GATE 2018 EC Q.45: solution and answer

GATE 2018 · ECQ.45 · 2 marks · NAT

A junction is made between $p^{-}$ Si with doping density $N_{A 1} = 1 0^{15} cm^{- 3}$ and $p$ Si with doping density $N_{A 2} = 1 0^{17} cm^{- 3}$ .

Given: Boltzmann constant $k = 1.38 \times 1 0^{- 23} J\cdotK^{- 1}$ , electronic charge $q = 1.6 \times 1 0^{- 19} C$ . Assume 100% acceptor ionization.

At room temperature ( $T = 300 K$ ), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be ____.

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Answer and solution

ANSWER

The magnitude of the built-in potential across this p⁻–p high-low (isotype) Si junction, at T = 300 K, is Vbi ≈ 0.12 V.

Official answer: 0.11 to 0.13Source: Official answer key, GATE 2018 EC (IIT Guwahati)

This is a p⁻–p homojunction (an isotype or "high-low" junction), not a p-n junction: both sides are p-type Si, with $N_{A 1} = 1 0^{15} cm^{- 3}$ on the lightly doped ( $p^{-}$ ) side and $N_{A 2} = 1 0^{17} cm^{- 3}$ on the heavily doped ( $p$ ) side.
With 100% acceptor ionization, the equilibrium hole concentration on each side equals the doping density: $p_{1} = N_{A 1}$ and $p_{2} = N_{A 2}$ .
For an isotype (p–p $^{+}$ ) junction, the built-in potential is set directly by the ratio of the majority-carrier (hole) concentrations on the two sides: $V_{bi} = V_{T} ln (\frac{p _{2}}{p _{1}}) = V_{T} ln (\frac{N _{A 2}}{N _{A 1}})$
Compute the thermal voltage: $V_{T} = \frac{k T}{q} = \frac{( 1.38 \times 1 0 ^{- 23} J\cdotK ^{- 1} ) ( 300 K )}{1.6 \times 1 0 ^{- 19} C} = 0.025875 V$
Compute the doping ratio: $\frac{N _{A 2}}{N _{A 1}} = \frac{1 0 ^{17}}{1 0 ^{15}} = 100, ln (100) = 4.6052$
Combine: $V_{bi} = 0.025875 \times 4.6052 \approx 0.1192 V$
Rounded to two decimal places, the magnitude of the built-in potential is $V_{bi} \approx 0.12 V$ .

KEY IDEA

This is a p⁻–p (isotype / high-low) homojunction, not a p-n junction. Since both sides are p-type, the built-in potential is set by the ratio of the majority-carrier (hole) concentrations on the two sides, not by an $n_{i}^{2}$ /minority-carrier expression: $V_{bi} = V_{T} ln (\frac{N _{A 2}}{N _{A 1}})$ , with $V_{T} = k T / q$ .

WATCH OUT FOR

Treating this as a standard p-n junction and reaching for $V_{bi} = V_{T} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})$ — that formula does not apply here since there is no donor density and no $n_{i}$ given (and none is needed): both regions are p-type, so the isotype-junction formula using only the two acceptor densities must be used instead.

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GATE 2018 EC Q.45: solution and answer

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