I want to drive the ADC AD7903 with the ADC Driver AD8476 at a throughput of 1MSPS. Is it a good choise, to take the AD8476, because in the datasheet I found: "Suitable for driving 16-bit converter up to 250 kSPS".
I would suggest to use the recommended driver on the AD7903 datasheet like the ADA4941-1, ADA4841-x, AD8021/22 etc. Please refer to the AD7903 datasheet.
The AD8476 is good amplifier but using this for a 1MSPS throughput rate on ADC like AD7903 would not be good. From the AD8476 datasheet the settling time is at 1us. This would mean that when the ADC samples 1MSPS(1us),the driver will not be able to settle in time for a full-scale step. But at lower throughput like 250ksps is applicable.
There is an article that you can check that discusses about the ADC front end circuitry. Please refer to the link.
Front-End Amplifier and RC Filter Design for a Precision SAR A/D Converter
thank you for your suggestion, I will heed it.
I have look at the link, but their are some things, which I do not understand. Maybe, you can help me?
Front-End Amplifier and RC Filter Design for a Precision SAR Analog-to-Digital Converter
I understand the equation U_step = (2*pi*f_IN * U_peak * t_conv * C_DAC) / (C_EXT + C_DAC ).
It is the voltage change at the ADC input in the conversion time with a sine wave input signal.
And with an external capacitor the voltage change is slower. It would be reduced by the factor C_DAC /(C_DAC + C_EXT). Ok.
And the euqation for the half LSB is clear, too.
I do not understand the euqation
tau = t_ACQ / N_TC and N_TC = ln( U_step / U_half_LSB ).
Can you explain it, please?
The Tau= T_acq/ ln( U_step / U_halfLSB ) was derive from the formula for the RC time constant.
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