Question 6.1: For the sample-and-hold circuit shown in Figure 5a, determin......
For the sample-and-hold circuit shown in Figure 5a, determine the largest-value capacitor that can be used. Use an output impedance for Z_{1} of 10 Ω, an on resistance for Q_{1} of 10 Ω, an acquisition time of 10 μs, a maximum peak-to-peak input voltage of 10 V, a maximum output current from Z_{1} of 10 mA, and an accuracy of 1%.
The expression for the current through a capacitor is
i=C\frac{dv}{dt}Rearranging and solving for C yields
C=i\frac{dt}{dv}where C = maximum capacitance (farads)
i = maximum output current from Z_{1}, 10 mA
dv = maximum change in voltage across C_{1}, which equals 10 V
dt = charge time, which equals the aperture time, 10 μs
Therefore, C_{max}=\frac{(10\, mA)(10\,μs)}{10\,V}=10\,nF
The charge time constant for C when Q_{1} is on is
τ = RC
where τ = one charge time constant (seconds)
R = output impedance of Z_{1} plus the on resistance of Q_{1} (ohms)
C = capacitance value of C_{1} (farads)
Rearranging and solving for C gives us
The charge time of capacitor C_{1} is also dependent on the accuracy desired from the device. The percent accuracy and its required RC time constant are summarized in
Table 1
For an accuracy of 1%,
C=\frac{10\,μs}{4.6(20)}=108.7\,nFTo satisfy the output current limitations of Z_{1}, a maximum capacitance of 10 nF was required. To satisfy the accuracy requirements, 108.7 nF was required. To satisfy both requirements, the smaller value capacitor must be used. Therefore, C_{1} can be no larger than 10 nF.
Table 1 : The percent accuracy and its required RC time constant
| Accuracy (%) | Charge Time |
| 10 | 2.3τ |
| 1 | 4.6τ |
| 0.1 | 6.9τ |
| 0.01 | 9.2τ |