The whole circuit is used for measuring voltage, but I am wondering what is the usage the circuit of the red section and how to analysis it? could anybody offer me some anlysis? Pleas ignore some notes in this paper…
Hi Gavin, the AD830 has a differential voltage range of +/-2 V. The circuit in the red frame attenuates the input signals by the factor (4,99 k + 10) / (499 k + 499 k + 4,99 k + 10) = 5 k / 1003 k = 0.004985. Thus the input voltage range is increased to +/- 401.2 V or roughly +/- 400 V.
The AD830 has an input capacitance of typically 2 pF, so with resistors but no capacitors we had two low pass filters with a cutoff frequency of about 1 / (2 * Pi * 5 k * 2 pF) = 15,9 MHz (at this frequency the signal would be attenuated by -3 dB). But the AD830 has only -0.1 dB at 15 MHz, and -3 dB at typically 45 MHz. In order to have a flat frequency response of the voltage dividers, a capacitor C needs to be connected parallel to R1 and R2; this way, the ratio of the impedances of R1 + R2 || C and R3 + R4 || input capacitance can be made independent of the frequency. However, the input capacitance of semiconductor devices isn't exactly the same for each device; therefore, the capacitor C has to be adjustable. But most trimmer capacitors aren't suitable for 400 V, so the designer put a trimmer parallel to the input capacitance where only a small voltage can occur.
The question remains why the capacitors aren't two orders of magnitude smaller. I guess 4.7 pF was the smallest value the designer could easily get. Since Xc = 1 / (2 * Pi * f * C), the ratio of the impedances of the capacitors equals the inverse ratio of the capacitances. Here we get (with trimmer adjusted to 79 pF and 2 pF input capacitance) 2.35 pF / 471 pF = 0.004989, so the voltage divider made of the capacitors corresponds to the voltage divider made of the resistors, and as a consequence the ratio of the whole voltage divider is independent of frequency.
The impedance of each input is now about 1 MOhm || 2.35 pF. Hope this helps.
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