How would you do this?

[Charles Linquis]

Isolation is usually the best policy. But sometimes isolation is not possible, or cost effective or...

In this case, ERMEGM (I hope that is an acronym for something!), has the grounds connected. In this case, if you added an opto, you would have both the Emitter and the LED Cathode connected to the processor GND. You would have some sort of resistor in series with the LED anode. The collector would be connected to VCC through a resistor.

Now, apply 12V to the LED through the resistor. The transistor will conduct, and the PIC pin will go to GND. Take away the 12V, and the PIC pin will go to Vcc - no higher. So far, so good.
If there is noise on the input, the opto (because it is slow), will not respond to really narrow 'glitches'. That is good, too.
A downside is that the LED has virtually no hysteresis. There will be very little noise immunity if the signal sits around some value
above the LED threshold (exact value determined by the resistor value).
If you apply a narrow 20KV spike to the Anode of the opto, you might probably blow everything - because most optos don't have an isolation voltage of greater than 5KV.
If you should accidently apply 50V continuously, you might burn the resistor in series with the LED, since the value of the resistor will have to be low enough to give it a few mA of drive at 12V.

On the other hand, if you use a divider you have the hysteresis of the PIC input pin. If you add a capacitor across the "lower" resistor, you block any narrow spikes, and you have noise immunity. Amazingly enough, this will also protect the PIC from those narrow
20KV spikes. How do I know? I have tested them. And if you should accidentally apply 50V continuously across the resistor network (7.5K/5K like I suggested), then you would have 44.5V across the "top" resistor (because the PIC would clamp at Vcc + 1 diode drop). I = E/R, so the current through the resistor would be ~7mA (6mA into the PIC pin and 1mA through the "bottom" resistor. Will the PIC care? The datasheet says it won't. A PIC input is good for at least 25mA in the "clamp" mode. No damage would result.

I'd say that this is a case where a divider is probably as good as anything.

[ERMEGM]

Thank you for your responses. As for my name, the name I wanted was already taken, so I used my kid's initials.

My original thought was to use a voltage divider setup with a filtering cap and a 5 volt zener in parallel with the cap going to ground. This way, the cap would filter out the ripple effects and the zener would protect the pic by regulating the input to 5 volts. As this is 12 volts from a vehicle, I'd like to protect the circuit as best as I can. This part of the circuit will not be detecting a momentary input. It's setup so that when it receives its input, it is constant, like a light switch. When the pic sees this, it goes into another routine until the input is removed.

With the optocoupler, if you have your resistor in place for the LED, any voltage spike would kill the LED. You could setup the same cap and zener across the LED portion of it, and this should provide the same protection, right? The zener would have to be a 12 volt zener though.

Tony

[Charles Linquis]

You can just feed the 12V into a small, 78L05-type regulator, and the output to the PIC.

[ERMEGM]

I'm looking at 4 inputs though. Two for a positive trigger and two for a negative trigger.

My thought was to pull the positive ones low and the negative ones high to give a distinct change.

[Charles Linquis]

I give up.

[ERMEGM]

Here are two circuits I've put together. One is a voltage divider and the other is an optocoupler setup. The positive inputs on the VD setup has a cap and zener for debouncing and has pulldown resistors. The negative inputs have a pullup resistor. The VD setup allows for only one input to trigger a new routine. The opto setup requires the LED circuit to be completed in order to trigger a new routine, so the setup has to work as a pair instead of individually.

In this case, is the VD setup the better setup to use? Disregard the resistor values and they were just thrown in there.

Thanks,
Tony

[amgen]

Charles,
Good point about clamp ability in pins (a few pins don't have that clamp diode), but why are you using such low val resisters ? Voltage dividers with 50, 100-200K value resisters make any voltage/power dissipation insignificant. I tried a touch sensor to a/d pin through a 10 MEG ohm R (and diode) which instantly brought a/d val to full, (5 volt).
Don

[Demon]

You can just feed the 12V into a small, 78L05-type regulator, and the output to the PIC.

If I may, Charles means something like this:




(for some reason, on-site attachments don't seem to work)

Robert

[ERMEGM]

Hi Guys,

Yes, I know what Charles is saying. What I'm asking though is another thing. This will be a production item, so adding a 7805 and caps to every board increases the board size and cost. My diagram from above I thought was pretty straight forward. Does it seem plausible? I think the diodes may be redundant.

Thanks,
Tony

[Demon]

The only comment I can make is "do you have experience with automotive electrical systems?"

Newer cars might be different, but the voltage used to vary a lot. I'd just make sure to have a circuit that can withstand up to 25V without breaking a sweat.

Oh yeah, I read a while back about some manufacturers wanting to use 24V and up, just a thought.

As for your circuit VS Charles divider or any other design, I'm not qualified to comment.

Robert

[ERMEGM]

Actually, I am very familiar with vehicle systems, voltage ranges and the spikes that can occur. However, when studying how other manufacturers have created their product to be installed in such a vehicle using the same features, I don't see much protection being used. Probably because the cost of the board and the likelihood of the malfunction and repair is not cost prohibitive. It almost seems like the opto circuit would be best in isolating one circuit from the other and maybe a 12 volt zener across the LED terminals would prevent damage to the opto. Any thoughts?