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Need advice about this AC Line Current Detector

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Avocado

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I have 4 motion sensor lights at my house (one at each corner of the building), and would like to have an audible indication when one of them is activated. I have found a circuit at Bowden's Hobby Circuits for an "AC Line Current Detector". A coil of 800 turns of magnet wire on a U-bolt is used as an inductive pick up. The description of the circuit mentions detecting 50 watts or more. With a pair of 75 to 150 watt floodlights in the motion sensor light, I will be presenting 150 to 300 watts to be detected passing the coil. Might this be a problem, and can I modify the number of turns on the coil to avoid any problem.
Also, rather then wind four of these coils by hand, (one for each of my motion sensor lights) I found a low frequency current monitoring transformer at Jameco Electronics. Can I substitute this for the inductive pick up?
Here is the info on the AC Line Current Detector and the low frequency current monitoring transformer.
Thanks for any insight you might offer.
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AC Line Current Detector
(schematic attached as AC Detector.gif)
This circuit will detect AC line currents of about 250 mA or more without making any electrical connections to the line. Current is detected by passing one of the AC lines through an inductive pickup (L1) made with a 1 inch diameter U-bolt wound with 800 turns of #30 - #35 magnet wire. The pickup could be made from other iron type rings or transformer cores that allows enough space to pass one of the AC lines through the center. Only one of the current carrying lines, either the line or the neutral should be put through the center of the pickup to avoid the fields canceling. I tested the circuit using a 2 wire extension cord which I had separated the twin wires a small distance with an exacto knife to allow the U-bolt to encircle only one wire.
The magnetic pickup (U-bolt) produces about 4 millivolts peak for a AC line current of 250 ma, or AC load of around 30 watts. The signal from the pickup is raised about 200 times at the output of the op-amp pin 1 which is then peak detected by the capacitor and diode connected to pin 1. The second op-amp is used as a comparator which detects a voltage rise greater than the diode drop. The minimum signal needed to cause the comparator stage output to switch positive is around 800 mV peak which corresponds to about a 30 watt load on the AC line. The output 1458 op-amp will only swing within a couple volts of ground so a voltage divider (1K/470) is used to reduce the no-signal voltage to about 0.7 volts. An additional diode is added in series with the transistor base to ensure it turns off when the op-amp voltage is 2 volts. You may get a little bit of relay chatter if the AC load is close to the switching point so a larger load of 50 watts or more is recommended. The sensitivity could be increased by adding more turns to the pickup.

Circuit found at Bowden's Hobby Circuits
(**broken link removed**)

--------------------------------
LOW FREQUENCY CSE187L
Jameco Electronics (www.jameco.com)

Description:
Designed to monitor current in low frequency applications. This Triad Magnetics part may be used to monitor current from .1 to 30 amperes at frequencies from 50 Hz to 400 Hz.

Technical Notes:
1. Turns ration: Primary to sense 1:500.
2. Suggested burden resistor: 60 ohms.
3. Typical output: 110 mV/Amp.
4. Primary DCR: 250µOhms max.
5. Sense DCR: 21 ohms max.
6. Constructed with UL recognized materials (Class B, 130¡C).
7. Hi-pot: 500 volts.
 

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The Jameco part is interesting, but it does require you to attach it in series with the AC line. If you can handle the safety issues, go for it.

I would much prefer using an iron-core inductor of up to 800 turns, and place the insulated AC line through the center. That's the advantage of the U-Bolt. I prefer powdered iron toroids. I would suggest something like the 12494 CH at www.mpja.com. But since it's only 72 turns, it might not be enough (unless you increase the op-amp gain). Search for similar inductors with at least 400, or more, turns.

If you can pass your AC line through the core multiple times, you can divide the necessary secondary turns by that number. Example, if you can pass your AC line through the 12494 CH about 11 times (Maybe 22 ga. wire, be sure to get 600V insulation) then the 72 turn inductor is enough.
 
mneary said:
The Jameco part is interesting, but it does require you to attach it in series with the AC line. If you can handle the safety issues, go for it.

I would much prefer using an iron-core inductor of up to 800 turns, and place the insulated AC line through the center. That's the advantage of the U-Bolt. I prefer powdered iron toroids. I would suggest something like the 12494 CH at www.mpja.com. But since it's only 72 turns, it might not be enough (unless you increase the op-amp gain). Search for similar inductors with at least 400, or more, turns.

If you can pass your AC line through the core multiple times, you can divide the necessary secondary turns by that number. Example, if you can pass your AC line through the 12494 CH about 11 times (Maybe 22 ga. wire, be sure to get 600V insulation) then the 72 turn inductor is enough.
The current transformer secondary needs to have more than 40uH inductance (as in 12494 CH) if it only has 72 turns. 72 turns will ideally yield 14mA of output current (into a short circuit) per amp of line current, multiplied by the number of input turns. Even with a 1 ohm burden (load), the output time constant is only 40usec (time constant=L/R). A 40usec time constant means the highpass corner is at 4kHz. This is over 6 octaves above 60 Hz, meaning 60Hz gets attenuated by over 36 dB. The corner frequency goes up (attenuation goes up) as you raise the burden resistance in an effort to get more voltage, so you gain nothing.
All this basically shows why current sense transformers have high-inductance secondaries. As you said, search for a transformer with the appropriate number of secondary turns.
Current transformers can give dangerous shocks if run without a burden resistor. The burden should always be connected before the primary is energized.
 
Thanks for the advice

Thanks for the advice. Sounds like I'll be spending my time winding four 800 turn coils.
 
I use 240/8 volts 1 amp transformers to monitor my outside lights, which are on a time clock.
I have LED indicators on them, and hour meters, and an alarm comes up when one lamp fails.

I have the 8 volts winding in series with each lamp which will provide enough flux to magnetise the 240 Volts winding.
Across the 240 Volts winding i put a damping resistor 470 ohms 1/2 watt to reduce the 240V side to 2 or 3 volts ac, to drive a led, and antiparrallel diode.

This system works well at my place for over 5 years now.
It saves you winding coils etc.
As long you fit the transformers in a well insulated box and adhere to the safety rules you can't go wrong.

If your lamp currents are over 1 amp you can use a 2 amp transformer or remove the secondary winding of the tx and put about 5 turns of 1.5mm² wire as current coil on it.
 
hi,
The method I use on my security lights is.

Inside the security light terminal block, where the incoming mains supply is connected, there is a 4 terminal connector block.

I connect an additional two core mains cable, the neutral of the two core to the neutral of the incoming mains and the live of the the two core to the terminal block that connects the lamp to the internal sensor relay.

At the free end of the two core cable I fit a 'free' mains socket, into which I plug in a plug type low cost 12Vdc psu.
[ most UK 'warts' have no earth pin]

When the security light comes ON, so does the 12Vdc psu from which I power other functions.

Eric
 
HI ! to all this is my 1st POST here !

I already use a method of RODALCO !

after some experiments it worked like I want & I believe it is the simplest

method.
 
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