Tutorial 3 - Photodiode

Any junction between an n-type material and a p-type material can act as a photodiode.  Indeed the junctions of any diode can be affected by photons of sufficient energy.  Diode junctions are usually encapsulated in opaque materials to prevent light getting at the junction.  However junctions can still be affected by X-ray and gamma ray photons.

 

A photodiode looks like this:

 

This particular one belonged to a defunct hard disc video recorder (HDD recorder).  Its purpose was to pick up the infra-red signals emitted by a hand-held remote controller.

 

The symbol for a photodiode is:

 

If the depletion zone of a junction is struck by a photon of sufficient energy, an electron-hole pair is generated. 

 

We will look at:

 

Photovoltaic Mode

The photons transfer energy to the semiconductor which results in electrons from the valence bands getting sufficient energy to jump to the conduction band.  They are free and are repelled by the electromagnetic force in the depletion zone.  They move to the n-type material.  When the electrons are removed, a hole is left, and this moves towards the p-type material as a result of the repulsive electromagnetic force in the depletion zone.  The field made by the depletion layer moves the holes towards the anode and the electron towards the cathode.  The electric field accelerates the electron, so they have a high drift velocity.  They cross the p-n junction without combining with the atoms there.

 

If the junction is zero biased, a small voltage is generated.  If connected to an outside circuit, a current will flow.  This is called the photovoltaic effect and can be used in a photovoltaic cell.  A photovoltaic cell uses this idea with large area p-n junctions.

 

The advantage of using the photodiode in photovoltaic mode is that the dark current is very low.  The disadvantage is that the response time is high. 

 

A photovoltaic effect can be observed by shining bright white light onto a yellow, orange, or green LED.  It doesn't work with a red LED.

 

 

Photoconductive Mode

The photodiode is always used in reverse bias.  When there is a reverse voltage on the photodiode, there is always a small current that passes through the diode.  It is independent of the light level and occurs even in the dark.  It is called the dark current, and often given the Physics code Il. The dark current has a typical value of about 0.5 mA (5 10-7 A).  When the reverse-biased photodiode is exposed to light there is a reverse current or photocurrent that depends on the light intensity, NOT the reverse biased voltage.  Therefore:

 

Total current = Dark current + photocurrent

 

 The dark current needs to be small for the photodiode to be sufficiently sensitive.

 

The pn-junction photodiode has now mostly been superseded by the PIN-junction photodiode.  Instead of just a p-type material and an n-type material, there is a piece of intrinsic semiconductor material placed between the p material and the n material like this:

While the n-type material has doping of Group 5 elements, and the p-type material has doping of Group 3 elements, the i-type material is the pure Group 4 semiconductor with no doping.  Therefore there are very few charge carriers in the conduction band.  Under reverse bias, n-type material is connected to the positive, and the p-type material is connected to the negative.  Therefore the holes will go to the negative end in the p-type material, attracted by the electrons.  The electrons in the n-type material will go to the positive end of the n-type material, attracted by the deficiency of electrons, like this:

 

This makes the depletion zone very wide indeed.  The majority carriers (electrons in the n-type and holes in the p-type) will not carry any current under these circumstances.  Minority carriers (electrons in the p-type and holes in the n-type) will carry a small electric current due to the externally applied electric field.  This is the dark current.

 

When photons of sufficient energy are applied to the depletion layer, electron-hole pairs are generated.  Due to repulsion within the depletion layer, the electrons go to the n-type material and the holes go to the p-type material.  Like the pn-junction photodiode the total current is:

 

Total current = Dark current + photocurrent

 

This might seem much more involved than the simple pn-junction, but it ensure that there are many more minority carriers to take the current.  There are advantages to this:

 

A further development is the avalanche photodiode.  A larger reverse-biased voltage gives more energy to the electron-hole pairs. 

 

  The higher voltage does NOT increase the number of electron-hole pairs.

 

 

 

 

The increased energy leads to a higher drift velocity, which will result in electrons being knocked off as the fast-moving free electrons collide with atoms.  These too are accelerated by the reverse voltage, and collide with other atoms, knocking off yet more electrons.  The result of this is an avalanche of electrons.  A snow avalanche is a useful way to think this through.  A small shock is applied to an unstable snowfield on a slope.  Bits of snow are dislodge, which in turn dislodge more snow.  Eventually thousands of tonnes of snow hurtle down the hillside and can do a lot of damage.  The advantages of this are:

The disadvantage is that there is a lot of noise (unwanted signals).

 

 

Measurements on Photodiodes

Like all electronic components, photodiodes have a certain number of operational parameters, or measurements that can be made:

The measurements can be made using a simple circuit like this:

 

 

Dark Current is the leakage current caused by the reverse-bias electric field when there is no light.  The number of free charge carriers can be increased with temperature.  This is because electrons acquire sufficient energy to go into the conduction band with a higher temperature.

 

Note that the vertical axis is not a linear scale, but goes up in decades. It is a logarithmic scale.  The dark current increases 1000 times as the temperature increases from 20 oC to 90 oC.  The implication of this is that photodiodes need to be used in a stable environment where the temperature is constant.  The slope of the graph will depend on what materials are used for doping the p-material and the n-material.

 

Be careful in the way that you read logarithmic scales. 

 

The point that is half-way between 0.1 and 1.0 is NOT 0.5; it is 0.316. 

 

Similarly the half way point between 1 and 10 is 3.16.

 

 

 

Responsivity is defined as the ratio of the photocurrent (A) to the power (W) of the light that is striking the photodiode.  If we know the intensity (brightness) of the light, we can work out the power by:

 

Responsivity has the physics code Rl and the units are A W-1.  The formula is:

 

 

Question 1

A photodiode passes a total current of 60 mA when light of intensity 10 W m-2  is shone on it.  The dark current is 1.5 mA.  The photodiode has a circular window that is 10 mm in diameter.

Calculate:

(a) the photocurrent;

(b) the power of the light;

(c) the responsivity.

Answer

 

The quantum efficiency reflects the number of electron hole pairs generated when photons land on the photodiode.  It has the physics code h ("eta" - a Greek letter long 'ē').  A typical infra-red photodiode has a quantum efficiency of 70 % at a wavelength of 950 nm.  You may need to revise what you did on photon energy in Particle and Quantum Physics.

 

Question 2

Use your answer to Question 1 b to calculate the number of photons falling on the photodiode every second, assuming the wavelength is 510 nm.

Hence work out the quantum efficiency, using your answer to Question 1 a.

 

Planck's constant, h = 6.63 10-34 J s

Electronic charge, e = 1.6 10-19 C

Answer

 

The typical response time of a PIN photodiode are in the order of nanoseconds (1 ns = 1 10-9 s).  For a typical IR photodiode, the rise time is 2.5 ns and the fall time is 2.5 ns.

 

Question 3

The rise and fall time of a photodiode are both 2.5 ns, and the diode conducts for 1 ns between the rise and the fall.  Calculate the frequency of the input signal.

Answer

 

Photodiode Response to Light

The response to wavelength of a photodiode is an important consideration for the electronic engineer.  If a photodiode is designed to be used in the infra-red region, it will not work at all if visible light is used.  Photodiodes have a range of wavelengths to which they are sensitive, and a wavelength to which they are most sensitive.  A typical example is shown on the graph:

 

 

The maximum sensitivity for this diode is about 950 nm.  This is the in the infra-red region.  Photodiodes are available for visible light as well.

 

Question 4

The maximum quantum efficiency of the photodiode that gave the graph above is 70 %.  Calculate the quantum efficiency if the wavelength is 1000 nm

Answer

 

We can plot a graph of the photocurrent against the light intensity:

 

This graph has decade or logarithmic scales on both the horizontal and vertical axes. 

 

Be careful in the way that you read these. 

 

The point that is half-way between 0.1 and 1.0 is NOT 0.5; it is 0.316. 

 

Similarly the half way point between 1 and 10 is 3.16.

 

If we measure the photocurrent against the reverse voltage at different light levels we get a graph like this:

 

 

Notice that the photocurrent increases with increasing light level.  The light level is in this case measured in lux, where 1 lux = 1 lumen per square metre.  It is not possible to do a direct conversion from lux into watts per square metre as there is a different conversion factor for every different wavelength.  The human eye has a maximum sensitivity to green light at a wavelength of 555 nm, at which point the conversion factor is:

1 W m-2 = 683 lux

 

Question 5

What is 2500 lux in W m-2 at 555 nm, green light?

Answer

 

For any given light level, the photocurrent does not increase that much as the reverse voltage increases.  This is because at a given light level, a maximum amount of electron-hole pairs are released, so the photocurrent is saturated.  An increased light level (brightness) more charge carriers are released.

 

A photodiode in forward bias can act as a very inefficient LED.  (Turn the voltage up high enough and there is a very bright flash and a puff of smoke. J)

 

 

 

How the Photodiode is used

Notice from the graph above that in forward bias, the photodiode acts as a normal diode.  It conducts as a forward biased diode at about 0.7 V.  Light levels make little difference in forward bias.  Therefore the photodiode is always used in reverse bias.  The circuit below is the very simplest way of using the photodiode:

 

The resistor R is there to limit the current through the photodiode which can only take a limited current.  It also provides a voltage which depends on the current.  So Vout is proportional to the total current (which is the sum of the photocurrent and the dark current).  To work out the value of resistor you need, you need to know what the light intensity will be and have a graph that will give you the current at that intensity.  We have seen this graph before:

 

Question 6

A photodiode is designed to work in a light intensity of 10 W m-2.  It is connected to a 3 V supply and a resistor.  When the reverse current is flowing, a voltage of 0.50 V is measured across the diode.  Calculate the value of the resistor.

Answer

  

In the Exam

The kind of question you are most likely to asked about the photodiode is one that asks you to lift data off a graph.  The sketch graphs I have drawn do not show the minor lines spaced out in a logarithmic manner.  In the exam, they will be.  You will get a mark for the correct reading of the graph.  A wrong reading will be penalised there, but the error will be carried forward for correct use of the incorrect value.

 

It is quite likely that you will be asked about the photodiode in the context of a wider circuit, like this:

This circuit contains an operational amplifier (op-amp) which we will look at later.

 

You will probably be asked a 1 mark question for a use of a photodiode.  Some are given below, but it's not an exhaustive list.

 

 

Uses for a Photodiode

There are many uses for photodiodes, including:

 

The diagram shows the idea for a fibre optic link:

The LED transmits pulses in the infra red region as optical fibre glass is particularly transparent to infra-red wavelengths.  An infra-red photodiode is used on the receiver.

 

 

Photodiodes with Scintillators

When a high energy particle such as a beta particle or gamma ray photon interacts with certain crystals, a flash of light is given off.  This is called scintillation.  A photomultiplier tube was often used to detect these flashes.  Nowadays photodiodes are more likely to be used to detect the flashes in the scintillators.  Here is a scintillator with a photodiode:

They are used where:

These types of photodiode detectors are not much good with low energy particles as the noise (unwanted signals) can swamp the scintillation signals.