GATE 2024 IN Q.23: solution and answer

Official answer: 1.6Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. What the op-amp does. There is no feedback, so the ideal op-amp works as a comparator. Its output goes to one supply rail or the other: $+5\text{V}$ when $V_{in}>V_{REF}$, and $-5\text{V}$ when $V_{in}<V_{REF}$. An ideal op-amp saturates exactly at its rails, so $V_{sat}=5\text{V}$.

2. Which state sets the limit. The Zener cathode is at $V_{out}$ and its anode is at ground. When the output is high ($+5\text{V}$), the Zener is reverse-biased and in breakdown it holds $V_{out}=V_Z=1.8\text{V}$, the logic-high level the 1.8 V CMOS gate needs. This is the only state where the Zener has to regulate. When the output is $-5\text{V}$, the Zener is forward-biased and clamps $V_{out}$ near $-0.7\text{V}$, so the breakdown current condition does not apply there.

3. KCL at the $V_{out}$ node. The CMOS input impedance is infinite, so the logic gate draws no current. All the current in $R_Z$ therefore flows into the Zener: $I_Z=I_{R_Z}$.

4. Ohm's law across $R_Z$ (output high, Zener in breakdown): $$I_Z=\frac{V_{sat}-V_Z}{R_Z}=\frac{5-1.8}{R_Z}=\frac{3.2\text{V}}{R_Z}$$

5. Keep the Zener in breakdown. $I_Z$ falls as $R_Z$ rises, so the condition $I_Z\ge I_{Z,\min }=2\text{mA}$ puts an upper limit on $R_Z$: $$\frac{3.2}{R_Z}\ge 2\times 10^{-3}\Rightarrow R_Z\le \frac{3.2}{2\times 10^{-3}}=1600\Omega$$

6. Result. $R_{Z,\max }=1.6\text{k}\Omega$. With any larger resistance the Zener current drops below $2\text{mA}$, and $V_{out}$ is no longer held at $1.8\text{V}$.