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View Full Version : How to monitor and ground a 5v-12v p-p signal?



sccoupe
09-13-2007, 04:40 AM
How do I condition a signal that varies from 5v p-p to 12v p-p with a 3.3volt input. I also need to ground this signal for 1ms over a given frequency. I am currently doing this with a BS2, but I wouldnt trust the input to hold up to that for a long time and wouldnt trust the propellers 3.3v input at all at 12vp-p. What kind of circuitry do I need to pull this off?

Thanks

Jason

Phil Pilgrim (PhiPi)
09-13-2007, 04:50 AM
You mentioned the p-p voltage, but not the common-mode extremes. What are the lowest and highest voltages w.r.t. ground?

-Phil

sccoupe
09-13-2007, 05:11 AM
I think what you are asking for is max: +12v to -12v and min: +4v to -4v with zero ground reference. Its been many years since ive had to deal with any of this so im not sure im using the right terms. I am tieing into a two wire sensor (one wire is ground) that measures when metal teeth pass buy it. This shows me a nice sign wave on the o-scope. I am using this to measure RPM and then grounding it when the RPM goes too high.ˇ at low rpm the voltage is fine but it increases with RPM. I thought it might just be a hall effect sensor, but those require a magnet to move past it.

Phil Pilgrim (PhiPi)
09-13-2007, 06:27 AM
It's probably an inductive pickup. That's the reason the voltage increases as the flux change increases. You might get by with a series resistor between the sensor output and the Propeller pin. 33K should work, if you're sure that 12V really is the maximum. This will effectively clip the negative peaks to -0.6V and the positive peaks to +3.9V, as the Propeller sees them, and keep the current through the protection diodes below the 500ľA limit. For extra protection — and possibly faster switching — you could also use the attached circuit. If it were me, I'd go for the extra protection, in case there are nasty noise spikes riding on that sensor line as well.

-Phil

PS. Your signals are actually 8V to 24V p-p.

Peter Jakacki
09-13-2007, 06:42 AM
Typically an inductive pickup will simply interface to a comparator such as an LM393, the output of which is an open-collector that you can feed straight into your logic input with an extra pull-up resistor. Also, the sensor could be referenced to ground on both sides via resistors and fed into complementary comparators (LM393's are dual) so that the outputs can be wired common-or and react on both the positive and negative phase.

Alternatively you could just use a simple NPN with the current limit resistor in the base to handle the signal and feed the collector to the prop with a pull-up resistor.

There is no need to preserve the actual voltage level as all you want is the timing.

Why do you want to ground the sensor only when the RPM goes too high?

*Peter*

Phil Pilgrim (PhiPi)
09-13-2007, 10:23 AM
If you go the comparator route, be sure it's powered by at least the voltages (positive and negative) that you expect to see on the inputs. Most comparators' input common mode ranges don't extend beyond the supply rails.

Peter, I like the NPN idea.

-Phil

wastehl
09-13-2007, 12:54 PM
This could be a VR sensor, if so, the voltage is determined by the speed of the passing gear tooth, and might exceed 12V. I solved this problem using a simple bridge rectifier from Radio Shack which generates a double pulse, then use a level converter to bring the pulse down to 3.3 vdc level for pin input...

Erik Friesen
09-16-2007, 04:54 AM
If the npn has a base emitter reverse breakdown voltage of 6v what is the practical limit of the negative acˇvoltage that you can feed it lets say with a 10K resistor?

Skogsgurra
09-16-2007, 05:13 AM
I do these things with nothing more than a resistor connected directly to the input. At 12 V and some prudent design, the resistor value can be something like 47 or 100 kohms. A small capacitor from input to ground doen't hurt. Make the 1/RC about equal to your highest expected input frequency (in rads/second). That will not only reduce interference but also keep yor input signal approximately constant (-20 db/decade from corner frequency - and your pick-up goes +20 dB/decade, so the result is a rather constant input voltage.

Edit: read 'lowest expected input frequency'

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Post Edited (Skogsgurra) : 9/15/2007 9:20:39 PM GMT

Phil Pilgrim (PhiPi)
09-16-2007, 05:17 AM
Erik,

Good point. I believe the answer would be 6V, since there's no current, hence no voltage drop through the base resistor. A simple protection device would be a separate diode between base and ground to limit the reverse voltage to 0.6V. The same technique is used to protect optoisolator LEDs that are subjected to AC voltages.

-Phil