Hedges
I See Electromagnetic Fields!
- Joined
- Mar 28, 2020
- Messages
- 20,493
Dirty surfaces or low torque on a nut could make a contact resistive.
In my AC breakers, I find the specified torque provides very little compression. If I just wiggle/rotate wire, it works loose and I can turn the screw further, so that's what I do.
Loose/bad connection would get hot, pushing breaker closer to tripping.
The breakers I'm used to can take 50% excess current for 10 minutes or so. I think 100% excess might trip in 1 minute.
Try to find current/trip curves for yours, see what would explain 1 minute trip.
Battery should supply approximately constant current. Capacitors in inverter should smooth out 60 Hz cycles, and the higher frequency PWM used to synthesize sine wave.
If the capacitors aren't enough to smooth out 60 Hz, each pulse of current will go through the breaker.
Try using AC Volts scale to measure voltage across breaker. Also across a shunt if you have one. Across a length of battery cable (you can look up resistance per foot and use it as a shunt).
Try DC Volts too.
The AC Volts would tell you how much ripple.
People say to derate the BMS, plan to use it for continuous current maybe half its rating.
If there is very high ripple current (zero to 200A instead of steady 100A) that would increase heating of both breaker and BMS (when back to using BMS in the circuit.)
Maybe added capacitors at the inverter would reduce ripple. It would also increase the current surge when closing DC breaker, so a precharge resistor might be a good idea.
My thinking is that invertor vendor might skimp on capacitors.
Bigger breaker/fuse would be a work-around. Just like getting a BMS rated twice what you need.
"I^2 R" or "I squared R" is the power dissipated in a resistor.
Pulses of 200A half the time would deliver same power as 100A all the time, but would cause more heating in the wires.
If wire, breaker, or another component was 0.001 ohm, power dissipated in it is:
100A x 100A x 0.001 ohm = 10W all the time
200A x 200A x 0.001 ohm = 40W half the time, 20W average.
Obviously if breaker dissipates 20W not 10W, it'll trip sooner. Can only carry 1/sqrt(2) = 0.7 times as much current
140A x 140A x 0.001 ohm = 20W half the time, 10W average
So 70A average current (140A half the time) would heat breaker same as 100A continuous.
In my AC breakers, I find the specified torque provides very little compression. If I just wiggle/rotate wire, it works loose and I can turn the screw further, so that's what I do.
Loose/bad connection would get hot, pushing breaker closer to tripping.
The breakers I'm used to can take 50% excess current for 10 minutes or so. I think 100% excess might trip in 1 minute.
Try to find current/trip curves for yours, see what would explain 1 minute trip.
Battery should supply approximately constant current. Capacitors in inverter should smooth out 60 Hz cycles, and the higher frequency PWM used to synthesize sine wave.
If the capacitors aren't enough to smooth out 60 Hz, each pulse of current will go through the breaker.
Try using AC Volts scale to measure voltage across breaker. Also across a shunt if you have one. Across a length of battery cable (you can look up resistance per foot and use it as a shunt).
Try DC Volts too.
The AC Volts would tell you how much ripple.
People say to derate the BMS, plan to use it for continuous current maybe half its rating.
If there is very high ripple current (zero to 200A instead of steady 100A) that would increase heating of both breaker and BMS (when back to using BMS in the circuit.)
Maybe added capacitors at the inverter would reduce ripple. It would also increase the current surge when closing DC breaker, so a precharge resistor might be a good idea.
My thinking is that invertor vendor might skimp on capacitors.
Bigger breaker/fuse would be a work-around. Just like getting a BMS rated twice what you need.
"I^2 R" or "I squared R" is the power dissipated in a resistor.
Pulses of 200A half the time would deliver same power as 100A all the time, but would cause more heating in the wires.
If wire, breaker, or another component was 0.001 ohm, power dissipated in it is:
100A x 100A x 0.001 ohm = 10W all the time
200A x 200A x 0.001 ohm = 40W half the time, 20W average.
Obviously if breaker dissipates 20W not 10W, it'll trip sooner. Can only carry 1/sqrt(2) = 0.7 times as much current
140A x 140A x 0.001 ohm = 20W half the time, 10W average
So 70A average current (140A half the time) would heat breaker same as 100A continuous.