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I'm totally new to electronics and I wonder why we need to put a resistor in series with a photoresistor to measure the variation of light? I mean, photoresistor is already a resistor, why do we have to decrease the voltage in the circuit with an additional resistor? Thanks in advance for your answers.

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How are you supposed to measure the voltage with only a single resistance? – Ignacio Vazquez-Abrams Jan 31 at 4:37
Because you're making a voltage divider. – brhans Jan 31 at 4:37
The input voltage of the circuit is 5V. If I have a single resistor in the circuit which is a photoresistor I am able to tell you the voltage difference by measuring the voltage between the photoresistor and the ground. Maybe I am missing something but I don't understand. – Moussamoa Jan 31 at 4:51
@Moussamoa If I have a single variable resistor between 5V and ground, does the voltage across it vary? – uint128_t Jan 31 at 5:33
up vote 19 down vote accepted

EDIT: Added example for calculating voltages in a voltage divider

Because if you want to measure the resistance of something, you need to apply voltage to it.
And if you apply voltage, you need to somehow measure that voltage, and by simply measuring between the photoresistor's terminal which is on the \$+5\;V\;(V_{cc})\$ and the terminal which is on \$GND\$, you get exactly \$+5\;V\$, there is no changing voltage, no matter how small or how large the resistance of the photoresistor is.


simulate this circuit – Schematic created using CircuitLab

You measure 5V in the schematic above.

You solve the problem by using a voltage divider:


simulate this circuit

Now you can measure the voltage drop on the resistor, and from that value you can guess the amount of light the photoresistor recieves.


In the second diagram you can see that the voltage is applied across a \$50\;\Omega\$ and a \$100\;\Omega\$ resistance. Because Ohm's law says that \$U=R\cdot I\$ and the current must be equal in a series circuit, the same amount of current flows through \$R_1\$ and \$R_2\$.
In a series circuit, current stays the same, but voltage is shared between the circuits.
We can write down the following equation:

\$U_{R_1}\$ = \$R_1\cdot I\$

You could ask how can we calculate the voltage if we don't know the current.
Well, we don't know the current, but we can calculate it using Ohm's law.
We write down the original Ohm's law equation differently:

\$U=R\cdot I\;\Rightarrow\;I=\frac UR\$

Because in this case the total resistance is \$R_1+R_2\$ (or \$150\;\Omega\$ in our example), the equation for the current will be \$I=\frac{U}{R_1+R_2}\$.

We can use this equation to substitute the single \$I\$ variable in the above mentioned equation.
So the equation for each of the resistors will be:

\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}\$

\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}\$.

If we have \$50\;\Omega\$ on \$R_1\$ and \$100\;\Omega\$ on \$R_2\$, then the voltages on them will be

\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}=50\;\Omega\cdot\frac{5\;V}{50\;\Omega+100\;\Omega}=50\;\Omega\cdot\frac{5\;V}{150\;\Omega}=50\;\Omega\cdot0,0\dot3\;A=1,\dot6\;V\$

\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}=100\;\Omega\cdot\frac{5\;V}{50\;\Omega+100\;\Omega}=100\;\Omega\cdot\frac{5\;V}{150\;\Omega}=100\;\Omega\cdot0,0\dot3\;A=3,\dot3\;V\$.

If \$R_2\$ will change (for example less illumination) and its resistance rises to \$150\;\Omega\$, the voltages will be

\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}=50\;\Omega\cdot\frac{5\;V}{50\;\Omega+150\;\Omega}=50\;\Omega\cdot\frac{5\;V}{200\;\Omega}=50\;\Omega\cdot0,025\;A=1,25\;V\$.

\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}=150\;\Omega\cdot\frac{5\;V}{50\;\Omega+150\;\Omega}=150\;\Omega\cdot\frac{5\;V}{200\;\Omega}=150\;\Omega\cdot0,025\;A=3,75\;V\$.

The more the resistance of the photoresistor rises, the more voltage will drop across it.

If we give the photoresistor more illumination and its resistance falls to \$75\;\Omega\$, then the voltages will be

\$U_{R_1}\$ = \$R_1\cdot\frac{U}{R_1+R_2}=50\;\Omega\cdot\frac{5\;V}{50\;\Omega+75\;\Omega}=50\;\Omega\cdot\frac{5\;V}{125\;\Omega}=50\;\Omega\cdot0,04\;A=2\;V\$

\$U_{R_2}\$ = \$R_2\cdot\frac{U}{R_1+R_2}=75\;\Omega\cdot\frac{5\;V}{50\;\Omega+75\;\Omega}=75\;\Omega\cdot\frac{5\;V}{125\;\Omega}=75\;\Omega\cdot0,04\;A=3\;V\$.

The lesser the resistance of the photoresistor gets, the less voltage will drop across it (and more voltage will drop across the other resistor).

As you can see, we moved from \$3,\dot3\;V\$ to \$3,75\;V\$ when the photoresistor's resistance rised then the voltage dropped to \$3\;V\$ when the resistance fell.

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In the first configuration the power supply and the photoresistor are in parallel. That means the voltage must be the same on both of them. If the photoresistor's resistance is smaller, then the same voltage still applies, with less resistance, it only means higher current. And the other way around. – domenix Jan 31 at 5:06
@Moussamoa the input voltage is effectively fixed, batteries and power supplies will provide whatever current is necessary at a fixed voltage. Alternatively you can use a fixed current source which will vary the voltage to keep the current constant. For dc circuit analysis you often assume that the supply voltage is a constant value regardless of load. In reality batteries and supplies have a current limit beyond which they no longer provide a constant voltage – crasic Jan 31 at 6:46
Why measure the voltage drop over the resistor, not over the photoresistor? – immibis Jan 31 at 10:03
@immibis Because you simply can't measure the varying voltage drop on the photoresistor. You measure a fix voltage drop, which is exactly what the voltage generator provides. It doesn't vary, no matter how hard you try. In the upper diagram I made you could simply move the multimeter's leads to the power supply's terminals. You would measure the same voltage, because those wires in the schematic are ideal, imaginary, zero ohm wires. The other answers explained in other ways why it shouldn't work as intended. – domenix Jan 31 at 14:59
@domenix Measuring the voltage drop over the photoresistor, rather than the resistor. This seems like it should be more convenient, because one terminal is at ground. – immibis Jan 31 at 19:21

It depends how you are using the photoresistor.

If you are using it manually on the bench, to measure light levels, then you only need to connect it to a multimeter on an Ohms range, and measure its resistance.

If you are using it as part of a circuit that responds automatically to light levels, then the circuit has to measure its resistance. There is no way it can do that without additional components. The simplest way to do it is to put another resistor in series and use the voltage at the point where they join.

While it may appear that an Ohms reading multimeter magically measures resistance, internally it has a whole bunch of extra components. On the Ohms range, the most significant of which is a resistor or current source in series with the thing that's being measured. Take a peep at the circuit board inside a multimeter next time you are changing the battery.

A popular way to measure the resistance with a microcontroller like PIC or Arduino is to put the photoresistor between an output pin and an input pin, with a capacitor from the input pin to ground. The output pin is toggled, and the micro counts how many clock cycles pass before the input pin follows. This is effectively using the logic swing on the output pin to define a voltage, and measuring the current into the capacitor as a time to charge up. No resistors here, but you are still using extra components to measure at least one of voltage and current.

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In a normal resistive series circuit, the voltage dropped by the circuit will equal the input voltage. If only one resistor is used, then the entire input voltage is dropped by it. A single Photoresistor will drop 9V if 9V is put across it. Simple Ohms law. V = I * R.

If more than one resistor is used, then the voltage drop is proportional across the resistors, based on their resistance. Resistors in series are a cumulative resistance, they simply add together. Again, ohms law, V = I * (R1 + R2 + Rn)

So a single Photoresistor, who's variable resistance is based on sunlight, will continue to drop the same voltage regardless the resistance. What changes is the current through it. V stays the same, r changes, so I changes.

By adding a fixed resistor, in relation to the photoresistor, you get a variable voltage across the photoresistor. The two resistances vary in proportion to the input voltage, causing a change in the voltage dropped against each. The total voltage drop across the fixed resistor and the photoresistor will be the same, but the actual drop against each will change.

This is the essence of a voltage divider.


simulate this circuit – Schematic created using CircuitLab

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To expand on domenix' great answer...

"Why measure the voltage drop over the resistor, not over the photoresistor?"

In the circuit (the second diagram in domenix' answer) that has a fixed-resistor (R1) in series with the photoresistor (R2), you can measure across either the fixed-resistor or the photoresistor for a voltage change when the light level (intensity) changes on the photoresistor.

The resistance of a photoresistor decreases with increasing light intensity.

This means that as the intensity of the light increases, the voltage you would measure across the photoresistor decreases, and the voltage you would measure across the the fixed-resistor increases.

So, the voltage across the photoresistor changes in the opposite direction as the change in the intensity of the light being detected. This may or may not be what you expect and may or may not be the behavior you want to see.

If you measure the voltage across the fixed-resistor, you will see that the voltage increases as the intensity of the light being detected increases.

Depending on your needs and the other components in your final circuit, you could look at the voltage across either the photoresistor or the fixed-resistor.

Also, keep in mind that if it will help in your circuit, you can swap the position of the photoresistor and the fixed-resistor. Then the voltage at the junction of the photoresistor and the fixed-resistor will increase relative to ground, as the intensity of the light being detected increases.

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