If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased, and (ii) reverse biased ?
.Dear Student,
In case of forward bias, p-type of the diode is connected to the positive terminal of the battery and n-type is connected to the negative terminal. When load is applied, negative terminal of the battery repels the electrons in the n side of the diode and they start moving towards the p-side.The more you apply voltage, more the width of depletion region is reduced and barrier potential reduces. At the same time, positive terminal also attracts the electrons moving from n side to p side. Once you cross the barrier potential(0.7 V in case of Si), current starts to flow in the opposite direction of the electrons. Here, current flows because of majority charge carriers( electrons in n-side and holes in p-side.
The polarity of this potential barrier is same as the polarity of voltage source applied during reverse biased condition. Now if reverse biased voltage across the pn junction is increased the barrier potential developed across the pn junction is also increased. Hence, the pn junction is widened. When positive terminal of the source is connected to the n-type region, the free electrons of that region are attracted towards positive terminal of the source because of that more positive impurity ions are created in the depletion layer which makes the layer of positive impurity ions thicker. At the same time since negative terminal of the source is connected to the p-type region of the junction, electrons are injected in this region. Due to the positive potential of the n-type region the electrons are drifted towards the junction and combine with holes adjacent to the layer of positive impurity ions and create more positive impurity ions in the layer. Hence, the thickness of the layer increases.
Regards
In case of forward bias, p-type of the diode is connected to the positive terminal of the battery and n-type is connected to the negative terminal. When load is applied, negative terminal of the battery repels the electrons in the n side of the diode and they start moving towards the p-side.The more you apply voltage, more the width of depletion region is reduced and barrier potential reduces. At the same time, positive terminal also attracts the electrons moving from n side to p side. Once you cross the barrier potential(0.7 V in case of Si), current starts to flow in the opposite direction of the electrons. Here, current flows because of majority charge carriers( electrons in n-side and holes in p-side.
The polarity of this potential barrier is same as the polarity of voltage source applied during reverse biased condition. Now if reverse biased voltage across the pn junction is increased the barrier potential developed across the pn junction is also increased. Hence, the pn junction is widened. When positive terminal of the source is connected to the n-type region, the free electrons of that region are attracted towards positive terminal of the source because of that more positive impurity ions are created in the depletion layer which makes the layer of positive impurity ions thicker. At the same time since negative terminal of the source is connected to the p-type region of the junction, electrons are injected in this region. Due to the positive potential of the n-type region the electrons are drifted towards the junction and combine with holes adjacent to the layer of positive impurity ions and create more positive impurity ions in the layer. Hence, the thickness of the layer increases.
Regards