In this article we will comprehensively study a transistor relay driver circuit and learn how to design its configuration by calculating the parameters through formulas.
Importance of Relay
Relays are one of the most important components in electronic circuits. Especially in circuits where high power transfer or mains AC load switching is involved, relays play the major role in implementing the operations.
Here we will learn how to correctly operate a relay using a transistor and apply the design in electronic system for switching a connected load without issues.
A relay, as we all know is an electromechanical device which is used in the form of a switch.
It is responsible for switching an external load connected to its contacts in response to a relatively smaller electrical power applied across an associated coil.
Basically the coil is wound over an iron core, when a small DC is applied to the coil, it energizes and behaves like an electromagnet.
A spring loaded contact mechanism placed at a close proximity to the coil immediately responds and gets attracted toward the energized coil electromagnet force. In the course the contact connects one of its pair together and disconnects an complementary pair associated with it.
The reverse happens when the DC is switched OFF to the coil and the contacts return to its original position, connecting the previous set of complementary contacts and the cycle may be repeated as many times as possible.
An electronic circuit will normally need a relay driver using a transistor circuit stage in order to converter it’s low power DC switching output into a high power mains AC switching output.
However the low level signals from an electronic which may be derived from an IC stage or a low current transistor stage may be be pretty incapable of driving a relay directly. Because, a relay requires relatively higher DC currents which may be normally not available from an IC source or a low current transistor stage.
In order to overcome the above issue, a relay control stage becomes imperative for all electronic circuits which need this service.
A relay driver is nothing but an additional transistor stage attached with the relay which needs to be operated. The transistor is typically and solely employed for operating the relay in response to the commands received from the preceding control stage.
Circuit Diagram
Referring to the above circuit diagram we see that the configuration only involves a transistor, a base resistor and the relay with a flyback diode.
However there are a few complexities that need to be settled before the design could be used for the required functions:
Since the base drive voltage to transistor is the major source for controlling the relay operations, it needs to be perfectly calculated for optimal results.
The base resistor value id directly proportional to the current across the collector/emitter leads of the transistor or in other words, the relay coil current, which is the collector load of the transistor, becomes one of the main factors, and directly influences the value of the base resistor of the transistor.
Calculation Formula
The basic formula for calculating the base resistor of the transistor is given by the expression:
R = (Us - 0.6)Hfe/Relay Coil Current,
Where R = base resistor of the transistor,
Us = Source or the trigger voltage to the base resistor,
Hfe = Forward current gain of the transistor,
The last expression which is the “relay current” may be found out by solving the following Ohm’s law:
I = Us/R, where I is the required relay current, Us is the supply voltage to the relay.
Practical Application
The relay coil resistance can be easily identified by using a multimeter.
Us will also be a known parameter.
Suppose the supply Us is = 12 V, the coil resistance is 400 Ohms, then
Relay current I = 12/400 = 0.03 or 30 mA.
Also the Hfe of any standard low signal transistor may be assumed to be around 150.
Applying the above values in the actual equation we get,
R = (Ub - 0.6) × Hfe ÷ Relay Current
R = (12 – 0.6)150/0.03
= 57,000 Ohms or 57 K, the closest value being 56 K.
The diode connected across the relay coil though is no way related with the above calculation, it still cannot be ignored.
The diode makes sure that the reverse EMF generated from the relay coil is shorted through it, and not dumped into the transistor. Without this diode, the back EMF would try to find a path through the collector emitter of the transistor and in the course damage the transistor permanently, within seconds.
Relay driver Circuit using PNP BJT
A transistor works best as a switch when it is connected with a common emitter configuration, meaning the emitter of the BJT must be always connected directly with "ground" line. Here the "ground" refers to the negative line for an NPN and the positive line for a PNP BJT.
If an NPN is used in the circuit, the load must be connected with the collector, which will allow it to be switched ON/OFF by switching its negative line ON/OFF. This is already explained in the above discussions.
If you wish to switch the positive line ON/OFF, in that case you will have to use a PNP BJT for driving the relay. Here the relay may be connected across the negative line of the supply and the collector of the PNP. Please see the figure below for the exact configuration.
However a PNP will need a negative trigger at its base for the triggering, so in case you wish to implement the system with a positive trigger then you may have to use a combination of both NPN and PNP BJTs as shown in the following figure:
If you have any specific query regarding the above concept, please feel free to express them through the comments for getting quick replies.
ketan says
Ketan Dhruv
Dear sir,
Which parameters should I have to take in account for same family but different functions digital ICS matching?
Swag says
Ketan, sorry I could not understand your question.
ketan says
Hello sir,
I am ketan and my question is:
I want to use 4013B and 74f74 for my different projects,as a trigger of relay driver circuits so which parameters of d flip-flop ICs should I take into account? for calculate output current of the ICs to be able to change the state of driver transistor to toggle as switch?
Swag says
Hello Ketan, the calculations are already explained in the above article, you can apply them in the following example applications
https://www.homemade-circuits.com/make-this-easiest-flip-flop-circuit/
https://www.homemade-circuits.com/build-these-simple-flip-flop-circuits/
Hamid Hussain says
Thanks for your explanation sir if the transistor base is connected to uncalculated base resistor what will be happen and is there any transistor driver to boost up current what’s should the main concern about choosing a transistor if I have a requirement is 30 to 100ma at collector of transistor what are main points should I noted in the datasheet and I have used a1015 PNP in the circuit but in the tsop 1738 its not working but bc557 a same PNP transistor is driving the led whats the reasonss
Swag says
Hi Hamid, you can connect by approximately judging the resistor values in circuits, it is not critical unless the value is referenced to some other dimension.
In relay circuits I normally use a 10K resistor as the transistor base resistor for all relays whose coil resistance is between 200 ohm and 400 ohms. If you calculate the transistor base resistor for a 400 ohm relay load, you will find it to be 56K, but I use 10k which does not make any difference expect a some mA more dissipation or wastage by the transistor and the relay.
Choosing transistor is also not so crucial, below 50mA load you can use BC547, and at 100mA and above you can choose 2N2222.
You have to basically consider the collector/emitter voltage and current rating of the transistor from its datasheet
Sriram Kp says
Hi, will the above formula appilcable for PNP like BC557B and S8550?
Because when I implemented the above formulae for BC557B with hFE=250, the result am getting is 95000 (95K). If the formula is not applicable for PNP means how to calculate the base resistor value for PNP?
Kindly help.
Swagatam says
Hi, the above formula is applicable for both NPN and PNP. the load current must be correctly estimated for getting the correct base resistor value. If you have calculated everything correctly and getting 95K then it may be the right value
Sriram Kp says
Hi, If I need to use two relays with single transistor, Can I change 800 instead of 400 in the above formula? So finally I suppose to use 110K resistor with BC547? Correct me if I am wrong.
Thanks
Swagatam says
Hi, since the two relays are in parallel will make the the result as 200 ohms not 800.
so please use 200 in the formula.
fahad alsharari says
Hi sir.
please which relay should I use to get the right temp. between 45 and 50 deg. in this circuit ?
And which components should I adjust.
If you give me the right values, it would be really appreciate it.
Swagatam says
Hi Fahad, which circuit are you referring to?
a relay can not be responsible for controlling temperature, it will only switch the load ON/OFF….the actual control would be done by the sensor and the connected circuit which will switch the relay ON/OFF according to the temperature levels.
SivaraJ P says
Dear sir, I need a dry run preventer cicuit for my open well, if the water goes down near to the submesible water bump the circuit will trip the motor if we want to run the motor we need to release the relay trip manually and the relay run through the no volt coil on the starter and also the circuit will not use of ic, please help me,
Swagatam says
Dear Sivaraj, I have one related circuit posted here, see whether it helps or not:
https://homemade-circuits.com/2013/07/underground-water-pump-motor-dry-run.html
Sriram Kp says
Hai, what is the minimum voltage required to activate a normal 12volt relay??
suppose If I give 10 or 11 volt means, will the relay get activate??
Swagatam says
around 11v to activate and 8V to deactivate.
Sriram Kp says
Thanks. I have some doubts.
1. suppose for a BC547 relay driver, how to activate the relay after 5 sec of switched ON the circuit?
2. and for a BC557 relay driver, how to activate the relay after 5 sec of switched ON the circuit?
3. I connected two 12v relays in collector of one 8550. Will the 10k resistor fine as base resistor?
Swagatam says
connect a 1000uF capacitor across base/ground and a 1N4007 between emitter and ground for NPN….and for PNP the capacitor should be across base and positive and the diode between positive and emitter.
yes 10k would be OK, although it can be solved using the formula as suggested above
MUJAHID SHAH says
I went through the post suggested by you so I got the formula to determine the value of a resistor to the base of a transistor thank you
can we use two transistors parallel to handle higher current drawing load.?
Swagatam says
yes we can use two transistors in parallel, but make sure to connect a small value resistor in series with the emitters of the transistors
Anish says
Sir, Can u suggest a simulator software which is able to do some testing, thanks.
Swagatam says
sorry, I have never used a simulator software, so can't suggest about them.
Anonymous says
Sir, Can you suggest a open end (freeware) software for simulation of Analog and Digital circuit
Swagatam says
Freewares are almost impossible to find, I don't think there's any good one online.
jay vyas says
Hi!!
if the base of the transistor is connected to an op Amp LM358 what would b its trigger voltage
Swagatam says
Hi,
Connect a proper resistor at the base of the transistor as per the given formula, voltage will get adjusted accordingly….
savyasachi says
Hi, Which is the software shown in this video ?.?
Anonymous says
Hi,
In the circuit shown in the video, what can we use to drop down 220V to 12V sufficient enough to supply the base of the transistor instead of 220-12V transformer.
Thanks.
Anonymous says
Is it possible to put 3 relays in 1 transistor?
Anonymous says
I'd like to use 3 relays at the same time, what transistor shall I use instead of BC557B?
Thanks…