AC Signal...

[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.

[Tear]

Hi,

What I am trying to accomplish is to build a delay device that turns on your amp a few secs after the radio is turned on. I have already built this using the amp turn on line from the stereo which is DC Voltage line. However for my brother's car, that I also would like to put this in, his stereo doesn't have an amp turn on. Due to this I need to use the speaker lead which is an AC voltage signal.

Orginally I thought that using a AC/DC converter on a microcontroller might do the trick, but on further inspection it turns out this is not the case. I am now thinking about building a simple PNP Transistor Switching Circuit where the AC Voltage signal (line) would be connected to the base and control the switching with a +5V DC voltage following to the PIC when the switch is closed. However, I am not sure if this will work.

Essentially the easiest way to characterize what I need to do is somehow have the PIC dectect when the AC signal is HIGH either directly or maybe through some circuit like a PNP Transistor switch. Once the PIC detects the line as being HIGH then it will delay and set the output HIGH.

I hope that this explains what I am trying to do a little better, and thanks for the help.

~Michael

[Melanie]

Hmmm not quite an 'AC Power Source' ...

So you're telling us that the AC is really a dinky Audio Signal from a car radio and not a High-Voltage 'Power Source' ?