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Low Volts High Current!

Discussion in 'Electronic Basics' started by NumanR, Nov 6, 2004.

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  1. NumanR

    NumanR Guest

    Hello everyone, I need some help,

    I have to convert voltage from a 12 volt car battery to 7.2 volts, BUT
    be able to draw 150 amps for a short time (approximately 2 or 3
    seconds). The voltage output must be accurate, can anyone help
    regarding design and recommended components please? As I understand it
    a potentiometer would be no good as it couldnt cope with the high
    current or am I wrong? Assistance please.

  2. John G

    John G Guest

    You are right about the potentiometer.
    Pray! Tell us what you need 150 AMPS for at a very precise 7.2 volts.
    This may lead someone to helping you solve your problem.
  3. You have that right. If the current was always exactly 150 amps, a
    fixed series resistor of (12-7,2)/150=.032 ohms would produce the
    correct output voltage (assuming the 12 volt input was also exactly
    right. But it is not generally a practical method.

    You need either an active resistance (big transistor of some type)
    that is controlled by a measurement of the output voltage (called a
    linear regulator) or you need a switching regulator that turns the 12
    volts on and off at a high frequency, and this pulse that is either 12
    volts or zero is them passed through a low pass filter to extract the
    average voltage of the pulse train (called a buck converter). The
    pulse duty cycle (% on time) also has ot be controlled by a circuit
    that measures the output of the filter. At 150 amps out, neither will
    be a simple or low cost circuit.

    How often will this 2 to 3 second load occur?

    For a quick course on linear and switching regulators, see:
  4. Terry

    Terry Guest

    Agree more info needed.
    If all the voltage including the 2 or 3 seconds at 150 amps must be
    maintained accurately to 7.2 volts then a very heavy, fairly complicated and
    expensive voltage regulating circuit will be needed. Also the response time
    of the voltage regulating should be known and designed for.
    A potentiometer is basically a preset variable resistance (R ohms); the
    voltage drop across it will be a function of the amount of current flowing.
    Voltage Drop (Volts) = Current (Amps) multiplied by Resistance (Ohms).
    So; if at one moment there is 150 amps the voltage drop will be 150 x R =
    Voltage drop.
    If a moment later the current drops to say 15 amps voltage drop will be 15 x
    R = One tenth of the voltage drop at 150 amps!
    Q? Must the device requiring 150 amps for short periods have its voltage
    regulated to 7.2 volts? Could it use the full 12 volts and other
    (electronic) apparatus kept at a reasonably constant 7.2 volts. If so what
    is the current demand of that other apparatus and is it steady or varying?
    Insufficient info; this question is rather like saying "I need to heat some
    water to 100 degrees!!!!!!!!!!!!!!!!!!"
  5. NumanR

    NumanR Guest

    Ok the reason I need 7.2 volts is because this is for a power supply
    for a racing motor dynometer. The voltage could be around 7.5 volts
    but needs to be stable as possible. The motor is placed in a machine
    and has a flywheel fitted and then is accelerated to full speed. The
    motor accelerates for approximately 3 seconds depending on power and
    then stays at full speed for another two. These results are then
    processed and then sent to my computer where they are graphed for
    comparison. It is the comparison that makes the voltage supply
    accuracy important. The normal supply is a nicad pack but this voltage
    drops quickly after a few runs.
    Thanks for your replies so far.
  6. NumanR

    NumanR Guest

    Oh and the load would be applied every ten minutes or so, I have been
    checking a few charts and the average current over the whole period is
    about 45-50 amps, but obviuosly it draws considerably more when
    starting the acceleration. Would it be cheaper to go mains electricity
    route as a starting supply? The supply voltage needs to be as smooth
    as possible.
    Thanks again, Mark.
  7. This application is very robust as far as high frequency ripple and
    noise, so a buck converter would work fine without much concern for
    filtering the output to a low noise and ripple content. All that
    matters is the average voltage over the PWM cycle (as long as this
    cycle is short compared to the mechanical time constant of the motor
    flywheel system and the motor electrical time constant).

    But at 150 amperes you are essentially constructing a golf cart speed
    control (a shoe box sized device).

    I would be thinking along the lines of a multi phase converter
    (several parallel buck converters that have active current sharing),
    just to keep the switching devices small and cheap and to reduce the
    problems with input and output ripple.

    This chips is not necessarily the one I would use to control such a
    beast, but its data sheet shows what would be involved in such a
  8. NumanR wrote...
    There is a commercial Zantrex power supply that nicely fills
    your needs. I'm thinking of their XFR 7.5-300, rated at 300A.

    The smaller XHR 7.5-130 is rated at 130A, but could be fudged
    with external programming to operate to 150A for a short time.
    I have several of these supplies, obtained on eBay. E.g.,
  9. Another alternative is to use a more suitable rechargable battery.
    A 7.x Volt lead-acid battery.

    Or rebuild a 12 Volt car battery, use only
    enough cells to produce 7.5 or so Volt.

    It is unnessary to build or buy a high power converter when it is so easy
    to find a battery with the right voltage.
  10. Unfortunately, a lead-acid cell is about 2.1 volts, so it will be
    pretty hard to make a 7.2 volt battery with them (and even if you
    could, the voltage would drop significantly when you try to draw 150

    I think the commercial power supply route that Winfield suggests is
    the best solution.

    Peter Bennett, VE7CEI
    peterbb4 (at)
    new newsgroup users info :
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    Vancouver Power Squadron:
  11. The other person said that the voltage did not have to be exactly 7.2
    Volt. He is testing motors, so the voltage has to be somewhere around
    that value.

    I think one of those values would suffice for his needs.

    There is also the possibility to reduce the voltage with one diode
    drop,by inserting a high current diode. It would take 1 Volt or so, which
    gives us 8.4 minus 1 is 7.4.

    If we use a 8.4 Volt battery I have a feeling that the voltage would be
    closer to 7.2 than 8.4 when we take out 150 Amps anyway.
    He is already using batteries, but they need to be recharged too often
    and drop too much voltage during the test. A heavy duty lead-acid battery
    would probably be a lot better than what he he using now.
    Probably good enough for his purposes.

    A lead-acid battery with a suitable voltage can be charged 24/7 with a
    charger which is a lot easier to build than a converter from 12V DC or
    the mains voltage. Building 2 Amp circuits is a lot easier and cheaper
    than building 150 Amp circuits.
    If he has the money, yes probably.

    If I had the same problem I would either get a lead-acid battery with a
    suitable voltage, or make a little change to a car battery so I can use a
    suitable number of cells.
    (Assuming that I could buy a car battery in good enough shape cheaper
    than a new heavy duty lead-acid battery with a suitable voltage)

    If one battery of a certain type is not enough it is easy to add another
    in parallell with it, but I doubt that it would be needed if we use a car
    battery or a similar battery with suitable voltage.
  12. Externet

    Externet Guest

    Hi Mark.
    Not a simple thing to do precisely.
    I would attempt other ways:
    Trick a high power switching supply by modifying the sense line with
    dividing resistors, to make it believe the output is at rated voltage
    but actually delivering 7.2V. I have modified a 12.0V 50 Ampere
    switching power supply and yields 13.8 V now.
    Buy a 8V automotive battery and insert a 200 Ampere diode in series,
    it will be close enough to 7.2V under load.
    Seven 200A diodes in series to 12V will drop ~5V.
    -Maybe one of these ways will be applicable for your application-
  13. Neil Preston

    Neil Preston Guest

    Is it practical to simply keep the battery on a charger between test runs?
    You might need to devise a high current charger (10-15 Amps or so), and be
    sure it shuts itself off to prevent overcharging the battery, but it would
    be a much simpler task......

    Neil Preston
  14. NumanR

    NumanR Guest

    The seven 200a diodes sounds the easiest option to me, where can i get
    them and how much would they cost?
  15. Doug

    Doug Guest

    I would recommend the 8 V battery. It's a common replacement for a six volt
    battery on a tractor and should easily be found at a battery suppier or
    farm store. In series with a diode it should give many runs at 150a.
  16. NumanR wrote...
    This won't work very well considering your varying current load.
    All silicon diodes have a change in forward voltage of 60 to
    100mV/decade of current. This adds up to as much as 0.7V change
    for seven diodes. Plus remember the forward-voltage temperature
    coefficient, which can be severe for a 100C junction temperature
    change at high currents. And more damaging, at high currents
    diodes exhibit a considerable internal series-resistance drop,
    up to 1V or more at their rated current, which is rather bad if
    experienced seven times in series! And not to mention all your
    wiring and contact drops. That's why commercial high-current
    power supplies have a pair of sense terminals, allowing you to
    nail the regulated voltage at your final destination.

    But I digress.

    The 1n1183 is a classic large silicon diode you might consider,
    According to fig 4 its forward voltage drop is 950mV at 10A
    25C, which drops to 800mV for a hot die temperature. The usual
    60mV per decade would predict 1000mV at 150A, but we actually
    see 1500mV, which tells us that it has about 3 to 3.5 milliohms
    of internal resistance. That sounds pretty good, but it means
    the voltage can range from 800 to 1500mV over these conditions.

    For currents over 50A you might consider a Schottky diode.
    The 100BGQ015 is a 100A part, with a drop of 580mV at 150A.
    Its forward voltage drop curve shows 250mV at 10A, with an
    ~80mV/decade slope. But even as low as 50A we see a resistive
    component pushing this higher, to 350mV. Unless it heats up,
    and drops to 280mV. So here we see a 280 to 580mV range over
    just the 50A to 150A region, not very good regulation either.
  17. Externet

    Externet Guest

    Hello Numan.
    Seven 200A diodes will cost you much more than a new 8 V battery. If
    you had them among your parts bin, fine, it would be an option to
    Seems you are in UK, just dial an industrial electric supplier and
    ask. Get a Nervo-Calm pill before asking the price.
  18. NumanR

    NumanR Guest

    Dont appear to be able to get an 8 volt battery in the uk, I cant find
    a supplier anywhere, any more ideas?
  19. I have taken the liberty of cross-posting this to s.e.d. because it
    can be nicely considered as a nontrivial electronics design problem.

    NumanR also wrote ...
    I suggested a commercial 1 or 2kW power supply, such as a Xantrex
    XHR 7.5-130, spec'd at 7.5V 130A (but can be externally programmed
    to operate to 150A for a short time, and I gave a $350 eBay link,
    That one finished, but the seller still has it, plus a few more...

    I guess Numan wasn't interested in spending $350 for a used eBay
    Xantrex, but hey, precision regulated 1.1kW power supplies aren't
    trivial things, that's why they are expensive.

    However, perhaps if one was to build one for himself it would make
    sense to start with a 12V auto or truck battery to simplify the
    task, as Numan envisioned.

    NumanR also wrote...
    A switching regulator would be most efficient, but a 1kW switcher
    is not a good beginning project for someone to attempt, so let's
    instead give Numan some guidance for making a linear 12V to 7.2V
    150A regulator. He delivers his kW power level for only a few
    seconds, and during that time dissipates 150A * 5V at most (likely
    less, due to battery and cabling voltage drops), which amounts to
    about 450 * 3 watt-sec = 1350J during that time. Unlike ordinary
    linear kW supplies, this won't present a serious heat sink problem.

    .. ______ ________
    .. | + | | |
    .. | o======| series |=======,
    .. | | | reg | |
    .. | | | +Vs |-------+
    .. | 12v | | | motor
    .. | batt | | -Vs |-------+
    .. | | |________| |
    .. | | | | |
    .. | o========|==+==/\/\====='
    .. |______| | 50mV I-meter shunt
    .. control

    Here's basically how it'll look. Note the two sense connections
    located near the motor terminals for the feedback control, and the
    on/off control input. Numan may also want a current-monitoring
    shunt, short-circuit protection, and some type of safety interlock.

    I'm going to sign off now, get breakfast, and shovel the snow. :>)
    Then I'll poke around and find an old post of mine to edit and put
    up as my entry for Numan's 150A linear regulator.
  20. The car battery you have there consists of a number of cells, which are
    connected in series. The voltage between the two outermost placed
    contacts is 12 Volt.

    In between them there are other connections which carry different
    voltages. If you can remove the plastiv cover on the battery, or drill
    through it, you can connect to any number of cells you like, in series.

    But I am pretty sure there are lead-acid batteries made for 7-8 Volt for
    sale in UK. A country filled with motorcycles of all ages and needing
    batteries of different voltages.

    The lead-acid battery technology is well suited to delivering very high
    currents for a short time. It will probably be the most cost-effective in
    this case.
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