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ECE 3455 -- Quiz #4 Shattuck Section MW 11:30AM - 1PM April 7, 2004
Keep this quiz closed and face up until you are told to begin.
1. This quiz is closed book, closed notes. You may use one 8.5" x 11" crib sheet, or its equivalent. 2. Show all work on these pages. Show all work necessary to complete the problem. If you go on to another page, indicate clearly where your work can be found. A solution without the appropriate work shown will receive no credit. 3. Show all units in solutions, intermediate results, and figures. Units in the quiz will be included between square brackets. 4. Do not use red ink. Do not use red pencil. 5. You will have 30 minutes to work on this quiz. ________________/20
Room for extra work
For the circuit given below, assume that the diodes can be modeled with a piecewise linear model with Vf = 1[V], rd = 2[k], and Is = 100[A]. Find VX. You are expected to be able to complete at least two complete tests, clearly stating your guesses, and showing how you tested your guesses, in the time period allowed.
+12[V]
R1 = 2.7[k]
VX R2 = 10[k]
1.5[mA]
-12[V]
-12[V]
-12[V]
Room for extra work
ECE 3455 -- Quiz #4 April 7, 2004 11:30AM 1:00PM MW Shattuck Section Solution For the circuit given below, assume that the diodes can be modeled with a piecewise linear model with Vf = 1[V], rd = 2[k], and Is = 100[A]. Find VX. You are expected to be able to complete at least two complete tests, clearly stating your guesses, and showing how you tested your guesses, in the time period allowed.
+12[V]
R1 = 2.7[k]
VX = R2 10[k]
1.5[mA]
-12[V]
-12[V]
-12[V]
We need to name the diodes, and define the names of our regions for the diode model. We do this in the diagram that follows.
+12[V]
iD 4 iD vD + vD -
R1 = 2.7[k] VD1 + R2 = 10[k] + VD2 D2 ID2 D1 ID1
3 2 1
VX
1.5[mA]
-12[V]
-12[V]
-12[V]
So, we will guess that diode D1 is in region 1, and diode D2 is in region 4. Now, we need to put the models in for the diodes. We put the two models in, and we get the circuit that follows.
+12[V]
R1 = 2.7[k] ID1 -0.1[mA] VD1 + + + VD2 -12[V] 2[k] ID2 -12[V] -12[V]
Now, we write KCL for the VX node, to get
VX 1[V] 1.5[mA]
R2 = 10[k]
VX - (-12[V]) V - 1[V] - ( -12[V]) - 0.1[mA] + 1.5[mA] + X = 0. 10[k] 2[k] Solving, we get VX = -13.5[V].
From this, we can get
VD 2 = VX - ( -12[V]) = -1.5[V]. Is VD 2 f 1[V]? The answer is no, so we made a bad guess.
Now, for this region, we have the test,
We need to guess again. This time, we will guess that both diodes are in region 1. We ha...