GATE 2022 EC Q.57: solution and answer

GATE 2022 · ECQ.57 · 2 marks · NAT

A circuit and the characteristics of the diode (D) in it are shown. The ratio of the minimum to the maximum small signal voltage gain $\frac{\partial V _{o u t}}{\partial V _{in}}$ is ________ (rounded off to two decimal places).

GATE 2022 EC question 57: Circuit: the input source V_in (+ on top, − at the bottom ground rail) feeds a series 2 k Ω resistor to node A. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

ANSWER

The ratio of the minimum to the maximum small-signal voltage gain is $0.75$ . The gain is $\frac{1}{2}$ with the diode ON and $\frac{2}{3}$ with it OFF.

Official answer: 0.70 to 0.80Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Diode model from the characteristic. For $V_{D} < 0.7 V$ , $I_{D} = 0$ , so the diode is an open circuit and its small-signal resistance is $r_{d} = \infty$ . At $V_{D} = 0.7 V$ the curve is vertical: the voltage stays at 0.7 V whatever the current, so $r_{d} = \frac{\partial V _{D}}{\partial I _{D}} = 0$ . For small signals it is a short.
Circuit structure. $V_{in}$ drives a series $2 k Ω$ resistor to node A, where $V_{o u t}$ is taken. Between A and B the second $2 k Ω$ resistor sits in parallel with D (anode at A). A third $2 k Ω$ resistor runs from B to ground. So $V_{o u t}$ is the voltage across ( $2 k Ω ∥$ D) in series with the bottom $2 k Ω$ .
Diode OFF. With $I_{D} = 0$ the same current $I = \frac{V _{in}}{6 k Ω}$ flows through all three resistors. By voltage division $\frac{\partial V _{o u t}}{\partial V _{in}} = \frac{2 + 2}{2 + 2 + 2} = \frac{2}{3} \approx 0.667$ This holds while $V_{A B} = \frac{V _{in}}{3} < 0.7 V$ , i.e. for $V_{in} < 2.1 V$ . That range includes every negative $V_{in}$ , where D is reverse biased.
Diode ON ( $V_{in} > 2.1 V$ ). Now D holds $V_{A} - V_{B} = 0.7 V$ . All the current entering A through the series resistor reaches B, through the upper resistor or through D, and then flows through the bottom resistor. So $\frac{V _{in} - V _{A}}{2 k Ω} = \frac{V _{B}}{2 k Ω} = \frac{V _{A} - 0.7}{2 k Ω} \Rightarrow V_{A} = \frac{V _{in} + 0.7}{2}$ Consistency check: $I_{D} = \frac{V _{B} - 0.7}{2 k Ω} = \frac{V _{in} - 2.1}{4} mA > 0$ for $V_{in} > 2.1 V$ , so the diode is indeed conducting.
Small-signal gain with the diode ON. From $V_{o u t} = V_{A} = \frac{V _{in} + 0.7}{2}$ , $\frac{\partial V _{o u t}}{\partial V _{in}} = \frac{1}{2}$ The same result follows directly from the small-signal model: $r_{d} = 0$ shorts the parallel pair, leaving the divider $\frac{2}{2 + 2} = \frac{1}{2}$ . The constant 0.7 V only shifts the operating point and adds nothing to the incremental gain.
Ratio of minimum to maximum gain. The gain is $\frac{2}{3}$ in one region and $\frac{1}{2}$ in the other, so $\frac{G _{min}}{G _{ma x}} = \frac{1/2}{2/3} = \frac{3}{4} = 0.75$

KEY IDEA

With a piecewise-linear diode, the small-signal gain is constant within each region. Below 0.7 V the diode is an open ( $r_{d} = \infty$ ). On the vertical part of the curve it is a short ( $r_{d} = 0$ ). Each case reduces to a plain resistive divider. The fixed 0.7 V drop only sets where the diode switches (at $V_{in} = 2.1 V$ ) and has no effect on either gain.

WATCH OUT FOR

Taking $V_{o u t}$ at node B (across the bottom resistor only) gives gains $\frac{1}{3}$ and $\frac{1}{2}$ and a ratio of 0.67, which is outside the key. The printed $V_{o u t}$ terminal is at node A, the diode's anode. Other slips are inverting the ratio (max/min = 1.33), or letting the 0.7 V drop enter the incremental gain when it only shifts the operating point.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 25 Diode circuits questions, with notes

GATE 2022 EC Q.57: solution and answer

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