How long does the PCI-6133 take to digitize a sample?

[MadMac_5]

I'm currently trying to acquire a fast (20 ns wide, 10 kHz) signal pulse from a shaping amplifier using my PCI-6133 DAQ card in LabWindows CVI 8.5.1.  When I try to acquire a triggered signal using a delayed constant-fraction discriminator, however, instead of measuring the pulse peak that happens at the exact same instant as my trigger (verified using an oscilloscope) I get a measurement that's considerably lower.  Moving the trigger further before or after the pulse doesn't change the measurement, so I'm wondering whether the DAQ card is unable to digitize the signal fast enough.  I'm currently using the trigger signal as my sampling clock to acquire a single sample at each event, and this SHOULD make the DAQ card sample the signal right at the same point in time that the trigger occurs.

The current setup I use for my DAQ hardware is in the attached .txt file; I've left out the rest of the C code, but can provide it if necessary.  With this setup, the way I think that it works is:

-Increment the counter (PFI 3) each time a rising edge trigger pulse occurs to keep track of number of events
-Use this same input as the sample clock for both the counter and the Analog channel
-Sample the analog channel each time the trigger occurs, since the sample clock "ticks" every time the trigger happens
-Read all gathered samples to a buffer array every 0.01 seconds using Timer ticks in the GUI

I read this thread (http://forums.ni.com/ni/board/message?board.id=250&thread.id=27238) to try and see if what was happening was due to a limitation in how the card samples data, and I've verified that this trigger mechanism works by digitizing a constant 3.0 V DC signal with the card.  It looks to me as though my card is sampling too late, but I'd really like to know whether it's a limitation of the card itself or whether I'm just doing something wrong in my DAQ config.

[dansch]

Hi MadMac,

I understand that you are trying to sample a 20ns pulse based on a clock which is synchronized with this pulse. Due to the very short width of this pulse and the bandwidth of this board, your signal will be highly attenuated. Additionally, besides bandwidth issues, the source impedance and input impedances will also play a role in attenuation of the signal.

Looking at the specifications, it appears that a 1V signal at 1.3 MHz will be attenuated by 3 dB. With a 1.3 MHz, 1 V sine wave, the maximum slope (assuming no attenuation) will be about 8.2 V/μs. Thus, in 20 ns, the signal could only change about 160 mV. As you can see, this is quite a bit of attenuation. In your case, you will not be able to see the signal with such a short amount of time. Hope this helps,

[MadMac_5]

Thanks for the answer, Daniel!  I managed to switch to a longer pulse for testing at the moment and it seems to be working fine.  I'll be getting a faster detector in the near future, though, so I'll bear this in mind when I switch to that new unit.