Can I power a 24V relay from +12V and -12V? - Page 1
Forum section: electronics / beginners

author: tarrryan
author_rank: Newbie
timestamp: March 15, 2015, 02:00:58 pm
content: I'm currently designing my first AC>DC supply to be built into another project later on (a function generator). I have a modest amount of experience with DC>DC supplies but this is the first time I'm going to use mains AC to power a project. It's also the first time in 8 years I have asked others for help, normally I would find the solution through trial and error or searching the internet, I'm not going to do that this time when mains AC is involved.I would like to use a relay at some point after the 7812 & 7912 regulators to switch the power on and off, I started off with a 12V relay switching the +/-12V rails on and off but realised that if the -12V fuse blows the +12V rail would not be terminated while the +12V fuse blowing would terminate the supply until it was replaced.After some though I wondered if I could power a 24V relay from the +/-12V rails, the contacts will open after a fuse blows letting it fail in a off state, or at least I think this is what will happen. I have attached a circuit diagram with the relay (K1) and fuses F2 & F3) in question as well as the rest of the circuit. please feel free to criticise any other aspect of the diagram/design you see, I'm perfectly happy dealing with AC or DC on their own but I'm still a little cautious about combining the two.I hope that makes sense and thank you for your help.

author: SeanB
author_rank: Super Contributor
timestamp: March 15, 2015, 02:14:53 pm
content: Power it from the unregulated rail. Most 24V relays will have a minimum pull in voltage of somewhere between 14 to 18V, and will operate with a 16V coil voltage. Depends on the relay, look on the datasheet for the minimum and typical pull in voltages, and see if your supply unregulated rail will deliver that. Most likely the negative rail, as those typicall are the lightest in load so the unregulated rail is the highest.

author: rs20
author_rank: Super Contributor
timestamp: March 15, 2015, 02:18:31 pm
content: You seem to be asking two different questions here:A. Can you power a 24V relay from +12V and -12V? Answer: Yes, that's a 24V differential; just be aware of how much current the coil is drawing, but apart from that, you'll be fine. (Although a 24V relay doesn't draw very much current, so you're probably fine).B. Will a 24V relay reliably turn off if I only feed 12V to it? Most probably not, check out atypicalrandomly chosen relay datasheet:https://www.components.omron.com/components/web/PDFLIB.nsf/0/0293ACA6B03161CD86257538007424DB/$file/G5LE_0813.pdfIn particular, the "Must release voltage" 10% min. of rated voltage (min.), or 2.4V in your case. So, the relay could be fed with 2.5V and it might still stay on while remaining in spec. Of course, there's likely a fairly large fudge factor in there, but you shouldn't rely on such a fudge factor, and to assume that it'll turn off at 12V is really pushing it. When the little electromagnet in the relay pulls the contact closer, it's easier to hold it close, just like a magnet snapping to a fridge, so that's why the holding voltage is so much less than the actuation voltage.EDIT: Looking at Sean's response, perhaps my randomly chosen relay is a bit of an outlier. I am open to correction. But certainly, check your relay's datasheet, everything you need to predict what'll happen is in there.

author: Richard Crowley
author_rank: Super Contributor
timestamp: March 15, 2015, 02:39:58 pm
content: It is not clear what benefit you derive from the addition of the relay? If you were designing a "bench supply" of some kind, that would be one thing. However you state this design is for a fixed circuit load (a function generator). Note that hard-switching the load downstream from the 3-terminal regulators, and leaving the regulators suddenly without a load, and the transients from switching that inductive load (the relay coil) are all issues I would find significantly negative vs. whatever value the relay may provide.And, as SeanB said, it is common to operate old-school "dirty" loads like relays, incandescent lamps, etc. from a SEPARATE bus, well isolated from the main regulated outputs. For example, see Gyraf's schematic for the SSL Mixbus Compressor Clone at:http://www.gyraf.dk/gy_pd/ssl/ssl_sch.gifYou can see that there is a separate regulator and output for the "light, relay" loads. The power is taken directly from the rectifiers, BEFORE entering any of the power filtering and regulation for the audio or control circuits.Honestly, your circuit seems like "overkill" and quite possibly a greater downside than any possible benefit.

author: tarrryan
author_rank: Newbie
timestamp: March 15, 2015, 04:14:14 pm
content: Thank you to everyone for your replies, I had a chance to read them and think about in on the way home from shopping.I'm pleased to hear I can power a 24V relay from a +/-12V supply, I thought this might be the case but i have no way of currently testing it. Also I don't think I was very clear on one point as I would never consider powering a 24V relay from anything other than its design voltage, thats just a bad idea in general.Now onto the messy business of my supply, I can see i have made something of a badgers arse of it. Firstly ignore the part about it powering the function generator for the moment, I believe I picked the wrong wording for it, my intention is to be able to supply 2A(ish) at +/-12Vdc with various signals and duty cycles for experimenting with, I dont just want to have a fixed low power output.I gather from your comments that powering the relay coil from the regulated side would be a bad idea and that I have two options, I can either power a 12V relay from the +15V unregulated side just after C2 or use a second 7812 to supply the relay coil. Of course I could just remove the relay entirely or replace it with a switch.I've also just noticed that it would be better to power the 7805 from the regulated side of the 7812 rather than the unregulated 15V, please correct me if im wrong.Thanks again.

author: SeanB
author_rank: Super Contributor
timestamp: March 15, 2015, 04:30:23 pm
content: You can run the 24V relay off the unregulated rail, provided the voltage there is over the minimum operate voltage. It will operate reliably.As to the 5V supply, yes running the 5V regulator from the 12V rail is better as it spreads the heat dissipation out over 2 regulator chips. Just note that you should use a regulator from a reputable supplier, as the no name ones might not have the ability to deliver the full current rating, and might fail with large voltage differences between input and output.As well look at the application notes and you will see a diode across the regulator, and another across the output, for reverse voltage protection. Please add these if you are going to use this as a benchtop power rail, as these protect the regulators against spikes and reverse input voltages. Without them there is a higher risk of blowing up the regulator from mistakes.

author: tarrryan
author_rank: Newbie
timestamp: March 15, 2015, 04:50:35 pm
content: THankyou for the heads up on the diodes, I had forgotten all about them, I'm planning to eventually have the power supply on one board and everything else for the sine/square wave control etc on another board with the two joined by a right angle 0.1 inch pin header, I like to keep my board broken up into modules if I can as I etch them all myself, Its much easier to replace or upgrade a module instead of an entire pcb with everything on it.

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2015, 05:00:28 pm
content: Quote from: tarrryan on March 15, 2015, 04:14:14 pmI'm curious about your hopes of getting 2 A from this supply. You don't have any pass transistors. I would think 500 mA is a more reasonable expectation.Also, 15 V unregulated at C2? If that is the peak smoothed voltage after the rectifier it would suggest you have a 12 V AC secondary on the transformer? I am not sure that will provide sufficient excess voltage to handle the drop-out requirement of the regulators.

author: tarrryan
author_rank: Newbie
timestamp: March 15, 2015, 05:25:22 pm
content: Quote from: IanB on March 15, 2015, 05:00:28 pmI'm unsure how i would have 12Vac at the transformer and then jump up to 15V at the cap? I think you might of made a typo or misread something.I'll have a 15Vac secondary, even taking into account the voltage drop from the rectifier diodes i should still have about 14.3V at the smoothing cap, with a dropout of 2V i will be cutting it a little close but it should still be functional, if needed I can always go up to 18Vac instead of the 15Vac if it proves to be too close.As for the 2A, I could use pass transistors but I have not had much trouble in the past as long as I heat sink the regulator, or are you suggesting pass transistors would be a more efficient way of doing this? Thanks for your input.

author: Seekonk
author_rank: Super Contributor
timestamp: March 15, 2015, 05:52:36 pm
content: I'd take a tip from almost every wall wart power supply. Power the relay from the raw 16V using an opto isolator. Opto likely will have enough drive current if not, just darlington a transistor. On the LED side use any zener over 12V - 20V and a resistor to set the current limit to led.

author: IanB
author_rank: Super Contributor
timestamp: March 15, 2015, 06:06:20 pm
content: Quote from: tarrryan on March 15, 2015, 05:25:22 pmBecause the 12 V AC on the transformer is the RMS voltage. If you rectify it and smooth it with a filter capacitor you get the peak voltage which is 1.4 times the RMS voltage: 1.4 x 12 = 17 V. (Less a bit from the diode drops.)If you start out with a 15 V transformer the 15 V will be the loaded voltage. Unloaded it will be more like 17 or 18 V, and multiply that by 1.4 you could have as much as 25 V on the filter caps.