How to calculate voltage drop

[Alex Hughes]

So I have two formulas for calculating voltage drop. The first is: Vdrop = Vin * R1 / ( R1 + R2...). This let's me calculate the voltage drop on the first resistor in a series of 2 resistors. You can also extend this to more resistors in series just by summing all the resistors. However, I do not find this useful in designing circuits because there is nothing to tell me what resistors to use. You sort of have to just keep plugging values in until you get the voltage drop you want. The second is: R2 = Vout * R1 / (Vin - Vout). This formula is more helpful to me because it allows me to calculate what resistor I would need (R2) to put in series with another resistor (R1) in order to achieve a certain voltage drop (Vout) if I have a given voltage from the source (Vin). The problem with this formula is I don't know how to manipulate it to allow me to calculate what resistors I would need to have for circuits where I would need more than 2 resistors in series. For example, if I had a 12v battery, and I wanted to drop the voltage to 8V, then 6V, then 3V, then 0. How would I calculate what resistors I would need to put in series to achieve this without just plugging in random resistors. Does anybody know a formula/method I can use? Thanks.

 

[phinds]

You need to study simple series and parallel circuits. They are quite simple and you are making more of the problem than is necessary.

 

[analogdesign]

You should look up Voltage Divider on Wikipedia: https://en.wikipedia.org/wiki/Voltage_divider

That is the kind of circuit you are describing.

You're missing one key bit of data and that is the total resistance you want. A high resistance level doesn't load the batter much but can't take much current, and a low resistance is the opposite. It's a trade off.

Once you know the total resistance, you have x equations and x unknowns and you can calculate all the resistance values.

 

[Hesch]

For example, if I had a 12v battery, and I wanted to drop the voltage to 8V, then 6V, then 3V, then 0. How would I calculate what resistors I would need to put in series to achieve this without just plugging in random resistors. Does anybody know a formula/method I can use? Thanks.

You must choose a current through all the resistors in series. The currents are the same, you know.
Let's say you choose 2A, then the resistor values will be:

1. (12V - 8V) / 2A = 2Ω
2. (8V - 6V) / 2A = 1Ω
and so on.

But other values can be used. Say that you choose a current = 2mA, the values will be:

1. 2kΩ
2. 1kΩ
3. . . . . . . .