Therefore:

The pn junction is in
about 5V reverse bias




W. Bergholz GEE 2 (Spring 2012)

12

01_Field Effect transistors

Pinch off:


This gives rise to the pinch
-
off effect, as
U
DS

is increased!




As the voltage goes up,



the space charge regions expand,
the channel gets narrower


hence the there is no increase in
current I
D
in spite of an increase in
the voltage U
DS

W. Bergholz GEE 2 (Spring 2012)

13

01_Field Effect transistors

Family of
characteristic curves

for the
J
-
fet:


J
-
fet stands for junction fet



For the bipolar transistor we had


I
C
vs. U
EC

for different I
B




For the J
-
fet we need something similar to use it as an
amplifier:




I
D

vs.
U
SD

for different
U
GS




W. Bergholz GEE 2 (Spring 2012)

14

01_Field Effect transistors

I
S

vs.
U
SD

for different
U
GS






W. Bergholz GEE 2 (Spring 2012)

15

01_Field Effect transistors

I
S

vs.
U
SD

for different
U
GS






W. Bergholz GEE 2 (Spring 2012)

16

01_Field Effect transistors


The parameter which characterizes the
„amplification“ of the BJT is the
current amplification



The corresponding paramter for FETs is
the

mutual conductance g
m






W. Bergholz GEE 2 (Spring 2012)

17

01_Field Effect transistors

Power comparision:



Mutual conductance 3 MilliMohs does not sound very promising, in
terms of amplification, but



What really matters is the power which is used in the input circuit



many sources have comparatively high voltages, but a very high
internal resistance, i.e. they cannot support „large“ currents (whatever
small or large means)

W. Bergholz GEE 2 (Spring 2012)

18

01_Field Effect transistors

Power comparision:


BJT: Ri = 100kOhm, U = 2V





I
base

= (2V) /(10
5

V/A) = 2 x 10
-
5

A





P
in

= 2 x 2 x 10
-
5

W = 4 x 10
-
5

W =
40 µW



not much, but far too much for many signal sources



J
-
FET: Ri = 10
9
Ohm (pn
-
junction in
reverse

bias!)





I
gate

= (2V) /(10
9

V/A) = 2 x 10
-
9

A





P
in

= 2 x 2 x 10
-
9

W = 4 x 10
-
9

W =
4 nW



This is much better, also much better than a typical op amp


W. Bergholz GEE 2 (Spring 2012)

19

01_Field Effect transistors

First application of an FET: Constant current source


Key idea:


2 circuit elements J
-
FET and R in combination


Feedback so that only one current is „permissable“



In detail:


I(U) curve of a resistor: Straight line, I = U / R


I(U) curve of a J
-
FET:




How does this help us?



W. Bergholz GEE 2 (Spring 2012)

20

01_Field Effect transistors


I(U) curve of a resistor: Straight line, I = U / R


I(U) curve of a J
-
FET:




Idea #1:


For R:


as U increases, I through R
increases


For the J
-
FET: as U increases, I
D

decreases



So: all we need to somehow „couple“ the 2 currents /
voltages that this sets up a kind of „equilibrium


how?




W. Bergholz GEE 2 (Spring 2012)

21

01_Field Effect transistors


Complete circuit: Power supply, [J
-
FET + Rs]=2
-
terminal device, load
resistor R
L


Numerical example:


Up = 8V, Uo=30V, Iop = 10mA, Idss=20mA, for this current

Ugs =
-
4.0V


Calculate the „right“ Rs: Rs = U
R
/Iop = 4.0V/10mA = 400 Ohm


Will this work for any load resistor?


R
L

U
L

W. Bergholz GEE 2 (Spring 2012)

22

01_Field Effect transistors


Answer: idea # 2 is to use
feedback


Feed the voltage drop over R back to the gate of the J
-
FET:








problem again: we have a non
-
linear circuit element


pratical method
: load line analysis


W. Bergholz GEE 2 (Spring 2012)

23

01_Field Effect transistors


Loadline analysis for J
-
FET + source resistor: U
GS

= U
R


Operating point:



only this current is „allowed“



I higher


higher Ugs


channel gets narrower



I lower


lower Ugs


channel gets wider

I = U
R
/Rs

W. Bergholz GEE 2 (Spring 2012)

24

01_Field Effect transistors


Loadline analysis for J
-
FET + source resistor: U
GS

= U
R


In Lab Gen EE2:


-

solve the system of equations:


Id = U
R
/Rs (1)


Id = Idss(1
-
Ugs/Up)


analytically


-

Solution for the 2 unknowns
I
dop

and U
gsop

represent the
operating point

I = U
R
/Rs

W. Bergholz GEE 2 (Spring 2012)

25

01_Field Effect transistors


NO !


Current in this example: 10mA


U
L

= R
L

x 10mA, i.e. the larger R, the larger the voltage drop over it, it cannot
be bigger than Uo, and even this is too much, since there must be „enough“
voltage left for the device to still operate in the linear region.


How much is „enough“?

R
L

U
L