USB-6009 counter input 5V

[Steve289]

Hello everyone,

I am trying to count pulses from an optical sensor.  This sensor outputs 20Vdc for a high value and 0.7Vdc for a low value.  I built a voltage divider that has a total resistance of 12.9Kohms.  As long as I don't connect the output of my voltage divider to the ground and pfi0 (counter) pins on the USB-6009 I get a signal of 4.9Vdc high and 0.175Vdc low which fall nicely into the TTL levels required.  When I make my connections to the USB-6009 my low signal rises to a level of 1.8Vdc and the high level stays the same at 4.9Vdc.  I believe the problem is that when I measure the voltage between the ground pin and the pfi0 (counter) pin of the usb-6009 with no signal wires attached I get a 5Vdc value, and from what I have found this is due to the internal 4.7Kohm pull-up resistor.  My question is, is there anyway I can measure this pulse signal using the USB-6009, I've tried install in 10Kohm pull-down resistor between pfi0 and ground but this doesn't do me any good.  What kind of sensor could handle having a 5Vdc signal present at its output connection?

Labview 8.5

Thank again for the support,

Steven M.

[tnek]

Hi Steven,

To isolate your voltage divider circuit from the internal circuitry of the USB-6009, you can use a voltage follower. This article gives a some additional information on this process with respect to DAQ devices. As for the 5V that you are seeing when there is nothing connected, this is because the pull-up resistor only contributes when the input is floating. When you actually connect a sensor, usually there is a much higher impedance associated with it and the pull-up resistor will have no effect. 

Regards,

Kent

Applications Engineer

[Steve289]

Hello again,

I gave the Op-Amp method a try.  I used a standard LM324 Op-Amp from Radio-Shack.  I have one op-amp on the positive output and one on the negative output.  I have a total impedance of around 17Mohms.  However when i connect the USB-6009 pfi0 input and digital ground my sensor is still getting saturated by the 5volts that is output by the pfi0 line.  I have the op-amps wired in the typical voltage follower method.  My signal line is connected to the positive side of the op-amp and the negative side of the op-amp is connected to the output of the op-amp.

Recap:

My voltage divider works perfect when it is not connected to the USB-6009 Digital Ground & PFI0.

When I connect my divider to the USB-6009 my output levels are saturated twords 5Vdc.

help.

Thank You,

[tnek]

Hi Steve,

You shouldn't need a voltage follower going to ground. I would try the setup again with the voltage follower only going to the digital input. 

Regards,

Kent

Applications Engineer

[Steve289]

Problem Solved . The voltage follower on just the positive leg worked.  However the the document listed above shows a resitor in series between the op-amp and the digital input.  My first attempt included this resistor (10Kohms) and my behavior was the same as if I did not use the op-amp.  I removed this resistor and connected the output of the op-amp directly to the PFI0 line and TADDA, works perfectly.  Two questions though, the specs of the USB-6009 state that the event counter can take an input from a gated inverter, could I have just connected the output of my voltage divider to the input of a basic IC inverter or maybe a AND gate with one input tied directly to 5Vdc.  Is the event counter just looking for a high impedance source, like that provided by an IC circuit, or is the op-amp required.  And finally when I am at a low state the output of my voltage divider and at PFI0 is 0.7VDC, if I connect a 330ohm resistor in parallel between the PFI0 and GND pins my low voltage goes to 0.3Vdc.  I can reduce this voltage down close to 0Vdc just by reducing the resistor value, however once I get to 10ohms then my HIGH value is effected, up till that point I get a perfect 5.1Vdc HIGH value.  The field wiring guide says I need to use bias resistors when measureing a floating or N.R signal, but they are talking about AI and I am doing a DI.  Is this parrallel resistor acting as my bias resistor and stopping current flow when I am at a LOW state.

I will post a diagram of my working circuit, and of the circuit that did not work later tonight.

Kent, thanks for the directions. 

Steve.

[tnek]

Hi Steve,

Glad things are working now. The digital input really only needed some isolation between the external voltage divider and the internal pull-up resistor. Anything which could provide some impedance should suffice. In addition, the resistor that you placed between the input and ground is acting as a pull-down resistor. Generally bias resistors are only used to remove excess charge that builds up on the input. 

Regards,

Kent

Applications Engineer