GATE 2023 EC Q.22: solution and answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2023 EC (IIT Kanpur)

1. What the options must mean. Connect the cascade so that the feedback is negative at low frequency. For an odd number of inverting CS stages, that means tying the output straight back to the input. With unity feedback ($\beta =1$), the loop gain is $L(j\omega )=\beta A(j\omega )=A(j\omega )$ and the closed-loop gain is $A_f=\frac{A}{1+L}$. Oscillation is decided by the denominator: $1+L(j\omega )=0$. At that point $A_f$ is unbounded, and a stable unity-feedback amplifier with large $|A|$ has $A_f\approx 1$. So the closed-loop gain is not the quantity any oscillation criterion uses. The options only make sense if "closed loop gain" is read as the loop gain $|L(j\omega )|$, and "phase shift" as the phase lag of $L(j\omega )$, on top of the $180^{\circ }$ inversion that makes the feedback negative. The official key awards marks to all candidates and gives no reason. The solution below answers the question as intended, under this reading.

2. Model the cascade. Assume $N$ identical stages. Each CS stage has a low-frequency gain magnitude $A_0=g_m{R}_D$ ($r_o$ absorbed into $R_D$) and one pole from its output-node capacitance $C_L$, at ${\omega }_p=\frac{1}{R_D{C}_L}$. $C_{gd}$ and its zero are neglected. Then $$L(j\omega )=\frac{A_0^N}{{\left(1+j\omega /{\omega }_p\right)}^N},|L(j\omega )|=\frac{A_0^N}{{\left(1+{\omega }^2/{\omega }_p^2\right)}^{N/2}},\varphi (\omega )=N{\tan }^{-1}\left(\frac{\omega }{{\omega }_p}\right)$$ where $\varphi$ is the phase lag of the loop. As $\omega$ rises, $|L|$ falls monotonically and $\varphi$ rises monotonically from $0$ towards $N\times 90^{\circ }$.

3. How many stages can reach $180^{\circ }$. At every finite frequency, each pole contributes ${\tan }^{-1}(\omega /{\omega }_p)<90^{\circ }$. So $\varphi$ reaches $180^{\circ }$ at a finite frequency only if $N\ge 3$. Call that frequency ${\omega }_{180}$: $N{\tan }^{-1}({\omega }_{180}/{\omega }_p)=180^{\circ }$. With $N=2$, $\varphi$ approaches $180^{\circ }$ only as $\omega \to \infty$, where $|L|\to 0$. A two-stage loop of this kind therefore cannot oscillate.

4. Oscillation criterion. A closed-loop pole on the $j\omega$ axis needs $1+L(j\omega )=0$, that is, $\angle L(j\omega )=-180^{\circ }$ and $|L(j\omega )|=1$ (Barkhausen). The extra $180^{\circ }$ of lag turns the negative feedback into positive feedback, and the signal comes back round the loop at its own amplitude. All open-loop poles are in the left half-plane and $\varphi$ is monotonic, so the Nyquist plot of $L$ crosses the negative real axis once, at $-|L(j{\omega }_{180})|$. If $|L(j{\omega }_{180})|>1$, the plot encircles $-1$. The closed loop then has poles in the right half-plane, and the oscillation builds up until the transistors' nonlinearity limits its amplitude. If $|L(j{\omega }_{180})|<1$, the loop is stable. The boundary case $|L(j{\omega }_{180})|=1$ is the constant-amplitude Barkhausen condition.

5. Restate it with the two crossover frequencies. Let ${\omega }_{gc}$ be the gain crossover, where $|L(j{\omega }_{gc})|=1$. Because $|L|$ falls and $\varphi$ rises monotonically, $|L(j{\omega }_{180})|>1⟺{\omega }_{180}<{\omega }_{gc}$. So the loop oscillates exactly when ${\omega }_{180}<{\omega }_{gc}$. In that case every frequency between the two has $|L|>1$ and $\varphi >180^{\circ }$, which is a negative phase margin: $180^{\circ }-\varphi ({\omega }_{gc})<0$. The loop is stable exactly when ${\omega }_{gc}<{\omega }_{180}$. In that case every frequency between the two has $|L|<1$ and $\varphi <180^{\circ }$.

6. (D): for $|L|>1$ and $\varphi >180^{\circ }$ to hold at the same frequency, we need $\omega >{\omega }_{180}$ and $\omega <{\omega }_{gc}$. That is possible only when ${\omega }_{180}<{\omega }_{gc}$, which is exactly the oscillating case. Conversely, every oscillating loop has such a band of frequencies. So (D) is the oscillation condition.

7. (A): $|L|<1$ together with $\varphi <180^{\circ }$ needs ${\omega }_{gc}<\omega <{\omega }_{180}$. That is possible only when ${\omega }_{gc}<{\omega }_{180}$, so this loop is stable, with positive gain and phase margins.

8. (B) and (C): $|L|>1$ with $\varphi <180^{\circ }$ holds at low frequencies, below both ${\omega }_{gc}$ and ${\omega }_{180}$, in every amplifier, stable or not. It is just the ordinary amplifying region. $|L|<1$ with $\varphi >180^{\circ }$ holds at high frequencies, above both crossovers, in every loop whose phase passes $180^{\circ }$, stable or not. Neither pair tells an oscillating loop from a stable one. If (C) is read as describing the $180^{\circ }$ point, it says $|L(j{\omega }_{180})|<1$. That is a positive gain margin, so the loop is stable.

9. Check with three stages (a ring oscillator). For $N=3$, $\varphi =180^{\circ }$ when each stage lags by $60^{\circ }$: ${\tan }^{-1}({\omega }_{180}/{\omega }_p)=60^{\circ }$, so ${\omega }_{180}=\sqrt{3}{\omega }_p$. At that frequency $$|L(j{\omega }_{180})|=\frac{A_0^3}{(1+3{)}^{3/2}}=\frac{A_0^3}{8}$$ Sustained oscillation needs $A_0=2$. With $A_0>2$, the loop gain is still above 1 where the phase lag reaches $180^{\circ }$, so the oscillation starts and builds up. This is the (D) condition.

10. Answer: (D), read as "the loop gain is greater than 1 while the phase shift exceeds $180^{\circ }$". As printed, "closed loop gain" is the wrong quantity (see the first step), and the official key awarded marks to all candidates.