GATE 2018 EC Q.30: solution and answer

GATE 2018 · ECQ.30 · 2 marks · MCQ

Q.30 Red (R), Green (G) and Blue (B) Light Emitting Diodes (LEDs) were fabricated using $p$ - $n$ junctions of three different inorganic semiconductors having different band-gaps. The built-in voltages of red, green and blue diodes are $V_{R}$ , $V_{G}$ and $V_{B}$ , respectively. Assume donor and acceptor doping to be the same ( $N_{A}$ and $N_{D}$ , respectively) in the $p$ and $n$ sides of all the three diodes.

Which one of the following relationships about the built-in voltages is TRUE?

(A)
$V_{R} > V_{G} > V_{B}$
(B)
$V_{R} < V_{G} < V_{B}$
(C)
$V_{R} = V_{G} = V_{B}$
(D)
$V_{R} > V_{G} < V_{B}$

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Answer and solution

ANSWER

The correct answer is (B): $V_{R} < V_{G} < V_{B}$ — the built-in voltage increases with the LED's band-gap, and blue LEDs have the largest band-gap.

Official answer: BSource: Official answer key, GATE 2018 EC (IIT Guwahati)

Photon energy (hence LED color) is set by the semiconductor's band-gap $E_{g}$ : among Red, Green and Blue LEDs, $E_{g, B} > E_{g, G} > E_{g, R}$ (blue light carries more energy per photon than red).
The built-in voltage of a $p$ - $n$ junction is $V_{bi} = V_{T} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})$ where $V_{T} = k T / q$ , and $N_{A}$ , $N_{D}$ are the same for all three diodes (given).
The intrinsic carrier concentration depends on the band-gap as $n_{i}^{2} \propto exp (- \frac{E _{g}}{k T})$ so a larger $E_{g}$ gives an exponentially smaller $n_{i}$ .
Since $N_{A} N_{D}$ is fixed, $V_{bi}$ grows as $n_{i}^{2}$ shrinks — i.e. $V_{bi}$ increases as $E_{g}$ increases.
Applying the band-gap ordering: $E_{g, B} > E_{g, G} > E_{g, R} \Rightarrow n_{i, B} < n_{i, G} < n_{i, R} \Rightarrow V_{bi, B} > V_{bi, G} > V_{bi, R}$ .
Hence $V_{R} < V_{G} < V_{B}$ , which is option (B).

KEY IDEA

A wider band-gap semiconductor has a much smaller intrinsic carrier concentration $n_{i}$ (since $n_{i}^{2} \propto exp (- E_{g} / k T)$ ), and since $V_{bi} = V_{T} ln (\frac{N _{A} N _{D}}{n _{i}^{2}})$ with the same doping in all three diodes, a smaller $n_{i}$ (larger $E_{g}$ ) gives a larger $V_{bi}$ .

WATCH OUT FOR

Assuming the built-in voltage is unaffected by band-gap because $N_{A}$ and $N_{D}$ are stated to be equal (choosing option C), forgetting that $n_{i}$ itself is exponentially dependent on $E_{g}$ and enters the same formula.

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 5 Optoelectronics questions, with notes

GATE 2018 EC Q.30: solution and answer

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