Voltage divider output is wrong

Question

working on a project with STM32F103 and need to measure couple of external voltages. I have a simple voltage divider setup like the attached circuit, but the output of the divider is surprisingly different that I can calculate..

I calculate the output at around 2.9 V, but on PCB this gives me only 1.9 V. Any idea, what can cause this? I have tried measuring the output of the divider by disconnecting the trace that connects to MCU, but the same result.

The resistance values are correct (measured with multi meter). What else should I look at to debug it?

The Zener datasheet is here https://evelta.com/content/datasheets/273-MM1Z3V3B.pdf

============ UPDATE =============

After further debugging, I see that the Zener is dropping the voltage. Zener reverse current? But datasheet says only 20 µA.

Answer 1

Zener diodes below 5 volts have soft knees as shown in the SPICE simulation shown below. The SPICE library is from ROHM's BZX84C3V3LY 3.3V zener diode.

The X-axis is the current flowing through zener diode U1, Y-axis is the voltage across the zener. The upper circuit is with your values. The current through the zener is about 65 uA which gives about 1.85 V across R1 and the zener diode.

The characteristics of the zener diode used in your circuit will be slightly different, however, this demonstrates the soft knee one would expect from a low voltage zener.

I'm guessing that you are trying to protect an input circuit from voltage spikes. If you don't need wide bandwidth, you can use a largish capacitor across R38 in your schematic (perhaps 1 uF). Another technique is using a biased diode as shown below. You may want to use Schottky diodes and tie D2 to the power supply of the input circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

Low voltage Zeners are horrible devices. See this answer of mine from a few years ago, and the graph of device I/V curves for various types of diode.

If you want to protect the input from over-voltage, then the usual configuration is a standard silicon diode to the positive rail. This has the disadvantage of limiting the voltage to a little above the rail. A series resistor between diode and device will limit any current flowing into the device.

If you really must have clamping at or below the rail, then take the diode to a voltage you generate a little below the rail, rather than rail.

Note from my graph that LEDs have a much lower leakage current at low voltages than zeners, but you won't quite meet the 2.9 V threshold even with a blue or white LED.

Here is a modification using only one extra resistor. It allows you to use a higher voltage, and thus lower leakage, zener diode.

^{simulate this circuit – Schematic created using CircuitLab}

I don't have any measurement for what the leakage current of a 5/6/7 V zener is, but it should be better, and it's easy to make the measurement.

Answer 3

You should also take the maximum allowed external impedance into account (apart from the issues coming from the Zener diode.)

Using lower resistor values (let's say 6.8 kΩ and 2.2 kΩ) would instantly improve your ADC reading. However, if you aim to maintain a very low current through the voltage divider, you might want to use a voltage follower.

From the Datasheet, chapter 5.3.18 - 12-bit ADC characteristics.

Example:

Answer 4

1μA through 1MΩ drops 1V.

Source impedance of your divider is about 20kΩ.

20E3 * 20E-6 = 0.4V

That’s not too far from the deviation you measure.

Given the relatively high impedance of the divider, there is not much need to “protect” the MCU. It would take much beyond 12V to cause any trouble. The MCU already has input protection diodes that go to 3.3V and 0V. You can put 1mA through those diodes without much harm unless the datasheet says otherwise.

Answer 5

An enabled pull-down resistors can also result in this behaviour, here a quick schematic:

Assuming in the following that you left RPD enabled by accident. This parallel resistive connection to R38 would decrease the lower half of the voltage divider. Let's look in the datasheet for RPD:

It gives a value of 40k +-10k, so let's assume 40k and calculate the combined resistance of R38 and RPD:

R38 || RPD = 22k || 40k = ~14k2

Now let's calculate the voltage of your voltage divider:

(14k2/82k2) * 12V = ~2,07 Volt

That is nearly what your voltmeter is showing, now take in account the leakage through the diode as mentioned by the other answers and the great tolerances of the internal resistors and it is plausible that you left the pull-down resistor on.