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Kerala Plus Two Electronics (DHSE) study materials
Kerala Plus Two Electronics (DHSE) study resources

Kerala Plus Two Electronics (DHSE): Basic Electronics and Semiconductor Devices

Kerala Plus Two Electronics (DHSE) · Work through your exam, one topic at a time.

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Basic Electronics and Semiconductor Devices

Basic Electronics and Semiconductor Devices tests how doped semiconductors, PN junctions, diodes and three-terminal devices behave in circuits. DHSE questions favour bias analysis, device characteristics and short numerical current or gain calculations.

Core concepts

  • Doping: pentavalent impurities such as P and As donate electrons and make n-type material; trivalent impurities such as B and Al create holes and make p-type material. Majority carriers dominate conduction; minority carriers are strongly temperature-dependent.
  • PN junction: diffusion leaves a depletion region and a barrier potential. Approximate cut-in values are 0.7 V for silicon and 0.3 V for germanium. Forward bias narrows the depletion layer; reverse bias widens it and leaves only a small leakage current.
  • Diode models: an ideal diode is a switch; a practical diode adds the cut-in drop when forward biased. A Zener diode is used in reverse breakdown for voltage regulation; an LED is forward biased; a photodiode is normally reverse biased.
  • BJT: two PN junctions form emitter, base and collector. In active mode it amplifies; in saturation and cut-off it switches. IE = IB + IC, α = IC/IE (less than 1), β = IC/IB, and β = α/(1−α). The CE connection gives current and voltage gain and inverts the signal by 180°.
  • FET: a voltage-controlled device with high input impedance. JFET and MOSFET operation depends on controlling the channel width; pinch-off and transconductance are key ideas.

Worked example

A silicon diode is forward biased in series with a 1 kΩ resistor and a 10 V battery. Using the practical model, the diode drops about 0.7 V, so the resistor voltage is 10 − 0.7 = 9.3 V. The current is I = 9.3 V / 1000 Ω = 9.3 mA. An ideal-diode answer of 10 mA is only an approximation; the practical answer is expected.

Traps that cost marks

  • Using 0.7 V for a germanium diode or ignoring the cut-in drop entirely.
  • Mixing up α and β; α is always less than 1, while β is usually much greater than 1.
  • Writing IE = IB − IC instead of IE = IB + IC.
  • Treating a Zener diode as forward biased in a regulator, or assuming reverse leakage is exactly zero.
  • Forgetting to convert mA, µA and kΩ consistently before calculating.