# Running a 24v relay from mains voltage?

• posted

I need to power a relay from 120 vac such that its action is delayed for a few seconds after power-on. I realize that this means getting DC voltage (via diode) and an R-C combination.

Ideally, a relay with DC-rated coil would be most appropriate, but I've got a nice new relay with properly-rated contacts sitting on the shelf with a coil rated at 24 vac, 330 ohms, 71 mA.

How complex would it be to run this relay with mains voltage? Is it as simple as a diode followed by a resistor? Would this resistor be 120/0.071/2 = ~500 ohms and 2.5 watts?

And a fundamental question: can a relay with an AC coil function properly & reliably on DC?

Since I need to delay the relay's turn-on, I'll be adding a capacitor, anyway, whether I use this relay or a true DC-rated type. What should the value of the capacitor be to give me a 1 or 2 second delay with the 500 ohm resistor (or, if I've miscalculated that value, with the proper value resistor)?

Any observations, suggestions, corrections (I'm getting used to these!!) are welcome.

Thanks,

• posted

(snip)

Unless you are doing this for the educational benefit, I suggest you just purchase a time delay relay.

These may help:

• posted

(snip)

Unless you are doing this for the educational benefit, I suggest you just purchase a time delay relay.

These may help:

• posted

On Thu, 20 Nov 2003 21:51:34 -0800, DaveC wrote (in message ):

It occurred to me that I didn't calculate the *dropping* voltage of the resistor, as I should have, and didn't use mains *peak* voltage, as I think I also should have:

((170/2)-24)/0.071 = ~850 ohms and 5 watts. Since this resistor value is not common, I'll have to use a more-common value. These more-common values will result in an applied voltage to the relay's coil of:

750 ohms: 32 vac - too high 820 ohms: 27 vac - max voltage allowed by relay mfr. 910 ohms: 20 vac - too low 1K ohms: 14 vac - way too low

And yes, John, I learn from all my electronics work, so you can always assume it's all for educational benefit :-)

Observations?

Thanks,

• posted

Dave:

Here is a little circuit that will do what you want, delaying the relay from latching for a few seconds. Its designed to pump up to 24V with a load of

330 ohms after about 2.5 seconds. It wastes a couple of watts in heat.

Don't use this in the bathtub...

:)

• posted

Lets set aside the other issues, and just address the delay time. In rough terms, a resistor in series with a capacitor will charge the capacitor to about 63 % of full voltage in a time equal to R times C. That figure is called the RC time constant. When R is only 330 ohms, you'd need a large C to get a few seconds delay - about

10000 uF to reach 63 percent of full voltage across the cap at about 3.3 seconds.

The rest of this post goes into some light theory and ends with a description of a nice circuit you can experiment with and possibly use in your amp.

You could get the same time constant with a much bigger resistor, and a much smaller capacitor. Make the resistor 33000 ohms, and the cap 100 uf (.00010 farads): 33000 * .0001 = 3.3. The trouble with that occurs when you connect the relay across the cap. All the time the 33000 ohm resistor is trying to charge the cap, the 330 ohm relay is discharging it. This will keep the cap at a very low voltage.

However, suppose instead of placing the relay coil across the cap, we place a very high impedance (many times higher than 33K ohms) across it? It won't have much of an effect on the time constant - but it will be able to "see" the voltage rising on the cap. Now, suppose we prevent that high impedance from seeing the voltage at all, until it reaches (or exceeds) some specific level? Well, we can do that by putting a zener diode between the high impedance circuit and the cap. Lets select a 9 volt zener, put it in series with the 33K ohm/ 100 uf cap junction, and stuff the other end of the zener into the base of a darlington transistor. Complete the darlington (NPN) circuit by connecting the emitter to ground and the collector to the relay coil, while the other side of the relay coil goes to + 12 volts.

What happens? When power (12 volts DC) is applied across the series RC made of the 33K resistor and the 100 uF cap, the cap charges to ~7.5 volts in ~ 3.3 seconds. In another second or so, it will reach 9 volts, and the zener will conduct. In turn this will cause the darlington to conduct and the relay will transfer. If you replaced the 33K ohm resistor with a 50K ohm pot, and added a 4.7K resistor in series, you would have a nice adjustable time delay.

This is a nice experiment, and the parts are cheap. You need to select (or make) a darlington that can easily handle the power that the relay coil needs and an appropriate zener and relay voltage. For a turn on delay for your amp, precision in the timing is not important, so even if the timing varies with temperature that should not matter. There are other alternatives to the circuit, but for initial experimentation with RC timing, it is hard to beat and it could also be a nice fit for your amp.

• posted

On Fri, 21 Nov 2003 19:29:06 -0800, Robert Monsen wrote (in message ):

What kind of caps are these? Electrolytic? Poly? Please specify each so I can know what to acquire.

Thanks for the circuit!

• posted

[snip]

The 4.7uF cap is a big fat non-polar 250V cap. A mylar cap should work fine. The voltage rating is marked on the cap, or you specify it when you buy it. Don't use anything with a lower voltage rating!

The 4700uF cap is a big electrolytic cap (you probably can't get anything else that big.) The minus on the cap (it'll be marked clearly) goes to the ground (which is the bottom line) in the circuit. Since you are going 24V, you need something that'll handle a high enough voltage, like a 25V cap. You should go 50V just to be safe and to extend the life of the cap.

You can get both of these, and a sandstone 10 ohm 5W resistor mailorder from

If you are going to build this, mark HIGH VOLTAGE clearly somewhere on inside the case, make sure NOTHING, including the relay, can be touched by users, and encase the wires and leads in heatshrink tubing (you can also get that at goldmine, I believe.) You could also use a cheapo hot glue gun to cover all the wires in goop. That should insulate them. Just make sure the goop is actually non-conductive.

One last thing, I'd put a 1MEG 1/4 W resistor in parallel to the 4.7uF cap just to be safe. This will keep you from shocking yourself when the circuit has been off for 2 hours and you pick it up. If you put the switch right before the cap, it could conceivably be charged up to 165V, which would mean

64mJ of energy stored across the cap, enough to wake you up.

Regards, Bob Monsen

• posted

In article , snipped-for-privacy@BulkingPro.com mentioned...

Rather than use that monster 4.7 uF capacitor, it might be cheaper to use a small power transformer. About the only place I can remember seeing a cap that size is for a fan motor, and it would have to be paralleled to get that much capacitance. And it was about the same price as a power transformer, maybe \$5 or so.

Among this list, the 7841 looks like it's a good choice; it's cheaper than that monster capacitor, and safer because it isolates you from the AC line voltage. When it is rectified and filtered, it will be about 24VDC. But your relay will take less than 24V if it's DC instead of AC.

• posted

Where are you getting these numbers? Did you measure 71mA or the 330 DC ohms? In either case the AC relay reactance is usually 10x the resistance if the relay is of any quality, and therefore all of your component planning will not come close. Measure the DC ohms Rdc and the total ac-RMS current, Irms. Then total reactance X=24VAC/Irms and inductive reactance XL= sqrt(X^2-Rdc^2). Then you solve for series R required at 120VAC operation, Rs, as |(Rdc+jXL)/(Rs+Rdc+jXL)|=24/120 etc...

• posted

A transformer would also work here. However, you can get two 2.2uF 250 caps for a buck at

and put them in parallel. That would also work (albeit without the isolation)

Actually, I was thinking this was a DC relay, but the OP clearly stated its a 24V AC relay, so... I'm not sure about the delays now. Aren't AC relays just DC relays with a diode? If so, the circuit should still work as advertised, since he measured DC current/resistance, and the circuit will supply DC current. If they are something else, all bets are off.

Regards, Bob Monsen

• posted

Many film caps in that range are not that big. I have several 3.3 uF caps rated for 250 or 350V, which are only about 1.5" by 0.5" by 1", and I saw

2.2 uF 250V caps that were about 1" by 1/4" by 1/2".
• posted

An AC relay is a DC relay with a shorted turn so that the current induced in the shorted turn (remember, this iron core is like a transformer when operated on AC), which keeps the magnetic flux around for a while while the AC cycle changes direction. The plunger also weighs more in most cases, so that it takes more time to accerate ftom the rest of the iron, helping prevent the relay from buzzing. When operated from DC, they often turn on at much lower voltages.

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On Sun, 23 Nov 2003 14:45:30 -0800, Robert Monsen wrote (in message ):

None of these numbers were measured; they come from the mfr's spec sheet:

(BTW, I goofed on the coil resistance; it's 303, not 330 ohms.)

According to Fred (posted later in this thread), an AC coil has one coil shorted.

Bob, how does this change your circuit values, if at all?

Thanks,

• posted

In article , snipped-for-privacy@privacy.net mentioned...

Actually, more like a half turn. That's that D shaped piece of copper that covers half of the end of the relay's pole.

• posted

According to your OMRON tiff, the thing 'must operate' at 75% of the rated voltage of 24V, so thats 18VAC. According to a different poster (Da Man) AC relays will operate at even lower values when operated with DC voltage.

Post the DC voltage that operates the thing, and I'll tell you what values to use to get it to operate after 2 seconds.

Regards, Bob Monsen

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On Mon, 24 Nov 2003 12:09:41 -0800, Robert Monsen wrote (in message ):

Bob, Since it will work within a range of voltages, shouldn't we aim for one that minimizes the size of components? Which would be max voltages, allowing for smallest dropping components. (But which, I am reminded, means larger components -- ie, working voltages -- for the others.)

Thanks,

• posted

If you really want a 2 second pause, you'll have to experiment with the values. Since I don't know what voltage the relay will close at, I can't choose the values for you.

If you are just looking for a delay near 2 seconds, then I'd guess (assuming it'll close at 15VDC) 3.3uF, 10R, 4700uF, which are standard values. I'd also put in a fuse for fire safety in case the 3.3uF cap fails.

I've updated the web page to reflect this and your new component values.

Regards, Bob Monsen

• posted

On Tue, 25 Nov 2003 12:46:00 -0800, Robert Monsen wrote (in message ):

Thanks, Bob.

What's R1 (303 ohm) for? Is that the relay's coil? There was no R1 in previous schematic...

Thanks,

• posted

Yes, its the relay coil's resistance. I used it to model the time delay.

Regards, Bob Monsen

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