Pull-up resistor & PLC input?????

Hey guys,

I have a Pepperl+Fuchs NBB4-12GM50-E0-V1 proximity sensor driving an input on a Telemechanique Twido PLC. The proximity switch is sensing 10 holes drilled into a 12" circum. wheel as an encoder configuration. The output of the prox did not drive the input of the PLC even though I was getting 24VDC(supply) out. So I put a pull-up resistor of 1K between the output and +24VDC. The machine works well up to 125 ft/sec and after that it seems the counting is getting off. Could this resistor be decreasing my switching frequency and/or increasing off decay-time? Thanks in advance for your advice.

C.

Are you performing the counting in the plc program or does the input have high speed counting ability.

If the input doesn't have high speed counting ability then you are limited by the scan time of your program.

If it does have high speed counting ability then read the documentation. I know that on AB M1000's you have to turn the filter off for those inputs to get the HSC ability.

10 Holes, on a 12" circumference wheel... 125 ft/sec... That works out to be about 1250 Hz.

That sensor has a rated switching frequency or 1000 Hz.

http://www.am.pepperl-fuchs.com/pdf/documents/sc04-095.pdf

And the pullup resistor probably isn't affecting your counting ability.

Well done rdrast.

Sorry, I meant 125ft/min.

125feet per minute = 2.083 feet per second x 10 =20.83 cycles per second or 20.8Hz

The 12" circumference with 10 holes means 10 "cycles" per foot which is why you multiply 10 x 2.083

Originally it was 125feet per second which meant multiply 125 x 10 to obtain 1250Hz.

To verify this 125ft per second is 60 times faster (60 seconds in a minute) than 125ft per minute...ie 1250/60=20.83

The size of the holes may also be of importance. This gives you the time the sensor goes to another state. This shouldn't be too short, otherwise the PLC may not detect it.

The 1000Hz for the sensor may be OK, but is the same true for the PLC?

Just some thoughts of mine.

Kind regards,