GATE 2020 EE Q.52: solution and answer

GATE 2020 · EEQ.52 · 2 marks · NAT

A non-ideal Si-based pn junction diode is tested by sweeping the bias applied across its terminals from $- 5 V$ to $+ 5 V$ . The effective thermal voltage, $V_{T}$ , for the diode is measured to be $(29 \pm 2) mV$ . The resolution of the voltage source in the measurement range is $1 mV$ . The percentage uncertainty (rounded off to 2 decimal places) in the measured current at a bias voltage of $0.02 V$ is ________.

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Answer and solution

ANSWER

At a bias of $0.02 V$ , the percentage uncertainty in the measured diode current is about $11.79 %$ .

Official answer: 11.50 to 12.00Source: Official answer key, GATE 2020 EE (IIT Delhi)

Use the full diode equation. The effective thermal voltage already includes the ideality factor, so $I = I_{s} (e^{V / V_{T}} - 1)$ The saturation current $I_{s}$ is not given with an uncertainty, so it is treated as exact. The only uncertain quantities are the bias $V$ and the thermal voltage $V_{T}$ .
Data: $V = 0.02 V = 20 mV$ , $δ V = 1 mV$ (the resolution of the source), $V_{T} = 29 mV$ , $δ V_{T} = 2 mV$ .
You cannot drop the $- 1$ here. The shortcut $I \approx I_{s} e^{V / V_{T}}$ only works when $V ≫ V_{T}$ . At this bias $\frac{V}{V _{T}} = \frac{20}{29} = 0.68966, e^{V / V_{T}} = 1.99303, e^{V / V_{T}} - 1 = 0.99303$ The exponential is only about twice the $1$ , so the $- 1$ roughly halves the current and has to be kept.
Partial derivatives: $\frac{\partial I}{\partial V} = \frac{I _{s}}{V _{T}} e^{V / V_{T}}, \frac{\partial I}{\partial V _{T}} = - \frac{I _{s} V}{V _{T}^{2}} e^{V / V_{T}}$ Divide each by $I = I_{s} (e^{V / V_{T}} - 1)$ . Both relative sensitivities then carry the same factor $k = \frac{e ^{V / V_{T}}}{e ^{V / V_{T}} - 1} = \frac{1.99303}{0.99303} = 2.00702$
Relative uncertainty in $I$ due to $V$ : $\frac{1}{I} \frac{\partial I}{\partial V} δ V = k \frac{δ V}{V _{T}} = 2.00702 \times \frac{1}{29} = 0.069208$
Relative uncertainty in $I$ due to $V_{T}$ : $\frac{1}{I} \frac{\partial I}{\partial V _{T}} δ V_{T} = k \frac{V δ V _{T}}{V _{T}^{2}} = 2.00702 \times \frac{20 \times 2}{2 9 ^{2}} = 2.00702 \times 0.047562 = 0.095459$
$V$ (set by the source) and $V_{T}$ (a property of the diode) are measured independently, so the two contributions add as a root-sum-square: $\frac{δ I}{I} = (k \frac{δ V}{V _{T}})^{2} + (k \frac{V δ V _{T}}{V _{T}^{2}})^{2}$
$\frac{δ I}{I} = (0.069208)^{2} + (0.095459)^{2} = 0.0047897 + 0.0091124 = 0.0139021 = 0.11791$
The percentage uncertainty in the measured current is $\approx 11.79 %$ .

KEY IDEA

Propagate the uncertainty through the full diode law $I = I_{s} (e^{V / V_{T}} - 1)$ . At $V = 20 mV$ the bias is smaller than $V_{T} = 29 mV$ , so the $- 1$ term matters. It multiplies both relative sensitivities by $e^{V / V_{T}} / (e^{V / V_{T}} - 1) \approx 2.007$ . The independent contributions from $δ V$ and $δ V_{T}$ then combine as a root-sum-square.

WATCH OUT FOR

The usual mistake is to use $I \approx I_{s} e^{V / V_{T}}$ , which is only valid for $V ≫ V_{T}$ . At $20 mV$ it gives $5.87 %$ (root-sum-square) or $8.20 %$ (linear sum), both far from the key. Two other slips: adding the two contributions linearly with the full equation gives $16.47 %$ , and taking half the source resolution ( $0.5 mV$ ) as $δ V$ gives $10.15 %$ .

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