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

Under forward bias the anode (p⁺) is positive, the junction barrier is lowered, and heavy carrier injection conductivity-modulates the lightly doped n⁻ drift region so a large current flows with a small drop.

[!ANSWER]

When forward biased the power diode conducts: carriers are injected into and conductivity-modulate the n⁻ drift region, giving a low resistance so a large forward current flows with a total drop $V_{AK}=V_j+I_F R_{ON}$.

[!TLDR]

Forward bias lowers the junction barrier and floods the n⁻ drift region with carriers (conductivity modulation), so the diode conducts a large current with a small drop $V_{AK}=V_j+I_F R_{ON}$.

A power diode has the layer structure p⁺-n⁻-n⁺, where the rectifying junction is the $P^+ n^-$ junction and the thick, lightly doped n⁻ region is the drift region that blocks reverse voltage.

Under forward bias the anode (p⁺) is made positive with respect to the cathode (n⁺). This reduces the built-in potential barrier at the $P^+ n^-$ junction, so majority holes are injected from the p⁺ anode into the n⁻ region and electrons are injected from the n⁺ cathode into the n⁻ region. Because the drift region is only lightly doped, the injected carrier concentration far exceeds its background doping. This heavy injection is called conductivity modulation: it greatly increases the number of free carriers in the drift region and so reduces its resistance $R_{ON}$ to a low value.

As a result a large forward current $I_F$ can flow with only a small voltage drop. The total forward drop is the sum of the junction drop and the ohmic drop across the drift region:

where $V_j$ is the junction voltage (about 0.7 to 1 V for silicon) and $I_F R_{ON}$ is the ohmic drop. Once the diode is fully conducting, the forward current rises almost linearly with the applied voltage because of the ohmic term.

[!ANSWER]

Under forward bias the $P^+ n^-$ barrier is lowered, carriers are injected and conductivity-modulate the n⁻ drift region to a low resistance, and the diode conducts a large forward current with drop $V_{AK}=V_j+I_F R_{ON}$.