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Determination of Boltzmann's Constant from the Current-Voltage Characteristic of a Transistor

Physics 310 – Thermal Physics

Introduction

In this experiment you will measure , Boltzmann's constant, which is the fundamental in thermal physics. The experiment is fairly easy to do, but it requires that you 𝐵𝐵 first become acquainted with some background𝑘𝑘 information about and transistors. The subject of electrical conduction in semiconductors is a complicated and fascinating application of thermal physics (take a glance at Chapter 13 in Kittel and Kroemer)— luckily we will need only the basic storyline and Boltzmann's law to understand this experiment.

Semiconductors

A semiconducting material (for example silicon or germanium) is a weak conductor of electricity. At room , most atoms in the are holding their valence electrons tightly, so these electrons are not free to move about the crystal and to serve as free charges to carry an electrical current. However, a few atoms are ionized at room temperature, contributing a free electron which can help carry a current. The probability that an atom is ionized at a given temperature is proportional to / , as given by the Boltzmann law, where is the ionization energy of the atom in the crystal.−𝐸𝐸𝑔𝑔 When𝑘𝑘𝐵𝐵𝑇𝑇 an atom is ionized, not only is a free electron created, but the ionized atom is left behind𝑒𝑒 as a positive charge. This positive ion can 𝑔𝑔 capture𝐸𝐸 the electron of a neighboring atom, leaving the neighbor ionized. The location of the positive ion can thereby migrate through the crystal: we call this a "hole" and we can treat it as positively-charged analogous to the electron. In a pure semiconductor, the of holes (denoted p) and electrons (denoted n) are equal, because each ionization process creates one electron and one hole.

Doped semiconductors

To make useful devices, one adds impurities, or dopants, to pure semiconductors. Let's take the example of silicon, which has 4 valence electrons. If one adds some arsenic (5 valence electrons) to silicon, the arsenic atom fits into the silicon crystal making bonds with the four neighboring silicon atoms, and holds its remaining electron rather loosely: this "extra" electron is readily freed. Note that the ionized arsenic atom does not function as a hole since it is not inclined to capture a neighboring electron: it is simply a fixed positive charge within the silicon crystal. Arsenic atoms are called donor atoms in silicon because they donate electrons, and arsenic-doped silicon is called an n-type semiconductor. The n-type semiconductor has a higher electron than its hole density, with the large majority of its electrons coming from the donor atoms. Alternatively, one can add some gallium (3 valence electrons) to silicon. The gallium fits into the silicon lattice, attempting to bond to the 4 neighboring silicon atoms, and readily captures the electron from a neighboring silicon atom. In other words, gallium atoms are electron-acceptors which contribute holes to the semiconductor. Gallium-doped silicon has a higher hole density than its electron density, with the large majority of holes coming from the acceptor atoms; it is called a p-type semiconductor. We will use the terms majority and minority carriers to refer to the electrons and holes: in an n-type material, electrons are the majority carriers and holes are the minority carriers; this is reversed in a p-type material.

pn Junctions and

The essential part of a semiconductor device is the junction between the p-type and n-type materials. The simplest device is called a , and consists of a p-type material and an n-type material with a boundary or junction between them. Near the junction, the density of electrons is much higher in the n-type material than in the p-type material, so random thermal motion leads to a net of electrons from the n-material to the p-material. These electrons recombine with holes in the p-type material, leading to a zone depleted of holes in the p-type material near the junction. This depletion zone has a net negative charge, since it still has the fixed negative ions (acceptor atoms) remaining in it. An analogous phenomenon leads to a depletion of electrons in the n-type material near the junction. The depletion zone is thus positive on the n- side and negative on the p-side; so there is an electric field resisting the further diffusion of majority carriers. In order to diffuse across the junction, a majority carrier would need to have enough to overcome the barrier potential Vo and according to Boltzmann, the probability of a given electron at temperature T having this energy is / . However, if an external voltage is applied across the device by means−𝑒𝑒 of𝑉𝑉0 a𝑘𝑘 𝐵𝐵battery,𝑇𝑇 this barrier potential can be changed. If a voltage V is applied with the positive polarity𝑒𝑒 connected to the p- type material, then the probability that an electron will have enough kinetic energy to diffuse across the barrier becomes ( )/ ; (1) the barrier is reduced, a current −can𝑒𝑒 𝑉𝑉0 flow,−𝑉𝑉 𝑘𝑘 𝐵𝐵and𝑇𝑇 the diode is said to be forward biased. If instead the positive polarity were connected𝑒𝑒 to the n-type material, the barrier would be increased, no current could flow, and the diode is said to be reverse biased. The diode is thus a one-way gate for current, allowing current to pass in only one direction.

Figure 1. E = electric field. Bipolar junction transistor

A bipolar junction transistor consists of three layers of doped semiconductor, either npn or pnp. In our experiment, we will use an npn transistor in the common-base mode, where the transistor functions essentially like an ideal diode described earlier. The three layers are called the emitter, base, and collector. With an external battery applying a forward-biasing voltage to the emitter-base junction, electrons from the battery enter the n-type emitter material and flow into the p-type base material. The base region is designed to be thin and weakly doped, so that most of the electrons diffuse through the thin base region, arriving at the base-collector junction where they are swept up the potential hill (since electrons are negatively charged) into the n-type collector. Thus the current of electrons entering the emitter is almost the same as that leaving the collector (a few electrons are lost by recombination with holes in the p-type base material), and this flow of electrons is controlled by the forward bias of the emitter-base junction. The transistor therefore is functioning as an ideal diode controlled by the voltage applied to the emitter-base junction. One might well wonder why we do not simply use a diode for our experiment rather than a transistor—the reason is a bit subtle. In a diode, in addition to the diffusion current of the majority carriers, there are significant contributions from a current of minority carriers and from surface current effects—these contributions are negligible in a transistor connected in the common base configuration, where the simple picture of the diffusion current described above is sufficient.

Figure 2: VM = voltmeter, AM = ammeter, E = emitter, B = base, C = collector.

Procedure

1. Record room temperature.

2. Connect the circuit as shown above. The potentiometer (pot) controls the voltage applied to the base-emitter junction which is measured by the voltmeter. The ammeter measures the 𝑏𝑏𝑏𝑏 collector current . Be sure to try the meters on the most sensitive scales for all ;𝑉𝑉 start the ammeter on the 200 µA scale (the numbers read in µA on this scale, but in mA on the 𝑐𝑐 other mA scales).𝐼𝐼

3. Starting at zero volts, gradually increase until you read a current. At lower voltages, the current is less than 0.1 µA. When you can read a current, record and , and then increase Vbe 𝑏𝑏𝑏𝑏 until approximately doubles and record the𝑉𝑉 new values. Continue increasing in steps that 𝑏𝑏𝑏𝑏 𝑐𝑐 make approximately double and record values, until reaches 𝑉𝑉around 4𝐼𝐼 mA. 𝑐𝑐 𝑏𝑏𝑏𝑏 𝐼𝐼 𝑉𝑉 𝑐𝑐 𝑐𝑐 4. When𝐼𝐼 you are finished, be sure to disconnect the battery𝐼𝐼 from the potentiometer and turn off the meters.

Questions

1. Since the probability that an electron at temperature T will have sufficient kinetic energy to cross the potential barrier is given by the Boltzmann factor (Eq. 1), the current is proportional ( to the same factor: = exp , where is a constant with units of 𝑐𝑐current, is the 𝑒𝑒 𝑉𝑉0−𝑉𝑉𝑏𝑏𝑏𝑏 𝐼𝐼 barrier of the pn junction (emitter and base), and is the voltage applied to 𝐼𝐼𝑐𝑐 𝐼𝐼0 �− 𝑘𝑘𝐵𝐵𝑇𝑇 � 𝐼𝐼0 𝑉𝑉0 the junction. Show that the above relation can be written in the form𝑏𝑏𝑏𝑏 = exp and 𝑉𝑉 𝑒𝑒𝑉𝑉𝑏𝑏𝑏𝑏 find . 𝐼𝐼𝑐𝑐 𝐼𝐼𝑐𝑐0 � 𝑘𝑘𝐵𝐵𝑇𝑇 �

𝑐𝑐0 2. Show𝐼𝐼 that if were plotted vs. ln , the result is expected to be a straight line. Find expressions for the slope and y-intercept of the line, in terms of the variables defined in the 𝑏𝑏𝑏𝑏 𝑐𝑐 previous question.𝑉𝑉 𝐼𝐼

3. Plot vs. ln . Using the numerical value of the slope, determine the value of Boltzmann's constant . Compare your answer with the accepted value (i.e. find the percent difference). 𝑏𝑏𝑏𝑏 𝑐𝑐 𝑉𝑉 𝐼𝐼 𝐵𝐵 4. Using 𝑘𝑘your plot, determine the value of . Can you determine the value of , the junction potential at room temperature? If so, find ; if not, explain why not. 𝐼𝐼𝑐𝑐0 𝑉𝑉0 𝑉𝑉0