How to detect when a device is using more than 200mA ? - Page 1
Forum section: electronics / beginners

author: tigrou
author_rank: Regular Contributor
timestamp: March 30, 2020, 11:36:56 pm
content: I have a small portable outdoor camera. It is powered by 8 x AA batteries.I would like to detect when the device is recording. It does not record all the time but only when something is near or when there is noise.Using a multimeter, I found out that the device use approximately 200mA when it's recording (with a peak of 1-2A at start). Then (once recording session is done) it fall back to 0.00A (the best I could measure with the multimeter). It's probably below 1mA.I have been thinking into using a current sense module like the ones based onMAX471orACS 712to measure current and trigger something.Is there a better or simpler way ? (eg : measuring voltage in a resistor using an opamp)My ultimate goal is to output a +5V or +3V signal when device is recording.The problem of the modules above is they produce a voltage which is a function of the current (eg : 100mv/A)

author: Manul
author_rank: Super Contributor
timestamp: March 31, 2020, 12:26:03 am
content: I would use a rail to rail type comparator hooked to a voltage drop resistor, and reference set by potentiometer, but it needs some work and design. In some old electronics I have seen overcurrent protection made by winding copper wire on a reed switch, thus making a low voltage drop current activating relay. I wonder how low activating current can be achieved with reasonable amount of windings.If you go for current sense resistor and some custom solution it may be useful to bypass a resistor with a reasonable size capacitor. It will reduce momentary voltage drop in case of high short startup current and also filter your circuit response a little. It depends how long are these high current pulses. Voltage drop should be minimized as possible to save energy and to not starve a camera when it wants to take high current.Keep in mind that current used may change a little as batteries are getting discharged.

author: magic
author_rank: Super Contributor
timestamp: March 31, 2020, 06:41:23 am
content: Possibly the crudest and cheapest solution (if it works at all):0.1Ω series resistor on the positive supply connection.A TL071 opamp with inputs connected to the ends of that resistor. Be sure to use a real one, not rebranded LM358Using the offset null function you should be able to set detection threshold to a few mV, which corresponds to tens of mA.

author: Ian.M
author_rank: Super Contributor
timestamp: March 31, 2020, 06:56:55 am
content: Wind a coil of magnet wire on a drinking straw as a former to give enough ampere turns to activate a reed switch.

author: tigrou
author_rank: Regular Contributor
timestamp: March 31, 2020, 08:08:38 am
content: Quote from: Ian.M on March 31, 2020, 06:56:55 amI like that solution (it's hard to make something simpler and it use no current at all). Since a reed switch usually need 10 and 100 AT to activate I will need lot of turns (50 to 500) for 200mA right ?Also since the device is now connected to an inductor, it will limit current somehow (at least for a short time). Maybe it will prevent device operating correctly ?

author: Ian.M
author_rank: Super Contributor
timestamp: March 31, 2020, 09:04:50 am
content: As the inductor doesn't have much iron in its core, its inductance will be fairly low. Put a few hundred uF of decoupling on the load side of the sense coil. A bigger problem is series resistance - due to the large startup current spike, its unlikely to tolerate more than a small fraction of an ohm so make sure that the gauge wire you use is thick enough.Hint: Work out the approx. length for 200 turns then consult an AWG table!

author: Zero999
author_rank: Super Contributor
timestamp: March 31, 2020, 02:43:02 pm
content: The reed switch idea might work. I don't see why there would be any need for additional capacitance, by the sounds of it, the camera already has large decoupling capacitors, hence the huge surge.How sharp does the 5V pulse need to be? Here's a crude circuit I've used in the past to switch an LED on, when it senses current flowing through a small motor. I've replaced the LED with a zener diode to give just under 5V out. Another transistor and some resistors could be added to give some positive feedback, hysteresis and a better pulse shape.Current switch.asc(1.87 kB - downloaded 60 times.)

author: tigrou
author_rank: Regular Contributor
timestamp: April 01, 2020, 09:15:17 pm
content: I just learned there is a device calledreed relaywhich contains areed switch with some winding around it.If I can found one that can be activated with a current that is low enough, that sounds like a easy solution.What is weird is there is two layers of copper wire on each side : one directly in contact with glass and one outside. Both seems to be separated by a thick layer.@Zero999: I have a question related to your schematic : how much current does it draw when there is no load ? (or something very low like 500 µA)

author: magic
author_rank: Super Contributor
timestamp: April 02, 2020, 05:58:00 am
content: Quote from: tigrou on April 01, 2020, 09:15:17 pmBe sure to also check if the coil's resistance is acceptable. This may prove problematic.For the opamp solution, a TLC271/TS271 could be better with 10µA idle current. The shunt resistor would go to the negative terminal this time.

author: SeanB
author_rank: Super Contributor
timestamp: April 02, 2020, 10:02:42 am
content: Your sliced in half reed switch has an inner copper tape wound around the reed glass, which is used as an electromagnetic shield, so that the coil current does not affect the small signals going through the relay contacts. in many applications this is not needed, but as the hermetic reed relays typically are used to switch tiny signals, it is needed, so it is there. Will be connected to the one pin that is also connected to the case, separate from both the reed contacts themselves and the coil windings.https://docs.rs-online.com/de05/0900766b81498d86.pdfLittelfuse reed relay with reasonable sensitivity, you will need 75 turns of wire wound around the glass body, probably on a plastic former, to guarantee your 200mA will close it, but some might work with only 50 turns for 200mA, or with some and 75 turns you might only need 130ma to close the coil. Drop out current, where the contacts open, will be around 100mA.0,75mm enamelled wire there will be around 2m required, 7. 5 layers, around 2R resistance, which will be around 0.4v drop in your circuit..https://docs.rs-online.com/e937/0900766b8129257e.pdfLinks for example only.

author: exe
author_rank: Unknown
timestamp: April 02, 2020, 10:50:32 am
content: Afaik, there are hall-effect current sensors, but they are not very precise devices, and, I heard, may drift over time. I found this one in Google, but there are more:https://www.monolithicpower.com/en/documentview/productdocument/index/version/2/document_type/Datasheet/lang/en/sku/MCS1802/As of reed relay, can parasitic inductance be a problem?

author: Zero999
author_rank: Super Contributor
timestamp: April 02, 2020, 11:21:36 am
content: Quote from: magic on April 02, 2020, 05:58:00 amThe TLC271 no, the TS271 yes.Quote from: exe on April 02, 2020, 10:50:32 amI doubt accuracy is important.QuoteThe camera probably already has lots of decoupling, so it might not be an issue.Quote from: tigrou on April 01, 2020, 09:15:17 pmThe standby current is equal to (V1-the voltage drops over D1 & D2) divided by R2 but, since the voltage drop across the diodes is quite small, they can be ignored giving just under V1/R2 = 12/100k = 120µA. R2 can be increased further to reduce the supply current more, but it will increase the sensitivity of the circuit, as less voltage will be dropped across the diodes and will reduce the base drive to Q1, so R3 will need to be increased, a lower current Zener used and it won't be able to drive such a low load impedance.

author: magic
author_rank: Super Contributor
timestamp: April 02, 2020, 01:30:18 pm
content: Quote from: Zero999 on April 02, 2020, 11:21:36 amWhat's the difference? I have never used the TI version but assumed they are about the same.

author: Zero999
author_rank: Super Contributor
timestamp: April 02, 2020, 03:33:58 pm
content: Quote from: magic on April 02, 2020, 01:30:18 pmI was mistaken. I looked at the section of the data sheet for high bias current mode and neglected the rest. The TLC271 would be fine.Anyway, an op-amp isn't ideal. If one wants to go down the IC route, a comparator should be used instead. The LMC7211 seems to be a suitable candidate.https://www.ti.com/lit/ds/symlink/lmc7211-n.pdf