# Low Volts High Current!

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

1. ### NumanRGuest

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.

2. ### John GGuest

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. ### John PopelishGuest

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:
http://www.national.com/appinfo/power/files/f4.pdf
http://www.national.com/appinfo/power/files/f5.pdf

4. ### TerryGuest

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. ### NumanRGuest

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

6. ### NumanRGuest

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. ### John PopelishGuest

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.
http://cache.national.com/ds/LM/LM2639.pdf

8. ### Winfield HillGuest

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.,
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=3850054721

9. ### Roger JohanssonGuest

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. ### Peter BennettGuest

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.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

11. ### Roger JohanssonGuest

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.

12. ### ExternetGuest

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

13. ### Neil PrestonGuest

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. ### NumanRGuest

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

15. ### DougGuest

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. ### Winfield HillGuest

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,
http://www.irf.com/product-info/datasheets/data/1n1183.pdf
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.
http://www.irf.com/product-info/datasheets/data/100bgq015.pdf
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. ### ExternetGuest

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

18. ### NumanRGuest

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

19. ### Winfield HillGuest

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,
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=3850054721
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. ### Roger JohanssonGuest

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.