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Input impedance of the ADC Pi

2131 Views - Created 19/10/2015


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I'm currently using the ADC Pi with various sensors, the setup is easy and it works pretty well.

As a voltage divider is included on the board, I'm not sure I've understood what is the actual input impedance and what voltage is measured by the board.

One of my sensors (http://www.omron.com/ecb/products/sensor/61/d6f-6.html) has an output impedance of 16.7 kOhm, which is quite high, so I read a lower voltage with the board than with a multimeter.

So here are my questions :

1) What is the input impedance of the ADC Pi  and what voltage is measured given the onboard voltage divider ?

2) Is it possible to increase this impedance ? (influence of PGA ?)

3) What is the coefficient I must multiply the measured voltage by to get the actual output voltage for the sensor I mentioned before ? (I suppose I will lose in accuracy though)


Thanks !


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The voltage divider on the ADC Pi inputs consists of a 10K and 6.8K resistor so the input impedance between the input and ground is 16.8K.  The MCP3424 ADC chip we use has a maximum input voltage of 2.048V so the voltage divider divides the input by 2.47 allowing the ADC Pi inputs to measure between 0 and 5.05V.

The only way to increase the input impedance would be to remove the voltage divider and connect to the ADC chip directly.  You could do this by removing the two input resistors and bridge the pads for the 10K resistor.  The downside is your input would then be limited to 2.048V.  According to the datasheet for the MCP3424 the input impedance for each ADC channel is 2.25MΩ/PGA.  As the input impedance on the ADC Pi is defined by the voltage divider changing the PGA value will have very little difference.

If your sensor has an output impedance of 16.7kOhm then that is being added to the 10K resistor on the input so the voltage divider now has 26.7K instead of 10K so if there is 1V at the ADC Pi input instead of seeing 0.405V the MCP3424 is seeing 0.203V.  If you multiply the measured voltage by 1.993 that should hopefully give you the correct value.

AB Electronics UK Tech Support


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Hello andrew, Thank you for your fast and comprehensive answer; I understand what was happening now. I'll stick to the software adjustment, simpler and less risky.

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