Hi,

I'm working with voltage dividers and I've noticed some very unexpected behavior when using high-resistance resistors.

I've created a voltage divider with 470k and 1M ohms resistors and hooked it up to a 9V battery (9.22V measured). The expected voltage would be 6.27V for 1M and 2.95V for 470k. Instead, I got 4.78V and 2.22V when measuring them individually, resulting in a total of 2.22V lost.

I've then recreated the test with 10x lower resistors - 47k and 100k. The loss was smaller, but still noticeable - at 0.32V lost compared to the power supply.

Could someone please explain what is going on? I'm a beginner in electronics and I'm sure there's something I don't know that would make me understand this phenomenon. Thanks in advance

 

Your voltmeter internal resistance will be the reason. Cheap voltmeters will have 1Meg of internal resistance.

 

Hi W6,
Welcome to AAC.
What is the tolerance of your resistors?

E

 

What is the resistance of the meter that you are using to measure the voltages?

 

What is the input resistance of your meter that you are measuring with. While many are 10 megaohm, it may be possible that some cheaper meters have a much less input resistance, and thus it is loading down the device you are measuring. That is, the 1M resistor and your meter are 2 "resistors" in parallel, possibly dropping the voltage even more. You can prove your meter is affecting the results by trying a 1k and a 470 ohm resistor, the calculated values should be much closer to what you measure.
Also, what is the % tolerance of the resistors? That can affect the values to a small degree as well.

 

Wow didn't expect that much engagement so quickly, thanks for the activity guys.

As for the multimeter, I've bought the cheapest one available, unfortunately, I can't find the information about its impedance anywhere. I don't know if I can post links here, here's its product page https://botland.store/universal-meters/18398-universal-multimeter-rebel-rb-830buz-5901890058063.html

So let's say it has a low impedance, why does that make it "suck up" more of the voltage as the resistance of the measured resistors increases? Shouldn't it take up more voltage when the measured resistors are smaller? I'm pretty comfortable with understanding the flow of electricity, resistance etc. but never really grasped the idea of impedance that well

 

Just realised I somehow added impedance to that while you were talking about internal resistance, not impedance. Must've gotten mixed up, it's been a long day. Just substitute impedance with resistance in my question and it'll still be valid

 

The situation will look like this:



And the output voltage will be equal to

Vout = Vin * (R2||Rvm)/(R1 + (R2||Rvm)) = 9V * (1MΩ||1MΩ)/(470kΩ + 1MΩ||1MΩ) = 9V * 500kΩ/(470kΩ + 500kΩ) = 4.639V

 

What kind of meter?? There are both digital meters,which have quite high input resistance and seldom affect the measured circuit, and also analog meters that have a much lower input resistance. There is also the very real possibility that connecting the resistors to the battery affects the battery voltage. One more thing that probably has an effect is that the resistor values may not be the exact amount shown by the color code.

 

So let's say it has a low impedance, why does that make it "suck up" more of the voltage as the resistance of the measured resistors increases?

Its a result of the Thevenin equivalent impedance of the two voltage divider resistors, which is equal to the two resistors in parallel.
This resistance is in series with the no load output voltage of the divider network, which forms an additional divider network between the Thevenin output impedance and the meter input impedance.

It doesn't "suck up" the voltage, it's just the simple IR voltage drop across the divider impedance due to the meter load current.

Make sense?

never really grasped the idea of impedance that well

Impedance is the term for the combined resistance and AC reactance in a circuit.
For a DC signal it becomes just the resistance.