a sample and hold (s/h) circuit, having a holding capacitor of 0.1nf, is used at the input of an adc (analog - to - digital converter). the conversion time of the adc is 1 microsec. and during this time, the capacitor should not loose more than 0.5% of the charge put across it during the sampling time.the maximum value of the input signal to the s/h circuit is 5v. the leakage current of the s?h circuits should be less than (a)2.5ma(b)0.25ma(c)25µa(d)2.5µa
Question
a sample and hold (s/h) circuit, having a holding capacitor of 0.1nf, is used at the input of an adc (analog - to - digital converter). the conversion time of the adc is 1 microsec. and during this time, the capacitor should not loose more than 0.5% of the charge put across it during the sampling time.the maximum value of the input signal to the s/h circuit is 5v. the leakage current of the s?h circuits should be less than (a)2.5ma(b)0.25ma(c)25µa(d)2.5µa
Solution
The leakage current can be calculated using the formula:
I = C * V * dV/dt
where:
- I is the leakage current,
- C is the capacitance,
- V is the voltage, and
- dV/dt is the rate of change of voltage with respect to time.
Given:
- C = 0.1nF = 0.1 * 10^-9 F,
- V = 5V, and
- dV/dt = 0.5% of V in 1µs = 0.005 * 5V / 1µs = 0.005 * 5 * 10^6 V/s = 25 * 10^3 V/s.
Substituting these values into the formula gives:
I = 0.1 * 10^-9 F * 5V * 25 * 10^3 V/s = 0.0125 * 10^-3 A = 12.5 µA.
Therefore, the leakage current of the S/H circuit should be less than 12.5 µA. The closest option to this value is (c) 25µA.
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