Relay driver and transient protection - Page 1
Forum section: electronics / projects

author: edmundopt
author_rank: Regular Contributor
timestamp: January 13, 2014, 11:14:38 pm
content: Hello againThis might be a beginners question, if so, feel free mister moderator to move this topic.I have a relays from Finder(12volts with 218ohm coil resistance) and I want to drive it via a 2N4401 transistor.I need some help to make the math on the diodes.First of all, are there any know values of the coil inductance? Can't find any on datasheet(Finder,Omron...).Second, the 2N4401 has a maximum collector emitter breakdown voltage of 40volts, just as little as 20Henry on the inductor and the voltage spike is near 30volts, what would be a safe value ?Do you have any sugestions for this ?

author: sleemanj
author_rank: Super Contributor
timestamp: January 13, 2014, 11:28:35 pm
content: This is an 8 henry inductor.How big is your relay to be 20

author: edmundopt
author_rank: Regular Contributor
timestamp: January 13, 2014, 11:40:18 pm
content: What can I say, one image equals one thousand words! Thank you.ok, the relay is a 12volts Finder 40.31 miniature PCB relay(10A,250VAC) , so, maybe I should consider like .02 henry? The fact is that I have no idea and no way to measure it, so I can't run the simulation.Anyway, considering the "new" value, a 0.5W 13volts zener will be enough!Do you agree with the transient protection ?

author: sleemanj
author_rank: Super Contributor
timestamp: January 13, 2014, 11:43:59 pm
content: Ok, properly speaking the inductor size isn't really a big factor in the inductance, but because more or less more turns are required, more or less larger inductance means larger inductor.A quick skim of some relay datasheets indicate 10 Ohm coils, around 0.5H coil inductance.

author: edmundopt
author_rank: Regular Contributor
timestamp: January 14, 2014, 12:19:32 am
content: I found something, using a diffrent google search, and this documenthttp://www1.futureelectronics.com/doc/TYCO/1393143-8.pdfhas the attached graph.Their 10series relay has a resistance fo 185ohms with the 12volts coil, the dimensions are larger.The fact that the graph is valid for all their models, means that, in this particular manufacturer of repays ,there is a certain linearity between the coil resistance and the inductance.So, with some false extrapolation, the inductance value of my relay would be arround 5Henry..Anyway, the supression circuit seems ok under the simulation.

author: Andy Watson
author_rank: Super Contributor
timestamp: January 14, 2014, 12:24:24 am
content: Quote from: edmundopt on January 13, 2014, 11:14:38 pmFrom the title, is it reasonable to assume that we are talking about the diode across the coil to catch the back-emf/kick-back (whatever you call it in your neck of the woods) ?QuoteInductance is not really meaningful - it will change when the relay pulls-in.The diode needs only to withstand the ON voltage of the relay, in this case 12V. Anything more is a bonus. The voltage spike that occurs when the relay is turned-off should be caught by the diode, i.e. the diode should be conducting to clamp the voltage spike. When the relay is turned-off the current decays, i.e. it is never more than the energising current, in this case 12V/218 R = 55mA.

author: edmundopt
author_rank: Regular Contributor
timestamp: January 14, 2014, 12:41:07 am
content: Yes, my concern was the Zener diode and the schottky diode parallel to the relay coil.I was concearned about inductance because of some simulations with large inductance values. They reached a voltage near the maximum collector emitter breakdown voltage of the transistor.Anyway, I have learned one or two things about it so I think the circuit will be ok.Thank you

author: vk6zgo
author_rank: Super Contributor
timestamp: January 14, 2014, 05:37:55 am
content: The Zeners are massive overkill-----normal commercial practice is to use a single diode (something like a 1N4007) connected across the coil,so that as far as the power supply is concerned,it is reverse biased.The spike from the relay is of opposite polarity to the supply,& is shorted out by the diode.Although the spike may be high voltage,it contains very little energy,so a power diode can easily handle it.

author: Kremmen
author_rank: Super Contributor
timestamp: January 14, 2014, 07:02:41 am
content: Guys. There will be no "spike" in the first place if you have the freewheeling diode in place.The voltage across a coil is proportional to the inductance and significantly, to the rate of current change; di/dt.If you break the coil energizing circuit the rate of current change would be infinite which would in turn cause an infinite voltage across the coil. This does not happen in practice but the resulting voltage excursion can be significant (kilovolts in a bad case, until something gives).Connecting the diode in antiparallel just provides a path for the coil current to continue. Thus the current will circulate, and slowly dissipates into heat in the resistance of the coil copper or other resistances in the freewheling circuit. No voltage spike because no abrupt di/dt.As already noted the diode needs only to withstand the coil working voltage. Unless the diode is minuscule, the freewheeling current is short in duration and thus the average diode current is low, placing no stringent requirements for the diode max current capability.

author: vk6zgo
author_rank: Super Contributor
timestamp: January 14, 2014, 07:12:47 am
content: Quote from: Kremmen on January 14, 2014, 07:02:41 amOf course,you're right!Bloody "Oldtimer's Disease" strikes again!I've described it to others years ago ,in very similar words to yours.

author: codeboy2k
author_rank: Super Contributor
timestamp: January 14, 2014, 12:54:49 pm
content: Quote from: sleemanj on January 13, 2014, 11:43:59 pmGood thing you said thatAs a kid (14-16) I used to build large inductors from silver plated copper tubing, like the one in the picture, which is actually a Harris 16uH inductor. It's about 12-14 inches long.