Watt's the wattage when used on a PIC?

[AvionicsMaster1]

A voltage divider with those resistors at that voltage should have just under 4.5 volts. If you're using that voltage to logic check or reference something you should use a zener or internal PIC voltage reference as a standard. There is some variation in voltage with engine running and off. Why not use some of the 5 volts from the power zener and voltage divide that to get around 4.5 volts? More stable and accurate. Though I still don't see how a voltage divider replaces an optocoupler.

[ERMEGM]

It doesn't replace the optocoupler. I would rather have the opto, however, after a recent string of failures, I am trying to minimize my repairs and use something more reliable like resistors.

So, if I use the Zener to regulate down to 5 volts, would I use a VD setup that brings the voltage down to say around 6 or 7 volts and regulate from there, or just use a single resistor for the load on the Zener?

My understanding is that the Zener should not be the only thing regulating the supply because it will blow. It WILL need to go through a resistor. Does this sound right?

Thanks,
Tony

[Charlie]

I'm not sure what your application is, but if it's an alarm or similar and needs to work when the vehicle is on or off, there is more to consider. You mention an input trigger, but not how you are powering the PIC. Inputs can not go higher than VDD, so this implies your voltage divider needs to be less that VDD when charging (as high as 14.4 V) and when the battery is quite discharged (about 10.8 V). You can do this with a voltage divider but there are a smaller set of values that will work. [R2/(R1+R2) = 2.85]. Finally, there's no need for zeners - a voltage divider will do it all. Zeners are useful where you need a constant voltage regardless of load. That is not your application - the load never changes if it's just the input of the PIC. You can make a zener work, but you are just wasting electricity (and of course money).

[ERMEGM]

Thanks for the replies. I don't work much with Zeners, so hence why I'm asking. I have been using optocouplers to trigger this part of the circuit, but they are expensive compared to a couple resistors, zener and cap.

My circuit is simple; main supply goes through a voltage regulator, so the supply to the PIC is always 5 volts. This other input is just to tell the PIC that this wire has been activated. 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 chose the higher values so the PIC would read them as ones and zeros. If it was lower, I would have to do an ADC, right? I've modified my circuit design from the opto to the VD. If I did an ADC, I would have to reroute the traces to the ADC pins, and that's a cumbersome redo of my board.

My comments about the load was stating that I just 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, or else the Zener would blow and then the PIC. Does that sound right?

Also, I am aware of the typical VD circuit where you have two resistors in series, supply is on one side of R1 and ground is on the other side of R2 and you tap in between R1 and R2 to get your divided circuit. I have found two such circuits with the same layout, but instead, Vin and Vout are reversed. The supply is applied between R1 and R2 and while R2 goes to ground, the other side of R1 is your output.

Trying to use the same formula but solving for the other unknown variable using algebra does not work. Is there a different formula for this method? In this instance, R2 is 10k and R1 is 100K.

All I want to do is trigger a pin on the PIC to let it know an input has power applied to it. There's optos, VD and transistors that can be used. Would a transistor be simpler to use in this instance?

Thanks,
Tony

[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]

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?