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 :
Both (A) and (R) are true and (R) is the correct explanation of (A).
Both statements are true, and the second is precisely the cause of the first.
The mobility figures for silicon at room temperature :
\(\mu_{n}\approx1350\ \text{cm}^{2}/\text{V}\!\cdot\!\text{s}\qquad \mu_{p}\approx480\ \text{cm}^{2}/\text{V}\!\cdot\!\text{s}\)
— electrons are about 2.8 times more mobile than holes, because a hole moves by successive covalent-bond exchanges while an electron travels freely in the conduction band.
Why that decides the frequency response. In an n-p-n transistor the carriers injected into the base are electrons; in a p-n-p they are holes. The time taken to cross the base is
\(\tau_{B}=\dfrac{W_{B}^{2}}{2D}\qquad\text{with}\qquad D=\dfrac{kT}{q}\mu\)
so a higher mobility means a larger diffusion constant, a shorter base transit time, and therefore a higher cut-off frequency:
\(f_{T}=\dfrac{1}{2\pi\tau_{ec}}\)
With identical geometry and doping, the only difference between the two devices is which carrier does the work — so the n-p-n wins by roughly the mobility ratio. The code is 1.
| n-p-n | p-n-p | |
|---|---|---|
| Carrier crossing the base | Electrons | Holes |
| Base transit time | Shorter | Longer |
| fT for the same geometry | Higher | Lower |
| Current gain β | Higher | Lower |
The same argument explains far more than this question. It is why n-channel MOSFETs are preferred over p-channel, why NMOS logic was built with n-channel devices, and why a CMOS inverter's p-channel transistor is drawn two to three times wider than its n-channel partner — the extra width compensates for the lower hole mobility so that rise and fall times match.
The exception worth knowing : in germanium the gap between the two mobilities is much smaller, which is why early germanium p-n-p transistors were commonplace, whereas in silicon the n-p-n has dominated ever since.
Hence, both (A) and (R) are true and (R) is the correct explanation of (A).
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) : 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 :
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:
