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.

Dear Joerg Fricke 

Very exact and full ansver.

Josef

Thanks Joerg - thorough analysis.