Once again, the TS is, in all likelihood, trying to solve a non-problem in a project when he has only a vague idea of what the project is.

Pinkyponky,

You will learn a lot more if you stop overthinking things and build some simple circuits, refining them as necessary.

Bob

 

Page 3 - Analogue input capacitance.

Only if you tell me the frequency of the amplifier output signal!
\(
I={V_{out}}{2 \pi f C}
\)
[Edit] apologies for taking three tries to get my TEX code right!

I did the calculations. But, may I know one thing is that how you evaluate the formula which is above. Moreover, Which do I need to be consider as a load current either Analog Input Leakage Current (according to datasheet in page 3) (OR) the value is calculating from the above formula? and also say why?.

 

I did the calculations. But, may I know one thing is that how you evaluate the formula which is above. Moreover, Which do I need to be consider as a load current either Analog Input Leakage Current (according to datasheet in page 3) (OR) the value is calculating from the above formula? and also say why?.

Both (add them together).
Because they are out of phase the answer will be
\(
I_{load}=\sqrt{i_{leakage}^2+({2 \pi V_{out} f C})^2}
\)
At a guess the current into the capacitor will be much greater than the leakage current (so that the leakage current could be ignored), and the total current will be insignificant compared to the maximum output current from the amplifier (so that the entire current into the input of the A/D can be ignored)

 

Both (add them together).
Because they are out of phase the answer will be
\(
I_{load}=\sqrt{i_{leakage}^2+({2 \pi V_{out} f C})^2}
\)
At a guess the current into the capacitor will be much greater than the leakage current (so that the leakage current could be ignored), and the total current will be insignificant compared to the maximum output current from the amplifier (so that the entire current into the input of the A/D can be ignored)

Thank you Ian0.

 

Thank you Ian0.

Full marks for thinking about it, though! There are many engineers who would always assume it to be insignificant, until, one day, their design doesn't work.

 

Both (add them together).
Because they are out of phase the answer will be
\(
I_{load}=\sqrt{i_{leakage}^2+({2 \pi V_{out} f C})^2}
\)
At a guess the current into the capacitor will be much greater than the leakage current (so that the leakage current could be ignored), and the total current will be insignificant compared to the maximum output current from the amplifier (so that the entire current into the input of the A/D can be ignored)

Where, f=20khz, Vout=5V,
According to datasheet, Ileakage=1uA, C=90pf.
Therefore, Iload=57uA

So, Can we say that load of the amplifier/ADC current consumption is 57uA and it would be negligible.

 

I think so - Most amplifiers (you didn't specify what the amplifier was) would be capable of 57uA output current.