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MHT-CET Physics · Semiconductor Devices

Special-Purpose Diodes: Zener, LED, Photodiode and Solar Cell

A Zener diode holds a fixed voltage in reverse breakdown and so regulates; an LED emits light in forward bias; a photodiode senses light in reverse bias; and a solar cell turns light into electrical energy with no bias at all.

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Why this matters

14 PYQs, none HARD. Two shapes: the Zener regulator — current through the Zener, or the series resistor its power rating allows — and matching each device to its bias and its job.

Concept 1 of 2: The Zener Regulator

In reverse breakdown a Zener's voltage stays fixed while its current changes, so a load in parallel with it sees a steady voltage. The series resistor drops whatever is left of the supply; the Zener takes whatever current the load does not.

Definition

  • Operating region: the reverse breakdown part of the I–V curve.
  • Load voltage =VZ= V_Z. Series current Is=Vin−VZRsI_s = \dfrac{V_{\text{in}} - V_Z}{R_s}; load current IL=VZRLI_L = \dfrac{V_Z}{R_L}; Zener current IZ=Is−ILI_Z = I_s - I_L.
  • Power rating: IZ,max⁡=PVZI_{Z,\max} = \dfrac{P}{V_Z}, so the least series resistance is Rs=Vin−VZIZ,max⁡R_s = \dfrac{V_{\text{in}} - V_Z}{I_{Z,\max}}.

Zener current

IZ=Vin−VZRs−VZRLI_Z = \frac{V_{\text{in}} - V_Z}{R_s} - \frac{V_Z}{R_L}

Worked example

An 18 V supply feeds a 12 V Zener through 200 Ω200\,\Omega; the load is 2 kΩ2\,\text{k}\Omega. Current in the Zener?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q19Moderate

Example 1 · Semiconductor Devices · Special Diodes — Zener, LED, Photodiode

A zener diode, having breakdown voltage 15 V is used in a voltage regulator circuit as shown. The current through the zener diode is

Giving the Zener the whole series current

The series current splits: the load takes VZRL\frac{V_Z}{R_L} and only the rest flows in the Zener. 20 mA through the resistor with 15 mA in the load leaves 5 mA, not 20.
Drill 5 more on the zener regulator

Concept 2 of 2: LED, Photodiode and Solar Cell

Light and a junction trade energy across the band gap. Forward-bias a junction and recombining carriers can give out photons — an LED. Shine photons on a reverse-biased junction and the extra carriers change a tiny current — a photodiode. Let the light drive current with no battery at all — a solar cell.

Definition

  • LED: always FORWARD biased; brightness set by the current; energy-efficient, long-lived, colours do not fade.
  • Photodiode: REVERSE biased; the reverse current depends on the minority carriers that light creates, so it rises with intensity.
  • Solar cell: light into electrical energy (not the reverse); best band gap about 1.0–1.8 eV.
  • A diode cannot AMPLIFY a signal — that takes a transistor.
  • Photon energy and band gap: Eg=hcλE_g = \dfrac{hc}{\lambda}, λ\lambda (nm) ≈1240Eg (eV)\approx \dfrac{1240}{E_g\,(\text{eV})}.

Band gap and wavelength

Eg=hcλ,λ (nm)≈1240Eg (eV)E_g = \frac{hc}{\lambda}, \qquad \lambda\,(\text{nm}) \approx \frac{1240}{E_g\,(\text{eV})}

Worked example

An LED has a band gap of 1.8 eV. Roughly what wavelength does it emit?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q21Easy

Example 2 · Semiconductor Devices · Special Diodes — Zener, LED, Photodiode

In the working of photodiode, the reverse current depends on

Detecting light in forward bias

In forward bias the large diffusion current hides the small change that light makes; a photodiode must be REVERSE biased, where the whole current is the light-made minority carriers.
Drill 7 more on led, photodiode and solar cell

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • The Zener Regulator

    Zener current

    IZ=Vin−VZRs−VZRLI_Z = \frac{V_{\text{in}} - V_Z}{R_s} - \frac{V_Z}{R_L}
  • LED, Photodiode and Solar Cell

    Band gap and wavelength

    Eg=hcλ,λ (nm)≈1240Eg (eV)E_g = \frac{hc}{\lambda}, \qquad \lambda\,(\text{nm}) \approx \frac{1240}{E_g\,(\text{eV})}

Watch out for (2)

Test yourself on Semiconductor Devices

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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