Match the following : Codes :List – I (Biasing of BJT) List – II (Functions) a. E-B junction forward bias and C-B junction reverse bias i. Very low gain amplifier b. Both E-B and C-B junctions forward bias ii. Saturation condition c. E-B junction reverse bias and C-B junction forward bias iii. High gain amplifier d. Both E-B and C-B junctions reverse bias iv. Cut-off condition
a-iii, b-ii, c-i, d-iv
Two junctions with two states each give exactly four regions of operation, and the question lists all four: a-iii, b-ii, c-i, d-iv — option 2.
| E-B | C-B | Region | Behaviour |
|---|---|---|---|
| Forward | Reverse | a. Active | iii. High gain amplifier |
| Forward | Forward | b. Saturation | ii. Switch fully on |
| Reverse | Forward | c. Reverse active | i. Very low gain |
| Reverse | Reverse | d. Cut-off | iv. Switch fully off |
The active region — why the gain is high. The forward-biased emitter junction injects carriers into the base; the base is made thin and lightly doped so that almost all of them survive to reach the collector, where the reverse bias sweeps them across. Since \(I_{C}\approx\alpha I_{E}\) with \(\alpha\) close to 0.99, only a small fraction returns as base current, and
\(\beta=\dfrac{\alpha}{1-\alpha}\)
is large. This is the region every linear amplifier uses.
The reverse-active region — and why it is nearly useless. Swapping the roles of emitter and collector shows immediately why the gain collapses: the collector is lightly doped and physically large, made for handling voltage and dissipating heat, not for efficient injection. Its injection efficiency is poor, so \(\beta_{R}\) is typically between 0.1 and 5 — hence "very low gain amplifier". The transistor is not symmetrical, and this asymmetry is deliberate. The one place the mode is used deliberately is the multi-emitter input of a TTL gate.
Saturation and cut-off are the two switching states. With both junctions forward biased the collector can accept no more current, \(V_{CE}\) falls to about 0.2 V, and the device behaves as a closed switch — note that \(I_{C}\lt\beta I_{B}\) here, so the usual active-region relation no longer holds. With both reverse biased only leakage flows and the device is an open switch. A digital circuit toggles between these two, passing through the active region only in transit.
Hence, the correct code is a-iii, b-ii, c-i, d-iv.
Similar Questions
The doping concentration of n-p-n transistor are
i. 5 × 1018 / cm3
ii. 1017 / cm3
iii. 2 × 107 / cm3Identify the regions in the above order i, ii and iii.
For the BJT to be biased in its linear or active operating region :
(a) the base-emitter Junction must be forward biased
(b) the base-collector Junction must be reverse biased
(c) the base-emitter Junction must be reverse biased
(d) the base-collector Junction must be forward biasedAssertion (A) : Two P-N diodes connected back to back cannot be used as a transistor.
Reason (R) : Fabrication of a transistor requires controlled doping of Emitter, base and collector regions.
Select your answer using the codes given below :
Assertion (A) : Considering two p-n-p and n-p-n transistors of identical construction as far as shape, size and doping are concerned, the n-p-n transistor will have a better frequency response.
Reason (R) : The electron mobility is higher than that of the hole mobility.
Select your answer using the codes given below :
Important Questions from Bipolar Junction Transistor
Find $I_C$, $V_{BE}$=0.7V.
- Voltage gain of common emitter amplifier with $R_E$ is given as :
The doping concentration of n-p-n transistor are
i. 5 × 1018 / cm3
ii. 1017 / cm3
iii. 2 × 107 / cm3Identify the regions in the above order i, ii and iii.
For the BJT to be biased in its linear or active operating region :
(a) the base-emitter Junction must be forward biased
(b) the base-collector Junction must be reverse biased
(c) the base-emitter Junction must be reverse biased
(d) the base-collector Junction must be forward biasedAssertion (A) : Two P-N diodes connected back to back cannot be used as a transistor.
Reason (R) : Fabrication of a transistor requires controlled doping of Emitter, base and collector regions.
Select your answer using the codes given below :
