Question

A sample and hold circuit is implemented using a resistive switch and a capacitor with a time constant of 1 $\mu$s. The time for the sampling switch to stay closed to charge a capacitor adequately to a full scale voltage of 1 V with 12-bit accuracy is ____________ $\mu$s (rounded off to two decimal places). 
 

Hint:

For an RC S/H settling to $N$-bit accuracy using {1 LSB} criterion: $t \approx \tau\,N\ln 2 \;(\approx 0.693N\,\tau)$.

Answer 1

For RC charging: $v(t)=V_{\text{FS}}\!\left(1-e^{-t/\tau}\right)$; residual error $=V_{\text{FS}}e^{-t/\tau}$.
12-bit accuracy $\Rightarrow$ error $\le$ 1 LSB $=V_{\text{FS}}/2^{12}$. Thus \[ V_{\text{FS}}e^{-t/\tau} \le \frac{V_{\text{FS}}}{2^{12}} \;\Rightarrow\; e^{-t/\tau} \le 2^{-12} \;\Rightarrow\; t \ge \tau \ln(2^{12}) = \tau\,(12\ln 2). \] With $\tau=1\,\mu$s, \[ t=1\,\mu\text{s}\times 12\ln2 \approx 1\,\mu\text{s}\times 8.3178 = 8.32\,\mu\text{s}. \] \[ \boxed{8.32~\mu\text{s}} \]