Chapter 17: The Bipolar Junction Transistor

The emitter

The arrowhead is always on the emitter lead, in NPN and PNP symbols alike, where it marks the base-emitter junction. The collector, the other outer lead, never carries an arrow.

It is the lead with the arrowhead

The emitter is always the lead with the arrowhead. Symbols may be drawn either way round, and the manual’s own NPN and PNP drawings put the emitter on opposite sides, so position tells you nothing.

A diode

The arrow marks the base-emitter junction, a PN junction with much the same properties as a diode: it conducts one way once its forward bias voltage is exceeded. A capacitor would block DC altogether.

Away from the base, outwards

The NPN arrow points outwards, away from the base, the way conventional current leaves by the emitter. An arrow pointing in, towards the base, marks a PNP transistor.

A PNP transistor

An inward arrow means conventional current enters at the emitter, which marks a PNP transistor. In the NPN symbol the arrow points outwards, away from the base.

Which way conventional current flows

As on a diode symbol, the arrow shows the direction of conventional current through the base-emitter junction. Electrons, being negative, drift the opposite way, so the close option has it backwards.

P-type

The letters give the layers in order: N-type emitter, P-type base, N-type collector. The base is always of the opposite type to the two outer layers, so an N-type base would make it a PNP.

The semiconductor types of the three layers

N and P give the semiconductor type of each layer in turn, emitter, base and collector, so an NPN has a P-type base between two N-type layers. The letters do not label the voltages on its layers.

Both P-type

A PNP transistor has P-type emitter and collector layers with a thin N-type base sandwiched between them. Two N-type outer layers would make it an NPN transistor.

A PNP transistor

The base is of the opposite type to the emitter and collector, so an N-type base sits between two P-type layers: P-N-P. An NPN transistor has a P-type base.

The emitter and the collector

In an NPN transistor the emitter and the collector are both N-type, with the thin P-type base between them, so the two outer layers always match. The base is never the same type as the emitter.

The base, in the middle

The base is a thin layer, in reality far thinner than drawings show, sandwiched between the much thicker collector and emitter. The emitter is one of those two thick outer layers.

They are much thicker

The base is the thin middle layer, and the collector and emitter on either side of it are much thicker. The thin base is what lets most carriers from the emitter reach the collector.

They carry on into the collector

The base is so thin that most of these electrons never meet a hole; they cross the collector-base junction and form the collector current. Only the few that recombine make up the base current.

Free electrons

The NPN emitter is N-type, so its carriers are free electrons, attracted across the junction by the positive base; the few that fill holes there make up the base current. Holes cross in a PNP.

Reverse-biased, with a depletion layer

With the collector positive, its N-type material meets the P-type base in reverse bias, so a depletion layer forms; current crosses only because carriers arrive from the emitter. The base-emitter junction is the forward-biased one.

A boundary between two of its layers

A junction is where two layers meet, P-type against N-type, as at the boundary between collector and base; a transistor has two, collector-base and base-emitter. Its leads are never joined to one another inside.

Two

Its three layers meet at two boundaries, collector-base and base-emitter, and each is a PN junction. Three would count the layers or the leads, not the junctions; a single junction makes a diode.

Positive with respect to 0 V

An NPN collector has to stay positive relative to its emitter, so with the emitter on 0 V the collector needs a positive supply. A negative collector suits a PNP transistor instead.

The collector is more positive

An NPN collector is held positive relative to its emitter, and in normal operation its collector-base junction is reverse-biased. A more positive emitter is how a PNP transistor is connected.

The emitter

In an NPN transistor the base and the collector are both held positive relative to the emitter, so the emitter is the most negative of the three. In a PNP the emitter is the most positive.

Practically none

With base and emitter joined, VBE is zero, well below the 600 mV needed, so no base current flows and IC = β × 0 = 0: the transistor is shut off. Shorting them never turns it on.

Its base-emitter junction must be forward-biased

Base current, and with it collector current, flows only once the base-emitter junction is forward-biased beyond its forward bias voltage, about 600 mV for silicon. Reverse bias keeps the transistor shut off.

Its collector current with its base current

β = IC ÷ IB: the collector current is β times the small base current, typically 100 to 500 times. Collector current over emitter current is a different ratio, always just under 1.

100 to 500

Ordinary small-signal transistors typically have β between 100 and 500, though it is poorly controlled; 10 to 50 is far too low. A figure just under 1 is collector over emitter current, not β.

6 mA

IC = β × IB = 150 × 40 µA = 6 000 µA = 6 mA. Dividing the base current by β instead gives 0,27 µA, less than the base current itself, which cannot be right.

80 µA

IB = IC ÷ β = 20 mA ÷ 250 = 0,08 mA = 80 µA. Multiplying by β instead gives 5 A, a base current far larger than the collector current it is meant to control.

200

β = IC ÷ IB = 9 mA ÷ 0,045 mA = 200, well inside the usual range. Dividing the base current by the collector current gives 0,005, the ratio turned upside down.

8 mA

IC = β × IB = 220 × 37 µA = 8 140 µA = 8,14 mA, so 8 mA is nearest. Dividing by β instead gives 0,17 µA, less than the base current, which cannot be right.

300

β = IC ÷ IB = 8,7 mA ÷ 0,030 mA = 290, so 300 is nearest. Dividing 8,7 by 30 without first putting both currents in the same unit gives 0,29, hence the 0,3 option.

It doubles

IC = β × IB, so with β unchanged twice the base current gives twice the collector current, a rise of β × 35 µA rather than 35 µA. Staying the same would need β to halve.

3 mA

IC = β × IB holds for changes too: 200 × 15 µA = 3 000 µA = 3 mA, a change 200 times larger than the base current’s. Dividing by β instead gives 0,075 µA.

Its current gain, β

hFE is not exactly the same thing as β, but close enough, so data sheets and exam papers may give hFE where β, the current gain IC ÷ IB, is meant. It is not a voltage gain.

β varies widely, even between transistors from one batch

β is not well controlled in manufacture, so two transistors of one type, even from one batch, can differ widely, and a circuit relying on one exact figure would behave differently with each. Ageing is not the reason.

The two values may be quite different

The manual warns that β is poorly controlled in manufacture, so even transistors from one batch can have quite different values. Matching within 1% would need careful selection from many samples.

So that it still turns fully on with a low-β transistor

β varies widely from one transistor to another, so a switch is designed for the lowest likely β; generous base current then turns any sample fully on. Extra base current does not raise β itself.

One designed for low-power work

Small-signal transistors are the ordinary types designed for low-power work, such as amplifying weak signals, and their β is typically 100 to 500. Switching large currents calls for bigger power transistors.

Amplification

Because a small base current controls a collector current many times larger, a transistor can make small signals larger: amplification. Rectification, turning AC into DC, is a diode’s job.

The voltage between the base and the emitter

V marks a voltage, and B and E name the two terminals it is measured between, base and emitter; in a conducting silicon transistor it is about 600 mV. A current would be written with I.

The voltage between the collector and the emitter

V marks a voltage between the two terminals named, here collector and emitter; in the manual’s NPN figure it is the supply that makes the collector positive. The collector current is written IC.

It is their sum, IE = IB + IC

In the manual both the base current and the collector current flow to the emitter, which therefore carries the two together. Subtracting would mean the base current flowed backwards.

25 µA

The emitter carries both other currents, IE = IB + IC, so IB = 4,525 − 4,5 = 0,025 mA = 25 µA. Adding the two given currents gives 9,025 mA, more than either of them.

50 µA

The emitter current is the sum of the other two, so IB = IE − IC = 10,05 − 10 = 0,05 mA = 50 µA. Adding the two currents gives 20,05 mA, more than either.

Collector and base both negative

A PNP transistor works like an NPN with every polarity reversed: its base must be about 600 mV negative of its emitter, and its collector negative too. Both positive is the NPN arrangement.

Lower than +12 V

A PNP collector has to stay negative relative to its emitter, so with the emitter on +12 V the collector sits lower, its load returning towards 0 V. A collector above the emitter suits an NPN.

Practically none

A PNP transistor needs its base about 600 mV negative of its emitter before base current, and so collector current, can flow; with no voltage between them it is shut off, as an NPN would be.

40 µA

A PNP transistor obeys the same rule as an NPN, IC = β × IB, so IB = 12 mA ÷ 300 = 0,04 mA = 40 µA. Multiplying by β instead gives an impossible 3,6 A.

6 mA

As for an NPN, IC = β × IB = 260 × 22 µA = 5 720 µA = 5,72 mA, so 6 mA is nearest. Dividing by β instead gives 0,085 µA, less than the base current.

It hardly changes

A conducting PNP transistor holds VBE near −600 mV whatever its base or collector current; only the currents grow. Ten times the voltage is how a resistor would behave, not a junction.

No base current and no collector current flow

Shut off means the base-emitter voltage is too low for base current, so with IC = β × IB no collector current flows either and the transistor acts as an open switch. It is not damaged.

In the same way, but with every polarity reversed

The manual says a PNP transistor operates like an NPN with the opposite polarity: base current still turns it on, with IC = β × IB, but its base and collector go negative of the emitter.

Every supply and signal polarity must be reversed

A PNP transistor works like an NPN with the opposite polarity, so its collector and base must go negative of its emitter: every supply and signal polarity must be turned round. The two types are not interchangeable.

The transistor stays off, so the relay stays released

With 0 V at the input the base-emitter junction is not forward-biased, so no base current, no collector current and no relay current flow. It takes the +5 V input to switch the transistor on.

150

β must be at least IC ÷ IB = 45 mA ÷ 0,3 mA = 150. Multiplying the two currents, 45 × 0,3 = 13,5, is not how a ratio is found.

90 mA

IC = β × IB = 180 × 0,5 mA = 90 mA, the most collector current, and so load current, it can pass. Dividing by β instead gives 2,8 µA, less than the base current itself.

The collector current

The coil is wired between the +12 V supply and the collector, so it carries the collector current. The emitter current also includes the small base current, which never passes through the coil.

1,23 mA

The base sits about 0,6 V above the emitter, leaving 3,3 − 0,6 = 2,7 V across the resistor, so IB = 2,7 V ÷ 2,2 kΩ = 1,23 mA. Forgetting VBE gives 1,5 mA.

5,6 kΩ

RB = (5 − 0,6) V ÷ 0,8 mA = 5,5 kΩ, so 5,6 kΩ is nearest. Leaving out the base-emitter drop gives 5 V ÷ 0,8 mA = 6,25 kΩ, nearer the 6,8 kΩ option.

5,4 V

With the transistor on, its base sits about 0,6 V above the emitter, leaving 6 − 0,6 = 5,4 V across RB. 5,8 V would take germanium’s 0,2 V, and 6 V forgets VBE altogether.

600 Ω

R = V ÷ I = 12 V ÷ 0,02 A = 600 Ω. Dividing 12 by 20 without first turning the milliamperes into amperes gives 0,6 Ω, a thousand times too small.

An NPN transistor

The relay driver uses an NPN transistor with its emitter on 0 V, turned on when the +5 V input drives current into its base. A PNP would need its base taken below its emitter.

4,7 kΩ

In this notation the letter k stands for kilohms and also takes the place of the decimal comma, so 4k7 means 4,7 kΩ, which is 4 700 Ω. Reading it as 47 kΩ ignores the comma the k marks.

It halves

The voltage across the base resistor, the input less VBE, stays the same, so by Ohm’s law doubling the resistance halves the base current. A larger resistor cannot pass more current from the same voltage.

Destroy the transistor

When the transistor cuts the coil current, the coil’s inductance produces a large back EMF that can exceed what the transistor withstands and destroy it. Without a diode the relay actually releases faster, not slower.

To protect the transistor from the coil’s back EMF

When the transistor turns off, the coil’s inductance tries to keep the current flowing and produces back EMF; D1 conducts and gives that current a safe path round the coil. The contacts are a separate circuit.

When the transistor turns off and the coil current stops

An inductor opposes any change in its current, so cutting the current suddenly makes the coil produce a large voltage spike. At switch-on its opposing voltage can never exceed the supply, so it does no harm.

It is reverse-biased by the voltage across the coil

D1’s cathode goes to the +12 V end of the coil and its anode to the collector, so with the transistor on it is reverse-biased; only the back EMF at switch-off drives the collector above +12 V.

Its inductance

A coil is an inductor, and an inductor opposes any change in its current; when the transistor cuts the current, that inductance produces the back EMF. Resistance only sets the steady current.

It carries the relay current while the relay is held on

D1 is reverse-biased, carrying nothing, while the relay is held on; it conducts only for a moment at switch-off, taking the current that the back EMF drives, so the transistor survives. The relay current flows through the transistor.

Turning a small DC fan on and off

The manual says transistor switches like the relay driver are usually used for DC voltages and currents; a high-frequency signal such as 7 MHz is normally switched with diodes or relays instead.

Diodes or relays

High-frequency signals are usually switched by diodes, such as PIN diodes, or by relays, while transistor switches like the relay driver mostly handle DC. A transistor may still work the relay’s coil.

A pair of diodes or a small relay

The manual says high-frequency signals are usually switched with diodes or relays, while transistor switches like the relay driver are kept mostly for DC. A capacitor or a fuse switches nothing.

So the weak 5 V signal can control the relay’s larger current

The output supplies only about 1 mA at 5 V, but the relay needs 20 mA at 12 V; the small base current lets the transistor switch the relay’s current from 12 V. It makes no voltage itself.

It can supply only about 1 mA, too little for the relay

The control output can supply only 1 mA at 5 V, but the relay needs 20 mA at 12 V, so a transistor must switch it. A 5 V output is far too weak to burn out the coil.