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I was reading the datasheet of ADS1115. It has an input capacitor stage which it continuously charges and discharges to measure the voltage between AINp and AINn.
Such a circuit is called a switched capacitor resistor. To see why it behaves as a resistor, consider the circuit in steady state. It is assumed that the switching frequency is low enough so that the capacitor has time to charge to full potential which is applied to it.
simulate this circuit – Schematic created using CircuitLab
SW1 is closed and SW2 is open: The capacitor was connected to voltage V2 so had initial charge \$q_i = CV_2\$. Now the capacitor has been connected to voltage V1 so has final charge \$q_f = CV_1\$. The amount of charge transferred from node V1 is: $$q_{tr} = C(V_1-V_2)$$ SW2 is closed and SW1 open: The capacitor was connected to voltage V1 so had initial charge \$q_i = CV_1\$. Now the capacitor has been connected to voltage V2 so has final charge \$q_f = CV_2\$. The amount of charge transferred to node V2 is: $$q_{tr} = C(V_1-V_2)$$
On average, a charge of \$q_{tr}\$ is transferred from node V1 to node V2 in one switching period. If the switching frequency is \$f\$, then average charge transferred per unit time or average current is equal to: $$i = fC(V_1-V_2)$$ $$\frac{V_1-V_2}{i} = \frac{1}{fC}$$ $$Z_{eff} = \frac{1}{fC}$$ Thus, the switched capacitor acts as an average resistor over the switching period with resistance \$\frac{1}{fC}\$.
(2) 0,7 is the typical forward voltage of a diode or a bipolar junction transistor (bjt) - So we can assume it's discharged with a bjt
(1) the internal capacitors are very small, so most circuits won't even notice a rise in capacity which just leaves us the internal resistance to deal with. If your circuit is also of low capacity and high resistance, it might be an issue. But a buffer amplifier in the ADC frontend will cure that issue with ease.
Re question 1 , the circuit uses a "flying capacitor" that basically transfers buckets of charge across. If we look at the circuit below, I have added a 6v , 1A light bulb , so if we apply 6v on the left hand terminals, the capacitor will be charged to 6v while facing left; if we then toggle the capacitor back and forth, it is discharged at say 2A into the light bulb for 2us, then back to the left to be charged at 2A for 2uS, so what your 6V source on the left "sees" is a 2A load for 50% of the time or 1A on average, so 6v x 1A = 6w, if you double the source voltage, you get double the current, hence it looks "resistive"
In the 1950's - 1970's mechanical vibrators and capacitors where used to converting a floating differential source to a single ended signal. You could also use "choppers" to convert floating DC into single ended AC, amplify that, and convert back to DC, back in the days of valves where DC performance was very poor.
