Connect with us

12 VDC Relays

Discussion in 'Electronic Basics' started by Conan Kelly, Nov 27, 2006.

Scroll to continue with content
  1. Conan Kelly

    Conan Kelly Guest

    Hello all,

    My background is not in electronics. It is in car stereo & security. So please dumb-it-down for me.

    What is the minimum voltage required to energize the coil of a standard 12 VDC relay? How can I wire 2 relays with resistors so
    that the output line will be switched off when the voltage drops below 9.6 Volts?

    I think that I might have a diagram on how to wire 2 relays and a momentary push-button switch (and maybe diodes) so that each push
    of the button toggles the output line between OFF and ON. I was thinking that there is probably some similar way of wiring 2 relays
    with resistors to accomplish what I'm trying to do.

    The reason that I'm trying to do this is because I have a Makita 12V cordless drill. I gutted an old Makita battery charger and
    wired it so I could drain the battery or power small 12V devices off of the drill battery. I read somewhere that the ideal
    discharge for NiCad batteries is 1.2 V per cell before recharging. (I'm guessing that there are 8 cells in a Makita 12V
    battery--12V / 1.5V per cell = 8 cells--8 cells * 1.2V per cell = 9.6 V ideal discharge) Right now, it is wired directly to the
    battery so the thing can be completely discharged if a cell phone cigarette lighter charger is left on it too long. I was hoping to
    find a way to use relays and resistors and what-not to get my contraption to shut off when it reaches the target voltage.
  2. John Fields

    John Fields Guest

    Usually 75% of the rated coil voltage at 25C, but the best bet would
    be to get the data sheet for the relay because it'll vary with
    temperature. And manufacturer.
    The short answer is, "You can't." because the relay release voltage
    varies all over the place. What you need is a comparator, a
    reference, and a voltage divider to determine when the voltage goes
    to 9.6V and then when that happens, open up the relay.
    Since a relay needs current to keep it closed it shortens the amount
    of time the battery can supply current to the load, so a better
    solution would be to use a P Channel MOSFET for the switch since it
    needs no current to keep it turned on. If you like, I can post a
    schematic for you which will do what you want, and I'm sure we'll
    all be happy to help you get it up and running if you're willing to
    get the parts and do the wiring and assembly. Interested?
  3. Conan Kelly

    Conan Kelly Guest


    Thanks for the feed-back.

    Sure I'm interested. Please go ahead and post your schematic.

    Thanks again for all of your help,

    Conan Kelly
  4. kell

    kell Guest

    remember this old thread

    A very similar question, and you even posted a spice simulation.
  5. John Fields

    John Fields Guest

  6. John Fields

    John Fields Guest

  7. John Fields

    John Fields Guest

    You're welcome, but just to clear up a couple of things, a NiCd
    battery, when fully charged, starts off at about 1.4V. Then, when a
    load is placed across it slowly falls to 1.2V, then very quickly
    falls to 1.0V, which is considered its 'cutoff' voltage.

    So, your 12V battery pack is probably made up of 10 cells, with an
    expected cutoff voltage of 10V, which is when you'd want to
    disconnect the load.

    This should do it for you:

    View in Courier

    VBAT>--+----------+-----------+-------S D--->LOAD
    | | | G
    | [150k] | |
    | | +---|---[1M]--+
    [412k] | | | |
    | +-[1k]--+--|+\U1A |
    | | | >-------+
    | | |
    | K|2.5V | +--|+\
    [137k] [LM385] | | | >
    | | | +--|-/U1B
    | | | |
  8. ehsjr

    ehsjr Guest

    Some general information not discussed in other replies:
    Nominal voltage for a NiCd is ~1.2V per cell.
    When fully charged, and immediately after being
    removed from the charger, voltage from the cell
    is ~1.43V. NiCd packs should not be discharged
    below ~.9V per cell. The "standard" 14 hour charge
    rate is ~C/10 where C is the Ampere-hour rating
    of the cell.

    When you have the specific recommendations from
    the manufacturer, they should be followed. Otherwise
    you can use the general information above.

  9. Conan Kelly

    Conan Kelly Guest

    John Fields,

    Thanks for the feedback.

    You where right, my drill's battery pack has 10 cells.

    Remember that I do not have a background in electronics, so I do have a couple of questions about your scematic.

    I'm not going to post them here because I have pix to go with them. Go to
    to see the pix/questions.

    Please confirm or correct my assumptions.

    Are these components that can be purchased at Radio Shack? ...Fry's Electronics?

    Thanks again for all of your help,

    Conan Kelly
  10. kell

    kell Guest

    You where right, my drill's battery pack has 10 cells.

    I don't want to steal John's fire, but I'd like to jump in here
    and answer your questions.
    Yes, the numbered components with no other information are resistors.
    K means 1000, M means 1,000,000.
    Yes, the + means connected and the other pictorial that looks like
    under-over means no connection. The positions of the components on the
    actual circuit boards might (probably will) be different in the actual
    execution than in that schematic, and the "wires" will actually be
    copper traces, if you fab an actual pc board for the circuit. Or you
    could do point-to-point wiring where you use wires soldered to the
    component leads. This you can accomplish in different ways, for
    1 wire wrap (I don't know how many people do it any more)
    2 "dead bug" construction (google it)
    3 or you can insert the components in perf board and run wires from
    lead to lead.

    The component you asked about, the LM385, is a voltage reference,
    similar to a zener diode. You can look up datasheets on websites like and, which are also good places to buy from.

    The U1A and U1B are both contained in a single chip, the LT1017 dual
    comparator. You only need one comparator, so you were right that U1B
    does not function in the circuit.

    As for how to change the voltage setpoint, look at the two resistors on
    the comparator's negative input, 412k and 137k. They take the battery
    voltage and divide it by a certain ratio for the comparator to look at.
    They form a voltage divider. You could make the circuit adjustable by
    replacing them with a potentiometer.
  11. John Fields

    John Fields Guest

    I don't know, but they're surely available from Digi-Key and
    probably also Mouser; check their web sites.
    You're welcome. Your questions:

    I’m assuming anything followed by a K or M is a resistor (412k,
    137k, 150, 1k, 1M)

    Correct. 'k' means that the value shown, in ohms, is multiplied by
    1000, while 'M' means that the multiplier is 1 000 000. Also, the
    resistors in the voltage divider feeding U1A- have a tolerance of
    +/- 1%

    I’m assuming when you see a plus sign in your diagram, this a
    connection where wires/components need to be soldered togther. And
    the other (-|-) is where wires/components cross over/under one
    another (I need to make sure that they are insulated and don't touch
    one another).


    I don't know what this component is.

    It's an LM385-2.5, a voltage reference.

    I don’t know what these 3 components are. It looks like the lower 2
    are similar. Is "LT1017" part of "U1A" or is it a separate component
    all together.

    They're both parts of an LT1017, a dual voltage comparator
    manufactured by Linear Technology. U1A is one of the comparators in
    the package and U1B is the other one.

    I know I’m probably wrong, but it doesn’t seem like the lowest one
    (U1B) serves much purpose. Is it necessary? What is it’s function?

    It's a spare and serves no function. It's not necessary, but it
    comes in the package anyway, so we spare it.

    The IRF4905L is a P channel enhancement mode MOSFET being used as a
    switch which allows the battery to be connected to the load when the
    battery voltage is greater than 10 volts, but disconnects the
    battery when its voltage falls to less than 10 volts.

    What changes would need to be made if I wanted to get this to cut
    off at 0.9 v per cell? You mentioned 1.0 v per cell as a cutoff
    voltage, someone else who responded to my post mentioned 0.9 v per
    cell, and I think I read somewhere ( Battery University or WikiediA
    ) that the cutoff voltage is 1.2 v per cell, so I’m not sure which
    one to use. In case I decide to go the 0.9-v-per-cell route, what
    changes would I need to make?

    Change the 137k resistor to 147k.

    looking at the circuit again, the LT1017 is probably a little pricey
    for your application, so for about 10% of its cost you could use an
    LM393, but the circuit would change somewhat and you'd have to add a
    pullup resistor:

    VBAT>-+----------+-----------+----------+--S D--->LOAD
    | | | | G
    | | | [10k] |
    | [150k] | | |
    | | +---|--[1M]--+-+----+
    [412k] | | | |
    | +-[1k]--+--|+\U1A |
    | | | >------+
    | | |
    | K|2.5V | +--|+\
    [137k] [LM385] | | | >---+
    | | | +--|-/U1B |
    | | | | |
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