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I am playing around with voltage drop resistor circuit, but the output voltage doesn't match the voltage I calculated.

This is the formula I am using

Vout = Vs*R2/(R1+R2)

schematic

simulate this circuit – Schematic created using CircuitLab

The battery I am using is 9V, it measures about 9.4V using voltmeter.

I am expecting an output voltage around 8.6V-8.9V, but instead I get 6V.

Is there something that could also be affecting the output voltage or am I calculating the output voltage incorrectly? I am a beginner so there is a good chance of this.

I have measured the resisters individually and they both measure the correct resistance.

Here is an image of my breadboard, although I think I got everything wired correctly. (Yellow wire is output V+.)

enter image description here

My end goal is to drop about 0.5Voltage between the input and output.

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    \$\begingroup\$ Measure the battery voltage itself when loaded with these resistors. You'll find it's a lot less than the off-load voltage. You'll then find the sums working. \$\endgroup\$
    – Neil_UK
    Jul 24, 2022 at 16:00
  • \$\begingroup\$ It appears that the battery is No-Name-Brand and Google has never seen one. I agree that the battery might be weak and drops its voltage when loaded. The connections on the breadboard also might be poor and reduce the voltage. \$\endgroup\$
    – Audioguru
    Jul 24, 2022 at 17:28

2 Answers 2

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When you load up your 9 volt battery with 230 Ω it seriously drops the output voltage seen on your battery and makes you think that your potential divider calculation is incorrect.

If you had a "strong" battery or used (say) 1000 Ω and 22 kΩ you would see a closer result.

The battery you are using is a zinc-carbon type according to this website and, according to this website it will have an internal resistance of about 35 Ω: -

enter image description here

So with 35 Ω internal series resistance and a 230 Ω load, the terminal voltage might drop from 9 volts to about 7.8 volts. If the battery is partially used or near it's end-of-life it may be even lower. Battery brands that do not provide a value for series resistance cannot be trusted in this situation; they may have a resistance even higher than the table above.

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    \$\begingroup\$ Agreed. Measure the battery terminal voltage when your circuit is powered up and you'll likely see it's not reading 9V anymore \$\endgroup\$
    – James Fotherby
    Jul 24, 2022 at 16:05
  • \$\begingroup\$ So the input voltage for the calculation should be the voltage once wired up, not the voltage of the battery without any load? \$\endgroup\$
    – Christopher Townsend
    Jul 24, 2022 at 16:06
  • \$\begingroup\$ @andy-aka This worked! Would you be able to point me to any resources online that explain why this is the case as I would really like to learn why 1000Ω + 22KΩ works but my initial 230Ω didn't. \$\endgroup\$
    – Christopher Townsend
    Jul 24, 2022 at 16:13
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    \$\begingroup\$ @ChristopherTownsend I have added a few notes to help you understand that. \$\endgroup\$
    – Andy aka
    Jul 24, 2022 at 16:13
  • \$\begingroup\$ Thanks @Andyaka. Am I correct in understanding that the higher resistance values you provided were better because higher resistance = less current, which in turn means less voltage is being used so the voltage drop wasn't as big as with my 230Ω ? \$\endgroup\$
    – Christopher Townsend
    Jul 24, 2022 at 16:19
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There are 2 possible situations:

1)Your voltmeter has very low resistance and it affects the reading

2)You are not taking into account the internal resistance of the battery.

To which is which , replace the resistors with x10 their value if the voltage between V+ and V- drops to approximately 0 then your voltmeter has stopped working.

If the voltage goes up then the internal resistance of the battery is big enough that in your original circuit it affects the voltage between V+ and V-.

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  • \$\begingroup\$ Just so you know that it wasn't me who d/v your answer, open this image. However, your paragraph under bullet point 2 is misleading. \$\endgroup\$
    – Andy aka
    Jul 24, 2022 at 16:48
  • \$\begingroup\$ The DMM used is a high quality instrument that apparently was also used successfully to measure the battery's open circuit voltage. So your situation #1 is impossible. \$\endgroup\$
    – PStechPaul
    Jul 24, 2022 at 23:13

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