I am using an ADC MCP3424.

After breaking one of the ADC inputs, I was saw this thread: ADC input protection?

I also read Microchip app note TB3013. Figure 3 in particular:

examples of input pin protection

The ADC's analog inputs need to be clamped at ± 0.3V of the supply rails.

The Vf of a schottky like BAT85 depends on the current flowing through it. So for a Vf < 0.3V, we would need a series resistance of 10k, to keep the current less than 1mA (assuming the Opamp Supply is +15V/-5V). Now the problem is wouldn't the 10k resistance slow down the charging time of the sampling capacitor and reduce my 18-bit accuracy? I am measuring slow moving DC voltages.

Is there any solution for the problem ?

  • \$\begingroup\$ Typically, vendors specify a maximum input voltage range but in reality it is the current that is destructive. A lot of vendors have started adding this to their datasheets. This particular device has ESD protection diodes clamped to +/-0.6V of the rail. It might be possible to just add a resistor if you can get Microchip to give you that destructive current. \$\endgroup\$ Oct 8, 2012 at 17:59

1 Answer 1


Putting a 10K resistor on the input will not change things to any great extent. Yes, technically it will slow down charging the sampling cap. It will act as a low pass filter where the cutoff frequency is 1/(2*pi*R*C). That works out to a cutoff frequency of 4.97 MHz. So any frequency above 4.97 MHz will be significantly attenuated.

This ADC has several modes, but the quickest mode it has runs as 240 samples per second. Meaning that it can handle frequencies less than 120 Hz. The RC filter caused by the input cap and the 10K resistor has a frequency response that is still 4 orders of magnitude higher!

In short, that 10k resistor on the input isn't going to do anything negative for you.

  • 5
    \$\begingroup\$ I heard 4.97 MHz === 5 MHz, have you heard that too? :-) \$\endgroup\$
    – stevenvh
    Oct 8, 2012 at 14:43
  • 2
    \$\begingroup\$ @stevenvh Only for sufficiently large values of 4.97. \$\endgroup\$
    – user3624
    Oct 8, 2012 at 14:53
  • \$\begingroup\$ ... in the limit as 1 approaches 0 \$\endgroup\$
    – vicatcu
    Oct 8, 2012 at 16:33

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