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Power at 240VAC relay with DC 312V

Discussion in 'Electronic Basics' started by Don, Mar 23, 2005.

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  1. Don

    Don Guest

    Hello all, I am just finishing my charger controller for my electric
    vehicle. I want to power a 240VAC coil relay with the power of the
    pack (312VDC nominal). The resistance of the coil is 3600 ohms and is
    rated at 4.0VA. This is a P&B T92 relay and the datasheet says it can
    take 75-120% of the rated coil voltage.

    What would be the DC voltage range in which I could drive this relay?
    Since I do not know how to convert a AC rating to DC, I am simply
    looking at the DC rated coils and guessing that the nominal value is
    70VDC, for a range from 52V to 88VDC.

    Given that my lowest pack voltage would be around 312VDC, and my
    highest pack voltage would be approx 380V (when charging at 4deg C in
    the winter), I think I can use a voltage divider circuit, but not sure
    if this is too wide of a range.


    If the current of the relay at 70 V / 3600 Ohms = 0.020 A, then power
    1.4W

    The resistor I would need is (380V-70V) / 0.020 A = 15.5 k Ohms and a
    power of
    6.2 Watts


    Will this work?

    thanks
    Don
     
  2. Chris

    Chris Guest


    Hi, Don. Not really. AC coils have lower resistance than DC for a
    given relay voltage, and are constructed differently. You'll probably
    smoke the coil.

    There's a reason why HV DC relays are becoming rare. Most automation
    stuff has a lower DC or AC voltage for the control circuitry. It's a
    safety issue.

    One thing you might want to do is making or buying a DC-DC converter to
    give you a 24VDC supply for the control circuitry. That will allow you
    to use the T92S11D22-24 , a 24VDC DPDT T92 series relay. They're
    great, I've used them many times for automation and lab projects.

    Good luck
    Chris
     
  3. John - KD5YI

    John - KD5YI Guest

    Hi, Don -

    I think your 70 V is a little high based on the following:

    Rated VA = 4
    Rated Voltage = 240 AC

    So, the rated current is 4/240 or .0167 amps.

    This means the coil is capable of dissipating .0167 * .0167 * 3600 = 1 W
    without overheating. The maximum voltage you should apply is sqrt(1 * 3600)
    or 60 VDC.

    The coil might be able to take more than this, but we don't know that.

    You may find that the armature will not pull in at that voltage but, if it
    does, the contact pressure may not be high enough. You can try it and see. I
    ran that test many years ago but I no longer remember the outcome.

    Good luck with your project.

    John
     
  4. John Fields

    John Fields Guest

     
  5. Rich Grise

    Rich Grise Guest

    He shouldn't need to do that - if it doesn't already have a separate
    control battery, he could just tap the main pile at 24V.

    And I agree that using over 300 volts in control circuity is Not A
    Good Idea. :)

    Cheers!
    Rich
     
  6. Bob Eldred

    Bob Eldred Guest

    Snip...

    120 Volts DC. The coil works at four volt-amps which is very near the
    actual power assumed to be four watts. There is a slight error here because
    of the unknown phase angle but not much because the DC will not dissipate in
    the copper shorting ring as AC will. Four watts is not excessive for a small
    relay coil. V = sqr(3600 * 4) = 120 VDC.
    Bob
     
  7. John - KD5YI

    John - KD5YI Guest

    Actually, Bob, you are correct that the phase angle is unknown. But, if
    there is a phase angle, then the power in the coil is not equal to the
    specified VA of 4 since they are related by the cosine of the phase angle.
    That is, P(ower) = VA * cos(angle).

    If the rated VA is 4 and the rated voltage is 240, then the rated current is
    I = VA/E which is 4/240 or 16.67 mA. The coil resistance is specified at
    3600 ohms. This means that the rated coil dissipation is P = I * I * R which
    is .01667 * .01667 * 3600 or 1 watt.

    Also, we know that applying 240 volts results in only 16.67 mA through the
    3600 ohm resistor. Well, that's not enough resistance to limit the current
    to 16.67 mA so there must be additional impedance involved and we can guess
    that the coil has indctive reactance. How much?

    We need an impedance of Z = E/I such that with 240 volts applied, the
    current is 16.67 mA. So Z = 240/.01667 or 14,400 ohms. We only have 3600
    ohms of resistance so the rest is made up of inductance. The impedance of a
    series RL is Z = sqrt(X * X + R * R). We can find the X (hence the
    inductance) with algebra. X = sqrt(Z * Z - R * R). So X must be 13,943 ohms.

    This gives us the impedance of the relay. It is 3600 + j13,943. In polar
    notation, the Z is therefore 14,400 at an angle of +75.523 degrees
    (arccos(3600/13943)). The current will lag the voltage by 75.523 degrees.
    Notice that the cosine of 75.523 degrees is .25 which brings us back to my
    original equation that the power in the power is p = .25 * 4 or 1 watt.

    Hence, it is important to limit the DC value so that the dissipation of 1
    watt is not exceeded. This will require a voltage of not more than E =
    sqrt(p * r) which is sqrt(1 * 3600) or 60 VDC.

    If I've made a mistake here, please let me know.

    John
     
  8. Don

    Don Guest

    Thanks for the help guys. Here is the situation. The charger I have
    for my electric vehicle has one major flaw. If no output is connected
    it will blow circuitry. I looked for other chargers but there are
    none, which can charge up to 400V at 30A which cost around $2000.

    Considering that I have already purchased the charger, and I
    definitely want to protect it from blowing up, I would like to
    develop a sense circuit which will ensure that no input power can be
    supplied if it is not connected to a pack.

    What comes to mind is the simple solution of a relay. But then again
    it does not sound so simple.

    Any ideas?
     
  9. John - KD5YI

    John - KD5YI Guest

    I meant coil.
     
  10. Chris

    Chris Guest

    Hi, Don. Back in days of yore (1980s), before Potter&Brumfield T92s
    ruled the earth, you would use a Potter&Brumfield PRD relay to switch
    that kind of current. They've got the same contacts, but are open
    frame relays with ugly black bakelite bodies. They still make 'em with
    110VDC coils, and that could provide one possible solution to your
    problem -- an AC coil won't cut it here.

    I'm guessing that your "battery charger" is a switching power supply
    with a series resistor and an in-line fuse. That would explain the
    "blow up on no load" phenomenon. But I'm not too sure about the 30 amp
    business. Let's see: 312V times 30 A = 9630 watts at 100% efficiency.
    Don't think that's something you're just going to plug into a 120VAC
    outlet, unless you've got an 80 amp service line somewhere. Or 240VAC
    times 40 amps, for that matter.

    But I'm hearing you say you don't want to turn on the charger, even by
    accident, unless there's a battery installed in the go-cart. If your
    charger is a simple one that applies a higher voltage than the battery
    voltage, and then either senses current or uses a timer to determine
    end of charge, or just stays on trickle charge until you unplug it, you
    might want to try this circuit (view in fixed font or M$ Notepad):

    '
    ' D1
    ' + >----o---->|-------------------.
    ' | 1K 1/2W T |
    ' | || ___ --- |
    ' o-----||-|___|-o---o o----o
    ' BATTERY | || R2 | SW1 |
    ' .-. CRY1A .-. |
    ' CHARGER R3 | | | | +|
    ' | | R1| | ---
    ' '-' 12K 5W'-' -
    ' L1 | | 312V|
    ' .-. C| |
    ' ( X ) RY1 C| |
    ' '-' C| |
    ' | | |
    ' - >---o--------------o----------'
    '
    ' ||
    ' >----------||----.
    ' AUX || |
    ' CONTACTS CRY1B |
    ' |
    ' >----------------'
    '
    RY1 P&B PRD-11DY0-110
    R1 12K 5WATT
    R2 1K 1/2W
    R3 100K
    L1 NE-51
    D1 85HFR60 (If=85A, PRV=600V)
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    I've made some assumptions here, which means that you can feel free to
    modify things as you choose. I've assumed your battery charger is a
    simple cheapie which applies a voltage higher than the battery voltage,
    and either has a timer or measures current to end charge, or turns into
    a trickle charger which can be left on until you're ready to use the
    go-cart. This won't work for a charger that uses battery voltage
    sensing. However, it shouldn't blow anything up, either. I've also
    assumed that, under normal circumstances, charging current can't exceed
    20A or so. (Make sure to heat sink the diode -- figure 1 watt per amp
    of charging current. The specified diode is probably overkill.)
    Another assumption is that you can turn on your charger with the set of
    AUX contacts. That may mean leaving the charger ON/OFF switch in the
    ON position, and just turning it on by switching on line voltage to the
    charger. Either that or, possibly, you can finagle the wires from the
    ON/OFF switch out to the auxillary contacts. Your call. I'm assuming,
    of course, that you're using the appropriate HV safety precautions. A
    312V battery can be fatal if handled improperly. If you don't know
    what you're doing, don't do it. If you don't understand what's going
    on here well enough to double-check the above and be confident it's
    right, don't do it, either.

    You can see from the circuit action that you have to press pushbutton
    SW1 to start the charge cycle. If there's no battery, the relay won't
    turn on. If the relay doesn't turn on, your charger won't turn on.
    But when the charger does turn on, it also keeps the relay on, and it
    will stay on until the charger is unplugged or otherwise turned off.
    You have to keep the pushbutton depressed until the charger is up and
    running.

    A simpler circuit might be possible at the cost of leaving the relay on
    after the charge cycle is complete, but that's a wasted 3.6 watts which
    will tend to discharge the battery. I would guess one big diode is
    worth it.

    Good luck, and play safe
    Chris
     
  11. John Fields

    John Fields Guest

     
  12. Bob Eldred

    Bob Eldred Guest

    That's all very nice but I doubt that one watt would pull the relay in
    unless it is a very small relay. A check of the PB T92 series shows a power
    of about 1.7 watts which means both of us are wrong, me too high, you too
    low. My calulations were an off the cuff estimation and yours, while more
    exact, neglected the losses in the shorting ring, a feature of AC relays.
    This loss will shift the phase ange more resistive than you have it and
    increase the dissipation. I stand corrected, 78 vdc should do it. I think
    you said near that above.
    Bob
     
  13. John Fields

    John Fields Guest

    ---
    If you run the relay on DC, the losses in the shorting ring vanish
    and, it seems to me, that in considering the _impedance_ of the coil
    while considering P&B's _Volt-Ampere_ rating for the coil he was
    right.
    ---
     
  14. Bob Eldred

    Bob Eldred Guest

    I did not bother to address your link, should I? Yes the DC losses vanish in
    the shorting ring, but the 1.7 watts is a DC rating, pure resistive for a 12
    or 28 volt coil. The ring losses explain, sort of, the lower power rating
    calculated at one watt by phase angle and VA, they have nothing to do with
    the DC rating only explain why the power number is low.
     
  15. John Fields

    John Fields Guest

    ---
    It's up to you, of course, I just think it curious that asked-for
    information was ignored.
    ---
    ---
    What you seem not to have considered is that the magnetic structures
    for the AC and DC relays are different (as explained in the
    application note) and that the currents required to pull the armatures
    down will be different for the different relays.

    In any case, I think you're going about the whole thing backwards.
    IMO, what you should be doing is sensing battery voltage and if it's
    below a certain level then using that information to keep the charger
    from being connected to the mains. That's why I asked you what sort
    of mains you had, which you also didn't address.
     
  16. Bob Eldred

    Bob Eldred Guest

    Cut........
    I'm going about it backwards??? why me? I'm not going about
    anything...Maybe you are a little confused.
    Bob
     
  17. John Fields

    John Fields Guest

     
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