Watt's the wattage when used on a PIC?

[ERMEGM]

Hi Guys,

As the titles alludes to, I am trying to figure out what size resistor I need when using a voltage divider on a PIC as an input trigger.

My circuit is being powered by a voltage regulator and I have been using an optocoupler to trigger my PIC, but am trying to minimize cost, so switching to a voltage divider setup. What I am trying to figure out is can I use two 1/8 watt resistors in my VD circuit along with a .1 uF cap and Zener, to make sure my circuit doesn't go above 5 volts?

For my 12 volt circuit, I have calculated using 10k and 4.7k, which gives me just under 5 volts. Is it safe to use two 1/8 watt resistors or should I bump the 10k to 1/2 watt or 1/4 watt? This is the only things that's stumping me. The input can either be momentary or continuous (if that matters).

Thanks,
Tony

[Normnet]

See the PIC data sheet for pin ma sink or source rating.
Ohms law is amps * volts = watts.


Norm

[richard]

or more appropriately


p= v*v/r ( v^2/R)
p= r*I*I ( R*I^2)

12 x 12 / 14700 = 0.009 watts for the total
I = V/R I= 0.8 Ma

P 4K7 Resistor = 4700 x 0.0008 x 0.0008 = 0.003 watts
p 10k R = 10000 x 0.008 x 0.0008 = 0.006 w

I don't think you can buy a resistor that would not be ok in that circuit even the smallest 603 smd could do it easily

[AvionicsMaster1]

I'm curious so here goes: Using those resistor values in a 12 volt circuit in series should yields around 3.8 Volts across the smaller resistor. That's well below 5 volts and could be considered to be in the ambiguous range if using it for a logic high. Maybe that's what you want but again I said I'm curious if that's what you really want. Those resistors can be of any size that will give you the voltage since you aren't really drawing current from them just a voltage level. Though I don't see how these will replace the opto-coupler as a trigger.

You say you want to use a zener with a cap and two resistors for VD clamping (may not be right term). Most zeners require a turn on current and I'm going to use 100mA as my minimum guess of that current. I'm assuming you're using the same 12 volts as the voltage divider for your source. The regulating resistor would have to drop 7 volts at .1A which means it would need to be around 70 ohms. So, I^2 x R would require it to dissipate .7 watts. You can use as many resistors as you wish to dissipate that heat but they must be at least capable of dissipating that heat continuously. You could use 6 each 1/8 watt 420 ohm resistors or one big 70 if you want. Either will work for dissipating heat.

If this isn't what you meant I apologize.

[MichelJasmin]

You say you want to use a zener with a cap and two resistors for VD clamping (may not be right term). Most zeners require a turn on current and I'm going to use 100mA as my minimum guess of that current.

With zener diodes the target is 10% of their maximum current. So for a 5V/1W zener it's maximum current is 200mA so the minimum is 20mA. 7V drop @20mA=350ohm.

If you just want to clamp the voltage at any input the PIC have protection diodes so a series resistor is sufficient. Read the datasheet to find a reasonnable current for protection diodes. Hint: check PBP manual for the SERIN command, there is a schematic to interface RS232 without level shifter...

HTH

[ERMEGM]

Thanks for the replies. My application is for a PCB being installed in a vehicle. The voltage is usually 13.8 volts. As I had some problems with some of my optocouplers being bad and had to replace them, I wanted to try another method (voltage divider) where you don't usually get bad resistors. I calculated 10k and 4.7k to bring it down to about 5 volts (or in my case, just under 4.5 volts). The Zener was to help maintain a max voltage of 5 volts to not damage the PIC, while the bypass cap helps to filter some of the noise in the line.

Do everyone's suggestions still stand? It seemed like a simple problem at first, but then the more I began thinking about all of the variables, I started second guessing myself. Is there anything I should change to the setup?

Thanks,
Tony

[grahamg]

Why do you not just use a 47k resistor in series with the input pin. The internal diodes in the pic will clamp the input voltage anyway. No need for zeners or voltage dividers.

[Dave]

Grahang, How are you going to limit the current?

[LinkMTech]

Why do you not just use a 47k resistor in series with the input pin. The internal diodes in the pic will clamp the input voltage anyway. No need for zeners or voltage dividers.

I've used this method too but also included a small filter cap to ground at the input pin for noise.
The DEBUG section of the PBP manual shows this as acceptable where +-25V signals are possible to the input.

[tumbleweed]

Bad idea.

See http://www.picbasic.co.uk/forum/showthread.php?t=15542 post #3 for the reasons.

[grahamg]

Dave with a 47k resistor the current at 12 volts is only about 255 Micro amps. I have used this method hundreds of times and never had any problems. The input resistor can be as high as 270K and stil works well.

[Badger]

If the input is only for a trigger (e.g.. turns on with the ignition), why don't you use a 5V reg (7805)?
I've used them and it works perfectly.

Badger

[Charlie]

If the input is only for a trigger (e.g.. turns on with the ignition), why don't you use a 5V reg (7805)?
I've used them and it works perfectly.

Badger

And it only cost about 100x as much as a resistor, and needs filtering so add another 10x, and is slow to respond, but sure a regulator and many other complex circuits will work to sense a level that doesn't change very fast. Grahamg has the cheapest solution with a single high value resistor, if the 5V VDD supply is capable of both sourcing and sinking current. A resistor divider pair will work if the supply can source only. No need for zeners or anything else.

[ERMEGM]

Dave with a 47k resistor the current at 12 volts is only about 255 Micro amps. I have used this method hundreds of times and never had any problems. The input resistor can be as high as 270K and stil works well.

So just one resistor from the input wire going to the input pin? Does this work in even the noisy environment of vehicles?

[AvionicsMaster1]

This certainly has gone tangential to your original question. Going back to the beginning, for the 10k and the 4.7k a 1/8W current rating would be enough.

I don't see where you said what PIC you're using. I looked at the 12F683 and see where it would be able to clamp 20mA at GPIO input. You may be able to use one resistor of sufficient size to take advantage of PIC internal circuitry to clamp that 12VDC but the datasheet shows only +.3V over VDD. To me a better way would be the voltage divider scheme you first posited.

I hope you finally got your answers.

[ERMEGM]

My original question asked if I had it right or if I needed to make the resistors bigger. Suggestions were made to make the voltage smaller. My question is that with 2.5 volts, will that trigger the input? For a digital signal, you need a 1 or 0, or in this case, 5 volts or 0 volts. Am I missing something?

I am using the 16F819. After reading the other article (link that was posted), I figured I would use the VD to a transistor to trigger the input. I just don't want to blow anything up.

Thanks for your help thus far. I just want to understand it better.

Tony

[Tabsoft]

Tony,

I did not see where you indicated which pin on the PIC you are using for input.
As you probably already know that which pin you use matters in the answer to your question about what voltage will be recognized as a 1 (high) or 0 (low) by the PIC.

The PIC you are using has 3 types of digital input pins, each with different thresholds for low and high. Standard Inputs, TTL Buffered inputs and Schmitt Trigger inputs.

Table 15.4 of the DS has the Min/Max for Input Low and Input High for each of the pin types.

You probably already knew this, but just in case.

Regards,