e011068.PDF

BASIC CIRCUITS

Relay Driver

Operation and design

By Karel Walraven

In many circuits the switching action is performed by a relay, which in turn

activates an external load. This part of the circuit is rarely covered in the

accompanying text, but there is more than meets the eye if you want to

design a relay driver properly.

We should really look at the relay driver as a

type of interface. After all, it forms the link

between the control circuitry or PC and the

outside world. The relay is used by the con-

trol circuitry to switch external appliances,

lamps, pumps, sirens, etc on and off.

Because they’re used so often, the relay

driver is as common as the integrated volt-

age regulator, an LED driver or a power-up

network. The description usually skims over

the operation of these basic circuits. We feel

we should look at the relay driver circuit in

more detail.

adapter should this be required.

Deriving this from a 6 V adapter

would be much trickier.

The manufacturer’s datasheet tells

us that the resistance of the coil of

the 12 V relay is 330 Ω (alternatively,

we could have found the resistance

using a multimeter). When we use

this 12 V relay, the current through it

will be 12 V / 330

Ω

= 37 mA.

Transistor

When the operating voltage of the

relay is 12 V, the transistor should be

able to cope with at least this volt-

age across the collector-emitter junc-

tion. To be on the safe side, we’ll

choose a value that is half as much

Supply voltage: Has to be equal to

the relay coil voltage and capable of

supplying sufficient current.

Just ﬁve components

Apart from the bypass capacitor, the circuit

for the relay driver contains only four or five

components, as shown in Figure 1 . It’s all

very simple really, but all components need

to be chosen according to their function. Next

we’ll look in detail at each of the components.

Relay: Consists of a coil and

switching contacts. The coil

voltage is usually 6, 12 or 24 volts.

The switching contacts are

usually rated at 50 or 240 volts.

Suppressor diode:

Prevents damage to the

transistor caused by

back-emf surges when

the relay is switched off.

Relay

First we should determine what load we’re

switching; for example a 100 W lamp operat-

ing at 240 V AC. A suitable relay would be

the E-Card relay (made by Siemens) which

we’ve often used before. This can switch

500 VA at a maximum current of 4 A and a

maximum voltage of 250 V AC.

But note that you can’t just multiply the max-

imum current by the maximum voltage

(4 A x 250 V = 1000 VA), since the manufac-

turer rates the relay at a lower power.

This relay is available for many operating

voltages: 5, 6, 12, 24, 48 or 60 V. The 12 V

relay is often used, since the required voltage

can be obtained from an inexpensive mains

adapter. It is also very simple to derive a sec-

ondary voltage of 5 V from the 12 V mains

Pull-up resistor:

Sometimes required

to ensure sufficient

base current.

Transistor: Energizes the

relay.Allows a large current to

flow from C to E when a small

current flows from B to E.

Base resistor: Prevents

excessive base current

(>5 mA).

005124 - 11

Figure 1. Circuit of the relay driver. The values chosen for the components depend

on the size of the load to be switched.

Elektor Electronics

1/2001

BASIC CIRCUITS

Table 1

Transistor

VCE0 (V)

Ic (mA)

Hfe (times)

BC546

100

110-800

BC547A

100

110-220

BC547B

100

200-450

BC548C

100

400-800

BC517

500

30000

BD139

100

1500

40-250

Table 1. Some characteristics of suitable

transistors

Figure 2. The relay driver is easily built on a UPBS-1 board.

Table 2

Logic family

again, which is at least 18 V.

The transistor also has to be capable

of switching the required current; in

our example that is 37 mA. In this

case we’ll use a safety margin of

100%, so choose a transistor which

can switch at least 74 mA.

In this example it makes sense to

use the smallest and most commonly

used transistor: the BC547. This can

switch 45 V and 100 mA.

resistor is calculated as follows:

Determine how much current is

required for the logic family used. If

we used LS logic the value for our

example would be

0.34 mA – 0.4 mA = –0.06 mA. The

negative result means that there is

enough current, so we don’t need a

pull-up resistor in this case. Had we

used a 6 V relay with an 80

Sink (mA)

Source (mA)

TTL

0.4

4000 (CMOS)

0.5

HC/HCT

Table 2. The source- and sink-currents of

several well known logic families

coil we

would need a base current of

75 mA / 110 = 0.68 mA. The differ-

ence of 0.68 – 0.4 = 0.28 mA would

then have to be provided by the pull-

up resistor. From our earlier rule of

thumb we know that the voltage

across the pull-up resistor is

5 V – 3 V = 2 V. This gives a resis-

tance of 2 V / 0.28 mA = 7140 Ω.

Again we choose the nearest value

in the E12 series, which is 6k8.

We just have to check that the dri-

ver IC can pull the base and pull-up

resistor to ground. The current ﬂow

would then be 5 V / 6800 Ω =

0.735 mA. This value has to be less

than the given sink current. In our

example for the LS-family that is the

case. But note that we couldn’t have

used a device from the 4000 series!

In this case we would have to

choose a transistor for T1 with a

higher gain, as that would reduce

the required base current.

Ω

that at the point when the transistor is

switched off the coil becomes a generator

and will supply a voltage with a reversed

polarity. That causes a voltage to appear at

the collector that is equal to the supply voltage

plus the voltage that is induced across the

coil. A risky business really, but the diode lim-

its the induced voltage to 0.7 V, which pre-

vents the transistor from being damaged.

Base resistor (R2)

Table 1 shows the most important

characteristics of several popular

transistors. The BC547 is made avail-

able in two versions; the A-type has

a gain (H fe ) of 110-220 and the B-

type 200-450. To be certain that the

circuit will work under all conditions

we’ll use the lowest gain (110) when

calculating the value of the base

resistor (R2). The relay current goes

into the collector and is 37 mA in our

example. The gain of the transistor is

110, so we’ll need a base current of

37 mA / 110 = 0.34 mA. As a rule of

thumb we’ll assume that the output

that drives the transistor can deliver

at least 3 V. The voltage across the

base resistor is then 3 V minus the

base-emitter voltage (0.65 V). The

value for the resistor is therefore

(3 V – 0.65 V) / 0.34 mA = 6910

Construction

Usually the relay driver will be part of another

circuit and will not be built independently.

But in the case when an existing circuit has

to be extended with a switched output, it

should be fairly easy to mount the five com-

ponents of the relay driver onto a piece of Ver-

oboard. We’ve ﬁgured out how to mount the

circuit onto a Universal Prototyping Board

size 1 (UPBS-1). As you can see from Figure 2

it went very well and you’ll be pleased to

know that the (unstuffed) UPBS-1 may be

obtained through our Readers Services.

We’ll round this down to the nearest

value resistor in the E12 series,

which is 6k8.

Ω

Flyback diode (D1)

The function of the ﬂyback (or ‘sup-

pressor’) diode is to protect the tran-

sistor. The relay coil has the property

(like all inductors) that it resists a

change in the current flow through

it. So when a voltage is applied

across a coil it will take a while for

the current to ﬂow, but the reverse is

also true: when the voltage across a

coil is suddenly removed the coil will

attempt to keep the current ﬂowing

at all costs. In practice the means

Conclusion

Pull-up resistor (R1)

T able 2 shows the source and sink

current capabilities of several popular

logic families. When the source cur-

rent is less than the required base cur-

rent, the difference has to be supplied

by a pull-up resistor (R1). As you can

see in Figure 1 it’s not connected to

the 12 V relay supply, but to the 5 V

logic supply! The value of the pull-up

These explanations and instructions make it

fairly simple for anyone to design a relay dri-

ver module to their own specification. We

would just like to point out that all calcula-

tions include a reasonable safety margin. The

ICs can usually provide a higher current, so

you don’t have to worry that something will

go up in smoke when the calculations have

been a bit tight.

(005124-1)

1/2001

Elektor Electronics

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