For the given circuit (shown in part (A)) the time dependent input voltage vin(t) and corresponding output vo(t) are shown in part (B) and part (C), respectively. Identify the components that are used in the circuit between points X and Y. [figure]
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Answer
(2) [figure]
Solution
Given branch (ideal diode + Zener, VZ=5 V):
⇒ When X is positive, the ideal diode is forward biased and the Zener is reverse biased.
⇒ It conducts once vin>5 V, so vo is held at +5 V (positive half clipped).
⇒ When X is negative, the ideal diode is reverse biased, so this branch is open.
The output is also held at -5 V in the negative half, so the X–Y branch must:
(i) stay open when X is positive, and
(ii) conduct from Y to X with a fixed 5 V drop once VY-VX>5 V.
Option (2): diode with cathode towards X, Zener with cathode towards Y.
⇒ X positive: the diode is reverse biased ⇒ branch open
⇒ Y positive: the diode is forward biased (0 V) and the Zener is reverse biased; it breaks down at 5 V ⇒ vo=-5 V
A resistor (options 1, 3) cannot hold the voltage constant, and in option (4) the Zener conducts forward at 0 V, so vo would be 0 V instead of -5 V.
Answer: (2)