Question

If the doping concentration in a Si-Zener diode is increased, the Zener breakdown voltage:

• A. Decreases
• B. Increases
• C. Remains unchanged
• D. Becomes broader

Hint:

More doping means a thinner depletion layer and a stronger field, so breakdown is reached at a lower voltage.

Answer 1

The Correct Option is A

Method: field-threshold reasoning.

Breakdown happens when the electric field inside the junction reaches a critical value large enough to rip valence electrons free by tunnelling across the narrow barrier. What matters is reaching that critical field, not a fixed voltage.

Heavier doping crowds more ionised donor and acceptor atoms near the metallurgical junction. This raises the space-charge density, which shrinks the depletion region needed to balance the charge. With the barrier now very thin, even a small reverse bias sets up an intense field $E \approx V/W$ across it.

Because $W$ has shrunk, the critical field is reached at a smaller reverse voltage $V$. In symbols, since $W$ falls with rising doping $N$, the voltage at which the field hits its critical value also falls: $$N \uparrow \;\Rightarrow\; W \downarrow \;\Rightarrow\; V_{\text{breakdown}} \downarrow.$$

This is the design rule used in practice: to make a Zener diode that clamps at a low voltage you dope it heavily; to get a higher clamp voltage you dope it lightly. Hence increasing the doping lowers the Zener breakdown voltage.

\[\boxed{\text{Breakdown voltage decreases}}\]