0
\$\begingroup\$

I plan to use LTC2641-12 DAC in some application and I simulate the analog part with LtSpice. For the DAC, I used a pulsed voltage source with this parameters:

PULSE(0 50m 10u 1n 1n 16u 32u)

Obviously, the rise/fall times cannot be 1nS but just to see the response. On output wave I can see some spikes due to very high given rise/fall.

enter image description here

Now the question is what will be the real (worst case) DAC rise/fall based on specifications:

Slew rate: 15V/uS Settling time: 1uS DAC Glitch Impulse: 0.5nV/S <-- what is this ? Digital Feedthrough: 0.2 nV/S <-- and this ?

My DAC will output up to 4.096, so according SR, will "jump" from 0 to 4.096V in 15/4.096 = 3.66 uS

This is the rise/fall time? What will be the worst case scenario?

Thanks for clarifications.

|improve this question
\$\endgroup\$
1
\$\begingroup\$

Actually it will "jump" in 4.096/15 or 273 nSec. Then add the 1 uSec settling time for 1.27 uSec.

"Glitch Impulse" is caused by variances in the R-2R network switching times for the resistors. Worst case is the switch from 7FFF to 8000 ("major transition"). Since all of the resistors get switched and they don't switch exactly at once, the output has a little "bump" or "glitch". It is specified as the absolute value of the area under the curve of voltage (compared to the final value) over the time of the glitch, hence the units are volt-seconds.

"Digital Feedthrough" is the glitch that occurs on the output when the input value bits do not change but one of the other digital inputs (clock, latch, etc) is changed. It is also specified as volt-seconds.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ I don't understand from where comes 273nS. The 15V/us reffer to full scale?? Then sum with 1uS? This means almost 785 KHz maximum if I intend to output square waves, right? I accepted for your clarifications, thanks. \$\endgroup\$ – orfruit Aug 7 '17 at 10:23
  • \$\begingroup\$ The slew rate is the slope of the rise or fall time, regardless of endpoints. So the bigger the change, the longer it takes to get there. If your output is 4.096 volts, it will take 2.73 nS to increase the output voltage, not including the settling time that takes care of overshoot or undershoot to the final value. A "square" wave at 785 kHz would certainly be possible with this rise time on the rising and falling edges and some small ringing. \$\endgroup\$ – John Birckhead Aug 7 '17 at 16:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.