Rd
Vdd
Rss
R2
Vdd
R1
T1
0
2
4
6
8
10
12
14
16
18
20
5
4
3
2
1
0
1
2
3
4
5
6
GatetoSource Voltage (V)
Drain Current (mA)
ECE 255
ELECTRONIC ANALYSIS AND DESIGN
Fall 2009
Homework 5 Solutions
Problem 1
Select element values for the circuit shown below to maintain
D
4
I
6 mA
≤
≤
for
transfer functions ranging from
2
GS
D
V
I
20
1
m A
4
=
+
to
2
GS
D
V
I
7.2
1
mA
1.44
=
+
assuming that V
DD
= 16 V.
Shown below are the two limiting transfer functions.
Since the only requirement was
D
4
I
6 mA
≤
≤
there are many possible solutions.
My first guess is
represented by the solid
red line
where V
G
= 4 V and R
SS
= 1 k
Ω
.
Another easy solution is given by
the broken
orange line
which goes through the point I
D
= 5 mA and V
GS
=  1 V and has V
G
= 6 V.
For the orange line
GS
SS
D
V
7V
R
1.4k
I
5m A
∆
=
=
=
Ω
∆
.
The limiting load line, shown as the broken
blue line
, can be determined from the extreme operating points which are V
GS
=  1.8091 V @ 6 mA
and V
GS
= –.3667 V @ 4 mA.
The load line running through these points has an intercept V
G
= 2.518 V
and slope corresponding to R
SS
= 721.2
Ω
.
To satisfy the design requirements, V
G
and R
SS
must be at
least as large (or larger) than the limiting values.
Completing the design using the
red load line
: given that V
DD
= 16 V we can let R
2
= 100 k
Ω
.
Then
R
1
= 300 k
Ω
to provide a gate voltage of V
G
= 4 V.
R
SS
was already assumed to be 1 k
Ω
.
The only
restriction on choosing R
D
is that it can not be so large as to put the FET into the ohmic region of
operation.
Let R
D
= 1 k
Ω
, then V
DG
≥
6 V and either transistor will be safely in the B.P.O. region.
+
V
i
–
+
V
o
–
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Problem 2
Show your calculation to determine the largest
value of R
D
that can be used in the circuit that you biased
in Problem 1 that will guarantee operation in the BeyondPinchOff region.
The limiting case is determined by the operating point with the highest current (assuming that this
operating point is also associated with the device with the largest V
P
, which is almost always the case).
The gatetosource voltage at the highest possible current is determined from
2
GS
V
6
20
1
4
=
+
,
therefore,
GS
6
V
4
1
20
=

±
⇒
V
GS
= –1.809 or –6.191 V.
Clearly the first value is correct.
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 Spring '08
 Staff
 Trigraph, Vgs, Grid bias

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