Potential Dividers

Potential Dividers
Introduction
Potential dividers are a very common part of even the most complex circuits and it is
therefore very important to understand what they do and how to do the associated
calculations.
A potential divider circuit is also known as a voltage divider circuit.
Basic Concepts
A potential divider is a pair of resistors
connected in series and connected to a
power supply or other source of EMF
(voltage).
The potential difference across ONE of the

resistors is a fraction of the applied EMF.
The total potential difference across the
resistors is divided between the two
resistors, each having its own potential
difference - hence potential divider.
In essence, a potential divider is a circuit with an input and an output. You apply a voltage to
the input (Vin) and get a smaller voltage at the output (Vout). The output voltage is a fraction of
the input voltage.
The output voltage depends on (a) the input voltage and (b) the ratio of the two resistors.
Note: The output voltage does not depend on the actual value of the individual resistors, it
depends on the ratio of the pair of resistors - they work together and the output of the
potential divider depends on their relative values.
Potential Divider Equation
The output voltage from a potential divider (Vout) is given by the potential divider equation:
Vout = Vin × R2 / (R1 + R2)

It is important to be able to use this equation.
Method of Ratios
The potential divider equation is very useful for calculating
Vout but it is less obvious when trying to calculate R1 or
R2.
It is usually much easier to use the ratios of the voltages

to work out what resistor values to use. The ratio of the
voltages is equal to the ratio of the resistances:
V1 ÷ V2 = R1 ÷ R2
or
V1 : V2 = R1 : R2
or
V1 : Vout = R1 : R2
Knowing Vin and Vout means you know V1 and V2 (V2 is Vout of course).
Once you know V1 and V2 you can choose R1 and R2 accordingly

An Important Consideration
The potential divider equation relies on the fact that the
current in R1 is the same as the current in R2 and no current
flows in the load (the load has an infinite resistance).
If this is NOT the case, the potential divider equation will not
give a correct value for Vout and neither will the method of
using ratios.
When calculating values of Vout, R1 or R2 it is always

assumed that no current flows from the output of the
potential divider
In reality, this is not possible to achieve. Any load such as a

voltmeter or subsystem connected to the output of the
potential divider will always take some current.
However, the theory is very nearly correct when the current

flowing from the potential divider to the load is much less
than the current through R1 and R2.
In practice, the values of R1 and R2 should be several orders of magnitude smaller than the
resistance of the load.
For example, if a voltmeter has a resistance of 1 MΩ then a potential divider made from a
pair of 10 kΩ resistors will work as expected but a potential divider made from a pair of
500 kΩ resistors will give a value of Vout smaller than expected.
The finite resistance of the load is effectively in parallel with R2 which reduces the overall
combined resistance of R2 and therefore reduces Vout.
Examples
Example 1
Given the two resistor values and the supply voltage, find the
output voltage.
The values given are non trivial so the best approach is to use
the equation for Vout
= 12 × 220 / (100 + 220)

Vout
Therefore:
Vout = 8.25 Volts
Example 2
Given Vout = 5 V, Vin = 9 V and R2 = 1 kΩ calculate R1.
The best approach when trying to find a resistor value is to use

the method of ratios.
The voltage across R2 is 5 V and therefore the voltage across R1

is 4 V
4 : 5 = R1 : 1000
Therefore:
R1 = 800 Ω
Example 3
Given Vout = 2 V, Vin = 6 V and R1 = 12 kΩ calculate R2.
Again, the best approach when trying to find a resistor value is to

use the method of ratios.
Voltage across R2 is 2 V therefore the voltage across R1 is 4 V
4 : 2 = 12k : R2
Therefore:
R2 = 6 kΩ
Effect of Changing R1 and R2
Where one of the resistors in a potential divider is variable - an LDR or Thermistor for
example - it is very important to know how Vout changes when either R1 or R2 changes (and
Vin remains fixed).
Case 1: R1 increases
R1 increases and takes a larger share of the input voltage
Therefore, the remaining voltage, Vout goes down
Case 2: R1 decreases
R1 decreases and takes a smaller share of the input voltage
Therefore, Vout goes up
Case 3: R2 increases
R2 increases and takes a larger share of the input voltage
Therefore, Vout goes up
Case 4: R2 decreases
R2 decreases and takes a smaller share of the input voltage
Therefore, Vout goes down

Potential Dividers with LDRs and Thermistors
Consider (and learn and understand) each of the following examples:
Thermistor as R2:
As the temperature increases the resistance of the
thermistor decreases.
As the resistance decreases, Vout decreases.
Therefore, Vout goes down as the temperature increases.
As the temperature decreases the resistance of the

thermistor increases.
As the resistance increases, Vout increases.
Therefore, Vout goes up as the temperature decreases
A similar situation occurs if an LDR is used in place of R2. If the light level increases, Vout
goes down and vice versa
Thermistor as R1
As the temperature increases the resistance of the
thermistor decreases.
As the resistance decreases, Vout increases.
Therefore, Vout goes up as the temperature increases
As the temperature decreases the resistance of the

thermistor increases.
As the resistance increases, Vout decreases.
Therefore, Vout goes down as the temperature decreases
A similar situation occurs if an LDR is used in place of R1. If the light level increases, Vout
goes up and vice versa
Website
https://www.electronicsteaching.com/Electronics_Resources/DocumentIndex.html
© Paul Nicholls
July 2020
Electronics Resources by Paul Nicholls is licensed under a Creative Commons Attribution

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Potential Dividers

A ​potential divider​ circuit is also known as a ​voltage divider​ circuit.

The potential difference across ONE of the

V​out​ = Vin​ ​ × R2 / (R1 + R2)

It is usually much easier to use the ratios of the voltages

Once you know V1 and V2 you can choose R1 and R2 accordingly

When calculating values of V​out​, R1 or R2 it is always

In reality, this is not possible to achieve. Any load such as a

However, the theory is very nearly correct when the current

​ ​ = 12 × 220 / (100 + 220)

V​out​ = 8.25 Volts

The best approach when trying to find a resistor value is to use

The voltage across R2 is 5​ ​V and therefore the voltage across R1

Again, the best approach when trying to find a resistor value is to

Voltage across R2 is 2​ ​V therefore the voltage across R1 is 4​ ​V

Therefore, the remaining voltage, V​out​ goes down

Therefore, V​out​ goes up

Therefore, V​out​ goes up

Therefore, V​out​ goes down

As the resistance decreases, V​out​ decreases.

Therefore, V​out​ goes down as the temperature increases.

As the temperature decreases the resistance of the

As the resistance increases, V​out​ increases.

Therefore, V​out​ goes up as the temperature decreases

As the resistance decreases, V​out​ increases.

Therefore, V​out​ goes up as the temperature increases

As the temperature decreases the resistance of the

As the resistance increases, V​out​ decreases.

Therefore, V​out​ goes down as the temperature decreases

Electronics Resources by Paul Nicholls is licensed under a​ ​Creative Commons Attribution

You might also like

A potential divider circuit is also known as a voltage divider circuit.

Vout = Vin × R2 / (R1 + R2)

When calculating values of Vout, R1 or R2 it is always

= 12 × 220 / (100 + 220)

Vout = 8.25 Volts

The voltage across R2 is 5 V and therefore the voltage across R1

Voltage across R2 is 2 V therefore the voltage across R1 is 4 V

Therefore, the remaining voltage, Vout goes down

Therefore, Vout goes up

Therefore, Vout goes up

Therefore, Vout goes down

As the resistance decreases, Vout decreases.

Therefore, Vout goes down as the temperature increases.

As the resistance increases, Vout increases.

Therefore, Vout goes up as the temperature decreases

As the resistance decreases, Vout increases.

Therefore, Vout goes up as the temperature increases

As the resistance increases, Vout decreases.

Therefore, Vout goes down as the temperature decreases

Electronics Resources by Paul Nicholls is licensed under a Creative Commons Attribution