Low Volts High Current!

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.

Mark.

Reply to
NumanR
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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:

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John Popelish
Reply to
John Popelish

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!!!!!!!!!!!!!!!!!!"

Reply to
Terry

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. Mark

Reply to
NumanR

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.

Reply to
NumanR

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.

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John G

Wot's Your Real Problem?
Reply to
John G

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.,

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 Thanks,
    - Win
Reply to
Winfield Hill

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 design.

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John Popelish
Reply to
John Popelish

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 amps.)

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

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Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
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Reply to
Peter Bennett

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. Or; Buy a 8V automotive battery and insert a 200 Ampere diode in series, it will be close enough to 7.2V under load. Or; Seven 200A diodes in series to 12V will drop ~5V.

-Maybe one of these ways will be applicable for your application- Miguel

Reply to
Externet

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.

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Roger J.
Reply to
Roger Johansson

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.

3*2.1=6.3 4*2.1=8.4 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.

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Roger J.
Reply to
Roger Johansson

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

Reply to
Neil Preston

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

Reply to
NumanR

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.

Reply to
Doug

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,

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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.

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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.

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 Thanks,
    - Win
Reply to
Winfield Hill

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 choose. Seems you are in UK, just dial an industrial electric supplier and ask. Get a Nervo-Calm pill before asking the price. Miguel

Reply to
Externet

Dont appear to be able to get an 8 volt battery in the uk, I cant find a supplier anywhere, any more ideas?

Reply to
NumanR

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,

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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 >

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.

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 Thanks,
    - Win
Reply to
Winfield Hill

I came across this old beast, wihch uses 75 pass transistors, for amusement. It's a 1.8V 250A quad Pentium linear regulator. :>)

Winfield Hill wrote,

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 Thanks,
    - Win
Reply to
Winfield Hill

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