Why my Relay short-circuited ?

Thread Starter

MrsssSu

Joined Sep 28, 2021
181
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I have a relay but rated at 3A, 30V DC only similar to this. When I connect the Pole and NC directly using a lithium ion battery (4V), a fire occur and it shortcicruicted. However, when i measure the resistance betwwen 2 pins (Pole and NC), it reads 16 Ohms. What should i do :) ?
Thank you for reading
 

ScottWang

Joined Aug 23, 2012
7,179
Pole(COM) is the middle pin and the NC is the Normal Close pin, the two pins should be connected to the Load or your device as switch and don't connected to the power like battery.

Normally you should connect the pole(COM) pin and NO(Normal Open) pin to the Load.

The power should be connected to the left side A, B pins.
 

panic mode

Joined Oct 10, 2011
2,113
yup, do not leave load out of the circuit. load is what limits the current. once the short circuit was formed, relay contacts are badly damaged (burned) which is why you read weird value.
 

crutschow

Joined Mar 14, 2008
29,521
What did you expect would happen when you connected the battery directly across the two relay poles that created a short-circuit across the battery?
 

Tonyr1084

Joined Sep 24, 2015
6,861
Redrawn from other schematics. I didn't feel like changing the battery description, but the point should be clear. The first is an instant short when connected to the battery. The second is a short as soon as you activate the relay. The third is the typical proper way of using a relay WITH A LOAD. No load is a dead short. I believe that's already been stated. You can use the relay to turn something ON when you activate the relay or you can use the relay to shut something OFF when it's activated. The point here is you didn't include a load to limit the current. A 4V battery (of any kind) should be able to deliver a lot of current, and cause a fire, as you experienced. To limit the current to some specific value, assuming the battery is forever constant voltage, to limit it to just 100mA (mA = Milli Amps; also equals 0.1A) you would need a load who's resistance (assuming a resistive load) would need to be 4÷0.1=40Ω. Lower the load resistance increases the current. Increasing the load resistance lowers the current. Of course a battery will drain, and as the voltage drops so will the current. But for sake of argument, a 4V source (battery or any other kind of source) delivering 100mA will occur when you have a (resistive) load of 40 ohms.
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