hello all. I have a simple question. What makes a 5v relay rated at 5v? Is it limited by the voltage or really by the current? I have a bunch of 5 volt relays that'd I'd like to use with 12 volts. is it possible to add a small resister in line with the terminals that switch the relay and have it turn on with a 12 volt supply? They will be one for a while, so I do not want to risk them heating up too much.
on the other hand, relays are pretty cheap. does it make sense to just get some proper ones? Either way, I'm curious to know! Thanks!
Yes, the relay is rated at 5v to turn on, and 250v for whatever it is running. I have 12v to the coil now, it is getting a bit hot. I'm not sure if they are supposed to get hot since they are such low resistance.
Can you tell me more about the power saving trick? Do I put a low value resister in series w/ the coil and put a cap in parallel with this resister? what value do you suggest?
BTW - are there any simple RC circuits to delay the turn on of the relay for a couple seconds?
You will burn the relay out if you do that. The coil is dissipating perhaps 5 times as much power as it should (a bit less than the theoretical 5.76:1 because the wire is getting so hot). Put a series resistor that is 7/5 = 1.4 times the nominal resistance of the coil and rated at least at 1.4 times the power dissipation of the coil.
For example, a 360mW 5V miniature relay might have a coil resistance of 69 ohms. If you put a 0.5W 100 ohm resistor in series you should be okay.
Use a 2N6028 and a 2N5064, four resistors and a capactor (see the data sheet for the former for circuit examples). Reduce the series resistor a bit to account for the 2N5064 forward drop.
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Best regards, Spehro Pefhany
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The source impedance in this case is less than 300 ohms. If we assume RC (63% pull-in), that's C ~= 6800uF. If it's a typical small 360mW power relay, the required capacitance will be more like 50,000uF.
The relay contact life may be significantly reduced by not switching it cleanly.
Best regards, Spehro Pefhany
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There are two voltage ratings for relays. One if for the contacts and is almost always higher than 5V. The other is the voltage rating of the coil. I assume that the coil's rating is what you have.
The coil has some resistance. It is usually specified by the maker and does not vary much from unit to unit. If you use that resistance and an external resistor to work out the yusual voltage divider design, you will have 5V on the coil when it is operating.
The coil has some inductance. This inductance will cause the voltage on the coil to be 12V very breifly when you first apply power. 12V isn't high enough to break down the insulation and won't last long enough to cause any other trouble so don't worry about it.
BTW:
There is a fairly simple trick that can save you some power. The relay takes a higher current in the coil to pull the armature in than it takes to hold it in. If you know the holding current, you can design the resistor so that you get a bit more than that in the steady state. You then will need to put a capacitor across the resistor to give an initial pulse of current to pull the armature in.
They are designed to be run at the specified voltage. Much more may cause heating, and poorer contact life (they bang together too hard).
Measure the resistance of the coil. Multiply by 12/5. Now, subtract the resistance of the coil. So, if 500 ohms, that's 1200 ohms - 500 ohms = 700 ohms.
You can add a R/C delay by adding this resistor, then putting a capacitor across the relay. (with the correct polarity)
Time constant is about R*C, so for 2 seconds, you might try 2/500 of a farad, or .004F (400uF)
It takes a particular voltage, usually something like 80% of the nominal coil voltage, to close a relay. However, it takes much less than that to keep it closed. You can take advantage of this by using an arrangement whereby turning it on generates a larger voltage.
Quick turn on will result in a 14V pulse across the relay coil, which then relaxes to 7V as the cap charges back to ground through the coil.
Assuming a coil resistance of R, this will save (12^2 - 7^2)/R watts.
The circuit will require some time to recharge the cap, so it isn't good for very quick close-open-close cycles.
The danger is that production coils will vary, possibly causing issues with some units. However, I believe that most mfgrs publish minimum voltages above which the relay is guaranteed to stay closed. If you make sure your voltage is above that, then this is ok.
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That is also true, but at turn-on there is zero volts across the cap. because you now have a series resistor, the cap slowly charges up (time constant = R*C seconds)
At turn-off the cap starts fully charged (5V if you picked your external resistor right) and will ring with the coil inductance at f =
I always used to use a very crude scheme - an npn transistor to drive the relay, with a base pullup resistor split in two, a cap to ground at the split, and a base pull-down resistor. That way Rb is perhaps 10k, so for 2s you would be looking at a cap around the 100uF mark. By suitably choosing the voltage divider ratio this can be reduced further, but is certainly a lot better than the 100mF or so required for the brute force approach. Were I not so lazy I would draw an ASCII schematic.
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