I wish to measure the phase voltages of an induction machine using Avago isolation amplifier (ACPL-790A).
(
http://www.avagotech.com/pages/en/optocouplers_plastic/plastic_miniature_isolation_amplifier/acpl-790a-000e/)
Figure 21 (attached) is copied from the datasheet and it is for current sensing. Note that the circuitry of R5 (10ohm) and C3 (47nF) forms an anti-aliasing filter.
The maximum phase voltage of the induction machine is expected to be not more than +/-50V. Since the recommended input voltage range of the ACPL-790A is only +/-200mV, a voltage divider is required to "step-down" the +/-50V to +/-200mV. To assure a current of not more than 1mA flowing through the "voltage divider" branch, R1=25kohm and R2=100ohm are selected (see attached figure).
According to datasheet (pp.14):
... The only restrictions are that the impedance of the divider be relatively small (less than 1kohm) so that the input resistance (22kohm) and input bias current (0.1uA) do not affect the accuracy of the measurement. An input bypass capacitor is still required, although the 10ohm (R5) series damping resistor is not (the resistance of the voltage divider provides the same function). The low-pass filter formed by the divider resistance and the input bypass capacitor may limit the achievable bandwidth.
Without the R5, and if R1=25kohm and R2=100ohm, what is the resistance of the voltage divider that will form the low-pass filter with the capacitor C3? Is it (R1*/R2)/(R1+R2)~=99.6ohm?
Do you have any idea to prevent the resistance of the "voltage divider" from affecting the bandwidth of the anti-aliasing filter formed with R5=10ohm and C3=47nF?
Cheers.