Option 1
Step 1: How to bias in reverse.
A junction is reverse biased when the external battery aids, rather than opposes, the built-in junction field. So join battery \(+\) to the n-material and battery \(-\) to the p-material, and put a sensitive current meter in the loop to read the small current.
Step 2: Depletion region response.
Because the external voltage pulls the mobile carriers back from the boundary, the carrier-free depletion zone grows thicker and the barrier height rises. Diffusion of majority carriers is now blocked, and the circuit carries only the thermally generated minority-carrier current, which is nearly independent of the applied voltage.
Step 3: Shape of the characteristic.
On an I-V plot the reverse branch lies in the third quadrant hugging the voltage axis (current a few \(\mu\)A). It remains flat as the reverse voltage is raised, until the breakdown voltage where it plunges steeply downward, showing a large reverse current for a nearly fixed voltage.
Step 4: Meaning of avalanche breakdown.
At large reverse voltage across a lightly doped junction the wide depletion region has a strong electric field. A minority carrier crossing it is accelerated so hard that on hitting a lattice atom it ionises the atom, releasing an electron-hole pair. Those new carriers are accelerated too and ionise further atoms, causing a chain (avalanche) multiplication and a sudden large reverse current. That is avalanche breakdown.
\[\boxed{\text{Chain impact-ionisation at } V_{br}\Rightarrow \text{steep rise of reverse current}}\]
Option 2
Step 1: Contrasting the two doped materials.
Doping silicon with a Group-13 (trivalent) atom leaves a shortage of one bonding electron, i.e. a hole; such p-type silicon conducts mainly through holes. Doping with a Group-15 (pentavalent) atom leaves one spare electron; such n-type silicon conducts mainly through electrons. Majority carriers: holes in p-type, electrons in n-type. Minority carriers are the opposite kind, and each sample stays neutral overall.
Step 2: Series loop reasoning.
When the diode conducts, going round the loop the source peak \(V_0\) must supply the fixed barrier drop \(V_b\) plus the ohmic drops across the diode's own resistance \(R_f\) and the load \(R_L\) (they carry the same current). Hence \(V_0=V_b+I_0(R_f+R_L)\).
Step 3: Peak current.
Rearrange: \(I_0=\dfrac{V_0-V_b}{R_f+R_L}=\dfrac{20-0.7}{10+500}=\dfrac{19.3}{510}\).
Dividing, \(I_0=0.03784\text{ A}=37.8\text{ mA}\).
Step 4: Load peak voltage.
Only the load resistance matters here: \(V_{L,0}=I_0 R_L=0.03784\times500=18.92\text{ V}\approx18.9\text{ V}\).
\[\boxed{I_0\approx37.8\text{ mA},\ \ V_{L,0}\approx18.9\text{ V}}\]