Powering a sensor wirelessly from a motor - Page 1
Forum section: electronics / projects

author: Annakin4
author_rank: Contributor
timestamp: July 29, 2016, 02:29:23 am
content: Hello everyone!Skip this paragraph if you don't care what the project isSo I am trying to get sensor readings off a industrial motor. The sensor will contain a ultra low power micro controller, and will need to sense-TemperatureRF NoiseCurrent drawThere is no power socket, light, thermal change, or anything to power the device. So I took a 120 volt motor and made a piss coil by hand and took the 1970s oscilloscope and I put the coil right put against the motor. I observed a 10mV AC current. So my plan is to have a coil next to a industrial motor (But 240+ volt industrial motor instead of my 120 volt fan motor) and have it go into a booster circuit which then feeds into a small (super) cap. Whenever a micro needs to take a measurement every hour it will wake up and take a sip off the cap. At the end of the day hopefully there will be enough power to send the recorded data quickly over LoRa to a base station. The motor will be contentiously running.Question-Can someone point me in the right direction to boost the milivolt current to 1.8v please? And if you read the above any input? And tomorrow I'm making a better, bigger coil and testing it out and I can post the results after I get to the lab._ ___ /^^\ /^\ /^^\__ _@)@) \ ,,/ '` ~ `'~~ ', `\._/o\_ _ _ _/~`.`...'~\ ./~~..,'`','',.,' ' ~:/ `,'.~,~.~ . , . , ~|, ,/ .,' , ,. .. ,,. `, ~\_( ' _' _ '_` _ ' . , `\_/ .' ..' ' ` ` `.. `, \_~V~ V~ V~ V~ ~\ ` ' . ' , ' .,.,''`.,.''`.,.``. ', \__/\ /\ /\ /\_/, . ' , `_/~\_ .' .,. ,, , _/~\_ `. `. '., \_< ~ ~ '~`'~'`, ., . `_: ::: \_ ' `_/ ::: \_ `.,' . ', \_\ ' `_ '`_ _ ',/ _::_::_ \ _ _/ _::_::_ \ `.,'.,`., \-,-,-,_,_,`'~~ `'~~ `'~~ `'~~ \(_)(_)(_)/ `~~' \(_)(_)(_)/ ~'`\_.._,._,'_;_;_;_;_;Thank you

author: Marco
author_rank: Super Contributor
timestamp: July 29, 2016, 04:23:49 am
content: Put a ferrite core in it, pulls in the field, or maybe trya standard ferrite core inductor?10 mV into a 1 MOhm scope means 100 pW ... that's not enough to work with, see what the voltage is when you load it with a resistor (say a kOhm).

author: Ian.M
author_rank: Super Contributor
timestamp: July 29, 2016, 04:46:24 am
content: Odds of success with a simple coil against the motor casing are extremely poor.However you'd need to add a current transformer on one of the supply leads anyway to collect the data you need and its quite likely that enough power could be harvested from the current transformer secondary in-between current readings by switching out its burden resistor so it drives a bridge rectifier feeding a supercap with a shunt Zener to prevent it overcharging.

author: Richard Crowley
author_rank: Super Contributor
timestamp: July 29, 2016, 05:23:58 am
content: 3. Get your coil CLOSER to the windings in the motor.2. Get your coil INSIDE so that you don't have the motor enclosure shielding the magnetic field.1. Put A LOT MORE windings on your coil.

author: LeonV
author_rank: Contributor
timestamp: July 29, 2016, 05:32:07 am
content: Sounds like you need better coupling...And i just thought that i'd let you know there are energy harvesting ICs that will tank tiny amounts of power.

author: Zero999
author_rank: Super Contributor
timestamp: July 29, 2016, 07:53:16 am
content: Increase the number of turns.Calculate the inductance.Add a capacitor in parallel with the coil, forming an LC with a resonant frequency equal to the mains in your area.

author: Ian.M
author_rank: Super Contributor
timestamp: July 29, 2016, 10:12:57 am
content: I ran the numbers and even with a 1/6 HP motor with no mechanical load, there should be about 2.5mW available for harvesting from the secondary of a typical 1000:1 current transformer, slipped over ONE of the motor leads. Larger motors or heavier mechanical loads will significantly increase the energy available.As the O.P's requirements include measuring current draw, he needs a current transformer anyway, and as long as the sample rate is considerably lower than the supply frequency, alternating sampling and energy recovery should have little effect on accuracy.

author: Annakin4
author_rank: Contributor
timestamp: July 29, 2016, 09:56:44 pm
content: Thank you guys for all of your suggestions. I have found that a coil with more winds (Still air core) brought 23mV AC. But I fallowed Ian.M and I made a piss poor coil around one of the leads and I got half a volt (Didnt measure wattage) which is a huge improvement and I'll continue in that direction to power my project.Quote from: Ian.M on July 29, 2016, 10:12:57 am

author: Ian.M
author_rank: Super Contributor
timestamp: July 29, 2016, 11:07:19 pm
content: The coil must not be wound around the lead - all that gives you is capacitive coupling. It must be wound on a core that totally encircles the lead e.g a reasonably large diameter ferrite ring

author: Marco
author_rank: Super Contributor
timestamp: July 29, 2016, 11:38:50 pm
content: Which brings up another option ... with both wires available you could tap them capacitively.

author: Ian.M
author_rank: Super Contributor
timestamp: July 30, 2016, 12:48:16 am
content: Quote from: Marco on July 29, 2016, 11:38:50 pmNot from outside the existing insulation!Lets consider a 6mm2cable to BS6004 with 0.8mm thick PVC insulation. Its O.D. is approximately 4.7mm.The dielectric constant for PVC insulation is in the range 6 to 8, so lets take it as 8 as the most optimistic result.If you wrapped a 1m length of it in foil, you'd have a cylindrical capacitor with I.D. 3.1mm. Converting to radii add using the calculator athttp://hyperphysics.phy-astr.gsu.edu/hbase/electric/capcyl.html, gives a capacitance of 1.07nF/m1nF @ 60Hz has an impedance of about 2.6Meg, so the max current to ground, assuming the cable is carrying 240V 60Hz would be 90uA. However if a VFD is in use, its chopper frequency is significantly higher and the pulses have sharp edges, so you are likely to get two to three orders of magnitude more current.I doubt its practical to cover more than 10cm or so of the cable with foil and your electrical inspector may insist on the cable being sleeved before foil is applied to it, so its marginal even if a VFD is in use.

author: Marco
author_rank: Super Contributor
timestamp: July 30, 2016, 01:51:58 am
content: Lets go a little lower than 10cm, lets say 1 cm giving 10 pF and 260 Meg impedance. That would give you 46 uW of recoverable energy, there are energy harvesting solutions which can work with that. In fact they go down to single digits before the leakage kills the ability to accumulate any energy.