EECE 310 Exam, Spring 2008
American University of Beirut · Electronics · 1 page
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Test 3
(Oct 22, 2007)
Two identical zener diodes are connected
back-to-back and in series with a 1.5 V battery as
shown. The diodes have VZ = 5.92 V at 30 mA and
rZ = 4 . When conducting in the forward direction,
they may be modeled by 0.7 V in series with 1 . If
vI is the rectangular waveform shown, determine the
upper and lower levels of vO.
Solution: For the zener diode, Vz = Vz0 + 0.004IZ.
Substituting Vz = 5.92 V and IZ = 30 mA, gives VZ0 =
5.8 V. During the positive phase of vI, the circuit will
be as shown, where Vz0 + VD0 = 6.5 V, and rZ + rD =
5 . It follows that
0.04 A, and
vO = 6.5 + 1.5 + 50.04 = 8.2 V.
During the negative phase of vI, the circuit
will be as shown. It follows that
0.1 A, and vO = 1.5 – 6.5 – 50.1 = – 5.5 V.
t, s
vI, V
10
-10
1
2
t, s
vI, V
10
-10
1
2
+
–
10 V
+
–
vO
45
1.5 V
6.5 V
5
I
+
–
10 V
+
–
vO
45
1.5 V
6.5 V
5
I
+
–
10 V
+
–
vO
45
1.5 V
6.5 V
5
I
+
–
10 V
+
–
vO
45
1.5 V
6.5 V
5
I
+
–
vI
+
–
vO
45
1.5 V
+
–
vI
+
–
vO
45
1.5 V