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 pn junction diode, depletion layer is formed due to
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 ?
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
