What is a Potential Divider or Voltage Divider Circuit

The potential divider or voltage divider is a very straightforward but very useful circuit. As a result the potential divider is found in a large number of electronic circuits.

It is very useful to see what this circuit is and how it works.

As the name implies the potential divider or voltage divider divides the potential or voltage at the input to give a lower output voltage. It divides it down according to the value of the resistors.

In its simplest form the potential divider consists of two resistors in series. As the same current flows through both resistors the voltage drop across them is in the same ratio as the value of each resistor.

This means the output voltage can be calculated from the formula Vout = Vin (R2 / R1 + R2). The output voltage is the value of bottom resistor in the circuit diagram divided by the total resistance of the chain times the input voltage.

In this way it divides the input voltage down to a lower value.

For the equation to hold true, the load resistance must be sufficiently high for it to be ignored, otherwise it must be included as a resistance in parallel with R2.

The circuit can also be used with a component like a light dependent resistor so that any resistance variations caused by light changes are converted to a varying voltage. It is found that if the LDR is placed in the lower section of the chain, then the voltage across it will fall with increasing light levels as the resistance of an LDR falls when the light increases.

The calculations for the voltages for the potential divider when light is shining on the LDR:
Vout = 10 x 5 / (5 + 10) = 10 x 5 /15 = 3.3333 V

The calculation for the output voltage of the potential divider when the LDR is in dark:
Vout = 10 x 200 / (200+10) = 10 x 200 / 210 = 9.5238V

If the LDR is placed in the top element of the potential divider and the 10k resistor is in the bottom position.
Then the light voltage will be:
Vout = 10 x 5 / (5 + 10) = 6.6V
and the dark voltage will be 10 x 10 / (200 + 10) = 0.476V

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[isuckatthisgame]

Nobody I've ever seen, until one of my favorite professors in college, explained the term Vout = Vin * (R2 / (R1 + R2)) as nothing else but Vout = I * R2 (a current multiplied by the resistance of any resistor you want to find the voltage drop over). Vin * (R2 / (R1 + R2)) is nothing else but the current flowing through both resistors that are connected in series. It's intuitive from the simple schematic but I know for a fact that many other students at Analog electronics class struggled even with simple stuff like this because it wasn't explicit immediately (as did I, but I managed to past that hell by working day in day out). Thanks for the explanation. I needed your source of information. Bless you, sir. 👍

[poptarts-fr]

Thank you! Simple and straight forward!

[challengesolvers]

Thanks, ,l have really learned a lot in this video

[UTubeIsTrackingYou]

Hi, newbie here. In the last diagram - with the photoresistor in the first position - shouldn't the red lead on the voltmeter be going to the connection between the two resistors, or am I missing something? Thanks for your videos!

[madjack]

this is totally different than the book i'm learning from. I have a voltage, two resistors in parallel and two more resistors in parallel with a load that has a voltage and current given. I'm