# Calculation of maximum input impedance

In the datasheet of the ADC of a microcontroller it was mentioned that: "The maximum recommended impedance for analog sources is 10k."

In a high voltage divider circuit such as this, how does one calculate the input impedance to the ADC? Is it 990k || 2k2 , or 2k2 or 990k?

The circuit node Vdiv is connected to the ADC.

• I see no valid ADC input in your circuit. Commented Dec 9, 2020 at 17:13
• Which MCU? My crystal ball tells me it's Atmega328P, or at least some other AVR, right? What is the purpose of the circuit? Commented Dec 9, 2020 at 17:50
• I don't see much point in creating a V/2 with R1 and R2 that has a 1.1 k impedance, then buffering it to zero ohms, then adding R6 to take it up to 2.2 k. Why not just make R1 and R2 4.3 k to the Vdiv node, lose U1 and R6, and move C2 to the power supply line? Commented Dec 9, 2020 at 18:01
• Is this part of an isolation scheme where the ADC is on the mains-side and there exists a galvanic boundary after the ADC's output? Or is there no isolation? If there's no isolation, you're asking for trouble. Commented Dec 9, 2020 at 18:57

As long as U1 is active and the output is stable (which may be negatively affected by the C2), the impedance looking into Vdiv is 2.2K || 990K || 100nF.

At 50Hz, approximately 2.2K.

As @Neil_UK suggests, maybe consider a circuit more like this one:

simulate this circuit – Schematic created using CircuitLab

Make sure the maximum peak voltage divided by the 990K does not exceed the minimum current draw of whatever is connected to the +5V supply unless your regulator can sink current (most can't). If necessary, you can add a 5mA power LED or whatever. Otherwise, a positive transient could lift the +5V rail and possibly damage something.

• @Neil_UK : This is a legacy circuit I am re writing the firmware for.In my case VDiv is connected to a PIC16F1969 through a 470Ohm resistor.I used LT1014 for a quick simulation,actually there is a LM324 over there.There is probably a current load of ca 30mA, LED, LCD with BL, etc.Thanks for the tip that I can eliminate the OpAmp, I am also putting a 10~47nF cap at the ADC input for better noise rejection. Commented Dec 11, 2020 at 17:54
• Hi Spehro,I went away too soon !! How does one solve your resistor network ? In my earlier circuit, I had solved for the node voltage by grounding the end of R6, getting the divided voltage value and then adding 2.5V. Commented Dec 11, 2020 at 19:47
• The voltage at the node (at low frequencies) is (5V/4.3K + VAC/990K)*(4.3K||4.3K||990K). Commented Dec 11, 2020 at 19:48

Since the input is fed to an ADC of a microcontroller that is extremely likely to be a sample-hold converter, the impedance needs to be considered on both DC and AC domains.

In AC domain, the 100nF capacitor alone has sufficient AC impedance during the sampling period to make a single measurement accurate enough, regardless of DC impedance. It could be biased with almost arbitrarily high DC impedance, as long as the sampling is done with large enough period to let the capacitor voltage to recharge before taking next sample.

In DC domain, the DC impedance is slightly less than 2.2k to begin with.

So there should be no issues sampling the voltage.

• Thanks much for replying.Its a PIC16F1xxx Commented Dec 11, 2020 at 17:55