Ttelmah
Joined: 11 Mar 2010 Posts: 19365
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Posted: Sat Nov 06, 2010 8:32 am |
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Start by reading the data sheet.
The INT pin does not require '5v'. it requires 0.8*Vdd.
Now, one answer then, if you are running slow enough, and don't 'need' 5v for anything else, would be to lower the PIC's supply voltage. Take Vdd down to 3.5v, and 2.8 or 3v, will trigger this input.
Beyond this, then you do need to buffer the signal. One very small solution, would be to set the PIC pin to interrupt on the signal going low, rather than rising, and connect your signal into the base of a transistor (with a resistive divider). Then connect the transistor's emitter to gnd, and the collector to the interrupt pin. Program the weak pull-up 'on', and the interrupt will fire when your signal turns the transistor on. If you (say) use a 10K resistor to the base, and a 3.3K to ground, this will happen between about 2, and 2.5v.
Best Wishes |
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