USB-6501 current sourcing/sinking capabilities

[parom]

I have a question about USB-6501, specifically about the Fig.5 of NI

USB-6501

User Guide and specification. Lets say my RI is 200 Ohms. If I follow your

procedure and I want to achieve 8.5 mA I have to set up Re to 423 Ohms.

This

will give me 8.5 mA to RI load. Accordingly to your specification the circuit connected to Port Pad is open drain transistor. If I operate that circuit my current from internal source will be 5V divide by parallel combination of

423

Ohm and 4700 Ohm resistors. The resulting current will be 12.8 mA which is

not allowed.

So by following your procedure I created self destructive circuit.

Am I missing something? Please send me explanation to this. Is 8.5 mA also limit of open drain transistor or is the limit coming from internal source capability. What is the current handling capability of transistor?

[K_Bull]

Hi there parom,
Is there a reason why you are choosing your RI to be 200Ohms? Also, the internal resistor is 4700Ohms, so to get the right value for the Re, you simply use the ammeter process described in the manual or Ohm's Law.
The source current can be driven up to 8.5mA and not higher.
If your external pull up resistor is of value 423Ohms, of course you will get 12.8mA but this doesn't happen as it exceeds the maximum sourcing current and you might end up burning some stuff.
From my calculations (not using the measure and choose method), you need an output current from P0.0 (ref-fig 5 in the manual) of 8.5 mA. So the combined Resistance for this between the internal and variable resistors would be 588.23 Ohms. Rounding the figures up, I make Re = 672.5 Ohms.
So stick a resistor of this value for Re and you can expect 8.5 mA given that it is 5V.
As the resistors act as potential divider, you need to select your resistor values as the manual suggests.
Best Regards,