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ECE 3254, Homework #1 Solutions (Due on 9/11/2006, 50 Points)
For vD = 0.6V, we have ( ) ( )
3 iD 0.2 10 I (^) S exp vD nVT 1 I (^) S exp vD nVT
− = × = ⎡⎣ − ⎤⎦≅
Thus, we determine that:
3 0.2 (^10 ) 1.950 10 A exp exp 0.6 2 0.
D S D T
i I v nV
− × − ≅ = = × ×
Then, for vD = 0.65 V, we have iD = I S ⎡⎣ exp 0.650 0.052( )− 1 ⎤⎦=0.523 mA
Similarly, for vD = 0.70 V, we find iD = IS ⎡⎣exp 0.70 0.052 ( )− 1 ⎤⎦=1.3687 mA
The load line equation is VS = R iS x + vx. Substituting values, this becomes 6 = 3 i (^) x + vx. Next we
plot the nonlinear device characteristic equation
3 ix = vx / 8 and the load line on the same set of
axes. Finally the solution is at the intersection of the load line and the characteristic as shown:
The diagram of a suitable regulator circuit is
We must be careful to choose the value of R small enough, so I (^) Z remains positive for all values
of source voltage and load current. (Keep in mind that the Zener diode cannot supply power.)
From the circuit, we can write, using KCL,
S L Z L
V V
I I
R
= −. Minimum I (^) Z occurs for
I (^) L = 100 mAand VS = 8 V. Solving for the maximum value allowed for R , we have
max
S L
Z L
V V
R
I I
.
Thus, we must choose the value of R to be less than R max (^) = 30 Ω. We need to allow some
margin for component tolerances and some design margin. However, we do not want to choose R
too small, because the current and power dissipation in the diode becomes larger as R becomes
smaller. Therefore, a value of about 24 Ω would be suitable. (This is a standard value.)
With this value of R , we have
,max ,max ,max
2 ,max ,max
,max ,max
208 mA 24
1.04 W
5 1.04 W
S R Z
R R
Z Z
V
I I
R
P I R
P I
