Simplest way to know if current is flowing in trace - Page 1
Forum section: electronics / projects

author: HSPalm
author_rank: Regular Contributor
timestamp: December 02, 2021, 03:02:08 pm
content: Hello,I am interested in finding a simple solution to have a digital signal to toggle when current is flowing or not in a wire- The wire is conducting 230VAC- The current will be around 90mA, but I'd like to widen the ranges of detection as much as possible- The detection circuit must sit "on the safe side", meaning galvanic isolationThere are lots of specialized hall effect or CT based ICs to measure the amount of current for an analog reading. Yes I can put this analog reading into a comparator, but looking for a simpler and cheaper solution.1. How much inductance would I need from a series inductor to create a magnetic field to be picked up by close by digital ouput hall effect sensor?2. Can I drop some voltage over a series resistor, and have the voltage drop power an AC photocoupler with digital output? I don't like putting series resistor, because I'm afraid of short circuit events.3. Those "wireless" voltage detectors that electricians use, how do they work? Must current be flowing for it to work?Thank you for chiming in with your thoughts!

author: CaptDon
author_rank: Super Contributor
timestamp: December 02, 2021, 11:20:05 pm
content: Those 'voltage detectors' do not need current flowing in the wire, only voltage to be present. You can use them to trace the hot lead of a wall receptacle even if there is nothing plugged in to the receptacle. I also like the two piece 'tracer' units. You can plug the little generator source into a receptacle and then use the detector part to sniff the fuse or breaker panel and find which circuit the receptacle is on. There are hall effect sensors made for PCB mount that have the current conductor built into them. We used some rated to over 30 amps but there are also more sensitive versions. They are available unipolar and bipolar. Of course bipolar is best for A.C., but then you also need a bipolar supply.

author: bob91343
author_rank: Super Contributor
timestamp: December 03, 2021, 03:36:51 am
content: I don't know what speed of response you are looking for, but consider a thermistor. Put a small series sensing resistor (if you even need it) next to a thermistor and see it change resistance with current. You can also put a second one in an area of no current and look at the differential.I have a DVM, HP3456A, that has temp readout with a 5k thermistor. I can set it for 6.5 digit temp readout and watch the millidegree change. This would work as a probe to look for traces that are carrying current, or not.

author: Berni
author_rank: Super Contributor
timestamp: December 03, 2021, 06:53:25 am
content: 90mA is not going to heat up a wire all that much. But temperature is a way of doing it. That youtube woodworker Mathias made a dataloger for a hot water heater using a I2C thermometer strapped to its mains cable. There the currents are large enough to make a easily detectable temperature difference.The easiest way to do this is a current transformer. You can buy them and they are just simply a toroidal inductor with a lot of turns. You pass the wire trough the middle of the toroid and this turns it into a transformer where your sensing wire is a single turn primary. If the toroid has 1000 turns this makes it a 1:1000 transformer so passing 1A trough the sense wire gives you 1mA on the output. This 1mA can then be sent into a 1 KOhm resistor to produce 1V. So you now have a completely isolated voltage that represents the current in the wire. All you need then is a comparator to detect it being above it. If you don't want to use a comparator then you can build your own out of a adjustable voltage reference like the TL431, or you can simply use a NPN transistor as that will turn on when you place about 0.6V on the base (Tho not quite as temperature stable as a reference).

author: fourfathom
author_rank: Super Contributor
timestamp: December 03, 2021, 07:03:19 am
content: Quote from: Berni on December 03, 2021, 06:53:25 amYes the current transformer is one way to go for good isolation, but of course you're going to get an AC voltage output which you will need to detect. The 50/60 Hz ones are fairly bulky as well.

author: gcewing
author_rank: Regular Contributor
timestamp: December 03, 2021, 08:03:15 am
content: If you want a really cheap solution, maybe you could put a low voltage light bulb in series with the circuit and use a photodiode or phototransistor to detect light from it. You might want back-to-back zener diodes in parallel with the bulb to protect it from excessive current, and a fuse in series with the whole lot in case of shorts.Still sounds a bit fragile, though. For robustness I think I'd go with the current transformer. Small ones don't seem to be terribly expensive, or you could probably make your own from a toroidal core if you wanted.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 03, 2021, 09:42:25 am
content: A makeshift current transformer is likely the way to go. There is no need for a super high turns ratio. Handwinding some 10 turns can be enough, more turns may help for high currents. The point is more that the secondary turns together should be thick enough to safely carry the current. So the number of turns times the cross section should be large enouugh, like 2 x the primary.Ideally there are 2 back to back diodes at the ouput of the CT to limit the worst case voltage and protect the circuit. For just a simple on/off the choice of the core is not so critical. For a current measurement the core should be high permeability (relatively large AL value), like the ones used in common mode chokes.

author: HSPalm
author_rank: Regular Contributor
timestamp: December 03, 2021, 02:13:58 pm
content: Quote from: CaptDon on December 02, 2021, 11:20:05 pmDo you know how these circuits operate or what they're called? Ideally they would only function when current is passing but it caught my interest. I'm talking about the touchless detector tool.Quote from: Berni on December 03, 2021, 06:53:25 amCurrent transformers can be quite bulky and expensive.. As mentioned by fourfathom. There are other single-chip solutions based on GMR (https://www.digikey.com/en/product-highlight/a/allegro-microsystems/acs70331-current-sensor-ics) or hall effect sensors. Ideally I would find one that respects isolation requirements if I go down that path. The ACS70331 does not.Quote from: Kleinstein on December 03, 2021, 09:42:25 amI see your point on current transformers, but I forgot to mention it's a high volume product, and I would like to keep the solution compacted to the PCB, with no wires or dangling transformers. I briefly mentioned above that the hall effect sensor can do the same job as the CT. Maybe also with isolation, I haven't done much investigation on them yet.Temperature is certainly a way to go. I wonder what response time I would get. Seconds is not an option. I'm pasting an image with my idea/concept on a borrowed schematic, which is ultimately based on light.

author: fourfathom
author_rank: Super Contributor
timestamp: December 03, 2021, 04:00:46 pm
content: Quote from: HSPalm on December 03, 2021, 02:13:58 pmThat circuit has a problem -- obviously the current-sense resistor needs to be a low value, but driving the optoisolator LED directly from that resistor will not work well as there is a very small zone between "LED off" and "LED overcurrent burnout". You might try replacing that resistor with a string of 1A diodes, then wiring the optoLED across those with a current-limiting series resistor. You would also need a reverse-polarity diode across the diode shunt. The voltage burden for the detector would only be a couple of volts. I haven't worked out the series resistor value or dissipation, as that depends on the optoLED characteristics.BTW, you really ought to have a fuse somewhere if you are adding power-dissipating elements (a current shunt or diode) to the circuit. Here's a circuit. The diodes are generic 1A 1N400X types, but if the current will exceed about 1/2A you will need something bigger. Regardless, you want a fuse in the circuit. That 1K resistor will provide somewhere between 0.7 and 2.2 mA to the LED, depending on temperature, component variations and current through the shunt. If you want a symmetrical voltage burden you can put three diodes in anti-parallel (I show just one). You can also adjust the lower current-detection threshold by putting a shunt resistor across the whole mess.

author: bob91343
author_rank: Super Contributor
timestamp: December 03, 2021, 06:29:05 pm
content: Berni said:"90mA is not going to heat up a wire all that much." How much is that? My Ohmmeter detects .001 degree, which is easily seen with only a few mA. You just need a sufficiently sensitive Ohmmeter.

author: Siwastaja
author_rank: Super Contributor
timestamp: December 03, 2021, 07:04:06 pm
content: Quote from: bob91343 on December 03, 2021, 06:29:05 pmYou can measure suchchangeon expensive precision ohmmeter, but two NTCs (trace measurement and reference outside of trace) do not match that closely, and worse, theywilldrift differently. Additionally, implementing that precision ohmmeter as a part of the circuit is costly and complex.Unless you deliberately add enough losses to make the temperature rise significant, in which case you could have just used series current sense shunt resistor, which indeed is one of the sensible ways to approach the problem, unlike yours. I hope we can forget this nonsense and get back on even remotely practical ideas.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 03, 2021, 07:38:39 pm
content: If the environment is reasonable quiet, not close to transformer or similar, one could still pi-up the AC field from a straight trace with a suitable SMD inductor on the other side. It may need a 2 nd inductor to compensate for an external field and some amplification.If you look around, there should be reasonable small, not too expensice current transformers. I have seen a high grade shop vaccum using a small CT that is like a small 15 mm dia toriod glued to the PCB and a single U shaped wire for the primary. So nothing to flumsy. They still sometimes break down, because they skmiped on diodes for protection.