Square Wave Voltage reduction

[jet_pack234]

How can I input a 24V square wave signal onto an i/o pin of a 16F84A? Could I use an external resistor on the input pin to pull up a voltage or would I have to use an op-amp. Thanks for any suggestions

[asmallri]

It depends on your source impedence. If it is low (output from a transformer / opamp/ amplifier etc) then the simplest way is to use two resistors and a zener diode. Ideally you would connect it to a Schmidt trigger input.

39K resistor from the +24v input to the PIC input pin

10K resistor from PIC input pin to ground (signal comon)

4.7volt 400mW zener diode. Cathode (bar) to the PIC input pin. Anode to ground.

Have a look at http://www.kpsec.freeuk.com/components/diode.htm#zener but add the second resistor across the zener.
_________________
Regards, Andrew

http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!!

[Humberto]

[Guest]

Thankyou guys, will have a look at both options.

[jet_pack234]

Why do I need to bother with the two resistors - which are acting as a potential divider (hence the chosen values?). I thought the whole point of a zener diode was so that if it is reversed biased it will maintain the breakdown voltage.

[asmallri]

Just playing it safe. The cheapest way of doing this is just to have the series resistor (not even the pull down resistor). In this case the protection diode clamp the input voltage on the PIC pin to 0.7 volts above the PICs VCC. However this is not really good practice.

The pull down resistor forms the voltage divider and is a cleaner solution than just the series resistor. If however you have a poor solder joint then the effective circuit is as the first scenario.

The zener is a clamp to ensure the input does not exceed the PICs VCC. In which case the zener conducts passing current to ground. You do not need the zener because, as in the first case, the input diode protection in the PIC will clamp the input to 0.7v above the power rail.

In other words the simplest solution is just the series resistor, a better solution is the voltage divider, the purist approach is the addition of the zener. You get to choose what level meets your needs.
_________________
Regards, Andrew

http://www.brushelectronics.com/software
Home of Ethernet, SD card and Encrypted Serial Bootloaders for PICs!!

[jet_pack234]

the only thing i dont understand now is if I put a multimeter across the input pin and ground, what would my input voltage be, assuming I am using the best approach of a reversed biased zener and a 10k resistor in parallel, Would the voltage drop be 5V from the potential divider or 4.7V from the zener?

Or is it (4.7 (from zener) + 0.19(drop across resistor) = 4.89V)

which would then stack up to give:
4.89V +19.1V (across 39K res) = 23.99V (original voltage)

I know it will work its just I like to understand whats happening before I put these things into practice!

Thanks so much for your help Andrew.

[jet_pack234]

Finally, may I ask you for one last piece of advice?

I am trying to program the following protocol.

http://www.sensorpulse.com/support/Documentation/MSP_Protocol_Theory.htm#Output_Delta

I am looking at the last section of the document (section 3.3)

What would be the best technique to capture this data. Using timers to measure the can rate or while loops etc etc. Just not sure how to approach the problem.

thanks

[asmallri]

[asmallri]