The document discusses junction field-effect transistors (JFETs). It describes JFETs as voltage-controlled devices with high input impedance that are commonly used as amplifiers and logic switches. The key types are n-channel and p-channel JFETs, which differ in whether electrons or holes serve as the majority current carriers. JFET operation involves controlling the channel resistance through a voltage applied to the gate terminal, allowing the drain current to be modulated.
2. Introduction (FET)
Field-effect transistor (FET) are important
devices such as BJTs
Also used as amplifier and logic switches
What is the difference between JFET and
BJT?
5. Types of Field Effect Transistors
(The Classification)
JFET
MOSFET (IGFET)
n-Channel JFET
p-Channel JFET
n-Channel
EMOSFET
p-Channel
EMOSFET
Enhancement
MOSFET
Depletion
MOSFET
n-Channel
DMOSFET
p-Channel
DMOSFET
FET
6. High input impedance (M)
(Linear AC amplifier system)
Temperature stable than BJT
Smaller than BJT
Can be fabricated with fewer processing
BJT is bipolar – conduction both hole and
electron
FET is unipolar – uses only one type of current
carrier
Less noise compare to BJT
Usually use as an Amplifier and logic switch
Introduction.. (Advantages of FET
over BJT)
10. N channel JFET:
Major structure is n-type material (channel)
between embedded p-type material to form 2 p-
n junction.
In the normal operation of an n-channel device,
the Drain (D) is positive with respect to the
Source (S). Current flows into the Drain (D),
through the channel, and out of the Source (S)
Because the resistance of the channel depends
on the gate-to-source voltage (VGS), the drain
current (ID) is controlled by that voltage
N-channel JFET
12. P channel JFET:
Major structure is p-type material
(channel) between embedded n-type
material to form 2 p-n junction.
Current flow : from Source (S) to Drain
(D)
Holes injected to Source (S) through p-
type channel and flowed to Drain (D)
P-channel JFET
16. JFET Characteristic for VGS = 0 V and 0<VDS<|Vp|
To start, suppose VGS=0
Then, when VD
Sis increased, IDincreases.
Therefore, IDis proportional to VD
Sfor small values
of VD
S
For larger value of VDS, as VD
Sincreases, the
depletion layer become wider, causing the
resistance of channel increases.
After the pinch-off voltage (Vp) is reached, the I
D
becomes nearly constant (called as ID maximum,
IDSS-Drain to Source current with Gate Shorted)
17. JFET for VGS = 0 V and 0<VDS<|Vp|
Channel
becomes
narrower as
VDS is
increased
21. JFET Characteristic Curve..
For negative values of VGS, the gate-to-channel
junction is reverse biased even with VDS=0
Thus, the initial channel resistance of channel is
higher.
The resistance value is under the control of VGS
If VGS = pinch-off voltage(VP)
The device is in cutoff (VGS=VGS(off) = VP)
The region where ID constant – The saturation/pinch-
off region
The region where ID depends on VDS is called the
linear/ohmic region
28. Transfer Characteristics
The input-output transfer characteristic of
the JFET is not as straight forward as it is
for the BJT. In BJT:
IC= IB
which is defined as the relationship
between IB (input current) and IC (output
current).
29. Transfer Characteristics..
In JFET, the relationship between VGS (input
voltage) and ID (output current) is used to
define the transfer characteristics. It is called
as Shockley’s Equation:
The relationship is more complicated (and not
linear)
As a result, FET’s are often referred to a
square law devices
2
V G S
ID = ID S S 1 -
V P
VP=VGS (OFF)
30. Defined by Shockley’s equation:
Relationship between IDand VGS.
Obtaining transfer characteristic curve axis
point from Shockley:
When VGS= 0 V
, ID= ID
S
S
When VGS= VGS(off)or Vp, ID= 0 mA
(off )
V
V
VP VGS
GS (off )
2
ID IDSS
1GS
Transfer Characteristics…
32. Exercise 1
ID
VGS = VP
1 -
2
IDSS
VGS
ID = IDSS 1-
VP
GS
V ID
IDSS
IDSS/2
DSS
I /4
0
0.3Vp
0.5Vp
Vp 0 mA
Sketch the transfer defined by
ID
S
S= 12 mA dan VGS(off) = - 6
33. Exercise 1
ID
VGS = VP
1 -
IDSS
Sketch the transfer defined by ID
S
S= 12 mA dan VGS(off)
= Vp
=- 6
IDSS
DSS
I /2
IDSS/4
2
VGS
ID = IDSS 1 -
VP
GS
V =0.3VP
GS
V =0.5VP
35. Exercise 2
ID
VGS = VP
1 -
2
IDSS
VGS
ID = IDSS 1-
VP
VGS ID
IDSS
IDSS/2
DSS
I /4
0
0.3Vp
0.5Vp
Vp 0 mA
Sketch the transfer defined by
ID
S
S= 4 mA dan VGS(off) = 3 V
36. Exercise 2
ID
VGS = VP
1 -
IDSS
Sketch the transfer defined by
ID
S
S= 4 mA dan VGS(off)= 3V
2
VGS
ID = IDSS 1 -
VP
VGS =0.3VP
VGS =0.5VP
VP
IDSS
DSS
I /2
IDSS/4