Physics 17 Lab Manual Planck’s Constant in Semiconductor Diodes Page 1

PLANCK’S CONSTANT IN SEMICONDUCTORDIODES

I. Introduction

This is a new lab. You will be the first Dartmouth students ever to try it out. Needless to say, theroad will not be perfectly smooth. In fact, I’m relying on your experience with the lab to help mesort 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 ann-doped region.

The vertical scale is not quite correct. For realistic materials, the acceptor and donor impurity levelsare about .01 eV. The diodes used in this lab are made from semiconductor compounds with energygaps on the order of 1.9 eV, so the impurity levels are less than 1% of Eg from the band edges. Ona diagram with a realistic vertical scale, the donor impurity levels are indistinguishable from theband edges.

The materials used to make these diodes are very heavily doped. That means the Fermi level EF isessentially at the band edge, and won’t move much as the temperature changes. (Actually, there isconsiderable evidence that these materials are degenerately doped, which means that EF is actuallyslightly inside the conduction (in the n-region) and valence (p-region) bands.)

In equilibrium the Fermi levels in the two regions must line up, so we get an energy diagram likethis:

Physics 17 Lab Manual Planck’s Constant in Semiconductor Diodes Page 2

Note that the contact potential eV0 is essentially the same as Eg . Remember in order to get fromthe 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 kBT ∼ eV0. The fraction of electrons that havethis kind of energy is proportional to

e−eV0/kBT

Since eV0/kBT 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 areduced barrier of height e(V0 − V ) where V is the externally applied voltage. Since the barrier islower, many more electrons have enough thermal energy to get over it, and there can be a substantialcurrent 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 theexcited state (conduction band) to the ground state (valence band). These diodes are called “light-emitting” because the energy given up by the electron as it relaxes comes out as a photon. (Truthsquad: even a “good” LED is lousy at this. Between 1% and 10% of the recombination events areradiative ones in which a photon comes out. The rest are non-radiative and just make the diodehot.)

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 emittedand the initial energy of the electron. You will measure the wavelength with a spectrometer.

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Since you have so far only used these spectrometers on atomic transitions, this will be a learningexperience. 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 theemitted photons will come from events in which an electron at the bottom of the conduction bandrecombines with a hole at the top of the valence band. Then hν = Eg and life is simple. But thefollowing things can happen: an electron can make a non-radiative transition to a donor level atthe top of the gap before dropping to the valence band. Or, it can drop (radiatively) to a donorlevel and then make a non-radiative transition the rest of the way to the valence band. Or it can doboth. All of these processes result in the emission of photons with hν slightly less than Eg (andtherefore slightly longer wavelength than “gap photons” with hν = Eg). It is also possible forthe 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 boththe electron and hole will be hot, making the photon even more energetic. The figure below is anattempt to illustrate these possible processes.

All these processes should be rare in comparison to recombination with hν = Eg , but when youlook through the spectrometer, it will be extremely difficult to judge the relative intensity of theemitted colors. You may have to resort to trying to match the color observed when directly viewingthe LED.

Having measured λ, if you knew Eg you could find h. Eg you determine from the I -V characteristicof the diode: measure the current through the diode as a function of the foward bias voltage. Asyou increase V , the current increases exponentially. This is because you are lowering the barrierheight for going from the n-region to the p-region, so the fraction of electrons that can get over isincreasing as

ee(V0−V )/kBT .

In other words, the forward voltage will eventually saturate at some value Vs. If you want to increaseV further, even by some tiny amount δV , the increase in current required will be enourmous—morethan your power supply can provide, and certainly more than the diode can tolerate.

When does this saturation occur? When the barrier height e(V0 − V ) has been reduced to the orderof kBT . Since kBT ∼ .03 eV, and since V0 ' Eg , the saturation will occur for V ' Eg . You won’tbe 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 lowervoltages and currents and try to estimate the saturation voltage Vs .

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III. Procedure

The device has six diodes and a current limiting resistor. To investigate, for example, Diode #1connect the positive terminal of the power supply to the left-most red terminal. Connect the negativeterminal 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 multimetereither on their way out of the power supply or on their way back to it. To measure the bias voltageapplied to the diode, connect the multimeter across the red terminal (for that diode) and the yellowone. 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 throughthe 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 exponentialincrease at lower voltages, and a steeper one at higher voltages. This second, steeper behavior isthe one you want. The lower voltage behavior is the evidence (mentioned above somewhere) thatthese diodes are degenerately doped. This current is associated with electrons tunneling directlyfrom the conduction band to the valence band.

What value for Planck’s constant do you get? With what uncertainty? Is this more or less accuratethan the photoelectric effect set-up?