Connect with us

using a 5v relay with 12v

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

Scroll to continue with content
  1. Matt

    Matt Guest

    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 Garth

    Lord Garth Guest


    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 Smith

    Ken Smith Guest

    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. Matt

    Matt Guest

    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 Stirling

    Ian Stirling Guest

    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)

    In inverse proportion to the coil resistance.
     
  6. 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. 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. Matt

    Matt Guest


    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. 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 Given

    Terry Given Guest

    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. Matt

    Matt Guest


    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 Stirling

    Ian Stirling Guest

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

    Terry Given Guest

    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. No, .004F = 4,000 uF
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-