GATE 2018 IN Q.15: solution and answer

Official answer: 0Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Assumptions. The op-amp is ideal: infinite open-loop gain $A$, zero input current, and output $v_o=A(V_{+}-V_{-})$, limited to its supply rails. The rails are not drawn. The transistor is an NPN with $\beta =20$ and conducts only when its base–emitter junction is forward biased.

2. Check the feedback sign before using a virtual short. $V_{-}=V_{+}$ holds only if the loop gives negative feedback. Follow a small rise in the op-amp output around the loop: $$v_o\uparrow \Rightarrow I_C\uparrow \Rightarrow V_C=5-(10.5\text{k}\Omega )I_C\downarrow \Rightarrow V_{-}\downarrow \Rightarrow (V_{+}-V_{-})\uparrow \Rightarrow v_o\uparrow$$ The disturbance reinforces itself.

3. What that means. The common-emitter stage inverts from base to collector. The collector goes back to the op-amp's inverting (−) input, which inverts a second time. Two inversions make the loop non-inverting, so the feedback is positive (regenerative). With the ideal op-amp's infinite gain the loop gain is far above 1, so the op-amp cannot stay in its linear region. The point $V_C=2\text{V}$ would give $I_C=3/10.5\text{mA}=285.71\mu \text{A}$ and $I_E=300\mu \text{A}$, but it is an unstable balance point. Any disturbance drives the output to a rail.

4. Find the saturated state that is self-consistent. Suppose the transistor carries no current. The (−) input draws no current either, so no current flows in the 10.5 kΩ resistor and $$V_C=5-(10.5\text{k}\Omega )(0)=5\text{V}$$ Then $V_{-}=5\text{V}>V_{+}=2\text{V}$, so $v_o=A(2-5)$ goes to the negative rail (or to 0 V on a single supply).

5. Check the assumption. With no current in $Z_L$ the emitter is at 0 V and the base is at $v_o\le 0$. So $V_{BE}\le 0$ and the transistor is cut off, as assumed. The circuit also starts in this state. At switch-on no collector current has flowed yet, so the collector sits at the +5 V rail, above the 2 V reference. The output swings negative, and the positive feedback keeps the transistor off.

6. Load current. In cut-off $I_B=I_C=0$, so $$I_o=I_E=I_B+I_C=0\mu \text{A}$$

7. Remark. 300 µA is what you get if the op-amp inputs are swapped (collector to (+), 2 V to (−)). Then a rise in $v_o$ lowers $V_{+}$, the feedback is negative and the virtual short holds. So $V_C=2\text{V}$ and $I_o=I_C(\beta +1)/\beta =285.71\times 21/20=300\mu \text{A}$. If the op-amp output sat at its positive limit instead, $I_o$ would depend on $Z_L$ and on the op-amp's rail and current limit. None of these is given, so that state cannot produce the single number the question asks for.