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I have a voltage divider as shown in the diagram below - 68k, 33k in series; output voltage Vs is measured across the 33k resistor. The arrangement is used to measure voltage from a 5V nominal source on a PCB. When the input is 5.1V, I measure Vs = 0.6V and the voltage across the 68k to be 4.5V using a digital multimeter, when I was expecting Vs to be 1.6V.

I have repeated the measurement after desoldering the Zener and the capacitors, but the result is the same. The capacitors are ceramic 0805, resistors are 1/8W 0805. I am wondering if there is some fundamental error or I am missing something obvious? Any pointers much appreciated!

enter image description here

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    \$\begingroup\$ What is the part number for D202. It is perhaps loading your voltage divider, and causing a lower output voltage. It there anything else the divider is connected to? \$\endgroup\$ Jun 15, 2022 at 22:00
  • \$\begingroup\$ What does your multimeter display when you measure a voltage of 5V through a 33k resistor? \$\endgroup\$
    – bobflux
    Jun 15, 2022 at 22:41
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    \$\begingroup\$ Vs=0.6 is suspiciously close to what you would get if the output was loaded by a forward-biased diode junction. Is perhaps an ADC input connected, with power not applied to the ADC? \$\endgroup\$ Jun 16, 2022 at 2:23

3 Answers 3

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You said that you disconnected the zener and capacitors, but you did not say whether or not Vs is still feeding some other part of the circuit. See if there is something else connected there.

If that is not the case then you probably have an incorrect resistor value.

If R207 is correct at 68K, R204 would be 9K.
If R204 is correct at 33K, R207 would be 250K.

Of course they could also both be wrong.

Verify the resistance values.

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  • \$\begingroup\$ Thanks, yes there is an analog MUX connected Vs which I had missed as mentioned in my answer. \$\endgroup\$
    – Mini23
    Jun 16, 2022 at 20:59
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Since all other components have been removed besides the two stacked resistors (68K and 33K), the lower than expected Vs voltage must be influenced by the voltmeter's internal resistance.

Solving for the voltage divider equation with the voltmeter resistance included results in the voltmeter having an internal resistance of 12.5K ohms. This seems unusually low for a digital voltmeter, so it may be defective.

I suggest doing the same voltage divider test with lower value resistors (say 680 and 330 using the same voltmeter. If the Vs voltage is near the expected 0.6V, then it is the voltmeter internal resistance causing the problem. The more accurate Vs reading is because the 12.5K internal resistance of the voltmeter is much larger than the parallel 330 ohm resistor.

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  • \$\begingroup\$ You wanted to say "near the expected 1.6 V" \$\endgroup\$
    – Jens
    Jun 16, 2022 at 0:43
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    \$\begingroup\$ The sum of the voltages measured (0.6 V + 4.5 V = 5.1 V) contradicts this theory. There must be a gap in the sum if the multimeter is a significant load. I assume one resistor is just bad or has a wrong value. \$\endgroup\$
    – Jens
    Jun 16, 2022 at 0:48
  • \$\begingroup\$ Thanks for the answer and the comments. The multimeter's impedance is ~10Mohm according to the datasheet so I had ruled out this factor. \$\endgroup\$
    – Mini23
    Jun 16, 2022 at 21:00
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Thanks all for your answers and comments. It was indeed what @Spehro Pefhany posted in their comment. Vs is connected to an analog MUX (I had missed this somehow) and I had not applied power to that. Once I did that, I measured ~ 1.5V at Vs which is close enough to the expected 1.6V. So it must have been a diode internal to the analog MUX that was loading the output.

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