Jump to content

As of July 17, 2015, the LabJack forums here at forums.labjack.com are shut down. New registrations, topics, and replies are disabled. All forums are in a read-only state for archive purposes.

Please visit our current forums at labjack.com/forums to view and make new posts. To post on the current forums, use your labjack.com login account. Your old LabJack forums login credentials have been retired. There are no longer separate logins for labjack.com and LabJack forums.


Photo

LJTickDAC reading commanded voltage

U3 LJTickDAC

  • Please log in to reply
1 reply to this topic

#1 Tom Kuiper

Tom Kuiper
  • Members
  • 10 posts

Posted 11 February 2015 - 12:10 PM

It seems it should be possible to ask what voltage a LJTickDAC channel has been set to.  I tried to use the examples of reading the cal constants and writing the volts to come up with an appropriate i2c command but I get back a list of 255s, which I interpret as an error response.  This was my attempt, in Python.

 

In [15]: lj[1].i2c(0x12,[48],NumI2CBytesToReceive=16,SDAPinNum=1,SCLPinNum=0)
Out[15]: 
{'AckArray': [3, 0, 0, 0],
 'I2CBytes': [255,  255,  255,  255,  255,  255,  255,  255,  255,  255,  255,  255,  255,  255,  255,  255]}
 
I also tried address 0x50.  Is it in fact possible to read these data?  If so, what would be the correct command?
 
Is there a fuller description of i2c communication for the U3?
 
Thank you.


#2 LabJack Support

LabJack Support
  • Admin
  • 8677 posts

Posted 11 February 2015 - 03:03 PM

For DACA and DACB on a LJTick-DAC, you can only set the output voltages and there is no address to read the currently set voltages. You could connect DACA and DACB to some analog inputs on the U3 and read the voltage values that way.

 

Low-level I2C communication documentation for the LJTick-DAC can be found here:

 

http://labjack.com/s...k-dac/datasheet

 

The U3's I2C command/response is documented here, and the Python i2c method is basically a wrapper for it:

 

http://labjack.com/s...rs-guide/5.2.19




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users