Need Help!- Vane Position Sensors

[frankg521]

After searching 11 pages I think I need help with this one I have these Vane Position Sensors need help wiring them to the BOB for the z limit on my mill the sensors have + - and output and the limit on the board has 5v and pin 12 if anyone could help i would greatly Appreciate it

Thank you
Frank

[frankg521]

has anybody veiwd this thread ???am i in the right forum??? or no one wants to help

[ViperTX]

Well the sensor is an open collector type....so, you need a pullup resistor...if you need the output to swing from near ground to +5 VDC then the pullup will go +5 VDC.

The input pin to the Bob is that a CMOS or a +5 VDC logic pin?

Do you know how to size the pullup resistor? Pick 20 ma

PullUp Resistor = (Pull Up Voltage -.4 volts) / 0.020

[frankg521]

thanks ViperTX .....but my knowloge of eletronics is limited this is the BOB im using http://www.cnc4pc.com/Store/osc/prod...products_id=45
what exactly is a pulup resister

Frank

[vger]

Frank, looks like you are using the same thing I used for limit / zero switches. As viper said, they will require a pullup resistor. The "Vane position sensor" is an optical device that contains an LED (light emitting diode) on one side and a phototransistor on the other. The phototransistor is similar to a standard transistor except the base (input) is powered by the light striking it. The phototransistor is like a switch that only passes current one way. Most comonly, phototrasistors are of the NPN type rather than PNP. The phototransistor has 2 leads, Collector and Emitter. The emitter is the negitive and the collector is the positive side. In this case, the collector is not connected to anything internally of the sensor and is left "open" and forms the output of the sensor. For the device to work, a resistor known as a "pullup" must be connected from the collector (output) to a positive power supply. When there is no light striking the phototransistor (sensor blocked) the phototransistor is switched off and the "pullup" resistor pulls the output to near the positive supply voltage. When light strikes the phototransistor, it switches on causing the collector (output) to be nearly grounded. A small current flows trough the pullup resistor and is wasted. The value of the pullup resistor is determined by the logic type the output it is connected to. If it's 5 V cmos 10K works well. If it's 5 volt TTL 4.7K does the trick. The positive supply voltage that the pullup is connected to should be the same voltage level that the logic input chip is powered by. This is most likely 5 volts.

This type of switch works well. On my little mini router, the repeatability of the limits are .001 or better.

Steve

[frankg521]

HI Steve thanks for the reply... this helps... the switch is magnetic but I think it’s the same principle, Do you have a schematic of this circuit?

Thanks again
Frank

[Al_The_Man]

The device looks like it has four connections, the outer are common and 5+ to 24+vdc supply, the centre two are either N.C. or N.O. , both open collector, so you can pick which one you want to use.
Which ever open collector you use, you wire a resistor to it and the +supply, and this collector will be your output, (input to BOB) if you have a BOB that has a source input, that is an input that has to be taken to common, then you do not need a pull up resistor.
This is a hall effect sensor so it is non-optical.
Al.

[frankg521]

Al the device has three wires Black(gnd) Green(output) Red(Vs)

[Al_The_Man]

Some of the literature suggested two outputs, if you only have one then that makes it simple, the same connections apply.

Personally I found the description of the BOB I/O conditions rather vague and ambiguous.
They do not show the nature of either the input or the output for the different pull-up, pull-down choices.
You want to use it with your sensor in the pull down or sink to common and the BOB in the 5+ source mode.
Al.

[frankg521]

Al if you could bare with me on this it does not seem to switch how do you wire the sencer on the bottom if the picture to the limit on the bob
Frank

[Al_The_Man]

It appears you need to set the jumper on the BOB 5+ 'common'
And take the vane output to any input, If the input is a opto isolator for e.g., you won't need a pull up resistor.
Al.

[vger]

Frank, my bad, hall effect not optical. Still the same output style. The vane must be low carbon steel and should be .040 inches thick and .4 inches wide. The switch is off when the vane is in the air gap meaning that the output (green wire) will be open (not grounded) with the vane in the air gap. Your BOB literature says "All inputs are outputs are tied to pull-down resistors" to keep noise from interfering with un-connected inputs. There is a jumper to select either +5 or ground to the common pin that is close by (physically near) the I/O pins. The switch will sink 20 milliamps so selecting a pullup of 470 ohms will be safe with 5 volt supply.

Sensor Black wire to ground on the BOB
Sensor Red wire to +5 V
Sensor Green wire to your input pin 12

Now for the pullup... obtain a 470 ohm 1/4 watt resistor from radio shack or other electronics supply. The color stripes on the resistor will be Yellow, Violet, Brown, with either a gold or silver after that.
Connect the resistor between the BOB pin 12 and the 5V connection right next to pin 12.

Make sure your power supply (5V 400ma) is connected to the BOB.

The pullup resistor just makes the input go HI when the vane is in the sensor. With the vane removed, the sensor makes the input go LOW.

Steve