1
\$\begingroup\$

I need to slow-charge a bank of capacitors that power a load which draws up to 150A at 12V, because I don't want to draw a massive amount of current when the battery is connected to the ~20000uF cap bank.

I thought I could do it by having a relay switch a high value resistor in series with the cap bank for a few seconds, then bypass this resistor for normal operation. However, such a relay would have to pass 150A in the normal operation mode. Many relays can't do this, because they are rated based on their interrupt current and not operational current. My application will only require 150A when a battery is plugged in, and when disconnected the relay will continue to be energised by a support circuit until the 150A load has shut down and draws nominal current. This will occur several seconds after the battery disconnect, so the actual interrupt current will be small, in the milliamp range. The 150A load is electronically controlled, and can be shut down very quickly, almost instantaneously. So, no arcing should occur on the relay contacts, which is what I believe gives them their normal current rating?

Are there any relays specifically optimised for this job?

\$\endgroup\$
4
\$\begingroup\$

You want a contactor, which is a heavy duty relay specially constructed for high currents.
Digikey lists more than a dozen of them capable of >150A (they're all either 500A or 600A devices, so the give you a lot of headroom).

\$\endgroup\$
0
\$\begingroup\$

You could dispense with the relay altogether and conceptually use a resistor in parallel with a diode between the 20 mF capacitor and the 12V line. The diode conducts in the discharge direction, which leaves only the resistor for charging.

The reason I said "conceptually", is because in reality there won't be just a single diode. You probably want a bunch of Schottky diodes in parallel, each with enough of a lead to provide just a little bit of series resistance. At high current, the voltage accross the diodes will go up anyway, so there will be some current sharing. Realize that the current sharing won't be very good and compensate by massively overrating the total current rating. It helps that the diodes will only be conducting for a short time, and that diode surge current rating is usually several times the continuous operating current rating.

\$\endgroup\$
  • \$\begingroup\$ I might be missing something, but won't I require a 150A diode? \$\endgroup\$ – Thomas O Sep 12 '11 at 14:35
  • \$\begingroup\$ @Thomas: Yes, conceptually you need a 150A diode. As I said, in reality this would be a bunch of Schottkys in parallel, each with a little bit of a individual lead. \$\endgroup\$ – Olin Lathrop Sep 12 '11 at 14:52
  • \$\begingroup\$ but that would be dissipating 40W-100W, which is a lot of power. \$\endgroup\$ – Thomas O Sep 12 '11 at 18:55
  • 1
    \$\begingroup\$ @Thomas: Yes, but for a short time only. Also keep in mind the system is delivering 1.8 kW during that time, so the 40-100 W is only 2-6% of the total. The fully charged capacitor holds less than 1.5 J, and the diodes will only dissipate a small part of that. Peak current is a issue, but not power dissipation of the diodes. \$\endgroup\$ – Olin Lathrop Sep 12 '11 at 20:03
0
\$\begingroup\$

Can you use an automotive starter solenoid? They are cheap, and it can handle 150 amps easily.

Not sure if they can be used for continuous duty however. Could be worth a try.

\$\endgroup\$
  • \$\begingroup\$ Why the -1????? \$\endgroup\$ – SteveR Sep 12 '11 at 19:53
  • 1
    \$\begingroup\$ I don't know either, so I gave it a +1 to cancel out the -1. \$\endgroup\$ – Olin Lathrop Sep 12 '11 at 20:05
  • \$\begingroup\$ @Olin - I'm not sure why someone posted a -1 (possibly because it's a hacky solution?), but a -1 vote confers -2 rep, while a +1 vote confers +10 rep. It's not cancelled out, it's a net gain of 8. \$\endgroup\$ – Kevin Vermeer Sep 16 '11 at 13:32
  • \$\begingroup\$ @Kevin: I wasn't concerned about the rep. I was trying to even out the score for the answer. A negative value means "bad answer", and I didn't think that was fair in this case. A starter solenoid essentially is a relay even though it's not called that, and it's meant to handle high current and run from 12V just like the OP asked about. \$\endgroup\$ – Olin Lathrop Sep 16 '11 at 15:17
0
\$\begingroup\$

Consider a solid state relay solution and an NTC inrush limiter to limit the charging current and provide the 150A path.

\$\endgroup\$
0
\$\begingroup\$

Is the load also charging the cap bank?

If not, why not just hard-wire the load to the caps, and then just switch the limiting resistor on the capacitor-bank input?

From the description, I do not believe your system is drawing 150A continuously. Therefore, your charging source is likely considerably lower-power then your load. This would make it much more approachable for a switch-limiter solution.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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