0195136047.pdf

7.4 FIELD-EFFECT TRANSISTORS 375

Solution

(a) ForvGS=1 V: Since it is less thanVT, the MOSFET is in the cutoff region so that

iD=0, which corresponds to the normally off state of the MOSFET,

vDS=VDD=20 V

(b) ForvGS=5 V: The MOSFET operates in the active region,

iD=

IDSS

VT^2

(vGS−VT)^2 =IDSS

(

vGS

VT

− 1

) 2

= 8 × 10 −^3

(

5

4

− 1

) 2

= 0 .5mA

vDS=VDD−RDiD= 20 −( 5 × 103 )( 0. 5 × 10 −^3 )= 17 .5V

The active MOSFET behaves like a nonlinear voltage-controlled current source.

(c) ForvGS=9 V: Since it is greater than 2VTso thatiDincreases whilevDSdecreases, the

MOSFET is presumably in the linear ohmic state when

|vDS|≤

1

4

(vGS−VT) and vGS>VT

The theory for this case predicts thatiD∼=vDS/RDS, whereRDSis the equivalent drain-to-source

resistance given by

RDS=

VT^2

2 IDSS(vGS−VT)

Here the MOSFET acts as a voltage-controlled resistor. For our example,

RDS=

42

2 ( 8 × 10 −^3 )( 9 − 4 )

= 200 

Since this resistance appears in series withRD,

iD=

VDD

RD+RDS

=

20

5000 + 200

= 3 .85 mA

and
vDS=RDSiD= 200 × 3. 85 × 10 −^3 = 0 .77 V
which is less than^1 / 4 ( 9 − 4 )V.
Figure 7.4.9 shows ap-channel enhancement MOSFET which differs from ann-channel
device in that the doping types are interchanged. Channel conduction now requires negative gate-
to-source voltage or positive source-to-gate voltage. WithvSG>VTandvSD >0,iDflows
from source to drain, as shown in Figure 7.4.9(a). Thep-channel andn-channel MOSFETs are
complementarytransistors, having the same general characteristics but opposite current direction
and voltage polarities. By replacingvGSandvDSwithvSGandvSD, respectively, the equations of
then-channel MOSFET apply to thep-channel MOSFET. However, ap-type channel does not
conduct as well as ann-type channel of the same size because holes are less mobile than electrons.
Consequently, smaller values ofIDSSare typical of thep-channel MOSFETs.

DEPLETIONMOSFETS

Figure 7.4.10 illustrates depletion MOSFETs and their symbols. Because the channel is built
in, no field effect is required for conduction between the drain and the source. The depletion