I have been searching like a maniac for as much info as possible before placing question here, obviously i failed...

The question is about voltage divider, i have an external source that gives me either 12-13V or 10-11V, now looking at voltage divider, if one would be implemented for this signal cable i should be getting out between 2-4V (max that i need is 5V so im gonna try stay below 5 for safety).

enter image description here

Looking at this picture, i saw somewhere if

  1. Vin = 12V
  2. R1 = 100K
  3. R2 = 10K

then i should get Vout at somewhere "around" 5V... now i tried that theory in practice, my supply is dynamic so i can change the Vin to 10, 12 or 15V or what ever, but the voltage never decreased more than 1-2V.

I also tried to chain up with more resistors, like multiple 1M, also with multiple 100K and 47K but the voltage never decreased and pleasurable amount to hook it into Arduino digital pins.

I have a hobby project that i want to finish up before saturday but this "issue" drives me crazy as im not even close to 5V.

Also worth to mention i was suggested to use a comparator (couldnt tho find any comparator that had pull-down output in my local store) but then later on some said a comparator would be "overkill" for my tiny project since i only need to compare this either 10V or 12V in arduino, where divider would be more suitable.

  • \$\begingroup\$ You need to learn Ohm's law. V= I*R. Calculate the current in your divider chain. (~0.1 mA) and then find the voltage drop across each resistor. (Since the current is the same in each resistor and the ratio is 10:1 the voltage drop is also 10:1.) \$\endgroup\$ Commented Sep 16, 2015 at 16:28
  • \$\begingroup\$ Are you measuring Vout across R2 - that is, between the Vout terminal and Ground? With a 10 volt supply, your 100K/10K divider should give about 0.9 volts across R2. \$\endgroup\$ Commented Sep 16, 2015 at 16:29
  • \$\begingroup\$ That is exactly what ive been trying to do, this famous V=I*R is always with me when dealing this issue. Tho im getting horrible confused when some say, okay you need 100K resistor, and a 10K one, and you can hook up to arduino for reading values... which obviously is loose ended responses \$\endgroup\$
    – Deko
    Commented Sep 16, 2015 at 16:30
  • \$\begingroup\$ Peter, actually no... think what ive done wrong, lol. I measured between Vin and Vout :S Gotta try that out now here.... \$\endgroup\$
    – Deko
    Commented Sep 16, 2015 at 16:32
  • \$\begingroup\$ Okay, i kind of worked in expected way, i got 12V with could of 1M resistors down few volts... think i could go for that ! \$\endgroup\$
    – Deko
    Commented Sep 16, 2015 at 16:39

2 Answers 2


Alright let's get the Math correct and then we can decide the Resistor values. The circuit above is a simple voltage divider circuit. Basically, the two resistors are connected in series and we try to measure (or read) the voltage from one of the resistors (here the one whose other end is connected to Reference (or lets say Ground). You can visualize the circuit as a simple series circuit the one shown below -


simulate this circuit – Schematic created using CircuitLab

We want to measure the voltage across R2 resistor. Since both the resistors are connected in series the current flowing through them will remain same. Therefore the voltage across R2 comes out to be -

V0 = R2* I ..... (1)

But we don't know the value of I. Well getting that is easy enough. The net resistance in series is the sum of the resistance connected in series. Therefore Rnet = R1 + R2. As a result the current the circuit comes out to be-

I = V1/Rnet ..... (2)

Using this value in (1)

V0 = R2 * V1/(Rnet)

=> V0 = R2 * V1/ (R1+R2)


In your case V1 = 10, 12 or 15 (you have a dynamic supply) and you want V0 to be around 5V.

Let's take R1 to be 10k ohm. Using the result above 5 = R2 * 15/(10k + R2)

Solving the above equation gives R2 as 5k ohm

So, you can use R1 as 10k and R2 as 5k (4.7k as 5k is not a standard value) and your input should be 5V. This should give you a voltage of about 5V across R2. I hope it helps.

  • \$\begingroup\$ thanks for very simple and good explanation! but still I'm confused about the resistor ohms. if I have vin at 12 and I add 2 1M resistors, why is the Vout still near 12. I think it only dropped .5V with each 1M. had to chain 12 resistors to pull 12V down to close to 5V. or are there different kinds of 10K resistor that divide current in various ways ? \$\endgroup\$
    – Deko
    Commented Sep 16, 2015 at 18:43
  • \$\begingroup\$ I have hardly seen this happening. If you are doing exactly as in the circuit diagram you mentioned. Then if both the resistors are of equal value (1M in your case) then the voltage across each one of them would be Vin/2 (around 6V in your case). I have a feeling that there might be some loose connection or the two resistors you are using are not of the same value. I have done practicals and I would say when working with resistors V=I*R generally holds true. \$\endgroup\$
    – Rahul Behl
    Commented Sep 16, 2015 at 18:51
  • \$\begingroup\$ I just tried your circuit live and absolutely, I got 4V , perfect, I'll try few more options with diff currents so it burns into my memory! thank you \$\endgroup\$
    – Deko
    Commented Sep 16, 2015 at 18:57

If the current into the Arduino's input is negligible, and your supply is a voltage source, then with R2 equal to 10k you'd drop 5 volts across it with 500 microamperes through it, since

$$ E = IR = 500\mu A \times10k\Omega = 5 \text{ volts.} $$

Then, with a source voltage of 12 volts, you'd have to drop the remaining 7 volts across R1 to account for the whole 12 volts and since current in a series circuit is everywhere the same, the 500 microamperes through R2 would also go through R1.

Such being the case, in order to drop 7 volts with 500 microamperes through R1, R1's resistance would have to be:

$$ R = \frac{E}{I} = \frac {7V}{500\mu A} = 14000 \text{ ohms.} $$

Next, since you need to know what Vout will be when Vin is 11 volts, you can use the voltage divider equation to determine that value, like this:

$$ Vout = \frac{Vin\times R2}{R1 + R2} = \frac{11V \times 10k\Omega}{14k\Omega + 10k\Omega} \approx 4.58\text { volts} $$


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