laser as z probe

dc42

Do you know what would happen if I feed negative voltage (-3.3V) into the Z probe input?

Probably nothing, because the ESD protection diode in the MCU will protect the input. But for safety, you could use a series resistor and a Schottky diode to ground.

Yes I was assuming that it is a time of flight sensor.

hurzhurz

@chilli Yes, I think so. Though, the needed resolution of the image sensor probably needs to be very high to achieve this crazy measurement resolution of 2-4µm... maybe they also use some clever algorithms to interpolate.
I just found this video with an illustration: https://youtu.be/g-EJ4FDB4qQ?t=34

chilli

I've been trying out a bit and it seemed to kinda work. I'm getting a signal in.
Can't work on it till probably next week, cause the z-axis clutch broke.

Another thing is, that the laser is giving 0V if the distance is 0. if it's more than 0 the voltage rises till 5V and stays there and same with negative, falling to -5V.
As the z probe input works with 3.3V i was thinking about using a resistor to ground and also in series a schottky diode for the negative voltage (thanks for that info @dc42).
I'm just not sure about what resistor and what diode to use. Any suggestions?

What current does the input draw at 3.3V?

maybe this schottky diode:
https://www.conrad.de/de/p/panjit-schottky-diode-gleichrichter-1n5817-do-41-20-v-einzeln-1304858.html

hurzhurz

Do you really need to deal with negative output voltage?

Pages 5&6 in the il1500 manual it look like you can set different output voltage range modes:

0V to 5V
-5V to +5V
1V to 5V

So I think you can simply use the first one together with a voltage divider?

Well, any maybe you could simply use the digital output "LOW/HIGH judgment output".
Here you could use a optocoupler for isolation between it and the duet.

chilli

@hurzhurz yes these modes are possible to set, but the problem is that the voltage stays after the set point is reached.
For example 0-5V: above 0 the voltage is 0V, the closer it gets to 0 the voltage rises (the point from which is the voltage rises can be set too) : 10mm 0V; 5mm 2,5V 0mm 5V.
If the distance gets negative the voltage stays 5V
So I thought it would be better that the voltage above and below 0 would change, so that it doesn't get too deep and still thinks it's at 0 distance

Not sure if I was able to explain it understandably...

The idea with the digital output and the octocoupler sounds possible, will try that!

hurzhurz

@chilli I hope I understood it correctly, but I would do it like this for the analog output way:
For your example values: 10mm=0V / 5mm=2,5V / 0mm=5V
I would define the middle (5mm) as my target trigger point and later set my probe height offset accordingly (G31 Zxxx).
Therefor I would adjust the probe trigger value (G31 Pxxx) so it matches the value that the duet measures at 2,5V (or 1,65V in case of an ideal voltage divider).
This way, you can see if you are below or above the trigger point.

Btw, we could also have a conversation in German if that helps...

JoergS5

I checked the optoNCDT sensor, they start used at 400$ (older models than the one mentioned in the video). Does someone have cheaper solutions with the same high resolution?

chilli

@hurzhurz yeah i think you got me right. that could actually work too.
In what unit would i put the trigger point value (behind the P of G32 Pxxx)?
Or is it just 0?
I think german would be easier, but english would be better for the others here

@JoergS5 Sorry, don't have any solutions for that

chilli

@dc42 What current does the duet draw at the 3.3V Z probe input?

hurzhurz

@chilli It is a number (between 0 and 1000 if I remember correctly) that represents the voltage:
0 for 0V and 1000 for about 3,3V (or the other way round, depends on the "I" parameter of M558). So you would enter something like G31 P500.
But I would try that out / look what G31 (without parameters) exactly tells when it is in the correct position.

You also have to consider that these 1000 steps are the available resolution for the probe input.
So if you map the distance of 10mm to the sensor output voltage range, you actually get a resolution of about 0,01mm at the probe input.
Because of this, maybe the digital output would give you a better repeatability.
I think you should try both variants...

For the probe input: @dc42 may correct me, but I think it shouldn't draw any noticeable current because it is a high impedance input.
There is also a 10k resistor from a RC filter which should limit it to 0,33mA (at 3,3V).