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Question

What happens to a diode, if the PIV rating of the diode is exceeded?

The correct answer is Diode gets destroyed

Diode PIV Exceeding Rating: Understanding the Consequences

A diode is a two-terminal electronic component that largely conducts current in one direction only. It allows current to flow easily in the forward-bias direction (when the anode is positive with respect to the cathode) but blocks current flow in the reverse-bias direction (when the cathode is positive with respect to the anode).

Peak Inverse Voltage (PIV) Explained

The term Peak Inverse Voltage (PIV) is a critical parameter for a diode. It refers to the maximum voltage that a diode can withstand in the reverse-bias direction without undergoing electrical breakdown. In simpler terms, it's the highest voltage that can be applied across the diode in the direction that normally blocks current, without causing damage.

  • When a diode is reverse-biased, only a very small leakage current flows through it.
  • As the reverse voltage increases, this leakage current remains very low until a certain voltage is reached.
  • This critical voltage, beyond which the diode's blocking capability is compromised, is known as the PIV rating or reverse breakdown voltage.

Diode Behavior When PIV is Exceeded

If the reverse voltage applied across a standard diode exceeds its PIV rating, a phenomenon called avalanche breakdown occurs. In this condition:

  • The electric field across the depletion region becomes very strong.
  • Free charge carriers (electrons and holes) gain enough energy to collide with atoms in the crystal lattice.
  • These collisions knock out more electrons, leading to a rapid and uncontrolled increase in reverse current.

This sudden surge of reverse current, often in the order of amperes, is not limited by the diode's internal resistance and leads to significant power dissipation within the diode. The power dissipated \(P\) can be expressed as:

\[ P = V_{R} \times I_{R} \]

where \(V_{R}\) is the reverse voltage and \(I_{R}\) is the reverse current. When \(V_{R}\) exceeds the PIV and \(I_{R}\) increases drastically, the dissipated power becomes very high.

This excessive power dissipation generates a large amount of heat. Since standard rectifier diodes are not designed to operate in this breakdown region (unlike Zener diodes), this extreme heat quickly damages the semiconductor material and the internal structure of the diode. The high temperature can melt the semiconductor material or its bonding wires, leading to a permanent short circuit or open circuit within the device. Consequently, the diode gets destroyed and becomes non-functional.

Understanding Other Options

  • Diode behaves as zener diode: While a Zener diode is designed to operate reliably in its reverse breakdown region (Zener breakdown) without damage, a standard rectifier diode is not. Exceeding the PIV of a standard diode causes irreversible damage, meaning it does not behave like a Zener diode which is built to regulate voltage in its breakdown region.
  • Diode conducts poorly: When PIV is exceeded, the diode does not conduct poorly; rather, it conducts excessively and uncontrollably in the reverse direction, which leads to its destruction.
  • Diode stops conducting: This is incorrect. Instead of stopping conduction, the diode experiences a massive surge of reverse current, leading to its failure. After destruction, it might effectively become an open circuit (stop conducting) or a short circuit (conduct poorly/excessively) depending on the failure mode, but the immediate event is uncontrolled conduction.

Therefore, for a standard diode, exceeding its PIV rating results in its permanent destruction due to thermal runaway caused by excessive reverse current.

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Important Questions from Diodes and Its Applications - Teaching

  1. What is the most distinctive feature of a tunnel diode's current-voltage ($I-V$) characteristic?
  2. The electronic circuit that converts AC to DC where the DC output peak value can be greater than the AC input peak value is -

  3. Which of the following statements are correct?

    A. Schottky barriers are established by depositing a metal, such as Tungsten, on a p‐type channel.

    B. The transfer characteristics of a depletion type MESFET are similar to those of a depletion type MOSFET.

    C. Maximum operating conditions are determined by the product of drain‐to‐source voltage and drain current.

    D. A complimentary MOSFET has negligibly small input impedance.

    Choose the correct answer from the options given below:

  4. Arrange the following in descending order of their switching times:

    (A) Schottky diodes

    (B) Power transistor (Darlington)

    (C) IGBT

    (D) Trine

    Choose the correct answer from the options given below:

  5. Arrange the following in decreasing order of the noise generated by them:

    (A) Diode

    (B) Transistor

    (C) Avalanche photo diode

    (D) FET

    Choose the correct answer from the options given below:

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