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According to Resnick, Walker & Halliday's Fundamentals of Physics (10th edition),

Figure 41-15 shows why a p-n junction operates as a junction rectifier. In Fig. 41-15a, a battery is connected across the junction with its positive terminal connected at the p side. In this forward-bias connection, the p side becomes more positive and the n side becomes more negative, thus decreasing the height of the potential barrier $V_0$ of Fig. 41-12c. More of the majority carriers can now surmount this smaller barrier; hence, the diffusion current $I_{diff}$ increases markedly.

The minority carriers that form the drift current, however, sense no barrier; so the drift current $I_{drift}$ is not affected by the external battery. The nice current balance that existed at zero bias (see Fig. 41-12d) is thus upset, and, as shown in Fig. 41-15a, a large net forward current IF appears in the circuit.

Another effect of forward bias is to narrow the depletion zone, as a comparison of Fig. 41-12b and Fig. 41-15a shows. The depletion zone narrows because the reduced potential barrier associated with forward bias must be associated with a smaller space charge. Because the ions producing the space charge are fixed in their lattice sites, a reduction in their number can come about only through a reduction in the width of the depletion zone.

Because the depletion zone normally contains very few charge carriers, it is normally a region of high resistivity. However, when its width is substantially reduced by a forward bias, its resistance is also reduced substantially, as is consistent with the large forward current.

My questions are as follows:

  1. According to the authors, the depletion zone becomes narrower when the diode is connected in forward bias because the height of the potential barrier, $V_0$, decreases. Height of $V_0$ decreases because the p-side is connected with the positive terminal of $V_{ext}$ and the n-side is connected with the negative terminal of $V_{ext}$, according to the authors. Can someone please explain more easily and clearly why the the height of the potential barrier, $V_0$, decreases when the diode is connected in forward-bias?
  2. If $V_{ext}$ is increased, will the depletion zone become narrower? Can $V_{ext}$ be increased so much that the depletion zone becomes non-existent? This question is a good segue to my 3rd question.
  3. Is the forward biasing voltage the cause of the dynamic resistance of the diode? If an AC source, $V_{{ext}_{ac}}$, is added with the DC source, $V_{ext}$, will the resistance of the diode vary with time? Is this resistance of the diode called the dynamic resistance?
  • $\begingroup$ Are you familiar with band diagrams? Have you seen a band diagram for a pn junction diode? What do you think might happen other than the energy barrier height decreasing under forward bias? $\endgroup$
    – Matt
    Jun 7, 2021 at 11:27
  • $\begingroup$ I am not familiar with band diagrams for pn junction diodes. Why doesn't the depletion zone/energy barrier height just remain constant? $\endgroup$ Jun 10, 2021 at 5:40

1 Answer 1


A block of semiconductor has some concentration of mobile charge carriers (electrons and holes) and their charge is exactly cancelled out by fixed charges in the form of ionized dopants (donor and acceptor ions)

The depletion region is, unsurprisingly, depleted of mobile charge carriers (electrons and holes). When depleted, what remains are the fixed ionized dopants. These charges create an electric field within the depletion region and this leads to the diode "built in voltage". That is the voltage difference between the p and n sides of the diode. This voltage is the potential barrier in the diode. When you apply an external voltage to a diode, you change this voltage difference and therefore change the barrier.

If $V_{ext}$ is increased, will the depletion zone become narrower?

Yes. The electric field is the integral of the charge in the depletion region, and the electric potential is the integral of the field. In order for the voltage to decrease you much reduce the total charge in the depletion region. When you do so, the depletion region narrows.

Can $V_{ext}$ be increased so much that the depletion zone becomes non-existent?

Yes in theory, but in practice no. If you were to apply enough voltage to actually completely remove the depletion region you would have an enormous amount of current. An amount that no physical diode would be able to support without burning, melting, or otherwise destroying itself.

Is the forward biasing voltage the cause of the dynamic resistance of the diode?

Typically you would think of dynamic resistance in situations where you have a small AC signal on top of a DC bias point. The dynamic resistance of a diode is the derivative of the I-V curve at your operating point. In doing so, your DC bias selects the effective resistance your AC signal sees. This assumes your AC signal is small enough to keep you in an approximately linear region of the diode's I-V curve. If you only have an AC signal, or it is a large signal, I wouldn't talk about dynamic resistance changing across your signal period, but some people might.

  • $\begingroup$ First of all, I'm very grateful to you for answering. I still have a question. If $V_{ext}$ is less than the barrier potential, will the depletion region still narrow? $\endgroup$ Jun 10, 2021 at 15:54
  • $\begingroup$ @user545735 Yes, any external bias changes the depletion region width. Reverse bias causes it to get wider, forward bias causes it to get smaller. Besides that though, the external forward bias is always smaller than the diode built-in potential for real diodes. They would burn up if you forward biased them enough to exceed the built in potential. $\endgroup$
    – Matt
    Jun 10, 2021 at 16:36
  • $\begingroup$ So, even though the external forward biasing voltage is less than the built-in voltage initially, upon being connected, the built-in voltage decreases and becomes less than the external voltage. Thus, current flows. Did I get it right? $\endgroup$ Jun 11, 2021 at 5:06
  • $\begingroup$ @user545735 The built in voltage is a constant. The external bias changes the barrier height. The barrier height equals the built in voltage when external bias is 0. Current flows under forward bias, even when there is a barrier, because electrons exist on an energy distribution and some have eneough energy to make it over the barrier. $\endgroup$
    – Matt
    Jun 11, 2021 at 11:51
  • $\begingroup$ So, the more the external biasing voltage increases, the more the barrier height decreases, and the more the barrier height decreases, the more is the probability that a free electron will be able to overcome the barrier height and enter the p-type semiconductor from the n-type semiconductor. So, current increases with the increase of the biasing voltage. Did I get it right now? $\endgroup$ Jun 11, 2021 at 17:04

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