Current limiting from power supply - Page 1
Forum section: electronics / beginners

author: JamesWedmore
author_rank: Newbie
timestamp: January 12, 2024, 08:47:37 pm
content: Hello,My problem that I am facing is that the power supply that we are using (12V 30A) is being used to power a 12V 10A linear solenoid. When attached directly to the power supply, it blew a 15A fuse. How would I go about limiting the current so that only 10A goes into the solenoid and not 30A?I don’t know if it is sending a constant 30A though and I need a current limiting circuit or if the in-rush current is just too much and I will need to use something like a PTC thermistor.

author: BeBuLamar
author_rank: Super Contributor
timestamp: January 12, 2024, 08:57:40 pm
content: Something is wrong with your solenoid. You do not need current limiting on the power supply. The solenoid should only draws what it needs. You may use an AC solenoid instead of DC which makes it draws a lot more power than it should.

author: JamesWedmore
author_rank: Newbie
timestamp: January 12, 2024, 09:42:12 pm
content: Sorry if this changes things, but we are using a 12V 30A battery to power the solenoid, not a power supply connected to the wall. My bad for not clarifying.

author: tooki
author_rank: Super Contributor
timestamp: January 12, 2024, 10:00:06 pm
content: What do you mean by a “12V 30A” battery? Battery specs can be more subtle than that.E.g. in a lithium ion cell, the maximum current it should be used at (the maximum current spec) is far less than the maximum current the cell is physically capable of providing (the short circuit current) — it’s just that doing the latter will damage the cell.Batteries don’t “send” constant amounts of current. They provide a constant voltage, which combined with the resistance of the load results in a current.A solenoid is an inductive load, so it doesn’t have an inrush current to deal with. (Inductors slow the rate of change of current flow.) In exchange they have inductive kickback when turning them off (the current wants to keep flowing but can’t, so causes a surge in voltage. We use diodes to clamp these spikes.) So I agree with BeBuLamar that there may be something wrong with the solenoid.Also, it’ll depend on the fuse: there are slow-blow, fast-blow, ultra-fast, and everything in between, and each value of each type from each manufacturer or series will have a different time-current curve.Can you please share the exact models of battery, solenoid, and fuse?

author: JamesWedmore
author_rank: Newbie
timestamp: January 12, 2024, 10:43:25 pm
content: The battery is I believe a Milwaukee drill battery that is 12V 3000mAh 36Wh rechargeable lithium ion battery. The solenoid is a Fielect DC 12V 80N 10A Push Pull Type Solenoid Electromagnet Open Frame Solenoid Electromagnet Linear Motion, JF-1683B-12V.I’m pretty sure we were using the fuse wrong by putting right after the battery so it blew immediately, but my group is worried that if we just connect the battery to the solenoid it might burn or damage the coil or wire in the solenoid. The other question I would have is how does the battery send the current, as in how much and how quickly it sends it through the wires in case directly connecting it to the solenoid would damage it.

author: Jwillis
author_rank: Super Contributor
timestamp: January 12, 2024, 11:06:45 pm
content: Quote from: tooki on January 12, 2024, 10:00:06 pmALL inductive loads have inrush current when switched on. Even chokes. Solenoids are either on or off and have an initial inrush current when activated. The amount of inrush current is dependent in the impedance of the coil. The lower the impedance the higher the inrush current.Soft starting solenoids with a PTC can slow the inrush currrent but can also slow the action. The 10A rating is the holding current. The datasheet should indicate an inrush current as well, which will be higher than the holding current depending on the impedance of the coil. You base the fuse on the inrush current and not so much the holding current.

author: tooki
author_rank: Super Contributor
timestamp: January 12, 2024, 11:16:39 pm
content: Quote from: Jwillis on January 12, 2024, 11:06:45 pmI cannot find a single instance of a solenoid datasheet listing an inrush current, and every source I found (both before writing, and after your response) says DC solenoids do not have inrush current. This is the best overview of inrush current types I found:https://electronics.stackexchange.com/questions/247297/inductor-in-rush-current

author: tooki
author_rank: Super Contributor
timestamp: January 12, 2024, 11:43:57 pm
content: Quote from: JamesWedmore on January 12, 2024, 10:43:25 pmThanks.OK, so a powerful battery and a no-name solenoid, so unfortunately no datasheet.How did you arrive at 30A for the battery? (That is a very plausible number for the 18650 cells that are likely within it, I’m just curious where you got the number since Milwaukee does not specify.)Quote from: JamesWedmore on January 12, 2024, 10:43:25 pmThe fuse is (or at least, is supposed to be!) in series, so its position in the circuit does not affect when it blows.Quote from: JamesWedmore on January 12, 2024, 10:43:25 pmAssuming the solenoid is in proper working order, it should not be damaged.Quote from: JamesWedmore on January 12, 2024, 10:43:25 pmThis is described by Ohm’s law, the most important fundamental relationship in electricity.Current flow is driven by voltage, which is basically electrical pressure. (More specifically, the difference in pressure between two points. For example, if we use a pneumatic analogy, if you fill two tanks of gas to the same pressure, and then connect them to each other with hoses, but with a tissue paper membrane separating their gas, the paper won’t break because the pressure on both sides is the same, even if it’s way higher than atmospheric pressure.) Think of your battery as a gas cylinder at pressure. Voltage uses the symbolVorUand uses the unit Volts (V).Next, your load has resistance — the degree to which it resists the flow of electricity. (This is akin to the size of the hole through which you let your pressurized gas escape: a big hole will let lots of gas flow, a tiny pinhole will let it flow very slowly.) Resistance uses the symbolRand uses the unit Ohms (Ω).The result of the above two factors is the current (the amount of flow). Current uses the symbolIand uses the unit Ampere (A).The relationship is:I = V/R.So suppose you have a load that has a resistance of 4 ohms. If you connect it to your 12V battery, the math says the the following current will flow:12V/4Ω = 3AGranted, inductive loads (coiled things like solenoids and motors) behave a little differently, but nonetheless ohm’s law is how current works.A key concept is to think of it as if voltage sources (like your battery or power supply) don’t push current, but rather that loads draw it. (I know I just explained above how pressure works, but electricity has to flow in a circuit, so until the load is there, it’s an open circuit and current can’t flow. So for basic practical purposes, think of it as loads pulling current.)As for “how quickly”, well, the answer is around 2/3 the speed of light in copper. But this doesn’t matter here.

author: Jwillis
author_rank: Super Contributor
timestamp: January 12, 2024, 11:53:53 pm
content: Quote from: tooki on January 12, 2024, 11:16:39 pmOH Geez. My apologies. I had my head in the AC cloud.

author: IanB
author_rank: Super Contributor
timestamp: January 13, 2024, 12:01:34 am
content: Quote from: JamesWedmore on January 12, 2024, 08:47:37 pmThe fuse you are using is probably too small or has an incorrect rating. Try using a 50 A fuse instead of a 15 A fuse.That was bad advice, stick with the 15 A fuse.Also note that according to the product information, this solenoid is for momentary use only, and should not be energized for more than 2 seconds at a time, or it will overheat and burn out.Is it possible you have previously tried to operate it for more than a second or two, and have overheated the internal windings, and therefore caused an internal short? That could explain an excessive current draw.

author: tooki
author_rank: Super Contributor
timestamp: January 13, 2024, 12:03:59 am
content: Quote from: Jwillis on January 12, 2024, 11:53:53 pmNo worries.(Meanwhile I wish my head were more in the AC cloud, by which I mean I need more experience and practice with AC theory and real-world circuits.)

author: tooki
author_rank: Super Contributor
timestamp: January 13, 2024, 12:08:14 am
content: Quote from: IanB on January 13, 2024, 12:01:34 amThat seems like terrible advice. It’s supposed to be a 10A solenoid, so there’s no reason it should be drawing 50A. And there’s no way that little battery is designed to deliver 50A. And a 50A fuse will pass 50Aindefinitely, with a quick blow happening at a significant multiple of the rated current, while a slow blow happens at something like 1.5-3x the rated current. There’s no guarantee that battery could quickly clear a 50A fuse, and letting 50A+ flow through 10A wire is asking for disaster.

author: JamesWedmore
author_rank: Newbie
timestamp: January 13, 2024, 05:24:14 am
content: For the 30A on the battery, I was told that you can basically take the mAh and that should be what it can output within a short instant.We ended up removing the switch we were using and just touched the wires together to see if the solenoid either works or burns up, and it ended up working correctly.

author: p.larner
author_rank: Frequent Contributor
timestamp: January 14, 2024, 08:02:23 am
content: so was the switch the issue?,what your saying makes no sense!.

author: BeBuLamar
author_rank: Super Contributor
timestamp: January 14, 2024, 10:20:24 am
content: Quote from: JamesWedmore on January 13, 2024, 05:24:14 amdo you still have the fuse in the circuit?

author: tooki
author_rank: Super Contributor
timestamp: January 14, 2024, 01:44:25 pm
content: Quote from: JamesWedmore on January 13, 2024, 05:24:14 amEh, not as a blanket statement, since it depends on the battery type. It’s not a horrible rule of thumb for lithium ion, but it’s a gross ballpark figure at best. And as I said in my first reply, there is a huge difference between the amount of current a batterycandeliver and the maximum current itshoulddeliver, because many batteries will be damaged by too high a current. For example, a standard 18650 cell is designed to be used at up to 10A, but the short-circuit current is somewhere between 50-100A. Some high-current 18650s are designed to be used at up to 35A, with a short-circuit current around 300-600A!! But drawing more current than it’s designed for will damage the cell: minor excess will just shorten the cell’s life, gross excess will cause catastrophic damage to the cell within seconds.Quote from: JamesWedmore on January 13, 2024, 05:24:14 am”within a short instant” - an instant is by definition short or definite.But did they mean “can deliver that high current for a brief moment”? Because “within” makes no sense; the battery doesn’t need to “build up” to the current or something, it’s delivered instantly.And just to be thorough, in the phrase “short-circuit current”, that means the current that flows if you were to short out the battery terminals with a very thick conductor. It doesn’t refer to length or time.