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Re: How would you do this?
Actually the picbasic pro manual shows under the serin command, you
can input 12 volts directly through a current limiting resistor. (22k in example)
But if you truly need a reliable trigger, without false triggers, I agree with aratti
and use optocoupler and caps.
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Re: How would you do this?
If the grounds are not isolated, can someone explain to me why an optocoupler is more reliable?
Charles Linquist
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Re: How would you do this?
If some kind of voltage spike comes along the circuits other than ground, less paths of that spike to the uP reduces chance of frying.
Don
Amgen
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Re: How would you do this?
I agree Charles. Isolated grounds would be best. But if thats not practical, than keeping circuit neighborhoods isolated would be next best. At least thats the way I understand it and have seen in practice.Originally Posted by Charles Linquis
Meaning of neighborhood. If you already know this than I apologize. One neighborhood would be the analog section. Another would be the digital. All neighborhoods should have their own ground that leads back to the point where power ground comes into the circuit. This point will have the lowest impeadance to power/earth ground. In other words, all neighborhoods share their ground at only one point. Doing this keeps noise created in one neighborhood from effecting another.
This does not work for all applications but using an optocoupler will keep this idea in place.
Also this idea relies on the use of the right capacitors in the right place. I still don't fully understand how caps fixed certain high speed trigger problems i had.
...maybe it was inductance or quantum interactions ... i don't know ???
Last edited by MikeWinston; - 13th September 2011 at 18:51.
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Re: How would you do this?
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.
Charles Linquist
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Re: How would you do this?
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
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Re: How would you do this?
You can just feed the 12V into a small, 78L05-type regulator, and the output to the PIC.
Charles Linquist
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Re: How would you do this?
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.
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Re: How would you do this?
If I may, Charles means something like this:
(for some reason, on-site attachments don't seem to work)
Robert
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