AC Signal...

[Melanie]

You obviously have a death wish.

The 16F84 has no A/D... and even if it did, it wouldn't help you.

The attached schematic might increase your life expenctancy (or that of your PICs). C1 is a 10nF 1kV Disc Ceramic (good for 110-240vAC). It's there to limit the current going through IC1 which is an opto isolator like Cosmo's K1010. D1 a 1N4007 is there to stop the reverse-bias voltage killing the opto's LED emitter. The PIC's pin needs a pull-up to +Vdd, or use one of the weak-pull-up's on PortB. Add your own Fuses (if required and as appropriate to your application) on the AC Input line. 50mA is more than adequate.

The PIC will see a pulse at the frequency of your AC supply. Check for it using COUNT or PULSIN or whatever method you prefer. No Pulse, No AC. I use this method (or variations of it) to detect AC Supply Phase Failure, Fuses Blown etc etc.

[Tear]

After posting this morning I was looking at the data sheet for the PIC16F84 and came to the same conclusion that the PIC16F84 doesn't have an A/D converter. However, even if I went to a pic that did have a A/D converter that still wouldn't do anything for me then? I am not sure how the A/D converter works and I am going to look into this this afternoon.

I am also having a little bit of trouble understanding your circuit schematic. From what I understand it looks like the diode goes across terminals 1 (L) and 2 (N) of the IC1. (L) I am guessing is the AC Voltage coming in through the capacitor and then into terminal 1 of the IC1. Pin 4 goes to the input pin of the microcontroller, and pin 3 goes to ground. What is the (N) or what I am guessing is pin 2 to IC1?

Thanks for the help and sorry for the confusion on my part.

Michael

[NavMicroSystems]

Tear,

before we continue our guessing session,
would you tell us a bit more in detail what you are actually trying to achive?

(At least to me t is not really clear)

Your initial posting on this thread sounds a bit like you would like AC to supply your PIC.

[Melanie]

The L and N are Live and Neutral from your high-voltage AC Supply Input.

[Warrier]

Ms. Newman,
don't you think you need to provide a DC path for the LEDs in the opto coupler to conduct? - even a 100k would be sufficeint as the voltage across the diodes is clamped to about +/-1.7volts.

I think Mr. Schmidt should be warned of potential shock hazards! If he is aware of it then he could even use a series resistor (1Meg or more) to any input as the internal clamping diodes will limit this to Gnd and Vdd.

But then Ralph's question is also valid... Is Mr. Schmidt trying to use an AC power source? Hope not since he has used the title "AC Signal" and refers to internal ADC for the F84!

-warrier

[NavMicroSystems]

...a 100k would be sufficeint as the voltage across the diodes is clamped to about +/-1.7volts.

Well, in general this is correct, but:
lets's assume the Opto's LED current is as low as 10mA,
even at that low current at a voltage of 120 or 240 V the resistor would have to be quite large (in size) and dissipate a lot of heat,
would you really like this?

[Warrier]

Ralph:

Sorry I didn't make it clear, I guess. When using the optocoupler, the resistor would be across the two back to back LEDs inside the opto coupler. So the voltage across the resistor would only be +/- 1.7volts (the LED drop). Wattage is thus not an issue...

The series resistor of say 1Meg is INSTEAD of the opto coupler and need not carry the LED currents. It would be connected in series with the input pin of the PIC and is to limit the current through the clamping diodes provided on all I/O pins with a max current limit of +/-20mA.

With a 1 M resistor, the I/O current is safely limited to 240uA and the wattage would be about 0.06watts for the 240VAC. My concern was that Mr. Schmidt may accidentally come in contact with the live AC line with a "shocking" experience**** There might be some fire-works also.

-warrier

-warrier

[Melanie]

>> don't you think you need to provide a DC path for the LEDs in the opto coupler to conduct?

No. Think outside the box. Look at what C1 is doing at the supply frequency.

>> the resistor would be across the two back to back LEDs inside the opto coupler

Again no. This particular opto only has one LED.

>> even use a series resistor (1Meg or more) to any input as the internal clamping diodes will limit this to Gnd and Vdd.

You really want to rely on a Resistor and the PICs clamping diodes to stand between you and oblivion?

In my circuit the LIVE AC is on the primary (input) side of the opto. No LIVE circuitry (nope not even 240uA of it) ever touches the PIC. If your Resistor failed, the PIC would become LIVE at supply potential. In my circuit, if either C1, D1 or the opto failed, you would still be insulated by the ISOLATION characteristics of the opto (usually 5kV breakdown or more).

This is not a 'Theoretical' circuit plucked out of the air... it actually works and is in use in thousands of products every day. In the snapshot appended, here is the very circuit where the PIC is monitoring the state of two totally separate but critical AC Supply fuses.