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Electronics: Class 12 Physics
Intrinsic and extrinsic semiconductors, p-n junction diode, rectification, transistor action and amplification, logic gates with truth tables, and the operational amplifier with its inverting and non-inverting uses.
Key points
A pure semiconductor such as silicon or germanium is called intrinsic. Each atom has four valence electrons and forms four covalent bonds. At absolute zero there are no free charges. At room temperature heat breaks a few bonds. Each broken bond frees an electron and leaves a vacancy called a hole, which acts as a positive charge carrier. In an intrinsic semiconductor the number of electrons equals the number of holes.
Doping means adding a small amount of impurity to a pure semiconductor to raise its conductivity. The result is an extrinsic semiconductor.
Types of extrinsic semiconductor:
- n-type: doped with a pentavalent atom such as phosphorus or arsenic (five valence electrons). Four electrons bond and one is left free. Electrons are the majority carriers and holes the minority carriers.
- p-type: doped with a trivalent atom such as boron or aluminium (three valence electrons). One bond has a missing electron, which gives a hole. Holes are the majority carriers and electrons the minority carriers.
- Both types are electrically neutral overall.
A p-n junction is formed when p-type and n-type material are joined in one crystal. Electrons diffuse from the n side to the p side and holes diffuse the other way. Near the junction they combine and leave behind fixed ions. This region is the depletion layer, which has no free carriers. The fixed ions create a barrier potential, about 0.7 V for silicon and 0.3 V for germanium, which stops further diffusion.
Biasing of a p-n junction diode:
- Forward bias: the p side is joined to the positive terminal. The applied voltage opposes the barrier, the depletion layer narrows, and current flows once the voltage is above the barrier value (the knee voltage). The current then rises quickly.
- Reverse bias: the p side is joined to the negative terminal. The depletion layer widens, and only a very small leakage current flows, due to minority carriers. If the reverse voltage is too large, the diode breaks down.
A diode conducts well in one direction only, so it can be used as a rectifier to change AC into DC. In half wave rectification one diode passes only one half of each cycle. In full wave rectification two diodes with a centre tapped transformer, or four diodes in a bridge, pass both halves, and the output frequency is twice the input frequency. A capacitor filter then smooths the output. A Zener diode is made for use in reverse breakdown and keeps a steady voltage, so it is used as a voltage regulator. LEDs give light when forward biased and photodiodes respond to light in reverse bias.
A transistor is a three layer device with an emitter, a base and a collector. In an n-p-n transistor a thin p base lies between two n regions. The emitter is heavily doped, the base is thin and lightly doped, and the collector is large. In normal operation the emitter base junction is forward biased and the collector base junction is reverse biased. The emitter current is IE = IB + IC.
The current gain in common emitter mode is β = IC/IB, and it is usually from 20 to several hundred. A small change in base current causes a large change in collector current, so the transistor can amplify. A transistor also works as a switch: with no base current it is off (cut off), and with a large base current it is fully on (saturation).
Logic gates and their outputs:
- NOT: output is the opposite of the input. 0 gives 1 and 1 gives 0.
- OR: output is 1 if at least one input is 1. Only 0,0 gives 0.
- AND: output is 1 only if both inputs are 1.
- NOR: the OR output inverted. Only 0,0 gives 1.
- NAND: the AND output inverted. Only 1,1 gives 0.
- NAND and NOR are called universal gates because any other gate can be built from either one.
An operational amplifier (op-amp) is a high gain integrated circuit with an inverting input (−), a non-inverting input (+) and one output. An ideal op-amp has infinite open loop gain, infinite input resistance and zero output resistance. With negative feedback the two inputs are at nearly the same potential. For the inverting amplifier the gain is −Rf/Rin, and for the non-inverting amplifier the gain is 1 + Rf/Rin. The output of the inverting amplifier is out of phase with the input by 180°.
Study tip: Learn all six truth tables and practise drawing them without notes. For the diode and transistor, draw the circuit and mark the biasing polarity clearly. In op-amp numericals, state which type of amplifier it is before choosing the gain formula.
Definitions and flashcards
Say the meaning first, then open the card to check.
Intrinsic semiconductor
A pure semiconductor in which electrons and holes are equal in number.
Doping
Adding a small amount of impurity to change the conductivity.
Hole
A vacancy for an electron in a covalent bond that acts as a positive charge carrier.
Depletion layer
The region near a p-n junction with no free charge carriers.
Barrier potential
The potential difference across the depletion layer that stops diffusion.
Rectifier
A circuit that changes alternating current into direct current.
Current gain β
The ratio IC/IB of a transistor in common emitter mode.
Universal gate
A gate (NAND or NOR) from which all other gates can be made.
Feedback
Returning part of the output to the input of an amplifier.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Transistor currents | IE = IB + IC | Emitter current is the sum of base and collector currents. |
| Current gain | β = IC / IB | Common emitter mode. |
| Inverting op-amp gain | A = −Rf / Rin | Rf is the feedback resistor. |
| Non-inverting op-amp gain | A = 1 + Rf / Rin | Output is in phase with input. |
| Full wave output frequency | fout = 2 fin | Frequency doubles after full wave rectification. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
What is the difference between n-type and p-type semiconductors? (3 marks)
An n-type semiconductor is doped with a pentavalent impurity such as phosphorus, so electrons are the majority carriers. A p-type semiconductor is doped with a trivalent impurity such as boron, so holes are the majority carriers. Both are electrically neutral.
What is the depletion layer? (2 marks)
It is the thin region on both sides of a p-n junction from which free electrons and holes have diffused away and combined. It contains only fixed charged ions and is almost an insulator.
Why does a diode conduct in forward bias but not in reverse bias? (3 marks)
In forward bias the applied voltage opposes the barrier potential, so the depletion layer narrows and majority carriers cross the junction as a large current. In reverse bias the applied voltage adds to the barrier, so the depletion layer widens and only a tiny leakage current of minority carriers flows.
Write the truth table of a NAND gate. (2 marks)
Inputs A, B and output: 0,0 gives 1; 0,1 gives 1; 1,0 gives 1; 1,1 gives 0. The output is 0 only when both inputs are 1.
Explain how a transistor acts as an amplifier. (5 marks)
In a common emitter circuit the emitter base junction is forward biased and the collector base junction is reverse biased. The signal is applied between base and emitter. A small change in the base current causes a much larger change in the collector current, because IC = βIB. The collector current flows through a load resistor RL, so the output voltage change across RL is much larger than the input voltage change. In this way a weak signal is amplified, and the extra power comes from the collector battery.
State the characteristics of an ideal operational amplifier. (3 marks)
An ideal op-amp has infinite open loop voltage gain, infinite input resistance so that it draws no current from the source, zero output resistance, infinite bandwidth and zero output when both inputs are equal.
Numericals with worked steps
In a transistor the base current is 40 μA and the collector current is 2.0 mA. Find β and the emitter current.
- IB = 40 × 10⁻⁶ A = 0.04 mA.
- β = IC/IB = 2.0 / 0.04 = 50.
- IE = IB + IC = 0.04 + 2.0 = 2.04 mA.
Answer: β = 50, IE = 2.04 mA
An inverting op-amp has Rin = 10 kΩ and Rf = 100 kΩ. Find the gain and the output for an input of 0.2 V.
- Gain = −Rf/Rin = −100 / 10 = −10.
- Vout = gain × Vin = −10 × 0.2 = −2 V.
Answer: Gain = −10, output = −2 V
A non-inverting amplifier has Rin = 2 kΩ and Rf = 18 kΩ. Find its gain.
- Gain = 1 + Rf/Rin.
- Rf/Rin = 18 / 2 = 9.
- Gain = 1 + 9 = 10.
Answer: Gain = 10
A full wave rectifier is fed from a 50 Hz supply. What is the frequency of the ripple at the output?
- A full wave rectifier turns both half cycles into the same direction.
- So there are two output pulses per input cycle.
- fout = 2 × 50 = 100.
Answer: 100 Hz
Common mistakes
- Saying that n-type material is negatively charged. It is neutral, with electrons as majority carriers.
- Mixing up the biasing: forward bias joins p to positive, reverse bias joins p to negative.
- Forgetting that IE = IB + IC, and wrongly saying IC = IE.
- Mistaking NAND for AND in truth tables. NAND is the inverse of AND.
- Dropping the minus sign in the gain of an inverting op-amp.