# DAC Fundamentals

#### lindula

Joined Sep 23, 2016
94
Hello, I'm revisiting DAC Fundamentals and need some help. I have a DAC that can go from +10V to -10V. It has a settling time of 20usec and an output slew rate of 5V/usec. Help me to understand how these figures are used to determine the fastest rail to rail signal I can generate please? I've see some people use 1/(2*20us) which is 25Khz, some use 1/20us which is 50Khz which us the settling time value.

Can anyone give an old guy some help please?

Thank you

#### nsaspook

Joined Aug 27, 2009
11,783

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#### MrChips

Joined Oct 2, 2009
29,236
You have two parameters to contend with, slew and settling time.

Slew of 5V/μs means that it would take 4μs to change 20V. This is shorter than the settling time of 20μs and that is good.

Settling time means that it takes 20μs for the output to settle. You can send a new DAC value faster than this but it means signal integrity will suffer. Your output rate should be 1/20μs = 50kHz or lower.

The article defines bandwidth = 1 / (2 x settling time) = 25kHz
This is correct because of the sampling theorem. You need at minimum 2 samples per wave cycle in order to satisfy the Nyquist limit.

In other words, if your output rate is 50kHz, the highest frequency signal that you can generate is 25kHz. In reality, if you want to preserve signal integrity, the highest frequency would be 10 times lower than this, i.e. 2.5kHz.

#### lindula

Joined Sep 23, 2016
94
You have two parameters to contend with, slew and settling time.

Slew of 5V/μs means that it would take 4μs to change 20V. This is shorter than the settling time of 20μs and that is good.

Settling time means that it takes 20μs for the output to settle. You can send a new DAC value faster than this but it means signal integrity will suffer. Your output rate should be 1/20μs = 50kHz or lower.

The article defines bandwidth = 1 / (2 x settling time) = 25kHz
This is correct because of the sampling theorem. You need at minimum 2 samples per wave cycle in order to satisfy the Nyquist limit.

In other words, if your output rate is 50kHz, the highest frequency signal that you can generate is 25kHz. In reality, if you want to preserve signal integrity, the highest frequency would be 10 times lower than this, i.e. 2.5kHz.
Hello, thank you very much for responding to my message. If you tried generating a 25Khz sinewave what you you expect to see on the o-scope? I assume that if you wanted 2.5Khz sinewave then it would look very good.

#### MrChips

Joined Oct 2, 2009
29,236
Assuming that your sampling rate is 50kHz, your samples are spaced every 20μs.
Therefore you have only two samples per cycle. Your output would be a 25kHz square wave.

If you want to generate a 25kHz sinewave you would need to output at least 10 or 20 points per cycle.
That is, you need a faster DAC with setting time of 2μs or shorter.

#### lindula

Joined Sep 23, 2016
94
I thank everyone for their comments. This has been very beneficial to me.

Thank you

#### MrChips

Joined Oct 2, 2009
29,236
I thank everyone for their comments. This has been very beneficial to me.

Thank you
You are very welcome.