Hello Colleagues,

I am planning to use DC-coupled differential output passive interface of DDS AD9958 to driving analog multiplier or mixer.

My question is:

How to calculate common-mode voltage at the outputs of a AD9958 DAC at a given load resistors ?

Sincerely yours,

Vlad Brusilovsky

Hello Vlad,

The DAC output is a current source and its full scale current is set by the Rset value. The full-scale current is inversely proportional to the Rset resistor as follows: Rset = 18.91 / Iout (full scale current) The maximum full-scale output current of the combined DAC outputs is 15 mA, but limiting the output to 10 mA max provides optimal spurious-free dynamic range (SFDR) performance. See page 18 for more details

To calculate the peak to peak voltage at the complementary DAC outputs, simply multiply the full scale current by the equivalent output load. Note, the DAC outputs are open drain and must be dc terminated for current flow. Whatever output network is used (termination resistors + filter + transformer, etc) the equivalent load must be used for the peak to peak voltage (Vpp) calculation. Again, Vpp = full scale current * equivalent load. For instance, say there's two 50 ohm output termination resistors at the DAC driving a differential filter with a 100 ohms at the output of the

filter. The equilvalent load at each DAC output is 25 ohms.

Since the AD9958's DAC outputs are referenced to AVDD (current sinking DAC), not ground (current sourcing DAC), the output common mode voltage for both outputs is (AVDD - Vpp/2). Note, if the DAC outputs are terminated into a transformer or inductors reference to AVDD, the DAC output will swing about AVDD. In that case, the common mode voltage is AVDD.

Hope this helps.