The 1 Meg Ohm in parallel with R2 191 K Ohm should not produce any noticeable loading effect, it would be negligible. The only way the Vout of the divider can be less than anticipated is if Vsupply is below 12.24 Volts. We already established the values of R1 and R2 to be correct and they are 1% tolerance precision resistors. The circuit current is negligible as to any heating effects. The meter used and the A/D converter agree so apparently 2.4 volts is what it is. To get that low the A/D imput impedance would have to be less than 750K. Something in all of this is wrong.

Ron

 

The 1 Meg Ohm in parallel with R2 191 K Ohm should not produce any noticeable loading effect, it would be negligible. The only way the Vout of the divider can be less than anticipated is if Vsupply is below 12.24 Volts. We already established the values of R1 and R2 to be correct and they are 1% tolerance precision resistors. The circuit current is negligible as to any heating effects. The meter used and the A/D converter agree so apparently 2.4 volts is what it is. To get that low the A/D imput impedance would have to be less than 750K. Something in all of this is wrong.

Ron

Off the cuff, 1 Meg Ohm in parallel with R2 191 K Ohm would be about 160k. That's not a insignificant change.

 

The 1 Meg Ohm in parallel with R2 191 K Ohm should not produce any noticeable loading effect, it would be negligible.

What?
How do you figure that 1 Megohm is negligible with respect to 191 kohms.
It will cause a about a 14% drop in the divider output voltage.

 

Resistance is measured at DC.
Impedance is frequency dependent and can be measured at a given frequency.

To measure the resistance or impedance, insert a resistance in series between the source and the receiving circuit.
Adjust the value of the resistance until the voltage at receiving circuit becomes one-half the source voltage.
The circuit resistance or impedance will be equal to the value of the series resistor.

 

That circuit will likely use more current than accurate voltage divider resistors.

Thanks a lot for the suggestions.

 

Basically a trade-off between resistor values with the resulting current versu measurement accuracy.

If you are using a micro to measure the voltage, you could always add a correction fudge-factor to the value to correct for the error due to the divider.
Thus if the measurement is 2.4V but the actual value is 2.9V, then multiply the measured value by a factor of 2.9 / 2.4 = 1.2.

Using a micro ADC from a high source resistance, for accuracy you need to make sure that the input is connected to the sampling capacitor for long enough for the capacitor to be fully charged.

 

What?
How do you figure that 1 Megohm is negligible with respect to 191 kohms.
It will cause a about a 14% drop in the divider output voltage.

My bad. Sorry. Thanks for pointing it out.

Using a micro ADC from a high source resistance, for accuracy you need to make sure that the input is connected to the sampling capacitor for long enough for the capacitor to be fully charged.

Something else to consider.

Ron

 

Another thought is to use the ACD input impedance as part of the voltage divider.
Thus, if the input resistance is 1 megohm, then a 236kΩ resistor for the bottom resistor will give the equivalent of 191kΩ.
You may have to experiment with the value to get the correct reading, depending upon the actual ADC input resistance.

Depending on the ADC acquisition time, you may need to add a small capacitor at the ADC input to get a proper reading.