I would like to know how to convert correctly dBc/Hz to pA/sqrt(Hz). The AD9744 is 50 pA/sqrt(Hz) for IOUTFS = 20 mA and the AD9707 is -161 dBc/Hz for IOUTFS = 5 mA and Fc = 175 MHz / Fout = 6 MHz, so what is the difference for the output noise current?

Regards,

Eric

Hi -

I have an excel spreadsheet that does this. It's attached.

Thanks,

Larry