Dynamic resistance of diodes and the depletion zone

Question

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?

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.