I would like to find out the RPM of an engine by capacitive measurement.
To do so I would wrap a wire around the inginition cable (to increase surface) and then connect that via a capacitor and resistor to the IO of the arduino. The grounds of the arduino and the engine would be connected.
The Ignition Impuls is somewhere around 20kV with a current somwhere in the low mA range.
I understand the arduino wont read much with this conficuration so in the drawing where I drew the dots it probably needs an op-amp curcuit of some kind.
But this question is more regarded to the protection of my Arduino.
-I dont really know what kind of signal to expect here, so how would I go about choosing the right Capacitor C1 in order to not blow the µC?
-Should I just start with a very small value(maybe 1nF?) and increase that until the signal gets strong enough?
-In case the signal is too stron anyways. Would two Schottky diodes connected to Vcc and Gnd be enough protection?
I dont really know what kind of signal to expect here
About 100 to 200V.
The capacitor does little to protect the Arduino pin it is the resistor and diodes that do the protection. You should use normal diodes not Zener ones.
I do not know how Zener diodes got into this discussion. The symbols are for Schottky diodes and the OP wrote about Schottky diodes. It is correct to NOT use Zener diodes for this application, but ordinary diodes and Schottky diodes are appropriate, with the Schottky diodes being slightly better assuming all else is equal.
I don't know about the capacitor, and the value of the resistor is unspecified. The diodes prevent the input voltage from going more than a (Schottky) diode drop above 5 volts or more than a (Schottky) diode drop below ground.
The diodes may work if they are not killed by excessive voltage or excessive current. I would be tempted (in this unspecified situation) to add more resistors, a couple of gas discharge tubes, and a couple of MOVs, and perhaps be prepared to replace the Arduino.
vaj4088:
I would be tempted (in this unspecified situation) to add more resistors, a couple of gas discharge tubes, and a couple of MOVs, and perhaps be prepared to replace the Arduino.
Grumpy_Mike:
That will kill the signal altogether, once triggered the impedance is very small and will swallow the signal.
That is sort-of the point. If the voltage is high enough to trigger a GDT, then it would have killed the Arduino. It does no good to allow the signal to come through if it is going to destroy the listening device or the other protection devices.
I suppose the OP can skip the GDTs so that the signal comes through but is limited by the MOVs and/or the diodes, but without further information it is possible that the MOVs are going to be destroyed. Hand wrapping an unknown wire (Does it have insulation? How much? What type? How many turns? etc. etc.) around an unknown ignition wire could work great or could be a recipe for disaster.
To clarify things a little:
At the moment this is more a theoretical question. My idea is to choose save values for capacitors and resistors and then from there work my way down until I can see a signal.
Same with the signal wire. I would start placing the wires close to each other and if that doesnt really give a good signal I would just start wrapping it more and more to increase the capacitive coupling.
Now the goal is to estimate where to start. Im trying to find out what is a safe circuit that will definately not blow my arduino.
Grumpy_Mike:
About 100 to 200V.
The capacitor does little to protect the Arduino pin it is the resistor and diodes that do the protection. You should use normal diodes not Zener ones.
Good to hear. I was thinking about Schottky diodes.
How do you estimate the 100 to 200V? This depends mainly on the area between the curled sensing wire and the sparkplug wire, right?
6v6gt:
I don’t see how the circuit in the OP offers any protection.
See here for some diode clipping circuits which clip both halves of the cycle: Diode Clipping Circuits and Diode Clipper
I may be wrong but I think the link you posted describes basically the same method that i use. If the voltage on the input exceeds 5V(+0.7V or something for the diode) it will be derived over the upper diode to Vcc. If it goes below 0V the diode from ground willhandle it.
vaj4088:
I don't know about the capacitor, and the value of the resistor is unspecified.
Its more of a theoretical question at the moment, thats why no values are specified.
I got the idea with the capacitor from this graphic.
I believe the capacitor is used to limit the current coming from the sensing wire. Once the capacitor is loaded. No more current can flow. No more danger for the circuit coming afterwards. Would that make sense?
Grumpy_Mike:
I connected the wire round the coil and there was enough voltage to fire a miniature neon lamp directly. Which you need over 80V to do.
Ah yes, I recall the old timing lights with a Xenon tube which you put in series with the spark plug leads - so they came with the same insulation as the spark plug leads.
