adccapacitormicrocontroller
I've learning about analog to digital converters. I came to read about the acquisition time of sample and hold circuit. But it confuses me. Can anyone help me to learn how to calculate the acquisition time of a sample and hold circuit ?
simulate this circuit – Schematic created using CircuitLab
Let's say we have a 8 bit SAR,
Resistor value = R;
Capacitor value = C;
How do we calculate the acquisition time ?
(Do not bother about the Op-Amp)
A peak detector is something of a sample and hold that samples all the time, and holds the peak:
Follow the input with this, and connect the output to your ADC. Make C significantly larger than the capacitance used by your ADC's sample and hold, or follow it with a buffer, so the voltage across C doesn't sag as you read it.
Trigger an ADC reading just after the event happens, and when it's done, close the switch to reset the peak, or make the switch a resistor forming an RC time constant significantly longer than the time it takes to measure the peak but significantly shorter than the interval between peaks to accept a little error but avoid the need to reset the peak detector.
There are two time constants here. One is when the input is driven and the other when it is open.
Probably the input is driven high, then left to float low. When the input is driven, you assume it has 0 impedance to ground. Note that this impedance is parallel to R1. What is R1//0? When the input is open-circuit, then it has infinite impdance to ground. The effective resistance seen at the top of C1 to ground should be obvious in that case.
Best Answer
The worst-case distance from the final solution in terms of voltage on C1 is (from the step response of an RC-filter):
$$ V(t)=V_{max}\exp(-t/(R_1C_1)) $$
Where \$V_{max}\$ is the maximum input swing (change per sample). Now you might want to settle a bit more than LSB. In terms of a relative deviation of (an arbitrary) 1/4th of LSB you can write:
$$ LSB/4 = \exp(-t/(R_1C_1)) \\ \log(LSB/4) = -t/(R_1C_1) \\ t = - (R_1C_1)\log(LSB/4) $$
Since you have 8 bits, LSB relative to 1 is \$2^{-8}\$. Then the above formula yields about 0.301 ms.