If the forward voltage in a semiconductor diode is doubled, the width of the depletion layer will
Become half
Become one-fourth
Remain unchanged
Become double
Formation of a Forward Bias Diode-
As the depletion region have no charge so the resistance is very high there so the applied voltage drops primarily across this region. The drop in voltage across the p and n side of the junction is relatively negligible. And the direction of the applied voltage (V) being opposite to that of the built-in potential (Vo) due to which the depletion layer’s width decreases and the barrier height reduce.
If the applied voltage is small, then the barrier potential is reduced marginally only below the equilibrium value. Then only small number of carriers crossing the junction, so the current is small. Similarly for a significantly high value of voltage, more carriers have the energy to cross the junction so, the current will be high.
One should also note that when the voltage is applied, some electrons cross to the p-side and some holes cross to the n-side. Under forward bias, this process is the minority charge injection process. Hence, the minority charge concentration which is electrons on the p-side are a minority and holes on the n-side are a minority, is significantly higher at the junction boundary.
Due to this concentration gradient, the injected electron diffuse from the junction-end to the far-end of the p-side. Similarly, injected holes diffuse to the far end of the n-side. This gives rise to current too.
The total diode forward current = Hole diffusion current + Electron diffusion current (mA)
So the answer is - It will become half
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