240V 1-lead only available LED-drive (current detect) - Page 1
Forum section: electronics / projects

author: Shakesword
author_rank: Newbie
timestamp: January 12, 2017, 09:51:05 am
content: I would like a LED-indication indoors when my water circulation pump motor outdoors is running.My woodburning stove heats water. It's connected to a 500Liter water tank. When the water temperature reaches >64 dgr C in the stove the water circulation motor is turned on.The switch that does this is an old(+20 years) but reliable gas-capillar-tube-relay-thermostat on the stove, it sits indoors.Inside the house is only one wire (240V )wire to the motor outside. The motor is max 50W.I would like a simple led to somehow draw its tiny current from this wire when the motor is running. Not using batteries or other power source.I was thinking if it would be possible to magneticaly induce enough current from the wire, to drive the led, when the motor is switched on. What would i need for that?.Perhaps using a ferrite ring as a transformer, but how many turns would I need for say ~10mA /~2.4 v ? Preferably very few turns on the 240V side....I tried an idea with 4 diodes in series in one direction in parallel with 4 diodes in series in the other direction. Over this i can get 4*0,7 VACIt worked ok...-ish, however the diodes got plenty hot because of the high current. And it does not feel safe for in-home use.Or any other smart non-mecanical , non-external-power, non-expensive solution?Any ideas?

author: capt bullshot
author_rank: Super Contributor
timestamp: January 12, 2017, 10:06:51 am
content: What you want to build is a current transformerTypical single phase induction motors have a PF of about 0.5, so calculate the RMS primary current: (50W / 230V ) / 0.5 -> 0.43AAssume there's about 0.4A ... 0.5A in the single wire, you'll need about one primary winding and about 40 turns secondary (0.4A / 10mA)But from my experience with such things, I doubt it'll work that simple. The core will saturate before you reach the secondary 2.4V. I may work with the secondary shorted You'll need a very large ferrite or maybe better iron core. Try 10 turns on the primary and 400 turns sec to get some more power out of the thing.I had useful results with a 1:700 turns current transformer designed to work on 50Hz lines driving an photocoupler LED from a single AC 230V wire carrying about 4A.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: January 12, 2017, 11:01:54 am
content: I'd look at using a mains transformer as a current transformer - find something with a low secondary voltage for minimum drop to the motor- connect the secondary in series with the motor, and LED with current-limiting and reverse-blocking diode to the primary.

author: Seekonk
author_rank: Super Contributor
timestamp: January 12, 2017, 03:04:09 pm
content: Just happen to have a split bobbin transformer handy that I cut out one of the windings and replaced with a couple turns of wire. They sell these CT/LED detectors ready made. The green CT is from a product I designed for a company that powered a 12V relay by itself. This is not a powdered core, but a wound length of flat steel. A core from a switching power supply will not work. If you are willing to remove some wire a transformer will work fine.

author: Shakesword
author_rank: Newbie
timestamp: January 12, 2017, 04:11:09 pm
content: Quote from: amyk on January 12, 2017, 11:50:09 amThis is not what i am after. I only have access to one wire. I need to somehow shunt off 2,4V/10mA while there is passing 0,5A thru the lead. This 'only' shows some good voltage indcators. But thanks for trying.

author: Shakesword
author_rank: Newbie
timestamp: January 12, 2017, 05:10:50 pm
content: The transformer idea seems safest so far but also probably difficult to make/get. I would like to get the theory on a tranformer solution before i try to make one. Preferably with less than say 50 turns on either side.How about a 5,5Ohm resistor ? 0,43A*5,5Ohm = 2,37 V ; 0,43*0,43*5,5 =1 WattsOr a coil that would need to generate 5,5Ohm reactance. How would that work if I put a LED in series with a current limeting resistor in parallel with such a coil. Would that work?Or can some clever high power semiconductor be used to solve this problem? Some sort of diac or something? That could generate the voltage over itself as it passes this ~0,5A thru it. This magical component should of course need to not heat up very much.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: January 12, 2017, 05:54:34 pm
content: Quote from: Shakesword on January 12, 2017, 05:10:50 pmA resistor is the simplest option. If current is 0.5A RMS, peak is about 0.7, so to drop enough for a red LED to light at the peaks, say 2.5v = 2.3 ohms. Call it 4.7 for some margin. Power dissipated 1.175watts - use a 5 watt wirewound to comfortably cope with startup surges.68R LED series resistance would give about 20mA peak current.Next option up would be a couple of inverse-series connected zeners, say 3.3v, to drop a constant-ish voltage regardless of load ( probably not an issue with a 50w motor) . Again use at least 5W ones for margin. A pair used to keep the voltage to the motor symmetrical - dc offsets can be bad.

author: Seekonk
author_rank: Super Contributor
timestamp: January 12, 2017, 06:27:02 pm
content: If you have access to something like an old 240 to 12V transformer type wall wart, that would give you a gain of about 20. Parallel that with a .47 ohm resistor in series and that would give enough voltage, no rewinding.

author: DBecker
author_rank: Frequent Contributor
timestamp: January 12, 2017, 07:49:05 pm
content: Current transformers are cheap on eBay. I recently bought 5 for $0.99.Even at that low pump current you could light an LED. If it's not bright enough, they are cheap enough to add more.Oh, and you'll almost certainly want a anti-parallel LED or plain diode for reverse voltage protection. A current transformer *really* wants that current to flow and will easily overwhelm the typical 5V reverse voltage of a LED.