I need help to detect an AC burned heater element

[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

[Nicmus]

One other approach will be to do a search in this forum for “transformerless” you will find ideas of simple circuits of powering up small loads (a LED part of the optocoupler in your case) .
Use one of the circuits in parallel with your switching device (relay, triac,…etc) and monitor its control signal and the optocoupler output.
If your heater is OK you will have a signal from your optpcoupler when the heater is not energzed. In case the heater is open you get no signal when the controlling device is OFF.

HTH

Nick