I need help to detect an AC burned heater element

[Archangel]

Thanks Archangel for reading my post and offer to help.
Here are the conditions:
I must check during the weld process, and i have to make sure that the current going through the wire is over 1 amp.
I do not think the other solution is feasible, due to the number of wires involved.
What i have in mind is to have a PCB for 16 heaters and bring the wire heater to it to one terminal, run a trace through some device to get the current, then get out on the next terminal and go to heater.
Current transformers is not a solution, as they are bulky, not to many models PCB mount and i still need to interpret the reading.
My option was for ACS712 , but if you look at the attached graph you will see that the output is + and -
I need somehow to convert it, or do something that i will get only the + portion, but not pulsating, i need it as a continuous line.
Also, i need that any voltage over 2.6 volts to be a logic 1 and any voltage from 0 to 2.6 to be a logic ZERO. In this case, i will have my inputs to PIC set to read 1 for a good heater, and 0 for a bad heater.
Thank you

Ok so I am thinking, use 2 diodes in parallel with their polarities reversed so there is a path for both phases, then only run the forward phase through the device.
0 amps = 2.5v 1 amp or more, is above 2.7v.
What do you think, will this work?

[Pic_User]

Hi igeorge,

If you just need an un-calibrated “Go - No Go” measurement, a simple “peak detector” could work.
If a silicon diode has too much junction drop, maybe a germanium or Schottky diode would work.
Just a diode to a capacitor circuit on the output of the ACS712.

-Adam-

[igeorge]

Hello Adam,
Thanks for reply but i am afraid, like i mentioned before about the power dissipation on the diodes.
I have to do some tests to see if it will work.
Thanks
Ion

[Pic_User]

Hi Ion,
You are right to be concerned with the several watts per diode dissipation with diodes passing Amperage..
But I am proposing using them in the 5V output circuit of the ACS712.
They would have almost no current. and would give your digital pin a 2.5V to 5V swing.
If the peak Voltage was a little too high for your logic levels then add a “Voltage divider (= two resistors).
-Adam-

[Ioannis]

Maybe a simple diode bridge with a optocoupler can give a low cost solution.

Put the ac pins of the bridge in series with the element. Short the dc pins together.

Put an optocoupler with a small series resistor in parallel to the ac pins of the bridge. Use a small integration of the pulses of the output optocoupler.

You are done!

Ioannis

[igeorge]

Adam and Ioanis, both of the solutions will works if i combine them. Optocoupler in series with heater is not feasible due to the amount of heat dissipation on the limiting resistor for the diode on the optocoupler. Also , as current varies from 0 to 5 amps, will be difficult to get a resistor value to light the led in the opto at everything over one amp, but do not fry the LED at 5 amps. Like Adam suggested, i will try to use the output from the ACS712. Here, i am going back to Ioannis advice to use an integrator to get a proper DC voltage out of the sine wave. From the DC voltage i need to use a comparator to get an output fro any voltage over 2.6 volts.
Ioannis, do you have any schematic for an integrator which will works on the voltage diagram posted above.
Let's say the Amps going through the heater are 5 A, so i will have a sine wave going up and down between +2vdc to +3vdc, where the 2.5 volts is the zero crossing point for zero amps. I need some kind of schematic to convert that sine wave to a dc line at 3 volts
Best regards
Ion

[Ioannis]

You did not followed exactly my thought.

Look at the attached schematic. You have to calculate the small resistor (I use SMD of 470 Ohms at 0805 size) according to bridge and Opto characteristics.

No need for power parts.

Light and dirty cheap.

The integrator or a simple delay will be done in software. When you have powered your load, current will flow through the bridge. A small voltage drop is then developed across the AC terminals of about 1,4-2 volts. The reverse voltage is small enough not to destroy the opto led.

So, when you have powered the load, normally the opto transistor will give a square waveform of line frequency. If the load is faulty, no current will flow and no pulses will be generated.

Ioannis

[igeorge]

Thanks Archangel,
I do not know if it will works, but i will try it. I am some how concerned about the power dissipation as they are 5 amps going through the diodes, and they will be 32 on one PCB for 16 heaters. I will test and let you know.
Thanks again
Ion