For an in-depth study regarding how a relay works please read this article. 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.
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 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. 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 basic formula for calculating the base resistor of the transistor is given by the expression:. 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. 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.
This is already explained in the above discussions. 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.
Normally, the supply voltage for a operating a relay is dimensioned to ensure that the relay is pulled-in optimally. However, the required retaining voltage is typically much lower. This is usually not even half the pull-in voltage. As a result the majority of relays can work without problems even at this reduced voltage, but only when it is ensured that at the initial activation voltage adequately high for the pull-in. The circuit presented below may be ideal for relays specified to work with mA or lower, and at supply voltage below 25 V.
Secondly, relays with higher voltage rating could be used with lower supply ranges. The circuit can be seen wired to a supply voltage capable of holding the relay perfectly. During the time S1 is open, C1 gets charged via R2 upto the supply voltage. The moment S1 is presed, the T1 base gets connected to supply common through R1, so that it switches ON and drives the relay.
The positive terminal of C1 connects to the common ground through the switch S1. Considering that this capacitor initially had been charged to the supply voltage its -terminal at this point becomes negative. The voltage across the relay coil therefore reaches two times more than the supply voltage, and this pull in the relay. Switch S1 could be, certainly, be substituted with a any general purpose transistor which can be switched on or off as required.
If you have any circuit related query, you may interact through comments, I'll be most happy to help! Your email:. Please T1 in the last circuit be any pnp transistor or Again what is the rational for the two 1k resistor in the npn and pnp transistor circuit above. Yes it can be any pnp transistor for relay coils higher than ohms. The two 1K resistors ensure that the base of the transistor is never held floating, raher always connected to some potential, either positive or negative.
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.? Hai, what is the minimum voltage required to activate a normal 12volt relay?? I have some doubts. I connected two 12v relays in collector of one Will the 10k resistor fine as base resistor?
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,.
And which components should I adjust. If you give me the right values, it would be really appreciate it. Hi, If I need to use two relays with single transistor, Can I change instead of in the above formula? So finally I suppose to use K resistor with BC?
Correct me if I am wrong. Kindly help. If you have calculated everything correctly and getting 95K then it may be the right value. 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 ohm and ohms. If you calculate the transistor base resistor for a 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.
Relay Driver :. In a low power circuit or an output from a Microprocessor is very low. It is sufficent for a LED to glow but to drive a high load you will need a Relay Electromagnet Switch , and to give proper voltage or current to a relay you will need a relay driver. Many times one transistor with a resistance is enough to make a Relay Driver. In this type of circuit Transistor is use as current amplifier and Relay does two things a they isolate current flow of electron this is important because high load appliances runs at different voltage potential difference thus Relay protects your sensitive electronics parts.
It is a type of machinical Switch which is pulled by a electro-magnet so its resistance is very low and thus it can control large power appliances. Now days In market Relay Driver Module is available which is generally combination of relay and a transistor.
When doing this, obviously the supplies need to be reversed, but also the diode polarity needs to be reversed. NPN emitter follower relay circuit Whilst the common emitter relay circuit will be the most popular, it is sometimes useful to use a common collector or emitter follower configuration for the relay circuit. This relay circuit just replaces the emitter resistor with the relay coil. Again the diode is incorporated into the relay circuit to prevent damage from the back EMF induced at turn-off.
The base resistor is placed in the circuit to limit the base current, although in many instances this may not be required. Like the common emitter circuit, this one too can use a PNP transistor, but with the diode polarity and supply reversed.
In some circumstances higher levels of current gain may be required. This issue can be solved using a Darlington transistor. However be aware that the base emitter voltage drop is twice that of a single transistor, i. Whilst it is important to design the circuits to drive the relays correctly, there are also points to note about the circuits that are being switched by the relays as well.
This is particularly important for reed relays where the contacts are more prone to damage. One of the key areas of importance is the current experience once the contacts are closed. Even when driving what may be thought to be low current circuits, the in-rush current caused by capacitors used for decoupling, etc can result in huge current spikes are switch-over.
This can significantly reduce the life of the reed relay because the inrush current can exceed the rated maximum current by many times.
Even comparatively small capacitors can use current spikes on many amps and this can significantly reduce the lifetime of the relay contacts, especially those of reed relays. The fact can be reduced by balancing the amount of decoupling and selecting the minimum value consistent with applying good decoupling on the voltage rails or lines that are switched.
It is also possible to use small series resistors to reduce the surge. Here the voltage drop across the series resistor needs to be calculated and if any current is carried, this needs to be retained within he acceptable limits. There are many different circuits that can be used with relays.
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