Q.Show that the ohmic drop makes the forward VI characteristics of a power diode more linear.

A power diode has a thick, lightly doped n⁻ drift region that adds an ohmic resistance $R_{ON}$, so its forward drop is $V_{AK}=V_j+I_F R_{ON}$; above the knee the current-proportional ohmic term dominates and the curve becomes a straight line of slope $R_{ON}$.

[!ANSWER]

Since $V_{AK}=V_j+I_F R_{ON}$, beyond the knee the forward voltage rises in direct proportion to the current, so the ohmic drift-region drop straightens (linearises) the forward V-I characteristic.

[!TLDR]

The series resistance of the n⁻ drift region adds a current-proportional term $I_F R_{ON}$ that turns the sharp exponential forward curve into an almost straight line.

Unlike an ordinary signal diode, a power diode has a thick, lightly doped n⁻ drift region inserted between the p⁺ anode and n⁺ cathode layers to block high reverse voltage. This drift region has a finite ohmic resistance $R_{ON}$. When the diode conducts in the forward direction, the total anode-to-cathode voltage is the sum of the junction drop and the ohmic drop across the drift region:

In a signal diode the forward current follows the Shockley relation and rises exponentially with voltage, so the characteristic shows a sharp knee and then shoots up almost vertically. In a power diode, once the junction is forward biased and $V_j$ is essentially fixed near the knee, any further increase in current mainly increases the linear term $I_F R_{ON}$, which is directly proportional to $I_F$. Hence above the knee the current increases in proportion to the voltage and the V-I curve becomes an almost straight line whose slope is set by $1/R_{ON}$, instead of the near-vertical exponential rise.

[!ANSWER]

Because $V_{AK}=V_j+I_F R_{ON}$, the current-proportional ohmic drop dominates above the knee and makes the forward characteristic a straight (linear) line of slope $R_{ON}$.