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:

The settling time of an RC circuit to n-bit accuracy is a classic problem. The required time is a multiple of the time constant $\tau$. The multiplying factor is $n \ln(2)$ if the error must be less than one LSB, or $(n+1)\ln(2)$ if the error must be less than half an LSB. Be aware of the convention used.

Answer 1

Correct Answer: 8.3

To determine the time for the sampling switch to stay closed, we start by understanding the capacitor charging equation: \( V(t) = V_f (1 - e^{-t/\tau}) \), where \( V(t) \) is the voltage across the capacitor at time \( t \), \( V_f \) is the final voltage (1 V in this case), and \( \tau \) is the time constant (1 µs).
Achieving 12-bit accuracy implies reaching within 1/4096 of full-scale voltage. Therefore, the error voltage should be less than \( \frac{1}{4096} \) V.
The required charging condition is:
\( 1 - e^{-t/\tau} \geq 1 - \frac{1}{4096} \)
Solving for \( t \), we have:
\( e^{-t/\tau} \leq \frac{1}{4096} \)
Taking the natural logarithm on both sides gives:
\(-\frac{t}{\tau} \leq \ln\left(\frac{1}{4096}\right)\)
\( \tau = 1 \) µs, substituting gives:
\( t \geq -\ln\left(\frac{1}{4096}\right)\) µs

Calculating:
\( \ln(4096) = 12 \ln(2) \approx 12 \times 0.693 = 8.316 \)
Thus, the time \( t \approx 8.316 \) µs, rounded to two decimal places is 8.32 µs.
This value should fit within the given range of 8.3, confirming its validity.
Therefore, the time is 8.32 µs.