Zener Diode Characteristics
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Direct answer
Reverse-bias a Zener diode and it holds its ground: past a sharp knee, the voltage across it stays pinned at the Zener voltage V_Z while the current through it swings widely — a controlled breakdown that is not destructive because a series resistor limits the current. In breakdown below roughly 5-6 V the mechanism is Zener (field-emission) tunnelling in heavily doped junctions; above that, avalanche multiplication — both reversible if the power rating is respected. Forward-biased, it behaves as an ordinary silicon diode with its 0.7 V drop. The flat reverse characteristic is what makes it a voltage regulator: wired in parallel with the load, it keeps V_Z across the load while the series resistance absorbs supply fluctuations.
What you must remember
- I-V shape: forward — negligible current below about 0.7 V, then a steep rise; reverse — tiny leakage until the knee at V_Z, then near-vertical current rise at almost constant voltage.
- Two breakdown types: heavily doped, low-voltage (below about 5-6 V) units break by Zener tunnelling; higher-voltage units by avalanche multiplication — both non-destructive under current limiting.
- Regulator wiring: supply V_in > V_Z in series with R_S; Zener in parallel with the load; the load sits at V_Z; R_S carries I_S = I_Z + I_L.
- Design equation: R_S = (V_in − V_Z)/(I_L + I_Z) chosen so the Zener always keeps some minimum current in breakdown.
- Power limit: the diode survives while V_Z × I_Z stays within its power rating; load removed, the Zener takes the whole series current and must still cope.
- Numerical anchor: V_in = 12 V, V_Z = 6 V, R_S = 300 Ω gives I_S = 20 mA; a 600 Ω load draws 10 mA and the Zener passes the other 10 mA.
- Ideal-versus-real framing: an ordinary diode is not designed to live in breakdown; the Zener is engineered for a sharp, stable V_Z — that is the entire distinction.
Regulating a supply, step by step
Build the standard circuit: 12 V supply, 300 Ω series resistor, a 6 V Zener across the load. The series current is fixed at (12 − 6)/300 = 20 mA. Attach a 600 Ω load: it demands 6/600 = 10 mA, and the Zener quietly passes the remaining 10 mA (dissipating 6 × 0.01 = 0.06 W) while holding exactly 6 V across the load. Now stress it. Supply rises to 15 V: series current becomes 30 mA; the load still takes 10 mA, and the Zener absorbs 20 mA — the output stays 6 V, which is the regulation promise kept. Finally overload it: drop the load to 300 Ω, demanding 20 mA — the entire series current; the Zener is starved out of breakdown, and regulation collapses. The Zener regulates only while it carries some current, and that boundary is the number every numerical turns on.
Where NEET sets the trap
The word "breakdown" baits the misconception that the diode is destroyed — the credited statement is that breakdown is non-destructive provided the current (hence V_Z I_Z) is limited by the series resistor. Circuit-identification items test polarity: in a regulator the Zener is reverse biased, its cathode to the positive side; options show it flipped and expect you to notice. The characteristic-curve question asks which quantity stays constant in the breakdown region — the voltage across the diode, never the current through it. Numerical traps: using V_in instead of (V_in − V_Z) when computing the series current, and forgetting that load and Zener share the series current — the dropout condition (load takes everything) is the concept the exam quietly tests. The 0.7 V forward drop and V_Z are different numbers and never interchangeable in options.
Frequently asked questions
What happens when a Zener diode is reverse biased past its knee?
The voltage across it stays essentially constant at V_Z while current varies widely — the flat region that does the regulating.
Why doesn't breakdown destroy a Zener diode?
The series resistor limits the current, keeping dissipation P = V_Z I_Z within the diode's power rating; only exceeding that rating causes permanent damage.
How does a Zener hold a 6 V output from a 12 V supply with a 600 Ω load?
With R_S = 300 Ω the series current is (12 − 6)/300 = 20 mA; the load takes 10 mA and the Zener the remaining 10 mA, pinning the output at 6 V.
What separates Zener breakdown from avalanche breakdown?
Zener (field-emission) tunnelling dominates below roughly 5-6 V in heavily doped junctions; impact-ionisation avalanche dominates higher — both are reversible mechanisms.
When does a Zener regulator stop regulating?
When the load current grows to equal the series current, the Zener drops out of breakdown — or when the supply falls to V_Z — and the output is no longer held.