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How much current can a USB-6009 sink?

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Hello,
 
I would like to use all 12 of the USB-6009 DIO pionts as open collector to sink the current required to turn on 12 MOC8020 opto-isolators.  The DIO point is connected to the cathode of the opto emitter (ie LED) and the anode is connected to 5V through a 470-ohm resistor.  The typical value for the forward-biased drop on the diode is @ 1V so I figure that the 6009 will have to sink (5V-4V)/470 = 8.5mA per DIO point.
 
(Please) What is the limit on current sink per DIO if all points could potentially be sinking simultaneously (I was unable to gather this info from the 6009 datasheet)?
 
Thanks,
Chris
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Hi Chris,
 
There is actually a KnowledgeBase (KB) article written on this issue.  See the following link:
 
 
The USB-6009 can only handle currents up to 1 mA.  However, you haven't taken into account the input impedance of the 6009 itself.  As the above KB mentions, the input impedance of the 6009 is 144 KOhms.  So, if for example you were to input the 5V directly, you could expect an input current of 5V/144000Ohms = 0.035 mA, which is well under the 1 mA limit.
 
Hope this helps!
 
Justin M
National Instruments
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Thanks for the reply Justin.  However, the knowledge base entry that you pointed me to seems to be for the AI not the DIO.  From the 6009 manual, I know that the 6009 DIO can source up to 8mA with an external pull-up, but I haven't really been able to decipher the sink capability.

The max sink per DIO line may also be 8mA, but I would expect it to be higher (15-25mA). 

Regards,

Chris

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Chris,
 
You're right, that KB is referring to the analog input.  Sorry about that!  However, we can still look at it in the same way.  Just as in the analog case, as long as you don't go over the maximum voltage specs, the USB-6009 will be able to sink the current.  For digital input on the 6009, the voltage must be within the range -0.5 to 5.8V.
 
Thanks,
 
Justin M
National Instruments
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Hi Justin M,

 

I know it's been a while (i.e. years) since this thread began; please bear with me...

 

Page 22 of the USB-6009 manual states that "The default configuration of the NI USB-6008/6009 DIO ports is open collector, allowing 5 V operation, with an onboard 4.7kOhm pull-up resistor."

 

1. I'm puzzled by this statement.  Can this onboard pull-up be disconnected through software or jumpers?  If not, then it would seem that the output is not really open collector, but is instead TTL (0V or 5V).  In other words, it appears that the USB-6009 digital outputs can not be used to convert from a TTL control signal to a +24V "industrial" control signal?.!

 

Best Regards,

Chris

 

 

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