--- title: "Reverse breakdown; Zener as voltage regulator" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:reverse-breakdown-zener-regulator" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/reverse-breakdown-zener-regulator site: BCT Engineering --- # Reverse breakdown; Zener as voltage regulator Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2073 Chaitra, 2+3 marks - Video: [Zener Diode as Voltage Regulator | Parameters & Example of Zener Diode as Voltage Regulator](https://www.youtube.com/watch?v=F6eGVaNcavg) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/reverse-breakdown-zener-regulator - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Define and explain reverse breakdown effect. Describe how Zener diode works as a voltage regulator. ## Short answer Reverse breakdown is a sudden sharp rise in reverse current when reverse bias reaches VBR, due to avalanche (high voltage) or Zener (low voltage) effects. A Zener regulator connects the diode in reverse parallel with the load and a series resistor; changes in input or load current are absorbed by the Zener current, keeping output voltage nearly constant. ## Answer **Reverse breakdown** is the sudden large increase in reverse current when the reverse bias voltage across a p-n junction diode reaches a critical value, the breakdown voltage VBR. Below VBR a reverse-biased diode carries only a very small leakage current (reverse saturation current). At VBR the current rises sharply while the voltage stays nearly constant. If the current is not limited by an external resistor, the heat produced destroys the diode. ```text I ^ | / | / forward -VBR | / ------+-------+--------/-------> V | small leakage 0.7 V | (reverse saturation current) | | steep rise in reverse v current at VBR ``` There are two mechanisms. #### 1. Avalanche breakdown - Occurs in lightly doped junctions with a wide depletion region, usually for VBR above about 6 V. - The minority carriers are accelerated by the strong field and gain enough kinetic energy to knock electrons out of the atoms by collision (impact ionisation). The new carriers cause further collisions, a multiplying chain like an avalanche. - Has a positive temperature coefficient (VBR increases with temperature). #### 2. Zener breakdown - Occurs in heavily doped junctions with a very thin depletion region, for VBR below about 5 V. - The electric field (about 3 × 10⁷ V/m) is strong enough to pull electrons directly out of the covalent bonds (field ionisation or tunnelling). - Has a negative temperature coefficient. Around 5 to 6 V the two effects cancel, giving a diode with almost zero temperature coefficient. #### Zener diode as a voltage regulator ```text Vin o----[ Rs ]-----+-------+----o Vo = Vz | | (Dz) [RL] | | GND GND ``` The Zener is reverse biased and connected in parallel with the load, with a series resistance Rs. - **Line regulation:** If Vin rises, the current through Rs increases. The extra current is taken by the Zener (Iz rises), and the drop across Rs increases by almost the same amount as the increase in Vin, so Vo = Vz stays almost constant. - **Load regulation:** If the load current IL increases, the Zener current Iz decreases by the same amount, so Is = Iz + IL stays constant and Vo remains nearly constant. Design condition: Rs = (Vin(min) − Vz)/(Iz(min) + IL(max)); Iz(min) must be at least the knee current Izk, and the Zener power (Vz Iz(max)) must not exceed its rating. Limits: it is only suitable for small currents (tens of mA), its efficiency is low, and the output changes slightly because of rz (ΔVo ≈ rz ΔIz). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement describes avalanche breakdown? A. Electrons are pulled directly from covalent bonds by the field in a heavily doped thin junction B. Heat melts the junction at low voltage C. Majority carriers recombine at the junction D. Minority carriers gain energy in the strong field and knock out more carriers by collision, multiplying like an avalanche; it occurs in lightly doped junctions at higher VBR Answer: D. Impact ionisation chain multiplication is typical for VBR above about 6 V. (Key point: Avalanche breakdown) 2. How do the two breakdown types differ in temperature coefficient? A. Avalanche is positive and Zener is negative; around 5 to 6 V they cancel B. Both are positive C. Avalanche is negative and Zener is positive D. Both are always zero Answer: A. VBR increases with temperature for avalanche and decreases for Zener breakdown. (Key point: Temperature coefficient) 3. How does a Zener regulator keep Vo steady when Vin rises? A. The Zener current falls so the load gets more voltage B. The extra current goes into the Zener, and the drop across Rs rises by the increase in Vin, so Vo = Vz stays almost constant C. The Rs drop stays the same and the load voltage rises D. The load current rises equally Answer: B. Iz absorbs the extra current while the Zener voltage stays nearly fixed. (Key point: Line regulation) 4. What happens to the Zener current when the load current increases? A. Iz increases by the same amount B. Iz stays the same and Vo drops C. Iz decreases by the same amount so Is = Iz + IL stays constant D. Is increases and Vo rises Answer: C. Rs current is fixed by Vin and Vz, so load increase is taken from the Zener. 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