Hi
sorry for my late response but yesterday I did some private work.
Back to your first response:
This is hardware part related. The 6040 card has a internal reference of +-10 Volt in bipolar mode. The DAC has a resolution of 12 bit. This will result in steps of 4.88 mV per bit. A sine with 5mV amplitude will give 0 mV, 4.88 mV, 0 mV, -4.88 mV which is more a rectangular wave than a sine. The best solution for this is to use external reference voltage. I do not have any experience which will be the smallest voltage you can use and the user manual, as far as I have seen, does not give any hint.
Next is the load. The user manual states that the card can drive -+5mA. If your load needs more current the signal will decrease. You say you
will have no signal when connecting the UUT I think there is a short circuit in the input of the UUT. Have you checked this?
Why have you measured 5 mV at the input of ACH0? The solution is the following: The last generated output will remain when the signal generation is ready. If your waveform ends with 5 mV then this voltage will stay after the genertion. The reading of the waveform will then read in this last sample.
You will not get any error by using AO and AI because both operations use different buffers.
PFI6 and PFI0 are signals also named WFTRIG and TRIG1. These signals are on the connector of the card. WFTRIG is set each time a waveform generation starts and TRIG1 will start a waiting waveform read.
For an example I have you to wait until tomorrow because I will test it on the PXI-System in the office.
I don't know in which time zone you life but I reside in Germany. Take this into account for "tomorrow".
Waldemar
Waldemar
Using 7.1.1, 8.5.1, 8.6.1, 2009 on XP and RT
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