How to connect a 12V pump to a relay? - Page 1
Forum section: electronics / beginners

author: Lucky-Luka
author_rank: Unknown
timestamp: July 03, 2020, 09:39:03 pm
content: Hi allI'm doing a little project involving Arduino.I have to control a 12V pump.https://youtu.be/iVRP8C9HSNEIt looks like it's working just fine.I am wondering if I need to add a flywheel diode or not.What do you think about it?Cheers

author: garethw
author_rank: Regular Contributor
timestamp: July 03, 2020, 09:59:57 pm
content: I would say that you would need freewheeling diodes.They will most likely be included on your relay control board already. You would have to add them yourself if you design a PCB for example.

author: JustMeHere
author_rank: Frequent Contributor
timestamp: July 04, 2020, 03:17:26 am
content: You should have one on the relay coil and one on the pump terminals. If the pump is AC then you need another type of snubber. I've never figured out the right answer for the AC snubber. MOVs are what I saw on a dishwasher controller. The pump one isn't as important but the relay's contacts will be damaged by the load dump over time.

author: Lucky-Luka
author_rank: Unknown
timestamp: July 05, 2020, 11:52:31 am
content: I have a 12V DC pump.Which flywheel diode should I use? 1N4007 (1000V 1A)?

author: garethw
author_rank: Regular Contributor
timestamp: July 05, 2020, 05:06:24 pm
content: 100V rated diodes would be more than sufficient. Even 50V ones would work. When you add the diode to the circuit the voltage cannot get above the voltage required to cause the diode to conduct (~0.6V)Lets, as an example, say that your relay requires 100mA to energise its coil. Your relay is switched on so you have 12V across your coil and 100mA flowing. Now you turn off the relay by breaking the circuit. The relay coil now wants to continue passing 100mA so its polarity flips and now looks like a current source. Your diode now becomes forward biased and starts conducting with around 0.6V drop across it and 100mA flowing through it. Your coil is quite happy draining its energy away without having to increase its voltage because the diode easily conducts until the coil has 'emptied'Now lets put a 1000 Ohm resistor in series with the diode. Now the coil has to pass its 100mA through the diode and the resistor. Ohms law gives us (neglecting the resistance of diode):100mA * 1000 Ohms = 100V.You relay coil would produce 100V to pass the 100mA. This would reduce very, very quickly as there is not much energy stored in the coil, but its still enough to cause sparks across switches or blow transistors.

author: Lucky-Luka
author_rank: Unknown
timestamp: July 12, 2020, 06:38:17 am
content: So it's better not to put a resistor in series with the diode, right?

author: Zero999
author_rank: Super Contributor
timestamp: July 12, 2020, 09:01:05 am
content: Quote from: Lucky-Luka on July 12, 2020, 06:38:17 amAdding a resistor speeds up the switch off time, but increases the back-EMF voltage. It's a trade-off. If your transistor is rated to 60V and the relay draws 100mA, you could add a 330R resistor in series with the diode, which would increase the back-EMF to just over 33V, which is added to the 12V supply, giving 45V, plus a diode drop.