# Lesson : Voltage Divider

In the past chapter, we looked at series and parallel circuit. One things you may have notice is that the series circuit can be used as voltage divider. We have already seen a small portion of the theory about the voltage divider in the series circuit lessons but we will now study in depth the voltage divider. We will also take a look at some real life applications of the voltage divider.

The voltage divider is actually quite simple. It is two resistors in series used to reduce the input voltage (in this case VDC) to a smaller voltage (Vout).

Below you will find all the equations for the voltage divider. All these formulas are found with Ohm’s law and the equivalent circuit:

$V_{out} = R_{2} \cdot I$

$I = \cfrac{VDC}{(R_{1}+R_{2})}$

If we replace I in the first formula, we get the formula for the voltage divider:

$V_{out} = VDC \cdot \cfrac{R_{2}}{(R_{1}+R_{2})}$

The voltage divider has multiple applications in real life. Below you can find one of the popular applications for it:

Voltage monitoring with microcontroller:

Often in personal computer and server, you have a microcontroller monitoring voltage, current, etc. These controllers often works at low voltage (< 3.3V) but needs to monitor voltage higher than this (12V, 5V). Sometimes the integrated analog to digital converter (ADC) works at even lower voltage than the microcontroller since they have internal reference. If you want to monitor the 12V rail, you will need to reduce the 12V to the appropriate voltage for your ADC. The voltage divider is perfect for this but you will need to use resistors which have a pretty low tolerance to get accurate result (<1%).

Example :

In the image below, we have 12V from the D.C voltage source. R1 = 100 ohms and R2 = 50 ohms. We want to calculate Vout. (Vout = V2 in the image below)

$Vout=Vin*\cfrac{R2}{R1+R2}$

$Vout=12V*\cfrac{50\Omega}{50\Omega+100\Omega}$

$Vout=4V$

We can also calculate V1 with the equation above. We only need to modify the equation for V1. We replace Vout with V1 and instead of R2, we have R1 :

$V1=Vin*\cfrac{R1}{R1+R2}$

$V1=12V*\cfrac{100\Omega}{50\Omega+100\Omega}$

$V1=8V$

The equation of the voltage divider could be modify for more than two resistors too if needed as long as the resistors are in series :

$Vx=Vin*\cfrac{Rx}{Rx+Rn1+Rn2+\dots}$

where Vx is the voltage across the resistor Rx and Vin is the voltage across all resistors in series.