Assertion (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 :
Both (A) and (R) are true, but (R) is not the correct explanation of (A)
Both (A) and (R) are true, but (R) is not the correct explanation of (A) — option 2.
The assertion is true. Two p-n diodes wired back to back give the right sequence of regions — p-n-p or n-p-n — and yet the arrangement has no transistor action at all. Connect them and you have two junctions that behave independently: whichever is reverse biased simply blocks, and no current reaches the far side.
The real reason is base width, and (R) does not mention it. Transistor action requires carriers injected by the emitter to reach the collector before they recombine. That imposes a condition on the middle region:
\(W_{B}\ll L_{B}\)
where \(W_{B}\) is the base width and \(L_{B}\) the minority-carrier diffusion length. In a real transistor the base is a fraction of a micron, so nearly every injected carrier crosses it and
\(\alpha=\dfrac{I_{C}}{I_{E}}\approx 0.98\ \text{to}\ 0.995\)
In two joined diodes the “base” is the whole thickness of two separate bodies plus the metallurgical joint — hundreds of microns. Every injected carrier recombines long before it arrives, so \(\alpha\approx 0\) and there is no gain.
| Requirement for transistor action | Real transistor | Two diodes joined |
|---|---|---|
| Base width ≪ diffusion length | Yes — sub-micron base | No — this is the fatal defect |
| Emitter doped far more heavily than base | Yes, for high emitter injection efficiency | Not arranged |
| Single continuous crystal | Yes | No — the joint is a barrier in itself |
Why (R) is true but does not explain (A). It is perfectly correct that a transistor needs controlled, graded doping — a heavily doped emitter, a lightly doped and very thin base, and a moderately doped collector sized for breakdown. But that is a statement about doping levels, whereas the decisive failure of back-to-back diodes is geometric: the base is far too wide. Two diodes could be doped to the correct profile and still not work, because the middle region would remain thick. (R) names a genuine requirement that is not the operative one, which is exactly the situation code 2 describes.
The lesson. In assertion-reason items, a reason can be a true and relevant fact and still not be the reason. Ask what would happen if the reason were satisfied but the assertion still held — here, correct doping alone would not rescue the device.
Hence, the answer is option 2.
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 :
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) : 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:
