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. \(n\cdot p=n_i^{2}\)
a-ii, b-iii, c-i, d-iv
Take the two easiest pairings first — both are one-line definitions.
d → iv, the law of mass action. In thermal equilibrium the product of the free electron and hole concentrations is a constant of the material, independent of doping:
\(n\cdot p=n_i^{2}\)
This is the definition itself, so the pairing is immediate. Every option except 1 already agrees, so it is a poor discriminator on its own — but it anchors the rest.
c → i, charge neutrality. Across the depletion region the total negative charge on the p-side must equal the total positive charge on the n-side. Writing x1 and x2 for the two depletion widths,
\(qN_Ax_1=qN_Dx_2 \Rightarrow N_Ax_1=N_Dx_2\)
This is the origin of the one-sided junction result: doping one side heavily makes its depletion width negligible, so the depletion region extends almost entirely into the lightly doped side.
b → iii, the built-in potential. Equating the Fermi levels on the two sides gives
\(\phi_o=\dfrac{kT}{q}\ln\!\left(\dfrac{N_AN_D}{n_i^{2}}\right)\)
Two features identify it on sight: the thermal voltage kT/q ≈ 26 mV at room temperature, which fixes the units as volts, and the logarithm of the doping product — a barrier of a few tenths of a volt for silicon.
a → ii, the junction depth. Whatever remains must be the depletion width, and the expression confirms it:
\(x_d=\sqrt{\dfrac{2\epsilon_{si}\phi_o}{qN_A}}\)
The square-root form is the signature of every depletion-width formula, and the presence of εsi (permittivity) with q and NA gives the dimensions of length.
A dimensional cross-check settles it without memorising anything. Item iv is a pure number density squared, item i is a relation between charges, item iii has kT/q so it is in volts, and item ii contains a permittivity under a square root so it is in metres. Only one arrangement matches "depth → metres, voltage → volts": a-ii, b-iii, c-i, d-iv.
Assemble. a-ii, b-iii, c-i, d-iv, which is option 4. The trap is option 1, identical except that it swaps c and d — so the whole question turns on distinguishing charge neutrality (a statement about the depletion charges) from the mass-action law (a statement about the carrier concentrations).
Hence, the correct matching is a-ii, b-iii, c-i, d-iv.
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 ?
In a pn junction diode, depletion layer is formed due to
Important Questions from PN Junction
The leakage current in a pn junction is of the order of:
The depletion region consists of:
In the P-N junction, the barrier voltage
- If the temperature increases by 10°C, the P-N junction barrier voltage
In an abrupt p-n junction if N A << N D, the barrier potential is:
