Physics 17 Lab Manual Planck’s Constant in Semiconductor Diodes Page 1
P LANCK ’ S C ONSTANT IN S EMICONDUCTOR D IODES
I.
Introduction
This is a new lab. You will be the first Dartmouth students ever to try it out. Needless to say, the road will not be perfectly smooth. In fact, I’m relying on your experience with the lab to help me sort out how it all works.
II.
Light Emitting Diodes
Why should we be able to get Planck’s constant out of a diode at all? Consider the diagram below, which shows the valence and conduction bands in a p
-doped region of semiconductor, and in an n
-doped region.
The vertical scale is not quite correct. For realistic materials, the acceptor and donor impurity levels are about .01 eV. The diodes used in this lab are made from semiconductor compounds with energy gaps on the order of 1.9 eV, so the impurity levels are less than 1% of
E g from the band edges. On a diagram with a realistic vertical scale, the donor impurity levels are indistinguishable from the band edges.
The materials used to make these diodes are very heavily doped. That means the Fermi level
E
F is essentially at the band edge, and won’t move much as the temperature changes. (Actually, there is considerable evidence that these materials are degenerately doped, which means that
E
F slightly inside the conduction (in the n
-region) and valence ( p
-region) bands.) is actually
In equilibrium the Fermi levels in the two regions must line up, so we get an energy diagram like this:

Physics 17 Lab Manual Planck’s Constant in Semiconductor Diodes Page 2
Note that the contact potential eV
0 is essentially the same as
E g . Remember in order to get from the n
-region to the p
-region, conduction band electrons need to somehow climb over this barrier, which means they need to have a thermal energy k
B
T ∼ eV
0
. The fraction of electrons that have this kind of energy is proportional to e − eV
0
/k
B
T
Since eV
0
/k
B
T is a large number, this fraction is very small.
To make the diode emit light, we need to forward bias it, as shown below. Electrons now see a reduced barrier of height e(V
0
− V ) where
V is the externally applied voltage. Since the barrier is lower, many more electrons have enough thermal energy to get over it, and there can be a substantial current of electrons from the n
-region into the p
-region.
The light emission comes from what happens to those electrons once they get over into the p
-region.
Since these electrons are a small minority surrounded by what is essentially their “anti-particle,” they will quickly find a hole to recombine with. Energetically, the electron “relaxes” from the excited state (conduction band) to the ground state (valence band). These diodes are called “lightemitting” because the energy given up by the electron as it relaxes comes out as a photon. (Truth squad: even a “good” LED is lousy at this. Between 1% and 10% of the recombination events are
radiative ones in which a photon comes out. The rest are non-radiative and just make the diode hot.)
The life cycle of an electron that recombines radiatively is shown below.
To determine Planck’s constant, we need to know two things: the wavelength of the photon emitted and the initial energy of the electron. You will measure the wavelength with a spectrometer.

Physics 17 Lab Manual Planck’s Constant in Semiconductor Diodes Page 3
Since you have so far only used these spectrometers on atomic transitions, this will be a learning experience. Atomic transitions are sharp. The emission from an LED is not so sharp. Determining
“the” wavelength will require some choices and have some uncertainty.
Here are some things to guide your thinking. Under ideal (for us anyway) circumstances, all the emitted photons will come from events in which an electron at the bottom of the conduction band recombines with a hole at the top of the valence band. Then hν = E g and life is simple. But the following things can happen: an electron can make a non-radiative transition to a donor level at the top of the gap before dropping to the valence band. Or, it can drop (radiatively) to a donor level and then make a non-radiative transition the rest of the way to the valence band. Or it can do both. All of these processes result in the emission of photons with hν slightly less than
E g (and therefore slightly longer wavelength than “gap photons” with hν = E g ). It is also possible for the initial electron to be “hot”—to have an energy well above the bottom of the conduction band.
If it recombines radiatively, then the photon with have a shorter wavelength than a gap photon.
Recombination with “hot” holes well below the valence band is also possible. Occasionally both the electron and hole will be hot, making the photon even more energetic. The figure below is an attempt to illustrate these possible processes.
All these processes should be rare in comparison to recombination with hν = E g , but when you look through the spectrometer, it will be extremely difficult to judge the relative intensity of the emitted colors. You may have to resort to trying to match the color observed when directly viewing the LED.
Having measured
λ
, if you knew
E g you could find h
.
E g you determine from the
I
-
V characteristic of the diode: measure the current through the diode as a function of the foward bias voltage. As you increase
V
, the current increases exponentially. This is because you are lowering the barrier height for going from the n
-region to the p
-region, so the fraction of electrons that can get over is increasing as e e(V
0
− V )/k
B
T .
In other words, the forward voltage will eventually saturate at some value
V s . If you want to increase
V further, even by some tiny amount
δV
, the increase in current required will be enourmous—more than your power supply can provide, and certainly more than the diode can tolerate.
When does this saturation occur? When the barrier height e(V
0 of k
B
T
. Since k
B
T ∼ .
03 eV, and since
V
0
− V ) has been reduced to the order
' E g , the saturation will occur for
V ' E g . You won’t be able to quite get all the way there, because the current through the diode under these “flat band” conditions would permanently cook the diodes. Therefore you have to work with the trend at lower voltages and currents and try to estimate the saturation voltage
V s .

Physics 17 Lab Manual Planck’s Constant in Semiconductor Diodes Page 4
III.
Procedure
The device has six diodes and a current limiting resistor. To investigate, for example, Diode #1 connect the positive terminal of the power supply to the left-most red terminal. Connect the negative terminal of the power supply to the GREEN connector. Getting that connection right is important!
To measure the current through the diode, the electrons have to pass through a Fluke 45 multimeter either on their way out of the power supply or on their way back to it. To measure the bias voltage applied to the diode, connect the multimeter across the red terminal (for that diode) and the yellow one. The back of the device is open making it easy for you to see how the connections go.
My expectation is that you will get a better determination of
λ if you try and keep the current through the diode to a minimum (in that part of the experiment). That should minimize hot electron effects, and make all the unwanted effects relatively fainter. But I could be wrong.
IV.
Analysis
You will not find a single exponential for the current. Rather you will find a slow exponential increase at lower voltages, and a steeper one at higher voltages. This second, steeper behavior is the one you want. The lower voltage behavior is the evidence (mentioned above somewhere) that these diodes are degenerately doped. This current is associated with electrons tunneling directly from the conduction band to the valence band.
What value for Planck’s constant do you get? With what uncertainty? Is this more or less accurate than the photoelectric effect set-up?