I'm apology if the title is kinda generic but I found a (on a AC) which use IC and a triac and there is some points I don't really get.

Image from this application note:

enter image description here

Datasheet used as reference.

  1. About the PTA0 pin :
    1. How can the pin accept 120v since the range of the ADC is VSS to VDD (Page 161). The maximum absolute voltage for input is VSS-0.3 to VDD+0.3 which is far away from 120v, (Page 149)
    2. What about -120v? The PTA0 shouldn't have a diode to be a voltage above 0v?
    3. Why using resistors for the zero-crossing on the PTA0 pin? If it is to limit the current I tought input pin handled them.
  2. Z1 shouldn't do a shortcut when it reach the breakdown voltage from C1 to Ground? (C1 -> D2 -> Z1) (I assume the top graph is the positive side)
    1. I'm trying to know why it need 0.5W, I don't think it may be the IC (25mA maximum for PTA* pins, but I don't find the power consuption overall (50mA could be the overall consuptions (including PTA* output pin I hope) ?)

Also, there is an exemple using a generic opto coupler where they connect the anode to a 100k which then go to the AC :

  1. I know LED work with current instead of voltage... but i'm kinda surprise they work at 120v. Is it really possible or this is only true for opto coupler?
  2. Should -120v destroy the led since it is much more higher than the regular 5v reverse input voltage?


  • \$\begingroup\$ Kind of distracted right now, but I think that circuit is actually floating on the AC line. So it's not like you just hooked up 120V to a micro connected to 0-5V. The whole floating power supply thing is kind of cool, but only usable when no one can come in contact with the circuit. It's just as dangerous as a live main. \$\endgroup\$ Oct 7, 2012 at 3:41

1 Answer 1



  1. It doesn't. Whether it's low or high, the resistors act as a pull-up while also providing the needed signal. The current, for a supposed logic low would be \$115\sqrt{2}/940k\Omega = 173 \mu A\$

  2. As Some Hardware Guy said, it's floating so the reference is 0V because its supply is mono-alternance and done through a capacitive impedance (reactance) of \$1\mu F\ = 2.65k\Omega \$ (60Hz) or\$ 3.18k\Omega\$ (50Hz).

  3. They're there for not actually putting the pin to 160V, your fear at #1.


  1. C1 has the mentioned reactance, acting as an "active" resistor, limiting current through Z1.
    1. I couldn't find the 0.5W you're talking about.

The rest:

  1. The LED will still have ~2V foreard voltage drop and the maximum current will be: \$ I_{max}=\frac{115\sqrt{2}-2}{100k\Omega} = 1.62mA\$

  2. There won't be -120V since its reference will be the IC's supply, therefore the bridge will clamp the voltage.

You're welcome, but you should really try those simple circuits that have proven to be good enough, instead of choosing this.


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