09-20-2014 04:02 AM
When I connect an input to DI channel of USB 6501 card from my board which has sourcing Output t, iIn the Off state it is not pulled down properly. The volatge on the port in is 2.1V -2.5V based on the pull down resistor on my board. I suspect that it acts as a voltage divider.
The state on the card is still show as "HIGH "state, even my sourcing output driver has a pull down resister. when I use this interface board for PCI mseries devices it works fine. Do the pull down resister on my sourcing output is not worth. Is this because of pull up resister 4.7k inside USB card 6501.
USB 6501 specifications shows the Digital input accepts both source and sink modes. By default the card is in sourcing digital input mode. How do we can convert this in Sinkind digital input mode.
Could anyone Please give your suggestions to overcome this issue .
Thanks you.
09-20-2014 05:30 AM
You really should not have both a pull up and pull down on the same line. Really weird things can happen. If you insist on having a pull down on your board, then I highly recommend putting a digital buffer chip between the DAQ and your board.
Some boards with DIO have a pull up, while others have pull down. You just have to watch the specs.
09-25-2014 05:00 AM
It seems as if the "thing" you connect to your input does not have a true TTL standard output. Obviously it has a so-called PNP transistor output, this switches the supply voltage to the output terminal when activated. If not activated, the output is in high impedance state.
A true TTL input line, on the other hand, requires that the output device connected to it can sink a certain amount of current to the GND line. If this does not happen, you will see a certain voltage level on the input terminal. What yousee falls in the range of voltage levels I observed on open TTL input lines. When a TTL line is open, it typically (but not necessarily) goes to a HIGH state.
In your case you will need a pull-down resistor on the input line to tie it to GND when the output connected to it is not activated. Another solution would be - as already suggested - using a buffer circuit.