Consider the following statements for a p-n junction diode : (a) It is an active component Which of the statements given above are correct ?
(b) Depletion layer width decreases with forward biasing
(c) In the reverse bias case, saturation current increases with increasing temperature
(b) and (c) only
(b) and (c) only — option 3. Statement (a) is the false one: a p-n junction diode is a passive component, not an active one.
(a) is false — the distinction that decides the question. The words active and passive have a precise meaning in circuit theory, and it is not about whether the device does something interesting :
| Active component | Passive component | |
|---|---|---|
| Definition | Can deliver power gain and needs an external supply to do it | Cannot supply net energy to the circuit |
| Can it amplify? | Yes — a small signal controls a larger one | No |
| Examples | BJT, FET, op-amp, vacuum tube, SCR | Resistor, capacitor, inductor, transformer, diode |
A diode rectifies, clips, clamps and switches, but it never delivers more power than it receives — there is no control terminal and no gain. Its two-terminal nature is the clue: control requires a third terminal, which is why transistors are active and diodes are not.
(b) is true. Forward bias opposes the built-in potential, so the barrier falls and the depletion region narrows:
\(W\propto\sqrt{V_{bi}-V_{F}}\)
Majority carriers can then cross in large numbers, which is exactly why forward current rises steeply. Reverse bias does the opposite — the layer widens, which is the basis of the varactor.
(c) is true, and strongly so. The reverse saturation current is carried by minority carriers, which are thermally generated, so it depends on temperature far more than on voltage:
\(I_{S}\propto T^{3}e^{-E_{g}/kT}\)
The practical rule is that \(I_{S}\) roughly doubles for every 10°C rise — about 7% per degree. This is why reverse leakage is specified at a stated temperature, and why germanium (\(E_{g}=0.7\) eV) leaks far more than silicon (1.1 eV) and is unusable at high temperature.
The shortcut. Statements (b) and (c) are standard results that few would dispute, so the question turns entirely on (a). Deciding that a diode is passive eliminates options 1, 2 and 4 together and leaves option 3 alone.
A caution on the word “passive”. Diodes are sometimes loosely called active devices because they are made of semiconductor and are non-linear. That is a different sense of the word — non-linearity is not gain. On the standard circuit-theory definition used in examinations, the diode is passive.
Hence, the answer is (b) and (c) only.
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 :
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
