The doping concentration of n-p-n transistor are i. 5 × 1018 / cm3 Identify the regions in the above order i, ii and iii.
ii. 1017 / cm3
iii. 2 × 107 / cm3
E, B, C
A bipolar transistor's three regions are doped in a strictly decreasing order — emitter heaviest, base intermediate, collector lightest — so the list as printed reads E, B, C, option 2.
| Region | Doping | Reason |
|---|---|---|
| Emitter | Heaviest, 1018–1019 | To inject carriers efficiently into the base |
| Base | Intermediate, ~1017 | Thin and lightly doped so few carriers recombine |
| Collector | Lightest | Wide depletion region to withstand voltage |
Why the emitter must be doped hardest. The emitter junction injects electrons into the base, but holes are simultaneously injected back from base into emitter, and that back-injection is wasted current. The ratio of the two is set by the doping ratio, giving the emitter injection efficiency
\(\gamma=\dfrac{I_{nE}}{I_{nE}+I_{pE}}\approx\dfrac{1}{1+\dfrac{N_{B}}{N_{E}}}\)
With \(N_{E}=5\times10^{18}\) and \(N_{B}=10^{17}\) the ratio is 50, so \(\gamma\approx0.98\). Doping the emitter more heavily raises \(\gamma\) and hence \(\alpha\), and since
\(\beta=\dfrac{\alpha}{1-\alpha}\)
a small gain in \(\alpha\) produces a large gain in \(\beta\).
Why the base is doped lightly and made thin. Injected electrons must cross the base without recombining. A light doping means few holes are available to recombine with them, and a thin base gives little time for it to happen; together these give a base transport factor close to unity.
Why the collector is lightest of all. The collector-base junction is reverse biased and must hold off the supply voltage. A lightly doped collector puts most of the depletion region on its own side, spreading the field over a greater width and so raising the breakdown voltage — which is why power transistors have very lightly doped collector regions.
A note on the printed figures. The third value, \(2\times10^{7}\) cm−3, is below the intrinsic carrier concentration of silicon (\(1.5\times10^{10}\)) and cannot describe a real collector; it is evidently a misprint for something like \(10^{15}\). The ordering it is meant to convey — smallest of the three — is unambiguous, and the answer is unaffected.
Hence, the regions in order are E, B, C.
Similar Questions
Match the following :
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 Codes :
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
Photo transistor is used for:
What happens if a voltage of about 0.7 V is applied across the base and emitter of the NPN transistor?
In a junction transistor, recombination of electrons and holes occurs in
A bipolar junction common emitter transistor is operating in saturation mode, identify the correct statement.
In BJT when both the junctions are forward biased, then its operating mode is called:
