Watt's the wattage when used on a PIC?

[Charlie]

You can make any of the methods you discuss work, but to reiterate, the simplest is a straight resistor divider. R1 goes from the ~12V high source to the PIC pin. R2 goes from the pic pin to ground.
Then: Vpin = vsource*(R2/(R1+R2)) As I stated above, the biggest Vsource you will see is 14.4V. The biggest Vpin you can stand is 5V.
so 14.4=5*(R2/(R1+R2)).
So 14.4/5 =R2/(R1+R2)
so 2.88 =R2/(R1+R2)
Now go hunting for the standard resistor values that when plugged into R1 and R2 = 2.88
Look for values between 1K and 100K
You probably won't find an exact match (43K & 15K might be close enough), although to be as safe as possible, err on R1 being just a bit higher than the ration implies.

[AvionicsMaster1]

I applied the Zener to the application since any kind of over-voltage situation will increase the voltage going to the PIC. I wanted to cap it off at 5 volts.

I'm just wondering what a voltage divider plus a zener would cost opposed to a voltage regulator to limit the input.

couldn't have the supply come in and go straight to the PIC and have the Zener regulate it to 5 volts. It has to pass through a resistor first

Yes, most zeners must have current limiting resistor. I say most cause as soon as I say all then someone will post one that doesn't.

The supply is applied between R1 and R2 and while R2 goes to ground, the other side of R1 is your output.

This sound like you're trying to make a logic gate for the PIC. There 's a good example in the PBP manual of how to do this.

Would a transistor be simpler to use in this instance?

You could use a transistor with the collector using your 5V supply and voltage divide the input to the base of the transistor. You'd still need a pull down resistor so you're not saving much. You'd end up with a higher parts count but not necessarily lower cost or reliability. Optos in essence are transistors and you've said your not having much luck with them. Though a 2n2222a and a couple of resistors aren't very expensive. If you wanted to go this route it's probably cheaper and easier to do the voltage regulator at the input.

Trying to use the same formula but solving for the other unknown variable using algebra does not work. I

I'm sorry to say math is math and it's been done here already.

Disclaimer: I've not seen your schematic and I'm often not good with word pictures. If I've botched something I apologize.

[Charlie]

To clarify - power supply for the PIC is not under discussion. This is all about scaling signals to not exceed VDD (5V).

[ERMEGM]

That is correct. The PIC will be powered. I want it to be able to see an input when 12 volts is applied. The program is busy doing a bunch of other things, but it will check that input like this; if input1 = 1 then do this, if input1 = 0, then don't do this.

[ERMEGM]

I'm just wondering what a voltage divider plus a zener would cost opposed to a voltage regulator to limit the input.


Yes, most zeners must have current limiting resistor. I say most cause as soon as I say all then someone will post one that doesn't.


This sound like you're trying to make a logic gate for the PIC. There 's a good example in the PBP manual of how to do this.


You could use a transistor with the collector using your 5V supply and voltage divide the input to the base of the transistor. You'd still need a pull down resistor so you're not saving much. You'd end up with a higher parts count but not necessarily lower cost or reliability. Optos in essence are transistors and you've said your not having much luck with them. Though a 2n2222a and a couple of resistors aren't very expensive. If you wanted to go this route it's probably cheaper and easier to do the voltage regulator at the input.


I'm sorry to say math is math and it's been done here already.

Disclaimer: I've not seen your schematic and I'm often not good with word pictures. If I've botched something I apologize.

Here is what I am speaking of. The left schematic is the one we are all talking about. The right schematic is one that I have seen a couple of times now. Still a high value on the top and smaller one on the bottom. What would be the formula for this bad boy?

[AvionicsMaster1]

For your current way diagram the output would be the voltage divided value at the point labeled out.

For your other way diagram, it depends. If you hooked a meter to the out without any other circuitry attached to the out you'd get whatever the +in voltage is applied. If there is other circuitry attached and there is current flow in the circuit you'd get some voltage dropped across the top resistor. It might be the whole +in but probably less than full voltage supplied. Even if the output is grounded you wouldn't get a voltage divided value across either resistor you'd get power divided across two resistors. Something altogether different.