The existence of cut-in voltage in a diode is mainly due to the presence of __________.
Depletion barrier potential
A p-n junction diode does not start conducting the moment even a tiny forward voltage is applied. Instead, current remains negligible until the applied forward voltage reaches a certain minimum value, after which current rises sharply. This minimum forward voltage is called the cut-in voltage (also known as the threshold voltage, knee voltage, or offset voltage) — approximately 0.2–0.3 V for germanium diodes and approximately 0.6–0.7 V for silicon diodes.
Why does this threshold exist? When a p-type and n-type semiconductor are joined, majority charge carriers diffuse across the junction (electrons from the n-side, holes from the p-side) until an equilibrium is reached. This diffusion leaves behind fixed, immobile ions on either side of the junction — positive ions on the n-side and negative ions on the p-side — creating a narrow region depleted of free charge carriers, called the depletion region. These fixed ionic charges set up an internal electric field directed from the n-side to the p-side, which corresponds to a built-in (barrier) potential, V₀, that opposes further diffusion of majority carriers across the junction.
Thus, the cut-in voltage is a direct consequence of the depletion region's built-in barrier potential, which must be overcome before appreciable conduction can begin.
The remaining options do not explain this phenomenon: a diode certainly does not have infinite conductivity (if anything, it behaves almost like an open circuit below cut-in and like a low-resistance path above it); the atoms in the depletion region are not simply “neutral atoms without charge” — they are ionised donor and acceptor atoms that create the barrier field precisely because they carry fixed charge; and no external magnetic field is involved at all in the operation of a basic p-n junction diode — its behaviour is governed entirely by electric fields and carrier diffusion/drift within the semiconductor.
At room temperature, the region in a PN junction diode where there are no mobile electrons or holes is called __________________.
Under forward-bias conditions, what is the primary mechanism that creates forward current across a PN junction?
Before breakdown, increasing reverse voltage causes reverse saturation current to __________.
What is the main disadvantage of a half-wave rectifier?
Consider an abrupt p-n junction under thermal equilibrium. If the acceptor doping concentration (NA) in the p-region is made significantly larger than the donor doping concentration (ND) in the n-region (i.e., $N_A ≫ N_D$), how is the depletion region physically distributed?
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 | ||||||
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| a. | Abrupt p-n junction | i. |
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| b. | Linearly graded p-n junction | ii. |
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| c. | Tunnel diode | iii. |
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| d. | p-i-n diode | iv. |
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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 temperature
Which of the statements given above are correct ?
At room temperature, the region in a PN junction diode where there are no mobile electrons or holes is called __________________.