6009OEM output high voltage versus current

[AJP1]

Hi all,

Just trying to make sense of: 

http://www.ni.com/pdf/products/us/20043762301101dlr.pdf 

On page 3 says:  Output high voltage (push-pull, I = -8.5 mA) minimum voltage = 2.0V maximum voltage = 3.5V  But where can I find out what current can it provide and still produce an output voltage over 4.2V?  (4.2V is the worst case scenario for the logic high threshold of a chip I want it to control).

Thanks!

J

[MikeParkin · Accepted solution]

Hi J,

As you mentioned the maximum voltage is 3.5V when you have an Output high voltage (push-pull, I = -8.5 mA). If you need more than 3.5V then you might need to change to an open drain output. As you can see from the link you provided this would enable you to receive a maximum voltage of 5V.

If you do use an open drain output you may need to combine it with a pull-up resistor.

I hope this helps.

[AJP1]

Hi Michael,

In the end I didn't look for/stumble across an option for open drain or push-pull.  When I wrote digital signals to the digital ports on the USB6009 OEM board they were around +5V for high, and 0V for low, so no problem there.

Thanks very much for the response any way.

James