Semiconductor Electronics: Materials, Devices and Simple Circuits — Questions & Answers

Sample questions from Semiconductor Electronics: Materials, Devices and Simple Circuits

1. Q1. A silicon p-n junction has a built-in potential of 0.72 V. If the doping on both sides is increased ten-fold, the built-in potential:

   • A.decreases, because the depletion width shrinks
   • B.is unchanged, being fixed by the band gap alone
   • C.increases, because V_bi depends logarithmically on the product N_a N_d✓
   • D.increases ten-fold
   Solution

   V_bi = (kT/e) ln(N_a N_d/n_i^2), so a ten-fold rise in each doping adds (kT/e) ln 100 = 26 x 4.6 = 120 mV, giving about 0.84 V. It can never exceed the band gap in volts, and the depletion width shrinks at the same time.

2. Q2. An emitter resistor R_E is added to a CE amplifier without bypassing it with a capacitor. The result is:

   • A.a more stable operating point with the gain unchanged
   • B.a higher gain and better stability simultaneously
   • C.a more stable operating point but a reduced voltage gain, approximately R_C/R_E✓
   • D.no change, since R_E carries no signal current
   Solution

   The unbypassed R_E provides negative feedback: any rise in I_C raises the emitter voltage and reduces the effective V_BE, stabilising the bias, but the same feedback cuts the gain to roughly R_C/R_E. A bypass capacitor restores the ac gain while keeping the dc stabilisation.

3. Q3. The I-V characteristic of an illuminated solar cell is drawn in the fourth quadrant (positive voltage, negative current) because:

   • A.it consumes power from an external source
   • B.it delivers power, the current flowing out of the positive terminal opposite to the sense of an applied-bias diode current✓
   • C.it behaves as a resistor
   • D.of measurement error
   Solution

   An illuminated cell generates current opposite in sign to the forward diode current while its terminal voltage is positive. The product V*I is therefore negative in the diode convention, which is the signature of a device delivering (not absorbing) power.

4. Q4. In the energy-band picture, doping silicon with arsenic introduces donor levels. These levels lie:

   • A.just above the valence band edge
   • B.just below the conduction band edge, about 0.05 eV from it✓
   • C.exactly at the middle of the gap
   • D.inside the conduction band itself
   Solution

   Pentavalent donors create a level a few hundredths of an eV below the conduction band edge, so at room temperature (kT = 0.026 eV) almost all donors are ionised and give electrons to the conduction band.

5. Q5. A photodiode with reverse dark current 10 nA shows 2.0 microamp when illuminated. Assuming photocurrent is proportional to intensity, the current when the intensity is halved is about:

   • A.1.0 microamp
   • B.1.0 microamp plus the 10 nA dark current, i.e. about 1.005 microamp✓
   • C.0.5 microamp
   • D.2.0 microamp
   Solution

   Total current = dark current + photocurrent. Photocurrent at full intensity = 2.0 - 0.01 = 1.99 microamp; halving intensity gives about 0.995 microamp, so the total is about 1.005 microamp.

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