ericgibbs

hi Krumlink,

Use a 1R resisitor is series with the 0V line.

Connect a LM3914 input across the 1R, drive 10 leds with the outputs from the LM3914.

You can calibrate the LM3914 to give say 0.1A or 0.2A steps on the leds.

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Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

kenjj1

Hi. Roll your own relay. Put enough heavy wire of suitable guage on some soft iron, like a heavy nail. Secure this coil so it can't move. Get a reed switch and, while pumping 1.5 Amps through the coil you just built, slowly move the switch towards the coil. When the switch finally engages, you've found the 'sweet spot'. Secure the switch in place. Now you use a processor pin to detect the switch's closure, or an ohmeter with a continuity buzzer to signal closure, or whatever you envisioned. Maybe you can even find a relay built to do just this.

END TWO CENTS...
kenjj

ericgibbs

hi,
I based this reply on your opening post, I see that the 'specification' changed as the thread developed.
So its now 120Vac mains, detecting a current window, as I cannot edit my previous post,
the solution I suggested [LM3914] will not apply.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

kchriste

Use the current sensing toroid transformer. If you get a split core clamp on ferrite core and clamp it over one lead of a zip line cord, you can do this fairly safely.
You can buy commercial ones for this purpose:

Or you can roll your own transformer from an empty core:

which might be easier and cheaper to find. You'll need quite a few turns of wire on the secondary to get a decent sensitivity (500-5000 turns) so the premade one may be better for you.

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qsiguy

Use a hall effect sensor, similar to what kchriste suggested, and a relatively simple comparator circuit. The hall effect sensor will output a voltage based on the current passing though it. Adjust the vRef on the comparator to match the voltage from the hall effect sensor when it reads 1.5 amps. You will, of course, need an ammeter to calibrate the circuit. Use the comparator output to drive a transistor which will drive a relay or whatever you need.

Diver300

A current transformer works very well with AC like that.

The thing to remember is that it is a CURRENT transformer not a voltage transformer, so you must not leave it open circuit. The output current is set by the input current and the turns ratio. There is a maximum output voltage that you can take. The transformer specification should say that.

It is quite easy to wire the output to a bridge rectifier, then a capacitor and a resistor.

You get much less power in the resistor for the same voltage drop, and it is much safer.

Krumlink

@Khristie: How would I do that? I have never used current sensing before.

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Mikebits

Her is a nice little Current sense. They work just like a transformer.
http://ww2.pulseeng.com/products/datasheets/P524.pdf

I think Digikey has some as well.

Krumlink

Mike that looks interesting, how could I use it?

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MechTronics

World of Warcraft is a disease

Mikebits

Well, you have your AC line, which is primary and essentially 1turn going through the xformer which is the secondary say the secondary is 50 turns. So if 1 amp is running through the primary 1/50 current will run through secondary. Have a resistor load on secondary and using ohms law Vsense = Isense x Rload (E=IR). For given voltage across Rload you know what I primary is. Check out this link on transformer equations. May help explain.
Transformer Formulas

Keep in mind the voltage would still be AC.

kchriste

I'll try and reply before this thread gets locked.
This datasheet shows how it is basically done at the bottom:
http://www.crmagnetics.com/pdf/3110.pdf
The I is the one wire of the AC lamp cord you pass through the core. It is also the amperage though that wire. They include a formula for you. So, if you put a 1K resistor on the output of the current transformer, you would get apx 0.48 volts RMS AC across the resistor for a current of 1.5Amps which agrees with the chart they provide you:
1.5A x 1000hm: / 3100 = 0.48V
You could amplify this voltage with an OpAmp (Or try a 10K resistor) and then rectify it to detect the 1.5A threshold.

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audioguru

The tiny transformer is made for high frequency SMPS and might not work at the mains frequency.

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ecerfoglio

No, its made for 50 / 60 Hz (look at the data sheet at http://www.crmagnetics.com/pdf/3110.pdf

Its a CURRENT TRANSFORMER. It needs a resistor in parallel with the secondary (or else an ammeter) so it doesn't develop a high secondary voltage.

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E Cerfoglio
Buenos Aires
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Mikebits

I think he was speaking about the one I suggested and I think he might be right. I should have looked more closely at the data sheet.