voltage level when measure two-edge time difference??

[ossoo]

I am using PCI-6229.

I have two signal, both are 20K Hz, amplitude are from -8V to +15V. Both the rising and falling edge of the two signal have a slight slop. I need to measure the time difference between the rising edge of the first signal and the falling edge of the second signal. The time diffence are different if measure at different voltage level.

I have two questions:

1. Do I need to do a level shift to the range of 0~5V? Because the counter channel are digital I/O channel?

2. At what voltage the Start and Stop channel of Counter start to trigger? Because the time difference are different if trigger at different voltage?

3. How can I trigger the Start and Stop at some voltage value I need, like 2.4V?

Because I do not quite understand the principle the counter works. Is anyone can help, I'll appreciate a lot.

Thanks

[John_P1]

Hi ossoo,

The digital lines of the 6229 use TTL Voltage Levels.  So to answer your questions:

1. Yes, you will need to shift the signal to fit within the input range of the card if you want to use the counters.  The 6229 Specifications page has information about the logic levels required:
   
                  
   
   
   
2. The transition from logic low to high will happen somewhere in between .8 and 2.2 Volts.
   
   
   
3. To trigger the start or stop at a specific voltage, you'll probably want to use a comparator.  This is especially true if the rise and fall time of your signal is slow--the comparator will give you a much cleaner edge.

So to summarize, I would look into using an external comparator that will output 0-5V depending on whether the signal of interest is above or below a reference voltage.  If the level that you want to trigger off of is within ±10 V then you can use the AO of the 6229 to set the reference voltage programmatically.

Once you get the signal into TTL form you can use the counters to measure the time difference using the Two Edge Separation method.

Best Regards,