Identification of Circuit Components Between Points X and Y

To determine the correct combination of components placed between points $X$ and $Y$, let us analyze the operation of the circuit during both positive and negative half-cycles of the input voltage $v_{in}(t)$.

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1. Analysis of the Output Waveform $v_o(t)$

• The input voltage $v_{in}(t)$ is a sinusoidal wave with a peak amplitude of $20\text{ V}$.
• The output voltage $v_o(t)$ is clipped at:
  • $+5\text{ V}$ during the positive half-cycle ($v_{in} > 0$).
  • $-5\text{ V}$ during the negative half-cycle ($v_{in} < 0$).

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2. Analysis of the Middle Branch in Part (A)

The middle branch connected in parallel with the source consists of:

• An ideal diode pointing downwards (anode connected to $X$, cathode connected to the Zener diode).

• A Zener diode ($V_Z = 5\text{ V}$) pointing upwards (cathode connected to the ideal diode, anode connected to $Y$).

• For $v_{in} > 0$ ($V_X > V_Y$):

  • The ideal diode is forward biased (acts as a short circuit with zero voltage drop).
  • The Zener diode is reverse biased. When $v_{in}(t) \ge 5\text{ V}$, it undergoes Zener breakdown.
  • This clamps the voltage across the parallel combination to $V_X - V_Y = +5\text{ V}$.

• For $v_{in} < 0$ ($V_X < V_Y$):

  • The ideal diode in the middle branch is reverse biased (acts as an open circuit).
  • Thus, no current flows through the middle branch during the negative half-cycle.

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3. Determination of the Components Between $X$ and $Y$

Since the middle branch is inactive during the negative half-cycle ($v_{in} < 0$), the circuit between points $X$ and $Y$ must be responsible for clipping the output voltage at $-5\text{ V}$.

To achieve this:

1. For $v_{in} > 0$ ($V_X > V_Y$): The branch between $X$ and $Y$ must act as an open circuit so that it does not interfere with the $+5\text{ V}$ positive clipping provided by the middle branch.
2. For $v_{in} < 0$ ($V_Y > V_X$): The branch between $X$ and $Y$ must conduct and undergo Zener breakdown at $V_Z = 5\text{ V}$, clamping $V_Y - V_X = 5\text{ V}$ (or $v_o(t) = -5\text{ V}$).

This requirement is met by a series combination of:

• An ideal diode pointing towards $X$ (cathode at $X$, anode towards $Y$), which is reverse biased when $V_X > V_Y$ and forward biased when $V_Y > V_X$.
• A Zener diode ($V_Z = 5\text{ V}$) pointing towards $Y$ (anode towards $X$, cathode at $Y$), which is reverse biased and breaks down at $5\text{ V}$ when $V_Y > V_X$.

This configuration matches Option (B).

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Conclusion

The correct component configuration between $X$ and $Y$ is given by Option B.