Dear Sir/Madam,

We are studying about your AD9280 8bit ADC.

I have a question.

We need the input-impedance for our simulation.

Impedance =Fs voltage / DC leakage current, we think, correct?

On your datasheet show DC leakage current =43uA when Fs=32MHz max.

Impedance mean 32MHz/43uA.

Our Fs(sampling freq (Conversion Rate)) is 2MHz.

So what is the DC leakage current Fs=2MHz?

Otherwise how should we think your input-impedance for this case?

Thanks Kaos

Hello,

The AD9280 is one of 1st CMOS ADC's (released in mid-1990's) by ADI thus one may consider latter generation of CMOS ADI ADC's to determine if they meet your applicaiton requirements. Note, the "clamp" function is unique to this device (for viideo applications) and not on latter generation standard ADC's. .

That said...........the AD9280 has a 'switched-capacitor" sample-hold input structure with its input impedance being frequency dependent as well as whether the device is in the "sample" or "hold" mode. Typically one is more concerned about the impedance in the "sample mode" where the differential sampling capactior (1 pf for AD9280) is connected to the driving source. DC leakage current (that is not sample rate dependent) need not be considered since this is quite low.

Please refer to application note AN-743 (on our website) on modeling an ADC input. This was written after the AD9280 was released but same concept applies. At low frequency , the track-mode impedance is quiite high with the "IMG" component dominated by the 1 pF sampling capacitor and the REAL part being quite high.

Regards.