0
\$\begingroup\$

Recently I had posted about a thread related to a bench DIY high current transformer 12VDC supply that I'm trying to built. But due to some transformer and diode rectifier limitations that project may take a little more time to be completed until I get my hands on a much beefier transformer or two transformers combined enough to do the job. Meanwhile I have a used spare 100 or 120A alternator (I don't remember the exact current rating) in my possession. I was thinking that if I mounted the alternator with an AC induction or a brushed DC motor like from a treadmill I can use it to generate the max current capability of the alternator to run high current 12VDC motors that pulls close to 80-90Amps for a brief testing like under 10 seconds. I've a treadmill motor in my possession, which I think its rated for 90 or 100VDC that is rated for 3150rpm I guess. I can use a pulley system to run it to obtain higher rpms at the alternator. The voltage output of the alternator doesn't really matter too much for me as long as its more than 12VDC, since my goal is not to charge a battery.

What do you guys think about this project? Is it a good idea? Also what is the RPM range required to obtain that max. current capability of an alternator? If you need a specific alternator model I can look it up and let you guys know later.

Edit: I'm aware that I'll need a 12V supply to excite the field windings. Planning to use a 12V transformer with rectifier for that job.

\$\endgroup\$
8
  • 3
    \$\begingroup\$ This won't work, as soon as you try to draw 90 amps from the alternator it will be hard to turn, and the treadmill motor will no longer be able to turn it. This isn't just about spinning it at a specific speed, it's about the power required to turn it. Just like if you hooked up a 90 amp load to your car, your engine would bog down, that's the engine taking the load... Using a pulley system to increase RPM will decrease the torque, making the situation worse. \$\endgroup\$
    – Ron Beyer
    Commented Apr 20, 2020 at 23:06
  • \$\begingroup\$ Also, alternators are not well suited to shouldering big loads by themselves, which is why batteries are involved and car-audio enthusiasts add extra batteries or capacitors. \$\endgroup\$
    – Ron Beyer
    Commented Apr 20, 2020 at 23:14
  • 1
    \$\begingroup\$ You will need at least a 1.5 HP motor to get 90 amps from the alternator. An automotive alternator MUST be connected to a 12 volt battery - otherwise the output voltage will be much much higher than 12 volts. The battery is a vital part of the voltage regulation system. \$\endgroup\$ Commented Apr 20, 2020 at 23:19
  • \$\begingroup\$ @Ron Beyer Looks like using an alternator system with heavy loads doesn't seem to be a good idea. At least with a single unit it seems from what you guys are saying. \$\endgroup\$ Commented Apr 22, 2020 at 19:40
  • \$\begingroup\$ @PeterBennett I would have added a small SLA battery like 12Ah if it needs it just for absorbing the spikes and just fo regulation . But I'm afraid that the alternator might pump higher current into the small battery and also it could be a problem when testing very high current 12V loads. \$\endgroup\$ Commented Apr 22, 2020 at 19:42

1 Answer 1

3
\$\begingroup\$

Alternators are designed to generate enough voltage to prevent the battery from discharging at the engine idle speed. Speed and field current determine the voltage. At idle, the regulator lets the maximum field current flow to the alternator. As engine speed increases, the regulator cuts back the current to prevent excess voltage. For your purposes, you would only need to drive the alternator at the speed it would run at the normal idle speed of the engine. You can operate the alternator without a battery, but you do need the proper voltage regulator and a 12-volt DC power supply for the field. If you operate the alternator at a voltage that is too high, the initial starter current will be higher and more difficult to provide.

Once the starter is at full speed, the current should drop somewhat, but probably not as much as might be expected with no load, because the field is connected in series with the armature. I would think that a proper bench would include measuring the idle current. There must be some guidelines about that someplace.

One thing that might help with the testing power supply would be to put a small value of resistance, perhaps 1/20 ohm, in series with the starter before connecting it to the power supply and then short the resistor when the motor has picked up some speed. You might be able to make the resistor from a piece of stainless steel.

\$\endgroup\$
4
  • \$\begingroup\$ One thing I really now is that this alternator setup might be good only for loads probably up to 35-40Amps. That's like a bike starter motor. One of the main reasons was I wanted to make this test bench to test rebuild car starter motors also. Even when the car starters are tested outside the car with any load and without the solenoid engaged (as the solenoid also pulls 35A suddenly and settles to 18-20Amps I have seen) it pulls above 60Amps.Though I might run it only for 3-4seconds, from your explanation the initial current draw by the starter will exceed 80A and may make the alternator stall \$\endgroup\$ Commented Apr 22, 2020 at 19:45
  • \$\begingroup\$ I'll try the resistor setup you mentioned to see how that performs. But it seems that using a pair of some very high current automotive boost charger 100A transformers or two or three from a high powered UPS seem to be stable and better choice for high current 12V test bench. \$\endgroup\$ Commented Apr 22, 2020 at 19:47
  • 1
    \$\begingroup\$ Yes. It would be the motor driving the alternator that would stall if it is not big enough, but the diodes in the alternator or even the windings could get too hot pretty fast too. \$\endgroup\$
    – user80875
    Commented Apr 22, 2020 at 20:09
  • \$\begingroup\$ The heating issue is another concern I had in mind. I might go for a transformer solution as I mentioned to BruceAbbott above. I seems to be the safer and easier solution. \$\endgroup\$ Commented Apr 22, 2020 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.