Calculate maximum data conversion rate in Wheatstone-Bridge

Question

A Wheatstone-Bridge is based pressure transducer that has a sensitivity of 10mV/kPa and it measures pressure within the range from 0 to 200kPa. So, the output signal from the transducer is buffered and fed within an analogue-to-digital conversion system.

The buffer amplifier utilized in the analog to digital converter can supply a maximum current of 10mA to a storage capacitance of 5pF within the sample-and-hold amplifier. When 25 percent of the acquisition time is utilized for binary encoding, calculate the maximum data conversion rate achievable.

Answer 1

The corresponding electrical circuit for the sample-and-hold amplifier is as demonstrated:

The maximum current will be drawn by the buffer amplifier while the output signals by the transducer this at its maximum and in that case storage capacitor is firstly discharged.

The maximum output voltage by the transducer is its output n full-scale conditions specified as:

Full-scale Output Voltage = Transducer Sensitivity x Full Scale Pressure

VFS = 10mV/kPa x 200 kPa = 10 x 10^-3 x 200 = 2V

In that case the resistance of the sampling transistor while ON is:

R = V_inmax/i_inmax = 2/(10x10^-3) = 2/10^-2 = 2x10² =200?

The time-constant for the sampling C-R network is then:

CR = 200 x 5 x 10-12

 = 103

 x 10-12

 = 10-9

 = 1ns

When duration of five times the time constant is acquired for charging in that case:

Sampling Time t_SAMPLE = 5ns

If 25 percent of the acquisition time is spent encoding, then the remaining 75 percent is spent sampling therefore:

t_SAMPLE = 0.75tACQ = 5ns

T_ACQ = 5 / 0.75 = 6.67ns

The maximum conversion rate is at that moment specified as:

Conversion Rate = f_CONV = 1/TACQ = 1/ (6.67 x 10^-9) = 10⁹ /6.67 = 150 MHz or 150 MSamples / s