Microelectronics Questions and Answers – Basic Physics of Semiconductor – PN Junction

This set of Microelectronics Multiple Choice Questions & Answers (MCQs) focuses on “Basic Physics of Semiconductor – PN Junction”.

1. If the donor or acceptor concentration are each increased by one order of magnitude, the built-in-potential changes by ______________
a) 120mV
b) Two order of magnitude
c) Half order of magnitude
d) 0

Explanation: The built-in-potential is proportional to the logarithm of the product of donor and acceptor concentration. One order of magnitude implies 10 times. Hence, if we increase both of them by one order of magnitude, the resultant change in the built-in-potential will be approximately 120mV. This can be obtained readily from the formulae Vbi=VT ln($$\frac {N_A*N_D}{n_i^2}$$).

2. What happens to the built-in-potential if temperature is increased?
a) It increases
b) It decreases
c) It remains unchanged
d) It increases linearly.

Explanation: The built-in-potential is directly proportional to the temperature. It would increase linearly with temperature.

3. What happens to the junction capacitance at zero bias if temperature increases?
a) It increases
b) It decreases
c) It remains constant
d) It increases linearly.

Explanation: The capacitance is inversely proportional to the square-root of the built-in-potential. The built-in-potential is directly proportional to the temperature. An increase in temperature would lead to a increase in built-in-potential which in turn would reduce the junction capacitance at 0 bias.

4. Under forward bias, what happens to the hole concentration in the n-side?
a) It increases
b) It decreases
c) It remains constant
d) It decreases non-linearly

Explanation: The hole concentration in the n-side increases a bit when the pn junction is forward biased. This is expected since some electrons would readily cross the barrier and flow into the n-side. It is also understood by the following equation pn-side=pp-side*[1/{exp(V0 – Vf.bias)/VT)}]. If Vf.bias increases, the concentration would increase since the denominator decreases.

5. If the cross-sectional area of the PN junction increases 10 times, how much should the diffusion length change so that the total reverse saturation current remains constant?
a) Ln and Lp should increase 10 times each
b) Ln and Lp should decrease 10 times each
c) Lp should increase 10 times each
d) Ln should increase 10 times each

Explanation: The equation for reverse saturation current is IS=Aqni2(Dn/(NALN) + DP/(NDLP)). We observe that if A increases 10 times, each of the diffusion length has to increase 10 times so that the total reverse saturation current remains constant.

6. What is the relation between reverse saturation current of Si and Ge?
a) ISi > IGe
b) ISi < IGe
c) ISi = IGe
d) ISi < 2 * IGe

Explanation: Si has a higher thermal equilibrium concentration than Ge. The reverse saturation current being proportional to the intrinsic carrier concentration, Si will have a higher reverse saturation current than Ge.

7. Why can we approximate the working of a diode as a constant voltage model?
a) The voltage is a weak function of current
b) The voltage is a strong function of current
c) The voltage is independent of current
d) We cannot do this approximation

Explanation: The voltage is related to the logarithm of the current. The voltage is seen to be fixed between 750mV-800mV for certain current levels and owing to the weak dependency, we use the constant voltage model.

8. The constant voltage model doesn’t take the reverse saturation current into account.
a) True
b) False

Explanation: Typically, the reverse saturation current is fixed and the voltage drop across a particular diode is modelled to be fixed after the input voltage crosses the threshold. The constant voltage model only allows us to reach a concluding remark about the status of diode in the circuit.

9. How can we keep the maximum electric field, in the depletion region, constant if the space charge region has somehow increased 2 times in the n-side?
a) Reduce the donor concentration
b) Reduce the acceptor concentration
c) Reduce the concentration of carriers
d) Increase the donor concentration

Explanation: The electric field becomes maximum at the junction of the PN-junction. The maximum value is proportional to the product of the donor concentration and the width of the space charge region in the n side. If the space charge region extends deep into the n-side, the donor concentration has to be reduced so that the maximum value of electric field remains constant.

10. If the width of the depletion region increases, what happens to the maximum value of electric field in the depletion region?
a) It increases
b) It decreases
c) It remains constant
d) It increases exponentially

Explanation: The maximum value of electric field is inversely proportional to the width of the depletion region. If the width increases, the maximum value of electric field will decrease.

11. If the donor and acceptor concentration increase 10 times, what happens to the depletion capacitance?
a) It increases 10 times
b) It decreases 10 times
c) It increases by √10 times
d) It decreases by √10 times

Explanation: The depletion capacitance is found to be inversely proportional to the square root of (1/ND and 1/NA). Hence, this capacitance would increase by √10 times if each of the concentration is increased 10 times.

12. If the potential across a junction varies like a ramp function, what happens to the potential energy of an electron as it travels through that junction?
a) It varies like an inverted ramp function
b) It varies like a ramp function
c) It oscillates
d) It remains constant

Explanation: The potential energy of a charged particle is positive and is equal to the product of the charge and the potential at which it is present. If the potential varies like a ramp function, the potential energy would also vary like a ramp function but it will get inverted since the charge of an electron is negative.

13. Why can’t an electron flow out of the potential barrier?
a) The kinetic energy becomes 0
b) The potential energy becomes 0
c) The total energy becomes 0
d) The barrier behaves as a wall

Explanation: After the depletion region is made, the kinetic energy of electrons become 0 as they reach the maximum width of depletion region in either side of the junction. The potential energy is not zero but it reaches a maximum. The electrons cannot invade further until a forward bias is applied.

14. Why doesn’t the energy level for an atom in a gas repeat for a semi-conductor solid?
a) The interatomic-distance varies considerably
b) The structure of the element varies
c) The behavior varies considerably
d) They are the same

Explanation: The interatomic distance in gas is far more than that in a solid. This is the main reason why energy levels in a gas and a solid are not similar. The periodic structure in crystals also contributes to the energy levels.

15. After crossing the diffusion length, the concentration of carriers decreases to____________
a) 1/e of it’s initial value
b) 3/e of it’s initial value
c) 2/e of it’s initial value
d) 4/e of it’s initial value

Explanation: It has been defined that the diffusion length is the product of Diffusivity of carrier times the minority carrier lifetime. The diffusion length indicates that after entering the extrinsic semiconductor as a minority carrier, the concentration has decayed to 1/e of it’s previous value after they travel a distance equal to the diffusion length.

Sanfoundry Global Education & Learning Series – Microelectronics.

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