# using a 5v relay with 12v

Discussion in 'Electronic Basics' started by Matt, Dec 5, 2004.

1. ### MattGuest

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!

-matt

2. ### Lord GarthGuest

Why not use a voltage regulator if you are going to run several 5 volt
relays
on a 12 volt supply. Use an LM7805 and a heat sink. If your total current
requirements are low, you may be able to use a zener diode and a limit
resistor.

Where is your 12 volts coming from? If it's a car, there are other
considerations.

3. ### Ken SmithGuest

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.

4. ### MattGuest

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?

Thanks!
-matt

5. ### Ian StirlingGuest

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

In inverse proportion to the coil resistance.

6. ### Spehro PefhanyGuest

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.

Best regards,

Best regards,
Spehro Pefhany

7. ### Robert MonsenGuest

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.

7V
+------+
| |
|< .-.
CTRL ---| | | 4.7k
|\ | |
| '-'
| |
o------|--------------------o-----\
| | | 12V
| | - Relay
.-. | || ^ Coil
10k| | o-----------||-------o-----/
| | | || |
'-' | V
| |/ C -
'----| |
|> |
| |
'--------------------'
GND

(created by AACircuit v1.28 beta 10/06/04 www.tech-chat.de)

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.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

8. ### MattGuest

I believe a cap across the relay will delay it turning off, not on, as the
charged cap will give voltage to the coil and leave it on.

9. ### Spehro PefhanyGuest

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

10. ### Terry GivenGuest

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 =
1/(2*pi*sqrt(L*C))

Cheers
Terry

11. ### MattGuest

Is it not really feasible to use an RC circuit to delay because of the low
resistance of the coil? sounds like it will take a large capacitance.

-matt

12. ### Ian StirlingGuest

It will.
But, it's probably easier than learning the proper way to do it.
(some sort of transistor/IC circuit)

13. ### Terry GivenGuest

ditto.

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.

Cheers
Terry

14. ### Terry PinnellGuest

No, .004F = 4,000 uF