Potential dividers and sensors

What you'll learn

How a potential divider produces a chosen output voltage from a supply.
How to calculate $VoutV_\text{out}$ using resistance ratios.
How LDRs and thermistors can turn light or temperature changes into voltage changes.
How to spot common exam traps, including the effect of a load.

Prerequisites: voltage, current and resistance

Before potential dividers, you need three circuit ideas.

Definition

Potential difference

Potential difference, or voltage, is the energy transferred per unit charge between two points in a circuit: $V = W / Q$ . It is measured in volts, V.

Current is the rate of flow of charge, measured in amperes, A. Resistance is how much a component opposes current, measured in ohms, Ω. For an ohmic component:

\frac{V}{I}

In a series circuit, components are connected one after another in a single path. The same current flows through each component, but the supply voltage is shared between them.

Key Idea

Series voltage sharing

In series, the same current flows through every resistor, so the resistor with the larger resistance gets the larger share of the potential difference.

Example

Finding voltage shares in series

A 12.0 V supply is connected across two series resistors: 2.0 kΩ and 4.0 kΩ. Find the potential difference across each resistor.

Add the series resistances:
$kΩR_\text{total} = 2.0\ \text{k}\Omega + 4.0\ \text{k}\Omega = 6.0\ \text{k}\Omega$
Calculate the current through the series chain:
$\frac{V_s}{R_\text{total}} = \frac{12.0\ \text{V}}{6.0 \times 10^3\ \Omega} = 2.0 \times 10^{-3}\ \text{A}$
Use $V = I R$ for each resistor:
$VV_1 = (2.0 \times 10^{-3}\ \text{A})(2.0 \times 10^3\ \Omega) = 4.0\ \text{V}$ $VV_2 = (2.0 \times 10^{-3}\ \text{A})(4.0 \times 10^3\ \Omega) = 8.0\ \text{V}$
Check the shares add to the supply voltage:
$V4.0\ \text{V} + 8.0\ \text{V} = 12.0\ \text{V}$

The potential divider

Definition

Potential divider

A potential divider is a series circuit used to produce an output voltage that is a fraction of the supply voltage.

Usually, two resistors are connected in series across a supply. The output voltage, $VoutV_\text{out}$ , is measured across one of the resistors. In the standard form below, $VoutV_\text{out}$ is taken across the lower resistor, $R_2$ .

Potential divider circuit with output voltage across the lower resistor

Because the resistors are in series, the current is:

\frac{V_s}{R_1 + R_2}

The output voltage is the potential difference across $R_2$ :

Vout=IR2=VsR2R1+R2\begin{aligned} V_\text{out} &= IR_2 \\ &= V_s\frac{R_2}{R_1 + R_2} \end{aligned}

Key Idea

Potential divider equation

If the output is taken across the lower resistor, use $Vout=VsR2R1+R2V_\text{out} = V_s\frac{R_2}{R_1 + R_2}$ . The resistor across which you measure the output goes in the numerator.

Example

Calculating the output voltage

A 9.0 V supply is connected across two series resistors. The top resistor is 2.2 kΩ and the lower resistor is 3.3 kΩ. Find $VoutV_\text{out}$ across the lower resistor.

Identify the resistor across which the output is measured:
$kΩR_2 = 3.3\ \text{k}\Omega$
Substitute into the divider equation:
$kΩV_\text{out} = 9.0\ \text{V} \times \frac{3.3\ \text{k}\Omega}{2.2\ \text{k}\Omega + 3.3\ \text{k}\Omega}$
Calculate the resistance fraction:
$3.32.2+3.3=3.35.5=0.60\frac{3.3}{2.2 + 3.3} = \frac{3.3}{5.5} = 0.60$
Find the output voltage:
$VV_\text{out} = 9.0\ \text{V} \times 0.60 = 5.4\ \text{V}$

Common Mistake

Putting the wrong resistor on top

Do not automatically put the lower resistor in the numerator unless the output is actually measured across it. If $VoutV_\text{out}$ is across $R_1$ , then $R_1$ goes in the numerator instead.

Adjustable potential dividers

A variable resistor has a resistance that can be changed. A potentiometer is a three-terminal variable resistor used as an adjustable potential divider. The middle contact, called the wiper, slides along the resistive track.

If a potentiometer is connected across a supply, the wiper can provide any output voltage from nearly zero up to the supply voltage.

Example

Using a potentiometer position

A 5.0 V supply is connected across a uniform potentiometer. The wiper is 40% of the way along the track from the 0 V end. Find the output voltage measured from the wiper to 0 V.

Treat the lower part of the track as 40% of the total resistance:
$RlowerRtotal=0.40\frac{R_\text{lower}}{R_\text{total}} = 0.40$
Use the same fraction for the voltage share:
$V×0.40V_\text{out} = 5.0\ \text{V} \times 0.40$
Calculate the output:
$VV_\text{out} = 2.0\ \text{V}$

Tip

End-point check

If the resistor across the output becomes very small, $VoutV_\text{out}$ should tend towards zero. If it becomes much larger than the other resistor, $VoutV_\text{out}$ should tend towards the supply voltage.

Sensors in potential dividers

A sensor is a component whose electrical properties change when its environment changes. In this topic, the important property is resistance.

Definition

LDR and NTC thermistor

A light-dependent resistor, LDR, has a resistance that decreases as light intensity increases. An NTC thermistor has a resistance that decreases as temperature increases; NTC means negative temperature coefficient.

Placing a sensor in a potential divider turns a change in light intensity or temperature into a change in output voltage. This is useful because electronic systems often respond to voltage signals.

LDR and thermistor potential divider circuits showing decreasing resistance trends

If the sensor is the lower resistor and $VoutV_\text{out}$ is measured across it:

Vout=VsRsensorRfixed+RsensorV_\text{out} = V_s\frac{R_\text{sensor}}{R_\text{fixed} + R_\text{sensor}}

So if the sensor resistance decreases, $VoutV_\text{out}$ decreases.

If the sensor is the top resistor and $VoutV_\text{out}$ is measured across a fixed lower resistor, a decrease in sensor resistance makes $VoutV_\text{out}$ increase instead.

Example

Finding how light changes the output

An LDR is used as the lower resistor in a potential divider with a 6.0 V supply. The fixed top resistor is 4.7 kΩ. The LDR resistance is 20 kΩ in dim light and 1.0 kΩ in bright light. Find $VoutV_\text{out}$ in each case.

Since the output is across the LDR, put the LDR resistance in the numerator:
$Vout=VsRLDRRfixed+RLDRV_\text{out} = V_s\frac{R_\text{LDR}}{R_\text{fixed} + R_\text{LDR}}$
Calculate the dim-light output:
$VV_\text{out} = 6.0\ \text{V} \times \frac{20}{4.7 + 20} = 4.9\ \text{V}$
Calculate the bright-light output:
$VV_\text{out} = 6.0\ \text{V} \times \frac{1.0}{4.7 + 1.0} = 1.1\ \text{V}$
Compare the outputs: as light intensity increases, the LDR resistance decreases, so the output voltage decreases.

Common Mistake

Assuming sensor output always increases

For an LDR or NTC thermistor, the resistance decreases when the input increases. Whether $VoutV_\text{out}$ increases or decreases depends on whether the sensor is above or below the output point.

Designing and calibrating sensor circuits

Definition

Sensitivity

Sensitivity means how much the output voltage changes for a given change in the physical quantity being measured.

A potential divider is most sensitive when the fixed resistor is similar in size to the sensor resistance in the range you care about. If one resistance is much larger than the other, the output may be stuck close to zero or close to the supply voltage.

In practical work, you might vary light intensity or temperature and measure $VoutV_\text{out}$ with a voltmeter or data logger. A calibration graph of output voltage against the measured quantity lets you convert voltage readings into physical measurements.

Example

Choosing a fixed resistor

An NTC thermistor is used as the lower resistor in a potential divider with a 5.0 V supply. At the required switching temperature, its resistance is 3.0 kΩ. You want $VoutV_\text{out}$ to be 2.0 V at this temperature. Find the fixed resistor needed above the thermistor.

Start with the divider equation:
$Vout=VsRTRfixed+RTV_\text{out} = V_s\frac{R_T}{R_\text{fixed} + R_T}$
Rearrange for the fixed resistor:
$Rfixed=RT(VsVout−1)R_\text{fixed} = R_T\left(\frac{V_s}{V_\text{out}} - 1\right)$
Substitute the values:
$kΩ(5.02.0−1)R_\text{fixed} = 3.0\ \text{k}\Omega \left(\frac{5.0}{2.0} - 1\right)$
Calculate the result:
$kΩR_\text{fixed} = 4.5\ \text{k}\Omega$

A nearby standard value such as 4.7 kΩ would usually be chosen.

Loading the potential divider

The simple divider equation assumes that the output is measured by a device with very high resistance, so almost no current leaves the divider at the output point.

A load is any component connected to the output terminals that draws current. If the load has a low resistance, it is effectively in parallel with the output resistor, changing the circuit.

Common Mistake

The divider formula assumes no significant load

If a load is connected across the output, first combine the load resistance in parallel with the output resistor. Then use the potential divider equation with this new effective resistance.

Example

Allowing for a load

A 5.0 V supply is connected across two 10 kΩ resistors, with $VoutV_\text{out}$ across the lower resistor. A 10 kΩ load is then connected across the output. Find the loaded output voltage.

Without the load, equal resistors would share the voltage equally, giving 2.5 V.
The load is in parallel with the lower 10 kΩ resistor. Two equal 10 kΩ resistors in parallel have an effective resistance of 5.0 kΩ.
Use the divider equation with the new lower resistance:
$kΩV_\text{out} = 5.0\ \text{V} \times \frac{5.0\ \text{k}\Omega}{10\ \text{k}\Omega + 5.0\ \text{k}\Omega}$
Calculate the loaded output:
$VV_\text{out} = 1.7\ \text{V}$

The load has pulled the output voltage down significantly.

Exam technique

In the exam

Mark exactly where $VoutV_\text{out}$ is measured before choosing the resistor for the numerator.
Keep resistance units consistent; kΩ can be used directly in ratios because the units cancel.
For LDRs and NTC thermistors, decide whether the sensor resistance increases or decreases, then apply the divider equation.
If a load is connected across the output, combine parallel resistances before calculating $VoutV_\text{out}$ .

Self review

Check yourself

If $VoutV_\text{out}$ is measured across the lower resistor, what happens to $VoutV_\text{out}$ when that resistor increases?
Why does an LDR potential divider not always give a higher output in brighter light?
Why should a voltmeter used to measure $VoutV_\text{out}$ have a very high resistance?

Potential dividers and sensors

What you'll learn

Prerequisites: voltage, current and resistance

The potential divider

Adjustable potential dividers

Sensors in potential dividers

Designing and calibrating sensor circuits

Loading the potential divider

Potential dividers and sensors Revision Guide