Understanding basic photoresistor circuit

Question

I'm new to electronics, so I just can't wrap my head around how this photoresistor circuit works.

I understand the following:

1. the photodiode can be understood as a variable resistor (i.e. works in principle the same as a photoresistor)
2. I want to measure the difference in voltage, that is caused by that variable resistor

Intuitively, I'd like to take a voltage measurement before (source: 5V) and after the diode. Can someone explain, maybe using an intuitive analogy (e.g. water current?), why this only works with an additional resistor (like below, copied from here)?

I built a working sensor using the arduino starter kit photoresistor (datasheet) using a 10kOhm resistor and a 5V source but want to understand it better.

Edit: afer helpful comments, this question is basically:
a) what is the difference between a photodiode and -resistor
b) why does a voltage divider need 2 resistors

Answer 1

The HW5P-1 is in fact a phototransistor.

This is the diagram of the example circuit from the datasheet:

The circuit symbol is clearly a phototransistor, regardless of what the poorly translated text says.

A phototransistor and a photoresistor (like a cadmium sulfide cell) both react to light, but they are fundamentally different in how they operate.

A photoresistor is a resistor whose resistance changes in response to light.

A phototransistor actually generates current from light in its base to emitter junction, which allows current to flow through the collector to the emitter. It is in effect a tiny solar cell connected to a single transistor amplifier.

You can use both to detect light, but you should familiarize yourself with how they work.

A short list of differences:

1. Photoresistors are slow. You can't really receive a modulated light signal with them. Phototransistors are much faster, and can pickup light signals modulated with signals that can reach megahertz frequencies.

2. Photoresistors keep changing resistance even after the light is gone. Datasheets specify the "dark resistance" for a time period in absolute darkness for this reason.

3. Photoresistors are usually most sensitive to visible light. Phototransistors are usually most sensitive to infrared. Results can be very different from what you expect if you just replace a photoresistor with a phototransistor without considering that difference.

4. They have different response curves. If you use the same circuit with both, they will react differently to the same light intensity. This might change your threshholds if you are using a simple level detector.

So, really, they aren't the same so don't use the terms interchangeably.

They can carry out similar tasks, but you have to use each as intended. Using one just like the other will lead to bad results.

Answer 2

The datasheet of this device isn't terribly clear. At some points it claims to be a drop-in replacement for a CdS sensor (i.e. a resistor whose resistance varies with light intensity) and at other points it seems to be a photodiode (i.e. a semiconductor device that creates a small current when illuminated with light). Some of the details in the datasheet suggest that it is a photodiode (and not a photoresistor) but I'll include both devices in my answer.

Photodiode (photoconductive mode)

If this device is actually a photodiode, you can use a very similar circuit, but the principle of operation (and some of the math) is different.

The photodiode acts similarly to a normal diode when reverse biased, meaning that it blocks most current and a little bit of current "leaks". By Ohm's law, the output voltage is equal to the leaking current times the resistor (labeled R in the schematic in your question). This means that there is a small output voltage even if the device is kept in complete darkness.

Due to device physics, when light hits the device, it allows more current to pass¹. The extra current causes the output voltage to rise, again due to Ohm's law.

A phototransistor acts similarly on the outside when properly biased (i.e. it allows a bit of leakage current and then much more when illuminated) but the working principle involves some different device physics.

Photodiode (photovoltaic mode)

The photodiode can also be operated like a tiny solar panel--it is not reverse biased with a voltage. This prevents leakage current from introducing background noise, but also makes the current smaller and harder to detect. A special circuit that makes use of an amplifier chip is needed to make such a setup work.

For reasons not discussed here, this circuit is also slower. You probably won't use it unless you have a specialty application that requires it. Additionally, the sensor you have might be unsuitable for this mode, since the manufacturer site mentions "Built-in micro-signal CMOS amplifier" possibly built into the device. The datasheet does not mention this, curiously.

¹ When the diode is reverse biased, a region known as a depletion region is formed. This region contains few electrons and few holes meaning that current has a hard time flowing. Light can create new pairs of electrons and holes, which create current right away.

CdS cell / photoresistor

If we consider this device as a drop-in replacement for a CdS (cadmium sulfide) photoresistor with the circuit you show, we end up with a voltage divider. I'm going to present this first because it's simplest, and a fair assumption for some cases.

^{simulate this circuit – Schematic created using CircuitLab}

Let's assume that the output is hooked up to an Arduino analog pin, meaning that it draws little to no current. Then, the current from the battery is given by \$\frac{5\,\text{V}}{R_1+R_2}\$ because the resistors are in series. By Ohm's law, the output voltage is:

$$ 5\,\text{V} + \frac{R_2}{R_1+R_2}$$

As you can see, this voltage depends on the value of R1, which varies with light.

Answer 3

It's not clear from the datasheet exactly what kind of device you have. I've answered based on the assumption that this is a photodiode or phototransistor.

First, a photodiode (or phototransistor) will act more like a light-controlled current source than like a variable resistor.

Second, consider what happens if you measure the voltage across R rather than across the photodiode.