24 VAC Switch

[ibwev]

Is there a component, preferably a chip, that can send a low VDC signal (to PIC16F690)when a 24 VAC signal goes high so the pic can do a series of test before allowing the 24 VAC circuit to be completed.

View attachment 65857

 

[ronv]

I'm guessing the 24 VAC comes from a transformer?? Can one side of the transformer be tied to DC ground? If so you could just voltage divide it rectify it and filter it a bit so you could sample it with the PIC. The a little triac switch or relay depending on how much current you need.

 

[4pyros]

No single chip that I know of.
You must rectify and filter the AC to DC and then regulatet it to a voltage the PIC can handle.

 

[ccurtis]

If you must use an IC something like this should work.

 

[dougy83]

If the AC signal shares a ground (or any reference) with the PIC, then a resistor from the AC signal into the PIC input can be sufficient to detect the 24VAC.

If it's isolated, then you can use an opto to detect the voltage, and an opto-fet or opto-triac or relay to switch the voltage

 

[4pyros]

If you must use an IC something like this should work.

Thats cool. Nice find.

 

[ronv]

What a cute IC. Didn't know they made such a thing.

 

[ibwev]

If the AC signal shares a ground (or any reference) with the PIC, then a resistor from the AC signal into the PIC input can be sufficient to detect the 24VAC.

1) The AC signal shares a common ground with the PIC. Using the above method, do I calculate the resistor size by V(supply)/I(max I/O from pic)
24V/.025A
960Ω

2) View attachment 67265I have the pic programmed to do a series of test once the Input Port goes high. Currently, I have it programmed to conduct the test once it receives a 5VDC signal. Since the new supply current is alternating, would I need to program the microcontroller differently to factor in the alternating current.

 

[alec_t]

960 Ohms is far too low a value. The input pin protection diodes would fry .
Try 100k - 1 Megohm. The protection diodes will then safely clamp the pin voltage between ~5.6V on the positive half-cycle and ~-0.6V on the negative half-cycle.

Since the new supply current is alternating, would I need to program the microcontroller differently to factor in the alternating current.

The micro can trigger on, say, the rising-edge of the waveform and ignore the falling-edge.

 

[ibwev]

Try 100k - 1 Megohm. The protection diodes will then safely clamp the pin voltage between ~5.6V on the positive half-cycle and ~-0.6V on the negative half-cycle.

Would a pull down resistor be needed (as in illustration)? or would it interfere with the circuit?

View attachment 67780

 

[alec_t]

No extra pull-down needed. The negative half cycle of the AC pulls the input down.

 

[MikeMl]

There are several MicroChip Application Notes that show exactly that circuit...