In many of my past projects I have made use of a circuit component (or set of components) called a voltage divider. I realized however, I’ve never really spent time talking about what these were or how they work. For that reason I wanted to spend some time showing this trick and how you can use it yourself.
So What Are They?
A voltage divider divides voltage! Brilliant analysis I know, but what does that mean? And why is it useful? Simply put a voltage divider is a way to get a smaller voltage from a larger one. For example lets say you are trying to power a chip which requires a 5V power source from a 9V battery. In this situation you could use a voltage divider at the power pin to lower your 9V supply to the required 5V. Another important use is signal attenuation. As we’ll soon see the output of a voltage divider is a linear function of the input voltage. That’s a fancy way of saying the voltage you get out will always be a specific fraction of the voltage you put in (Based on the resistor values). That means if you have a waveform or audio signal you can reduce its amplitude while still maintaining the original waveform.
Note: Voltage Dividers are a cheap and easy way to reduce voltage however they are not highly stable or accurate. If you require a stable supply it is better to use a voltage regulator IC or Buck Converter
Math Time!
Voltage dividers are an excellent illustration of Kirchoff’s Loop Law and we can use this along with Ohm’s Law to calculator our output voltage given the values of R1, R2 and the input voltage.
To clarify how these calculations are done I have drawn the divider as a loop with a voltage source V1. We know from Kirchoffs Laws that the Sum of the voltages around any closed loop is zero. This means V1 minus the voltage lost across the two resistors is 0. This also means the voltage V2 will be equal to V1 minus the voltage lost across R1:
To determine the voltage dropped across R1 we can rely on Ohm’s Law. Ohm’s Law states that V = IR. From this we know the voltage dropped across R1 is equal to the product of the resistance and the current. Unfortunately we still need the current for this equation. Looking at the loop as a whole though we can solve for it:
Now that we have an equation for current we can substitute this into the equation for the voltage drop through R1:
Finally we can input this new function into our original equation to solve the ratio between V1 and V2:
We can divide both sides by V1 to clean this up a bit:
For good measure we can do a little more simplification to come to the final voltage divider equation:
And there you have it! using this equation you can take any input voltage and any desired output voltage and calculate the values of R1 and R2 you will need to accomplish that reduction.
