Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Engineering Physics 3F3 Experiment #3 Fabrication and Characterization of a CdS Photoconductor Objectives: Single crystal CdS will be used to make a photoconductor by doping it with indium and attaching ohmic indium contacts. The spectral response and electron lifetime of this CdS:In photoconductor will then be measured. By knowing the gain of the device, it will be possible to determine the value of mobility of the material. Background: CdS is a highly resistive material in its pure form, due to its bandgap of 2.5 eV. This means that not many valence band electrons are excited into the conduction bands at room temperature (kT = 0.026 eV << Eg.). If illuminated, however, electrons may be promoted to the conduction band as was found in Experiment #2. In order to measure the flow of current due to these electrons, one may attach two electrodes to the CdS crystal and attempt to measure current when a voltage is applied across the CdS from an external source. This is known as photocurrent. The best available CdS crystals are only pure to the parts-per-million level. This causes electrons that should be mobile to become trapped at impurities or defects. This quenches the photocurrent, thereby making the material unsuitable for a photoconductor. However, if an ntype dopant is added to the CdS, many electrons exist in the conduction band and the traps will be filled. Additional electrons may now be photoexcited into the conduction band and the current will rise when light is incident on the material. The CdS in this experiment is doped with In, which replaces Cd and acts like an n-type dopant since it has one extra electron in its valence shell. Doping was carried out using diffusion from a gaseous source. If In is heated in nitrogen, it will vaporize and an equilibrium partial pressure of indium will be achieved. The partial pressure may be found from the vapour pressure curve found in Figure 1. Note that in equilibrium, some indium will remain in liquid form. Nitrogen is used to prevent formation of indium oxide. Pressure (Torr) Indium 1.00E+00 1.00E-01 0 1.00E-02 1.00E-03 1.00E-04 1.00E-05 1.00E-06 1.00E-07 1.00E-08 200 400 600 800 1000 1200 1400 Temperature (C) Figure 1. Vapour Pressure of Indium The CdS crystal will become doped when it is placed in this atmosphere containing Indium. Indium atoms in the vapour collide with the CdS surface and may replace a Cd site as they become trapped in the semiconductor. Due to thermal vibrations, exchange of atomic positions will now take place, and the In atoms will penetrate inside the CdS. The depth of indium in CdS is determined by its diffusion coefficient given by: D = 2 × 10 −1 exp − 1.82 (1) kT where kT is measured in eV. The diffusion coefficient, D, has units of cm2/sec and is used in the equation for Fick's Law which defines the flux in particles per unit area per second as: φ=D ∂C (2) ∂x where C is the concentration of diffusing atoms at any point, and ∂C is the concentration ∂x gradient in the x-direction. The surface concentration Cs of In will be, for our purposes, equal to the concentration of In in the vapour. It can be obtained from the ideal gas law, PV= nRT, where P is found from the vapour pressure curve in Figure 1. The depth of diffusion may be characterized by using Fick's second law, namely: 2 dC d C =D (3) 2 dt dx Solving this equation leads to solutions of the form C(x,t) = C s 1 − erf (4) (x) 2 Dt where t is the length of time of diffusion and erf is the error function. This solution is plotted in Figure 2. 1 C C s 0.75 0.5 0.25 0 0 0.5 1 1.5 2 2.5 3 3.5 4 x Dt Figure 2. Penetration curve for unidimensional diffusion into a semi-infinite medium and constant surface concentration Cs By measuring electrical conductivity in CdS, the number of electrons in the conduction band increasing with illumination intensity can be observed. The relationship is linear if the number of electrons excited is small compared to the number of valence electrons excited to the conduction band states available. Each time an electron is excited to the conduction band, a hole is created in the valence band. These holes have very low mobility in CdS and, therefore, contribute little to the flow of current. If a bar of CdS with two ohmic contacts is connected across a voltage source, an electric field is generated in the material and the electrons flow in one direction. As each electron leaves through the positive contact another electron may enter from the negative contact to maintain charge neutrality. Current continues to flow until a hole and electron recombine. By this time, thousands of electrons may have flowed. This results in gain defined by: G = τ /t, where τ is the free electron lifetime and t is the transit time between the electrodes. The gain may also be written as: G= τv τµε L = L = τµV L 2 (5) Here v is the electron velocity, µ is the mobility, ε is the electric field, L is the electrode separation and V is the voltage between the electrodes. The free electron lifetime can be determined by suddenly removing illumination and then observing the decrease in conductivity in a photoconductor. The time it takes, t1/3, for the voltage signal pulse to decay to 1/3 of its peak value, as measured using an oscilloscope, can be used to determine the free electron lifetime, τ, using: V1 / 3 1 = = exp(−t1 / 3 / τ ) (6) V peak 3 Procedure: Pieces of CdS which have already been doped will be given out one to each group. The CdS was doped at 525°C for 1 hour; this is the diffusion time to be used to find the concentration. Indium contacts have been attached to the doped CdS and the sample has been mounted in a sample holder with electrode clips. 1) Measure the electrical resistance of the mounted sample using a digital multimeter, using the attached indium electrodes. Note that indium makes good ohmic contact to CdS. See if room light has an effect on the resistance. 2) The TA will demonstrate the same measurement using an undoped sample. What is the measured resistance, with and without illumination? 3) Measure the doped CdS sample thickness, a common piece from the same crystal will be used. Also note the width of the indium electrodes and the distance between them. 4) Position the CdS sample holder such that light from the monochromator is incident on the entire sample area. Set up the lenses and position the sample as in Experiment 2. 5) Using the bias box, supply a voltage to your sample with a current-sensing resistance circuit as shown in Figure 3. Record the values of Vs and R. Figure 3. Circuit for photoconductor Note that V = iR, and hence i may be determined since R is known and V is measured. 6) Check that V, measured with a digital multimeter, changes when light is incident on the CdS. Then measure V as a function of wavelength every 5 nm from 400 nm to 700 nm. 7) At the wavelength where the maximum V was found, measure the value of V as a function of the illumination intensity. In order to determine intensity, position the lamp dial to 10 settings that give 10 equally spaced readings of brightness as measured using the commercial (Thorlabs) photodetector on the optical bench. Remove the CdS sample to do this and record the voltage out from the commercial detector on the voltmeter. Then replace the sample, and using the same 10 lamp dial settings, measure V for the CdS sample for each setting. 8) Remove everything from the optical bench. Now, mount a 10 cm focal length lens 22 cm from the exit slit of the monochromator and observe that the light focuses at a point 17 cm behind this lens. Place the chopper such that the blade is at this focal point. The motor should be between the lens and the blade. Place a 5 cm focal length lens 25 cm from the first lens and place the CdS sample 8 cm beyond the second lens such that it is fully illuminated. Using an oscilloscope, measure V(t) while the chopper is running. Use the 5 ms/cm time scale and determine the length of time it takes for V to decay to 1/3 of its maximum value. Note that there is a very slow decay with a time constant of several minutes, in addition to the faster decay. Ignore this slow decay in your measurement. Do not change the wavelength. Since the sample holder has a hole which allows some of the light to travel through the back of the holder, this light can be measured by the Thorlabs detector, which has a fast response time, and be used to externally trigger the oscilloscope. 9) Note down the chopper speed in RPM, and the radius of the disc where it intercepts your light beam. Note also the width of the light spot through which the blade passes. Report: 1) Calculate the diffusion coefficient D using (1) for the conditions used in this lab for In in CdS given a temperature of 525°C. Remember to make sure kT is in eV. 2) Determine the equilibrium vapour pressure of In in the furnace during your diffusion from Figure 1 and find the surface concentration, Cs in atoms/cm3 using the ideal gas law. Hint: don’t forget about Avogadro’s number to convert mols to atoms. 3) Graph C vs. x where C is the indium concentration in the CdS, and x is the depth into the CdS. Take your diffusion time to be one hour and use the values from Figure 2. 4) If each indium atom contributes one electron to the conduction band of the CdS, estimate the electric conductivity in (Ω ⋅ cm) −1 and hence the resistance R in Ω of the CdS sample. Note that: i) σ = µNe µ = 340 cm2/V.s, the mobility of CdS N is the electron density (#/cm3), in this case the density of In e is the electron charge ii) R = l / σA ; l = conduction path length, i.e. the separation of the electrodes A = cross-sectional area of conduction path, i.e. the thickness of the sample multiplied by the width of the indium electrodes. 5) Compare this calculated R to your experimentally measured R and discuss sources of error. 6) If we wished to achieve almost uniform In doping throughout the CdS sample, how might we prepare the sample? Calculate the length of time a diffusion would take at 525°C that would achieve this. Limit the variation in C between surface and bulk to ± 10%, i.e. C/Cs = 0.9, and use the corresponding value of x / Dt in Figure 2 and the measured value for the CdS thickness for x to find t. You will find that such a diffusion would take an extremely long time, so in your discussion, comment on using higher temperatures and other methods to achieve this doping. 7) Plot V vs. wavelength, correcting for the tungsten lamp response you found in experiment #2. State the peak wavelength and comment on how Eg as obtained from experiment #2 is relevant here. 8) Plot the conductance 1 of the CdS photoconductor as a function of brightness, noting R that: RCdS = V CdS = V s − V i i 9) Calculate the gain G of the CdS photoconductor for each lamp intensity setting using: G = electrons conducted in one second / number of photons absorbed in one second The numerator is obtained from the voltage measured for the CdS sample in part 7. Divide the voltage by the circuit resistance (including the multimeter in parallel) and e = 1.6 x 10-19 C to convert this measurement to electrons per second. To obtain the denominator, it is necessary to convert the reading in volts from part 7, as measured by the commercial photodetector without the CdS sample in place, into photons per second. To do so, use the detector sensitivity value for the wavelength you used, found in Figure 5 of Experiment 2, and equation (2) from Experiment 2 to convert the voltage into optical watts. Remember that the terminating resistance, Rload, is 50 Ω. To convert to the number of photons per second, simply divide this power in Watts by the photon energy E = hc / λ, in Joules, where λ is the wavelength of the light you are using. Ignore reflection or transmission losses. 10) Now solve for the mobility µ in your CdS for each intensity setting using the gain values you just calculated and equation (5). Use the voltage V measured from the CdS sample, the length L between the electrodes and the free electron lifetime, τ, determined by equation (6) and the voltage pulse decay time (to 1/3 from the peak voltage) measured using the oscilloscope in part 8. How do these values compare with the literature value of µ = 3.4 x 10-2 m2/V s for CdS. Comment on sources of error. 11) A high speed photoconductor that is able to respond very quickly to changing light levels has a gain which is small. Why must this be so? Comment on how you would design a photo-conductive detector for (i) high sensitivity and (ii) high speed. 12) Using data from part 8 and 9 of the procedure show that the value of t1/3 you measured is much larger than the length of time the chopper takes to turn off the light beam incident on the CdS sample. Use the measured disc radius, r, and the chopper speed, ω, of 100 RPM or 1.667 Rev/s, to find the angular velocity of the disc given by: 2πωr. Now use this velocity to find the time required for the disc to block the beam, using the measured width of the beam spot.