0
\$\begingroup\$

So I got this surge test I need to do some research on and need some help as my electrical background minimal.

The test i'm working to setup is as follows: I have to surge test this big switch (just shoot 12V DC, 2000 AMP to the switch terminal for 1 second)

We only have a 500 amp power supply but we got tons of 300 Watt load banks From what I understand, I can connect the 500 amp power supply to one load bank(master) and than chain a bunch of other load banks (slave) in parallel to add up the currents until I get 2000 AMPs.

The final load bank at the end of my parallel chain positive lead will connect to the positive terminal on the switch and the first load bank (master) negative lead will connect to the ground terminal on the switch?

Am I doing this correctly? I believe my math is correct, that I can use my 12VDC/500A power supply to create a 12VDC/2000A output via chaining x amount of load banks in parallel I'm not sure if my wiring is correct thou..also how will the resistance from the wires affect my output? I believe i need a 3/0 AWG wire at the minimum..(probably safer to go with something even bigger)

Any help or suggestions for better methodology would be appreciated! Thank you!

\$\endgroup\$
  • 12
    \$\begingroup\$ If you manage to get the 2000A, then make videoes. Have the survivors post them here for our ̶e̶n̶t̶e̶r̶t̶a̶i̶n̶m̶e̶n̶t̶ education. \$\endgroup\$ – JRE Oct 20 '16 at 15:46
  • 2
    \$\begingroup\$ The load banks are just resistors, right? The current, one way or another, has to come from your power supply. You could get 2000A from a bunch of lead-acid batteries but it wouldn't be very well controlled. \$\endgroup\$ – Spehro Pefhany Oct 20 '16 at 15:49
  • 1
    \$\begingroup\$ A power supply supplies power, a load bank consumes power. It will not give you an extra 1500A. \$\endgroup\$ – Steve G Oct 20 '16 at 15:49
  • 6
    \$\begingroup\$ Wait a minute, this guy is posting under the username "BPIntertek." I now have significantly less faith sending my products to Intertek for testing. \$\endgroup\$ – Peter Oct 20 '16 at 17:57
  • 3
    \$\begingroup\$ @PeterK At the very least archive this question for future strong-arm purposes when you need discounts (read: instant approval). \$\endgroup\$ – Asmyldof Oct 20 '16 at 18:06
5
\$\begingroup\$

Car batteries are rated by the sustained current from full charge at 12.5V to 7.5V or a 5V drop at low temp to yield a cold cranking amperage CCA rating to start cold engines which have more friction.

If a Battery has a CCA rating of 750A at 0'C, it will be higher CCA with temperature due to battery chemistry.

  • This 750 CCA battery has equivalent ESR = 5V(drop)/CCA(rise)
  • To perform a current surge test at low voltage, it makes no different if the voltage is 12 V or 7.5V on the switch equivalent ESR and resulting power dissipation which results in a high temperature rise in 1 second. This will be a value in Joules per second or Watts and the ESR of the switch will result in a measureable Tjcn rise which then translates into an accelerated aging MTBF value that may be researched and calculated as to how many such surges can be performed in it's life time.

    • As in any electronics or mechanical switch, for low loss and high efficiency, we use conservative values and opt for a ratio of switch ESR or RdsOn or Rce to load resistance such that we desire ESR/R(load) < 2%
    • This translates into the voltage drop (I*ESR) and the power drop (I^2*ESR) both a linear proportion with ESR

Conclusion

  • You only need ;
    • a battery bank with a cumulative CCA >> 2000A. (3 car batteries)
    • 50 mV current shunt at 2000A or a suitable ammeter
      • it must dissipate 100W in 1 sec. (many shunt wires of Nichrome )
      • signal wires must be shielded with twisted pairs
      • the sense wires must be coming off at 90deg to reduce induction
      • it could be on ground side to read full scale current of 2000A
    • a calibrated load to yield 2000A at 8 to 10V or 20kW of heat
      • such as many shunts of nichrome wire in water (NiCr 10x better tempco than copper.)
    • **edit* Test Phase 2 Inductive load test
    • since we know the inertia of motors translates into stored energy as a generator, we must consider not just the initial load current but the stored inductive energy and mechanical inertia of the motor now acting as a generator with an contact arc that is sustained until the arc is quenched similar to the holding current threshold of an SCR.
    • so the test results of a resistive load are far better than a switched inductive DC load and the stored energy in that load.
    • thus the surge current load test must define this stored inductive and rotational stored energy as part of test #2
    • although Relays often rate contact currents for Inductive AC and DC loads much lower than Resistive loads
    • define your real target loads by Inductance and Rotational stored energy then measure contact temp rise and get a spec for contact max temp rise as a test criteria... ( more later)

      I have used this basic method to monitor currents greater than 100,000 A with Voc<5V (no load) for Zirc/Steel diffusion bonding machines when I was at Bristol Aerospace Canada in Winnipeg. Zirconium shims were the interface material of choice for welding two Monel Steel 4" tubes together for Nuclear Reactor steam. There were a lot of fireworks in the welding process with forced water spray cooling and the current had to be increased as the pipes were rotated between two solid copper wheels (Inner and Outer) acting as rolling pressure contact as inner steel tube was inserted to another and rolled between the 2 solid copper wheels carrying the current. The weld naturally lowers the ESR of the joint as the shunt resistance drops as it rotated in welded circumferential resistance. **

This high current resistance weld, called "diffusion bonding". ~1979

more Opinions

What may happen to your switch makes contact may be less important than when it breaks contact with the arc power generated by the stored energy in an inductive motor generator load. SO an inductive load will ALWAYS BE more damaging than a Resistive load of sparks until the >>5000'K heat burns off the surface oxidize creates a lower contact resistance by melting impurities.

What happens during long term exposures (I*t) and the thermal properties of the metal alloy contacts is a matter of great research in chemistry and physics of higher temperature exotic metal alloys. If you are new to this look at the contact materials used by companies like OMRON that use the best metal alloys for the contacts required.

\$\endgroup\$
  • \$\begingroup\$ on 2nd thought I agree with you , water is also easier to clean and less messy too H2O=0.609 W/mK Oil, transformer 0.136 W/mK . I forgot Oil is better for hiVac cooling due to dielectric constant of 2 vs water=80.. I'll delete reference to oil \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 20 '16 at 16:55

Your Answer

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