18490 Exam 2 Solutions
1
Solution Problem 1 (5 points)
From inspection we can see that
VGS = 5V
(1)
VDS = VDD − ID RD .
(2)
iD = K(VGS − Vto )2 = 0.4 · 10−3 (5 − 1.5)2 = 4.9mA.
(3)
VDS = 5 − 4.9 · 10−3 · 103 = 0.1.
(4)
VGS < VDS + Vto .
(5)
Part 1
Let’s see if the transistor is in saturation:
Using Equation (2) we get
If the circuit is in saturation then
However, this is not the case since VGS = 5V . Therefore the hypothesis of saturation is not correct.
Part 2
Let’s see if the transistor is operating in the triode region. If this is the case, then using the value VGS = 5V
the current becomes:
2
2
iD = K 2(VGS − Vto ) VDS − VDS
= 2.8 · 10−3 VDS − 0.4 · 10−3 VDS
.
(6)
Using Equation (2) we get the quadratic equation
2
0.4 VDS
− 3.8 VDS + 5 = 0
(7)
VDS = {7.92, 1.58}.
(8)
The solutions of this equation are:
To pick the right one we need to fulfill the triode condition
VDS + Vto < VGS .
Only the solution VDS = 1.58 fulfills the condition and therefore is the solution.
1
(9)
2
Solution Problem 2 (5 points)
There’s no current going into the gate, so vGS = VGG = 7V.
KVL around the right loop gives the load line,
−VDD + R iD + vDS = 0
(10)
Zeroing vDS in Equation 10 gives the y-intercept:
iD = VDD /R = 2.0A1
(11)
Zeroing iD in Equation 10 gives the x-intercept:
vDS = VDD = 10V
(12)
Thus, drawn on the characteristics graph, the load line looks like this:
From the intersection between the load line and vGS = 7V, iDQ = 0.8A, and vDSQ = 6.0V. The transistor
is in saturation.
1 Realistically, R
D should be something like 5 kΩ, not 5 Ω, so iD should really be 2mA. As it appears on the test, though,
it’s 2A. Both answers got credit.
2
3
Solution Problem 3 (5 points)
3