**Voltage dividers are a good and simple way to drop voltage for things that require almost no current like for example small signal for GPIO pin or other such.**

Many will just take the circuit, decide total sum of R1 and R2 and then find the proportion between them to get correct voltage drop, like for example:

3.3 / 5 = 0,66 for R2, making R1 get portion of 0.34

And then decide total value of the resistors should be 300 Ohm *(See bellow why making random choice of the total might not be good idea)* making R1 = 0.34 * 300 = 100 Ohm and R2 = 0.66 * 300 = 200 Ohm.

This is all good and well if and only if your sure the current is small enough to not distort the equation much since this equation fails to take into account the current used by the resistors and the current used by the output that connects to the 3.3V

**Taking into account the current:**

We use Ohms law to calculate:

*I = V / R*

I=V/R = 5 / (100000 + 200000) = 16.6667µA *(Total current used by R1 and R2)*

By turning I=V/R around to V=R×I then we can calculate the Voltage drop over R1

100000 × 0.000016667 = 1.6667V

Therefore voltage in the middle is 5V – 1.6667V = 3.333 V

If the current drawn in the middle is close to nothing like in small signals then this will work well. But if you put any kind of load in the middle then this will not work.

**Lets look at why this cannot work with any kind of load in the middle:**

We put a device at the middle that needs 150 mA, and we recalculate the voltage drop

100000 × (0.000016667 + 0.15A) = 15001.6667 V

Given the voltage drop is much higher than the 5V that we started with then you would get zero in the middle at this current.

**Choice of resistors in the voltage dividers matters**

If you use resistors with high values in the voltage divider then you will waste less current. But bigger resistors will make your circuit more senstive to the current usage in the middle.

If we recalculate the example from above using smaller resistors:

I=V/R = 5 / (100 + 200) = 0.01667 A* (This is the amount of current that will be wasted over the resistors)*

If we put the same 150 mA device in the middle then the equation will look like this:

100 × (0.01667 + 0.15A) = 16,667 V (voltage drop)

The voltage drop is far less than before but would still lead to zero voltage in the middle if applying big load such as the 150 mA in this example.

**Conclusion is you should only use voltage dividers where current is close to nothing, and size of the resistors may need to reflect the current that you will draw.**

**If needing to drop voltage for example from 5V to 3.3V to run a device then better option would be to use Voltage regulator.**