In a pn junction diode, depletion layer is formed due to
recombination
Setting up the physics. When p-type and n-type silicon are joined, there is a huge concentration gradient across the metallurgical junction: holes are the majority carriers on the p-side, electrons on the n-side, with typical densities differing by many orders of magnitude.
Step 1 — diffusion. Driven by that gradient, majority carriers diffuse across the junction: electrons move from the n-side into the p-side, holes from the p-side into the n-side.
Step 2 — recombination. As soon as an electron enters the p-region it is surrounded by holes (and vice versa), so the two recombine and both mobile carriers vanish. The region immediately around the junction is therefore swept clean of free carriers.
Step 3 — the space charge that is left behind. Recombination removes only the mobile carriers; the dopant atoms that supplied them are fixed in the lattice and are now un-neutralised — negative acceptor ions (NA−) on the p-side, positive donor ions (ND+) on the n-side. This carrier-free, charged region is the depletion region (also called the space-charge region).
Step 4 — the barrier that stops the process. The exposed ions set up an internal electric field, and hence a built-in potential
\(V_{bi} = V_T \ln\!\left(\dfrac{N_A N_D}{n_i^2}\right)\)
(about 0.7 V in Si, 0.3 V in Ge). This field opposes further diffusion, so the depletion layer stops widening and equilibrium is reached.
Why the other choices are not the cause. Doping supplies the carriers and sets the width of the layer but does not by itself remove carriers. Barrier potential is a consequence of the exposed ions, not the mechanism that creates them. Biasing only modulates an already existing layer — reverse bias widens it, forward bias narrows it.
Takeaway. Diffusion carries the carriers across, recombination destroys them, and the leftover immobile ions constitute the depletion layer; its width varies as \(W \propto \sqrt{V_{bi} - V}\) and is larger on the more lightly doped side.
Hence, the depletion layer is formed due to recombination of the diffusing majority carriers.
Similar Questions
In a p-n junction diode
1. In reverse bias mode the bulk resistance is dominant.
2. The junction impedance is variable.In a MOS transistor, if n+ region is diffused in p-type substrate, the type of pn junction generated towards channel and drain is:
Match the following lists :
List – I List – II a. Abrupt p-n junction i. p+ n+ b. Linearly graded p-n junction ii. p+ n n+ c. Tunnel diode iii. p π(ν) n d. p-i-n diode iv. p+ p n n+ Correct codes are :
Consider the following statements for a p-n junction diode :
(a) It is an active component
(b) Depletion layer width decreases with forward biasing
(c) In the reverse bias case, saturation current increases with increasing temperatureWhich of the statements given above are correct ?
Match the following lists :
List – I List – II a. Junction deptd i. \(N_Ax_1=N_Dx_2\) b. Junction built in voltage ii. \(\sqrt{\dfrac{2\epsilon_{si}}{qN_A}\phi_o}\) c. Charge neutrality condition iii. \(\dfrac{kT}{q}\ln\left(\dfrac{N_AN_D}{n_i^{2}}\right)\) d. Law of mass action iv. \(n\cdot p=n_i^{2}\)
Important Questions from PN Junction
Diode junction breakdowns above 5 V are caused by:
The leakage current in a pn junction is of the order of:
For a PN junction, we have
A. Width of depletion layer
B. Junction barrier voltage
C. Reverse leakage current
Which of the above parameters will decrease when the temperature of the junction rises?
The depletion region consists of:
In the P-N junction, the barrier voltage
