Watt's the wattage when used on a PIC?

[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

[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