Building a current limiting circuit - Page 1
Forum section: electronics / projects

author: Henning S
author_rank: Newbie
timestamp: January 04, 2014, 08:50:35 pm
content: Hey there folksI´ve read an article about shunt resistors and thought whether i could use it myself to measure current coming out of an PS and limit it to e.g. 3AThe idea is not to drop the voltage if a current of 3A is exceeded but to cut the voltage completely with a relay.After drawing some stuff i created a project in Eagle to not harm you with my drawing skills.A screenshot of the circuit should be attached below.Question: Does Does the Relay cut the Voltage if a current of 3A is exceeded, with this configuration?I´m aware of the fact that the Relay would connect again if it disconnected because the current would be 0 A again but adding a "memory circuit" afterwards is not a problem for me.

author: tszaboo
author_rank: Super Contributor
timestamp: January 04, 2014, 10:24:04 pm
content: The MAX907 will blow up with 12V power supply. So after you changed it, take care that the common mode voltage close to zero is not causing problems. After the (oh shit) realise, that the input offset voltage will make your circuit absolutely unreliable, as it is half of your full scale input voltage. And relays are PITA.So read about circuits which are designed to for current sensing (caution, self advertizing ahead):http://nandblog.com/current-shunt-monitors/and, after you understood how they work, look at the datasheet of thehttp://www.ti.com/product/ina200,figure 31.Hey, it is also cheaper if you do it the proper way!

author: papo
author_rank: Regular Contributor
timestamp: January 04, 2014, 10:47:26 pm
content: HelloQuote from: Henning S on January 04, 2014, 08:50:35 pmThe working principle is correct, but there is a number of things that will prevent this particular circuit from operating properly:First of all, I'd say that using a comparator instead of an op-amp was a sensible decision. However, the MAX907 is not rated for 12V. Hence you will most likely fry that part. Also, you should consider decoupling the supply rail with a 0.1uF cap close to the compoarator.Then, the MAX907 is quite a weak driver. Have a look at the center plot on page 5 of the datasheet, "output high voltages vs. source current". Note that this plot assumes a 5V supply rail, but you must expect output voltages of less than 4 V with no load, and if a load is attached it gets even worse. A general purpose relay won't do it with such a low driver. If I were you and I were to use a relay, I would look for another comparator or use the comparator to drive a transistor controlling the relay.This might be just some fuzziness while drawing, but you have a 300 Ohms trimmer and the threshold at 3 A is 3.6 mV which you get when you max out the trimmer. Ideally, you would have room in both directions. Make sure to take into account the tolerances of the 12V supply, of the 1Meg resistor (what if the 1Meg is off by 300 Ohms in either direction? Is this within the tolerance of the resistor and what are the consequences?), and especially of the shunt. 1.2 mOhm is terribly small. This results in voltages that are really small, too. And noise might trigger the relay accidentally? Maybe you can go for a slightly larger shunt resistance? Keep in mind though that the shunt will "lift" the negative terminal of your load. E.g. with a 1 Ohm shunt will have a 3 V drop across it at 3 A and the load will only see 9 V. I assume this is not acceptable, so you'd have to go for something inbetween. High enough to create a significant drop to measure but sufficiently low not to disturb the load.Also, you have to decide whether you want to trigger the relay on spikes > 3 A or whether you want to do so after the over current condition maintains itself for a brief delay.Quote from: Henning S on January 04, 2014, 08:50:35 pmThat is very true and important to recognize. Not only would the relay turn on again if no latch is present, but the overcurrent condition would trigger the relay again, leading to a sad relay emitting machine-gun-like sounds.A last comment: Don't short the shunt. If you short the shunt, an overcurrent condition will not be recognized as such due do the lack of a drop. For example, if your 12V supply is not floating, do not attach the ground clip of an oscilloscope to the negative terminal of the load, which might look like "ground" of the load. (What is ground, anyway...)I hope that helpsBest regardsMatt

author: Henning S
author_rank: Newbie
timestamp: January 04, 2014, 11:17:26 pm
content: QuoteOk that´s what i wanted to hear.I haven´t thought of any comparator in particular, just took one of the list available in Eagle. Power for this Chip would add just another little regulator.QuoteWell 1.2 mOhm was the spec of one shunt i´ve found online. Don´t think the go as high as 1 Ohm when they are rated for 50 A. There must be a way of using a shunt with a resistance this low otherwise they would not be offered.QuoteA delay circuit could be added between the comparator and they relay same as the latching circuit. Thanks for the tip.QuoteI would have used a floating PC PSU for the supply and also my USB Osz. is powered by a floating PSU so this shouldn´t be a problem. Watched Dave´s Video about that too.

author: papo
author_rank: Regular Contributor
timestamp: January 04, 2014, 11:58:57 pm
content: HelloQuote from: Henning S on January 04, 2014, 11:17:26 pmOK then. Also note what NANDBlog said above whose points I would second. In any case, there's many ways to achieve a particular goal. Some people would not go for the regulator but either with a comparator that can take the 12 V, with a current monitor etc. It's always interesting to discuss possible options.Quote from: Henning S on January 04, 2014, 11:17:26 pmWell if you pump the max rated current of 50 A through that shunt you end up with a drop 60 mV. This is more than 10 times what you are trying to measure and I would argue that in terms of measurements, 60 mV requires considerably less precautions than 3.6 mV. If you think about it that way, there clearly is a way to use that shunt.In your case, a 10 mOhm shunt would lead to a 30 mV drop and a 0.1 Ohm shunt to 300 mV. Since I don't know what the load is, I don't know whether this is acceptable or not, but it would make a few things easier for sure.Quote from: Henning S on January 04, 2014, 11:17:26 pmRight, you can probably even combine the two. When sizing the delay, you may want to look at the current when you turn the load on. Some loads have quite high in-rush currents during turn-on. And since you are putting some circuitry between the amplifier and the relay anyway, that would also be the place to ensure that you can drive the relay into pickup (i.e. with something stronger than the comparator's 3 V at 1 mA or whatever it was).Quote from: Henning S on January 04, 2014, 11:17:26 pmNice. Just wanted to have that cleared. By the way, I just moved to a new place and have not yet moved my gear back from work to the "home lab", so I had to use a PC supply for a quick test with a low side shunt. And guess what, that supply was not floating and I ran *exactly* into what I described above. 30 mA of current just magically disappeared.Regards and have funMatt

author: pmbrunelle
author_rank: Regular Contributor
timestamp: January 05, 2014, 05:13:12 am
content: 1.2 mOhm is a very low resistance. You need Kelvin sensing (4-terminal measurement). There had best be nothing but the resistor between the voltage measurement points, i.e. no solder joints. For the voltage sensing points, solder flying leads directly to the (uninterrupted) resistor leads.Look at "dual-coil latching relays".