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Question Bank (5 marks) · Q2

Q.Explain the operation of power diode under the reverse biased condition.

Karnataka PUCTextbookLong· 5mImportance★★★★★est
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[!TLDR]

Reverse bias widens the depletion region into the lightly doped n⁻ drift region; the diode blocks with only leakage current until avalanche at VBRV_{BR}.

Under reverse bias the anode (p⁺) is made negative with respect to the cathode (n⁺). The P+n−P^+ n^- junction is now reverse biased, so its depletion region widens. Because the n⁻ drift region is very lightly doped (few donor ions), the depletion region spreads deep into this drift region in order to expose enough fixed charge to support the applied voltage. The lightly doped drift region therefore withstands most of the reverse voltage, and the electric field is maximum at the junction.

Only a small reverse leakage (saturation) current flows, carried by thermally generated minority carriers. As the reverse voltage is increased the depletion region widens further: in a non-punch-through diode it stays within the drift region (triangular field), while in a punch-through diode it reaches the n⁺ layer (trapezoidal field).

If the reverse voltage exceeds the device's rated reverse breakdown (avalanche) voltage VBRV_{BR}, the high field causes avalanche multiplication, and the reverse current rises sharply, which can destroy the diode. Therefore in normal operation the diode is kept below VBRV_{BR}, where it acts as a blocking (OFF) device carrying only leakage current.

[!ANSWER]

Under reverse bias the depletion region spreads into the n⁻ drift region, which supports the reverse voltage while only a small leakage current flows; conduction breaks down (avalanche) only if the reverse voltage exceeds the rated VBRV_{BR}.

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