03-26-2009 05:51 PM
Hello,
I beleive I have exhausted the search options before posting this question although someone has asked it before:
http://forums.ni.com/ni/board/message?board.id=250&message.id=20558&query.id=168227#M20558
The reply claims that as long as you are below the voltage specs (5V input) then the port should be able to sink the current. if this is true then maybe I am doing something wrong:
I am using a DIO port on the 6009 to drive the gate on an APD module (SPCM-AQRH-13) low. The gate on the module is internally connected to its 5V power supply with a 50ohm pull up resistor so in order to turn the module off, one needs to sink 100mA by driving the gate with a low ttl signal. I haven't been successful using the 6009 to do this and I think it is becasue the DIO port cannot sink this 100mA.
How much current can the 6009 sink? it apparently can source 8mA.
Thank you for your time and I apologize if this has been answered somewhere else
Aaron
Solved! Go to Solution.
03-27-2009 02:03 PM
you're right this isn't listed in any specification anywhere. however there is a knowledge base article that addresses this issue.
It is listed as 1 mA, hope this helps.
03-27-2009 02:10 PM
Thank you for your response Doug,
I have looked at that knowledge base article. It seems to be referring to the Analog Input current limit but maybe that spec is the same for both AI and DIO.
Is that true?
thanks again
Aaron
03-27-2009 02:25 PM
03-27-2009 02:42 PM
Thank you Dennis,
So this is exactly where I am unclear. I would think that the open collector mode would be able to sink more current than active drive. Are you saying that this 8.5mA limit is for all modes of operation?
Thanks,
Aaron
03-27-2009 04:49 PM
You could probably use a transistor and a couple resistors to provide a higher sink drive to the APD.
-AK2DM
03-27-2009 08:39 PM
That's what I ended up doing
thanks AK!