Voltage divider not making any sense

Question

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).

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.

Answer 1

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.

Answer 2

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:

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