Driving Electromagnetic Relays
DISCLAIMER
This
information is provided as a guide only; no responsibility is taken for any
omissions, errors, inaccuracies or any damages arising directly or indirectly
from the use of this information.
Why use a Relay?
There are situations where you may want to control switching a high power load
or an AC load from a low voltage or low current circuit (for example from a
microcontroller or a PC parallel port); in this situation it is hard to find a
simpler or cheaper solution than a relay. But what if your output can’t provide
enough power to drive the relay? This short article aims to give you enough
information to design your own simple transistor drive relay circuit to suit
almost any situation. You should however be aware that they are a
mechanical device and, as such, have a limit to how fast they can switch off
and on. You aren’t going to get 22,000 switch cycles per second out of a normal
electromagnetic relay.
Relay Basics
An electromagnetic relay consists of a small electromagnet which pulls the
contacts of a switch open or closed. The switch contacts are normally
electrically isolated from the electromagnet giving the controlling circuit a
reasonable amount of protection from the switched circuit.
When selecting a relay there are three basic considerations:
-
The switch contact load rating – how much current the relay can safely switch
(don’t try to control a 240v, 2400watt fan heater with a relay rated for 1 amp,
unless you really like sparks and flames)
-
The relay coil voltage – this is the minimum voltage required to overcome the
resistance of the coil to pull in the switch contacts
-
The relay coil resistance – this can be used to calculate the current required
to pull in and hold the switch contacts
Hopefully, these three things are labeled on the relay, but at the very least
the coil voltage and maximum switch load current should be visible. If you only
know the coil voltage, you can always measure the coil resistance with an
ohmmeter (or multimeter) and use the coil voltage with the measured resistance
to determine the coil current.
Selecting a Transistor
The first thing to determine when selecting a transistor is whether you want the
relay coil turned on when your control input is high (positive voltage) or low
(negative or zero voltage).
If you want the relay coil turned on when your control input is high (also
known as positive logic) it is easiest to use a NPN transistor. If you want the
relay coil turned on when your control input is low (also known as negative
logic) it is easiest to use a PNP transistor. See Figure 1 below.
Having decided which transistor type to use (NPN or PNP) the next thing to do is
to determine the collector-emitter current that will be required, which just
happens to be the coil current of our relay. If your relay has this detail
written on it, use the value they give you, otherwise you can calculate it with
Ohm’s law:
Coil Current = Coil Voltage / Coil Resistance
Don’t forget, Ohm’s law works in Volts, Ohms and Amps. Not millivolts, milliamps
or kiloohms. It is a common mistake to forget to convert to or from these units
and end up with a completely incorrect figure.
Our selected transistor must be able to “pass” at least this much current and
preferably a bit more to allow for variation in coil resistance (and hence
current) due to temperature or manufacture variations. This is known as the
collector current and is usually noted in the transistors specifications as
“Ic”.
Since we should only ever have our transistor in the “off” state or the “fully
on” (saturated) state our transistor should actually never need to dissipate
much power and so should never heat up. This is due to virtually no current
passing in the off state (so power, which is current times voltage, is near
zero) and the transistor having near-zero voltage across it in the saturated
state (again, power is near zero). This means we can virtually ignore the power
rating of our transistor as long as we ensure it is either off or saturated,
which brings the next question; how do we ensure our transistor is saturated
when on?
To ensure our transistor is saturated, we need to ensure that the current being
supplied to the base of our transistor is greater than the collector current
(coil current) divided by the transistor’s minimum DC current gain which is
normally noted in the transistor’s specification as “hfe” or “min hfe”. This
figure is used to calculate the value of R1 (Fig. 1) using ohm’s law and the
highest voltage you are going to put into your control line.
For example:
We want to drive the following relay from a fictional positive logic CMOS output
which has an output voltage of 4.7v when on and a maximum output current of 5mA
(0.005A);
Relay coil Voltage: 12v
Relay coil Resistance: 400ohms
Relay coil Current = 0.03A or 30 Milliamps (= coil Voltage / coil Resistance)
We’ll use positive logic and so a NPN transistor will be used. Scrounging
through a few datasheets, we find a suitable transistor;
bc548 –
Ic: 100mA - we only need 30mA, so this is plenty.
hfe min: 125
To ensure the transistor is saturated, we need to supply at least 0.03A / 125 =
0.00024A. It is generally a good idea to supply double that figure (0.00048A)
to be on the safe side. Note that since our fictional CMOS output can provide
up to 0.005A we can easily achieve this. If our output could not provide such a
current, we would need to look for a transistor with a higher hfe.
Now we need to calculate the value of R1 using ohms law:
Given our maximum output voltage of 4.7volts and our required current of
0.00048A we get:
R1 = 4.7 / 0.00048 = 9791.66666 ohms
Or 9.791k which is close enough to the preferred value of 10k, especially since
I’m going to use nice, cheap 20% tolerance resistors.
Figure
2 shows the example circuit with the components we have selected substituted
in. This might be a good time to introduce the protection diode and the
pulldown resistor (R2).
The protection diode is there to protect the transistor from reverse voltage
spikes from the relay coil when it transitions between on and off. The relay
coil is, after all, an inductor. As the magnetic field collapses at the moment
it is turned off, it induces a current in the coil which can result in a
reverse-polarity spike that will not make the transistor happy. The diode is
there to short out that spike (which is why it is installed “backwards” – we
only want to short out that negative spike, not our relay driving current) and
prevent it from reaching the transistor. Selection of the protection diode is
not critical except that the reverse-breakdown voltage must be higher than the
supply voltage for the relay. The breakdown voltage is the voltage at which the
diode begins to conduct, even in the wrong direction, and the supply voltage
for the relay is usually the same as the coil voltage. For 12v or lower, a
1n4001 or 1n4004 should be fine.
The pulldown resistor is required when your control output is an open collector
output. Open collector outputs don’t set the output voltage to a specific level
when the output is not on. Instead, they let the output level “float” as if it
is disconnected. In this situation the base (control) of the transistor can
pick up stray electrical noise which can be quite a problem (overheating
transistors, mysteriously chattering of the relay and unexpected appearances of
avocados!). The pulldown resistor’s job is to ensure that the base is pulled
down to ground when no signal is present so that this doesn’t happen. The
actual value of the resistor isn’t critical but it has to be high enough to
ensure the transistor still saturates when required. As a rule of thumb, R2
should be about 10 times R1, so in our case R2 would be 100k. Note that if your
control output is not open collector, R2 can be omitted completely.
In the case of the negative logic circuit you should note that the pullup
resistor is connected to Vcc, NOT the relay supply voltage. Vcc should be about
the same voltage as the maximum voltage of your control output and is often
tied to the supply of your control circuit (e.g. if your output is from a CMOS
chip, Vcc would be tied to the positive supply of that CMOS chip).
Well, I think that’s about it for now. Have fun, and don’t blow up your computer
(or your house).