Semiconductors, p-n Junction and Transistors
The security light on a house in Galway comes on only when it gets dark, with no one touching a switch. Inside, there is no moving part: a tiny slice of silicon decides when to let current through. How can a material that barely conducts be turned into the switch behind every phone and processor? By the end of this page you will be able to explain it.
Semiconductors and doping
Metals conduct well because they have many free electrons. Insulators have almost none. A semiconductor such as silicon sits in between.
In pure silicon each atom has four valence electrons, and all four are shared in bonds with neighbouring atoms. At low temperature none is free. At room temperature, thermal energy frees a small number of electrons from their bonds. Each leaves behind a gap called a hole, which behaves like a positive charge carrier because a neighbouring electron can move into it. The result is a small number of free carriers, so pure silicon is a poor conductor.
There are two types:
| n-type | p-type | |
|---|---|---|
| Dopant | Group 5: phosphorus, arsenic | Group 3: boron, aluminium |
| Valence electrons of dopant | 5 | 3 |
| Effect | one extra electron per atom, free to move | one missing electron per atom, creating a hole |
| Majority carriers | electrons | holes |
| Minority carriers | holes | electrons |
Doping can raise conductivity by several orders of magnitude, because each dopant atom adds a carrier, and there are far more dopant atoms than thermally freed carriers.
Temperature
In a metal, heating makes the lattice ions vibrate more, so electrons collide more often and resistance increases. In a pure semiconductor, heating frees many more electrons from bonds. The number of carriers rises sharply, so resistance decreases. This is why a thermistor (an NTC thermistor) is made from semiconductor material.
Silicon is a semiconductor which can be doped. a) What is meant by doping? b) Name one element used to make n-type silicon and one used to make p-type silicon. c) State…
Silicon is a semiconductor which can be doped.
- a) What is meant by doping?
- b) Name one element used to make n-type silicon and one used to make p-type silicon.
- c) State the majority charge carriers in n-type and in p-type material.
Pure silicon has a small number of free charge carriers at room temperature. A sample of silicon is doped with phosphorus. a) Explain how a phosphorus atom, which has…
Pure silicon has a small number of free charge carriers at room temperature. A sample of silicon is doped with phosphorus.
- a) Explain how a phosphorus atom, which has five valence electrons, contributes to conduction in silicon (four valence electrons).
- b) Explain why the doped sample is still electrically neutral overall.
- c) Explain why the resistance of pure silicon decreases as temperature rises.
A student claims: "Doped semiconductors are charged because they have extra electrons or holes." a) Evaluate this claim. b) Explain why doping increases conductivity by…
A student claims: "Doped semiconductors are charged because they have extra electrons or holes."
- a) Evaluate this claim.
- b) Explain why doping increases conductivity by several orders of magnitude.
- c) Compare, with reasons, the effect of raising temperature on the resistance of a metal and of a pure semiconductor.
- d) Explain why transistors in a processor are made from semiconductors rather than metals.
The p-n junction and the diode
Join a piece of p-type and a piece of n-type material and you have a p-n junction.
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Electrons are plentiful on the n-side and scarce on the p-side, so electrons diffuse across to the p-side. Holes diffuse the other way.
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Near the junction, electrons fill holes and both carriers disappear. This leaves a region with almost no free carriers: the depletion layer.
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The n-side loses electrons, so its fixed ions become positive. The p-side gains electrons, so its fixed ions become negative. This sets up an electric field pointing from n to p across the layer.
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This field opposes further diffusion. The process stops when the field is strong enough to balance it. The potential difference across the layer is the potential barrier: about $0.7\ \text{V}$ for silicon.
Biasing the junction
| Forward bias | Reverse bias | |
|---|---|---|
| Connection | p-side to positive terminal | p-side to negative terminal |
| Depletion layer | narrower | wider |
| Barrier | reduced | increased |
| Current | flows once the supply exceeds about $0.7\ \text{V}$ | practically none (tiny leakage) |
In forward bias the supply pushes carriers towards the junction, shrinking the layer so they can cross. In reverse bias it pulls carriers away, widening the layer, so nothing can cross.
A p-n junction with two terminals is a diode: it conducts in one direction only. It is not ohmic, because its I–V graph is not a straight line through the origin.
Rectification
An alternating supply pushes current one way, then the other. A diode only lets current through when it is forward biased, so it blocks the negative half-cycles. In a half-wave rectifier the output is a series of positive pulses separated by gaps: it is direct current in direction, though not steady.
a) What is the depletion layer in a p-n junction? b) Explain in terms of charge carriers how it forms. c) State whether the depletion layer gets wider or narrower when…
- a) What is the depletion layer in a p-n junction?
- b) Explain in terms of charge carriers how it forms.
- c) State whether the depletion layer gets wider or narrower when the junction is reverse biased.
A silicon p-n junction has a potential barrier of about $0.7\ \text{V}$ across its depletion layer. It is connected in series with a $330\ \Omega$ resistor and a…
A silicon p-n junction has a potential barrier of about $0.7\ \text{V}$ across its depletion layer. It is connected in series with a $330\ \Omega$ resistor and a $5.0\ \text{V}$ supply in forward bias.
- a) Explain how the depletion layer changes in forward bias.
- b) Calculate the current in the circuit, assuming the diode drops $0.7\ \text{V}$.
- c) Calculate the power dissipated in the resistor.
In a p-n junction, electrons diffuse from the n-side to the p-side and holes in the opposite direction. a) Describe the formation of the depletion layer and the…
In a p-n junction, electrons diffuse from the n-side to the p-side and holes in the opposite direction.
- a) Describe the formation of the depletion layer and the resulting electric field direction.
- b) Explain why diffusion does not continue indefinitely.
- c) Explain why the junction conducts in forward bias but practically not in reverse bias.
- d) Explain why a diode can be used to convert alternating current into direct current, sketching in words the output of a half-wave rectifier.
The transistor as a switch
An n-p-n transistor has three terminals: the base (B), the collector (C) and the emitter (E). A small current into the base controls a much larger current from collector to emitter.
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If the base-emitter voltage is below about $0.6\ \text{V}$ the transistor is off: practically no collector current flows.
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Once it is above about $0.6$ to $0.7\ \text{V}$ the transistor is on: current flows from collector to emitter.
The collector current is roughly $I_C = h_{FE} \, I_B$, where $h_{FE}$ is the current gain. When used as a switch, the base current is made large enough to switch it fully on (saturation).
A dark-activated switch
A potential divider has two resistors in series, and the base is connected to the midpoint. The position of the LDR decides whether the circuit responds to light or to darkness.
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With the LDR in the upper half (between the supply and the base), the fixed resistor is between the base and 0 V. In the dark the LDR's resistance rises, so it takes a larger share of the supply and the base voltage falls. The transistor switches off, so this arrangement gives a light-activated switch.
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To make the LED light in the dark, put the LDR in the lower half, between the base and 0 V, with the fixed resistor above it. In daylight the LDR has low resistance, so the base voltage is low and the LED is off. In the dark the LDR's resistance rises, it takes a larger share of the supply, and the base voltage rises above about $0.6\ \text{V}$, so the LED lights. Always check which resistor is across the base.
The rule to remember: the base voltage is the voltage across the resistor connected between the base and the 0 V rail. When that voltage exceeds about $0.6\ \text{V}$, the LED lights.
A transistor is used rather than connecting the LED directly to the divider because the divider can supply only a tiny current. The transistor lets the small base current switch the much larger LED current from the supply.
A transistor can act as an electronic switch. a) Name the three terminals of an n-p-n transistor. b) State which terminal is the input that controls the switch. c)…
A transistor can act as an electronic switch.
- a) Name the three terminals of an n-p-n transistor.
- b) State which terminal is the input that controls the switch.
- c) Describe what happens to the current between collector and emitter when the base voltage is below its threshold of about $0.6\ \text{V}$.
A transistor is used in a circuit to switch on an LED in a night-light when it becomes dark. A light-dependent resistor (LDR) and a fixed resistor form a potential…
A transistor is used in a circuit to switch on an LED in a night-light when it becomes dark. A light-dependent resistor (LDR) and a fixed resistor form a potential divider connected to the base.
- a) Explain how the voltage at the base changes as light level falls, if the LDR is in the lower half of the divider and its resistance rises in the dark.
- b) Describe how this turns the LED on.
- c) Why is a transistor used rather than connecting the LED directly to the divider?
An n-p-n transistor switches a relay coil of resistance $120\ \Omega$ from a $12\ \text{V}$ supply. The base is connected to a $5.0\ \text{V}$ source through a resistor…
An n-p-n transistor switches a relay coil of resistance $120\ \Omega$ from a $12\ \text{V}$ supply. The base is connected to a $5.0\ \text{V}$ source through a resistor $R_B$. The base-emitter voltage is $0.7\ \text{V}$ when conducting.
- a) Calculate the current through the relay coil when the transistor is fully on (ignore the collector-emitter voltage).
- b) The transistor's current gain is $h_{FE} = 100$. Calculate the minimum base current needed.
- c) Calculate the largest value of $R_B$ that will ensure full switching on.
- d) Explain why a diode is often connected across a relay coil.
Back to the start: the security light uses a light sensor to set the base voltage, and the transistor switches the lamp circuit.
Summary
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A semiconductor has few free carriers when pure. Resistance falls as temperature rises, the opposite of a metal.
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Doping adds impurities: n-type (group 5, majority electrons) or p-type (group 3, majority holes). Doped material is still neutral.
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A p-n junction forms a depletion layer with a barrier of about $0.7\ \text{V}$ in silicon. Forward bias narrows it and current flows; reverse bias widens it and almost none flows.
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A diode conducts one way only, so it can rectify a.c.
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An n-p-n transistor (base, collector, emitter) switches on when $V_{BE} \gtrsim 0.6\ \text{V}$; $I_C = h_{FE} I_B$.