actuatorcurrentrelay
I need to design a circuit to drive a relay and I have some doubts.
Unfortunately, I don't have current specification neither a relay model. My doubt is about functioning, I'm afraid to fry the relay or the transistor.
simulate this circuit – Schematic created using CircuitLab
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This question has probably already been answered 100's of times. But here is maybe the 201'st time.
Use a circuit that looks like this:
In the schematic shown below, the entire secondary of the transformer has been used, two rectifiers have been eliminated, and C1 is used to smooth the full-wave rectified 16.5 volts.
U1 is a 12 volt linear regulator which drops the 16.5 VDC down to 12VDC for the relay and whatever else the 12 volts is being used for, R6 representing both the relay coil and the remaining parallel load, if there is one.
U2 is a voltage comparator with an internal 400 millivolt reference. The voltage divider R3 R4 is used to set the voltage at the non-inverting input of U2 to slightly over 400mV when the output of U1 is at 12 volts. That will force the output of U2 high, turning Q1 ON and energizing the relay coil.
If, for some reason, U1 OUT should fall below 12 volts, the voltage at U2+ will fall below 400mV, U2's ouput will go low, and the relay will become de-energized, opening its normally-open contacts.
U2 is supplied with about 6.5 millivolts of internal hysteresis and if more is needed it can be applied externally as shown on page 10 of the data sheet.
Best Answer
You only need a resistor in series with your relay coil if your are applying a voltage that is different than what your relay coil is rated for. At the rated voltage, the coil resistance will be sufficient to limit the current to the levels necessary to energize the coil.
However, if you apply a larger voltage than what the coil is rated for, the coil resistance is insufficient to limit the current to safe levels. In that case you must add enough series resistance so that the rated voltage appears across the coil and the rated current runs through the coil, even though you are applying a larger voltage.
This is easy enough to calculate using V=IR along with your applied voltage, and two of the following: the coil's rated voltage, current, and resistance from the datasheet
If you are applying a voltage less than the relay's rated coil voltage, you do not need a resistor but the relay will also not switch since there is not enough current.