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I'm building a voltage divider to use for feeding into an ADC to determine the battery voltage of a device I'm designing. Is it important to consider the input impedance in the ADC and to account for the voltage drop caused by that extra load or should I not even worry about it?

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4 Answers 4

Yes, you definitely need to consider the input resistance and/or bias current requirements of the ADC input.

The usual technique is to keep the Thévenin resistance of your divider low enough that the offset caused by the ADC input is "low enough" (e.g., less than 1% or 1 LSB), depending on your system requirements.

For example, suppose that your ADC input resistance is 1 MΩ and you want to keep the error caused by this to less than 1%. Therefore, the Thévenin resistance of your divider should be less than 10 KΩ.

So, now you have two equations, one for the divide ratio:

$$Ratio = \frac{R1}{R1 + R2}$$

and one for the resistance:

$$R_{TH} = \frac{1}{1/R1 + 1/R2}$$

and two unknowns, R1 and R2. It's straighforward math to plug in your known values for Ratio and RTH and solve for R1 and R2.

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Can you share the method for accomplishing this? I'm just not 100% sure what to do, though I am familiar with Thevenin equivalent resistance. – Pugz Apr 17 '14 at 14:52

Yes, you do need to consider the input impedance and probably more importantly you'll need to put a capacitor across the raw input to the ADC - if you are planning on using high impedances, a lot of ADCs that are built into MCUs will take a little glitch of current that will affect the accuracy of the reading taken so, put a 100nF capacitor to ground on the ADC input.

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would you expand a bit the "current glitch part"? Is it about the sampling cap, so that putting a 100nF near the imput pin shorts the input spike current to ground? – Vladimir Cravero Apr 17 '14 at 14:18
@VladimirCravero that's exactly it. – Andy aka Apr 17 '14 at 14:52
Sounds like it to me. – Pugz Apr 17 '14 at 14:53
@Andyaka Can you elaborate on how to accomplish this in a circuit? – Pugz Apr 17 '14 at 14:54
You put a 100nF capacitor in parallel with the resistor at the bottom of your "voltage divider" - the resistor that connects to 0V. – Andy aka Apr 17 '14 at 14:58

It would really help to know what ADC you are using. In any case, you need to consider it, but the good news is that most microcontroller ADCs have high impedance inputs so a divider made with a few tens of kilo-ohms won't be affected by it very much.

How much accuracy do you need? Beyond a certain point you will need to calibrate anyway, so the impedance of the ADC can be compensated for at that point anyway.

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It's the ADC in the Atmel SAM4S chips... 12-bit. At what point would you say I need to calibrate and what's the reasoning behind this? – Pugz Apr 17 '14 at 15:38

ADC input impedance is generally much higher than ESR of the battery, but if too high then imbalanced stray capacitance can inject CM noise. CMRR must be considered over entire BW of the circuit. Calibration is key to verifying Vfref noise and linearity glitches from Digital noise getting into analog measurement, so guarding, grounding , nyquist filtering and shielding improves the results. Then ratio of resistor divider voltage should be linear and accurate. Otherwise, circuit will inject aliasing noise , CM noise and offset input result in inaccurate results. Use a good signal or DAC sawtooth voltage to calibrate linearity , gain and offset.

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