How shall I power the safety cutoff relay?


Nice, but at this point, I have no space left.


So you say I should lower the current after it closed? How could I do that?


someone know this type of switch, I want to use it! but I do not know if it’s good? for battery from 6S to 16S Lipo, 100A / 200A.


That was my idea last night, you could add a series resistor, but this would lower your losses only a bit as the resistor produces some heat then.
I would try to use one of those cheap 1€ buck converters / step-down converters from ebay and apply a lower voltage to the relays actuation coil. The relays coil is in static use like a resistor, it will draw current according to Ohm’s law. The the first buck converter I found is from china, but you should be able to find one from europe with reasonable shipping times.
But this is only an idea now, I don’t know how much you can reduce your temperature by this. But 25% * 12V = 3V and your coil has a resistance of 9,5V/1A=9,5Ohm. It should lower your losses to about 1W compared to 9,5W. You will have to test at which voltage your relay really drops off though.

Edit: You could get the same 3V across the relay and the same ~0,32A current with a series resistor of ~20Ohms but it will produce about 2W of heat.


You are using arduino to close it?
Why not use PWM signal?
Start with 100% Duty and after it is closed you can test with how low you can go, but I guess even 50% Duty should be enough…

Edit: only works if your relay closer relay is not mechanical.


That’s an even better idea, just a little more complicated than adding a second relay if you don’t love electronics :slight_smile:
@MaxMaker if you remember my schematic in Pacificmeister Build Info and CAD Sources you should be able to do this.


I don 't that much yet in relay but what about to use a normally closed relay and then apply power to open it and cut the battery?


Maybe even a simple transistor (like the 2N2222A) can do the job. Depending on the current you need for your coil…


Well, I can turn the UBEC down to 8V. And then make a relay divert the current through a buck converter at maybe 5V after the relay was closed. But with that I can only change the voltage. Will the coil not automatically draw more current at a lower voltage?

I could also take out the winding and double the number of loops with thinner windings.


The relay coil acts like a resistor to DC voltage. More voltage more current, U=R*I basically. Just some nasty spikes from the inductance when switching off.
Don‘t modify the coil, it‘s not worth the effort.


Would that not mean that more voltage equals less current?


I just did some testing at the lab power supply.

Relay Closes at 8.5V and draws 0.5 A.

At 9.5V it draws 0.56A, but I measured 1A with the UBEC at the same Voltage. Is th UBEC a current source???

At 5V it draws 0.29A
At 3V it draws 0.17A
At 2V it draws 0.12A
Below 2V it opens up again.

So Flo is right, I could save a lot of energy. But why does it consume less current with the lab power supply than with the UBEC? 1A vs 0.5A.


no, R is constant so I=U/R
less voltage, less current

how do you measure it? and where, before BEC or after?


Ok today I briefly tried again an measured .5A. Maybe the coil drew more A at 100°C?

Ill do more tests when I have time.


Well, I guess the windings of your coil are copper, a̶n̶d̶ ̶y̶o̶u̶ ̶a̶r̶e̶ ̶r̶i̶g̶h̶t̶,̶ ̶c̶o̶p̶p̶e̶r̶ ̶i̶s̶ ̶a̶ ̶N̶T̶C̶ ̶(̶H̶e̶i̶ß̶l̶e̶i̶t̶e̶r̶)̶ ̶s̶o̶ ̶R̶ ̶g̶o̶e̶s̶ ̶d̶o̶w̶n̶ ̶w̶i̶t̶h̶ ̶t̶e̶m̶p̶e̶r̶a̶t̶u̶r̶e̶,̶ ̶a̶n̶d̶ ̶a̶c̶c̶o̶r̶d̶i̶n̶g̶ ̶t̶o̶ ̶I̶=̶U̶/̶R̶ ̶=̶>̶ ̶I̶ ̶w̶i̶l̶l̶ ̶r̶i̶s̶e̶…but the coefficient is something like 10^-3 per degree (page 5). So I guess that might have had an effect, but nothing like doubling (thats just guessing, to really proof it you have to measure R at some given temperature, then you can calculate how the resistance will change and so how the current changes (given that you are in a steady state))


Hi, Copper is a PTC in principle, a “Kaltleiter” in german. After switching on the constant voltage, the coil is heating up to a certain temperature difference to ambient and will remain there if you do not change the voltage, by heating up the resistance increases and the current decreases.
Max, check your measurement equipment and your setup, something is wrong. Your UBEC has unexpected features.
Circuits used to decrease the voltage applied to coils after switching has been done are also called “Stromabsenkung” in german, current reduction in english.


There are some solenoids that actually use dual relays to power on, then once “on”, the other one drops out, thereby reducing the current by half.


oh, yeah sure, you are right. Alpha is positive for copper so R increases…
I will correct it above.


Thats interesting, do you have some example which could help us?


I found the mistake. Two cables were switched and the relay actually received 30V at 1.5A which is 45W. Now at 9.5V and 0.6A it only reaches about 40°C. Thanks for your help.

I could still attach one more relay that would connect it to a lower voltage circuit after it closed, but I don´t have physical space left or pins on my Arduino.