diodes and semiconductors

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timer Asked: Apr 13th, 2017

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four la questions

MORGAN STATE UNIVERSITY DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING EEGR.215 Materials and Devices Spring 2017 Credits – 4 LAB DUE DATE: April 14, 2017 Lab #4: Introduction to Semiconductor Diode Characteristics and Parameter Extraction Objective: This is the first of a group of labs on basic semiconductor components. The student will study semiconductor characteristics (I-V, rectification) and some of their applications, particularly modeling. This particular lab will study diodes. The student will learn how to measure and extract parameters to simulate the diode’s I-V characteristics in ADS. The student will graph the diode’s characteristics in EXCEL. Background: Diodes and transistors are made from semiconducting materials: typically crystalline silicon. Pure silicon, at room temperature, has a finite number of free electrons and holes and, therefore, is quite resistive. To increase its conductivity, the silicon is normally “doped” with other elements. Some dopants, like phosphor, arsenic and antimony, easily give up one of their electrons to the now impure silicon. As these donated electrons are free to move about the silicon, its conductivity increases dramatically. Other dopants, like boron, indium and aluminum, when substituted for silicon atoms, possess incomplete bonds. These “dopant” atoms grab or accept electrons from the surrounding silicon atoms, leaving positively charged silicon ions behind. In turn, these now positive silicon ions try to lessen their charge by grabbing electrons from their neighbors…the net result is that there are regions of “positiveness” floating around the crystal lattice. Such “absences of electrons” are called holes. Amazingly, holes behave almost exactly like positively charged electrons; they move, respond to electric fields, and appear to have an effective mass close to that of an electron. A doped semiconductor with more mobile electrons than holes is called an “n-type” semiconductor; conversely, a doped semiconductor with more holes than mobile electrons is called a “p-type” semiconductor. If doping’s only effect was to increase semiconductor conductivity, semiconductors would be obscure, little-used materials. The utility of semiconductors comes from the remarkable effects of placing p and n-type materials next to each other. Such juxtapositions are called “pn” junctions. An isolated pn junction makes a semiconductor diode. Other semiconductor components are made from more complicated arrangements; bipolar npn transistors, for example, are made by sandwiching a p layer in between two n layers, hence the name npn. The current through an ideal pn junction is given by the diode equation. ID(VD)= Isat exp( qVD)/(n k T) - 1 , (1) where VD is the applied voltage drop across the junction, Isat is a constant called the saturation current and depends on the temperature and the particular geometry and material of the junction, q C is the charge of an electron, k J/K is Boltzmann’s constant, and T is the temperature in Kelvin. The constant n, is the ideality factor also known as the emission coefficient that varies between 1 and 2 depending on the particular diode, but is typically equal to 2 for discrete diodes. For an ideal diode n is equal to 1. Notice that the diode’s response is directional and highly nonlinear. When forward biased, (VD positive) enormous currents can flow through the diode because of the exponential dependence of ID on VD. When reversed biased, (VD is negative), the current approaches -Isat . Since Isat is typically very small (picoamps are not uncommon), very little current flows. Thus, the diode acts like a one-way valve; current can only flow in one direction. When forward biased, the positive end of the diode is called the anode, and the negative end is called the cathode. Equipment Needed: Power Supply Resistor (1 kΩ) Silicon Diode (1N4007 or 1N4001) Voltmeter Ammeter ADS Part 1. 1. Using the diode equation, plot the I-V characteristic for a diode having n=1.2, Isat=2*10-9 A and VThermal = 26mV. Limit the current range to 0-50mA. (Note: VThermal is kT/q) 2. For this diode, determine the incremental resistance (rd) at ID = .5, 2.5, and 50mA. The incremental resistance (rd) is VD/ID. 3. If this diode were used in the following circuit ( Figure 1.), and the +10V supply voltage varied by 20%, how much would the output voltage vary? 4. Determine the line regulation of the diode circuit (change in output voltage / change in input voltage). Express the results in percent and mV/V. 5. If the supply voltage is constant, but a 500Ω load is added across the diode, how much will the output voltage drop? 6. Determine the load regulation (change in output voltage / change in load current). Express the results in mV/mA. R1 =2k R= 1kΩ Ammeter + ¯ Vdc Vdc = 10V Figure 1. Vout Vdiode PART 2. Measuring the Diode Characteristics Build the diode circuit shown in Figure 2. Let R = 1kOhms R= 1kΩ Ammeter + ¯ Vdiode Vdc Figure 2. 2. Measure the diode voltage Vdiode (1N4007) as the power supply (Vdc) is varied from 0 to 10 volts using the steps given in Table 1. Record the diode voltage, Vdiode, the resistor voltage, VR, and measure or calculate the diode current, ID, in mA. Table 1 Measurement results for simple diode circuit: Source Volage, Vdc (Volts) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Diode Voltage Vdiode (Volts) Resistor Voltage VR (Volts) Measured ID (mA) 2. Measure the diode voltage Vdiode (1N4007) as the power supply (Vdc) is varied from -10 to 0 volts using the steps given in Table 2. Record the diode voltage, Vdiode, the resistor voltage, VR, and measure or calculate the diode current, ID, in mA. Table 2 Measurement results for simple diode circuit: Negative Vdc values Source Volage, Vdc (Volts) -10.0 -9.0 -8.0 -7.0 -6.0 -5.0 -4.0 -3.0 -2.0 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 Diode Voltage Vdiode (Volts) Resistor Voltage VR (Volts) Measured ID (mA) For each of the voltage steps above, you can check your measured diode current, ID, by measuring the voltage drop across the 1kOhm resistor and calculating ID. Since the value of the resistor is known, the current flowing through the resistor can be easily calculated. This is the same current flowing through the diode. 3. Using Table 1 data, plot the forward I-V characteristic for the diode in EXCEL. This plot will have the Diode Voltage (Vdiode) on the horizontal (x) axis and the Diode Current (ID) on the vertical (y) axis. 4. Using Table 2 data, plot the reverse I-V characteristic for the diode in EXCEL. This plot will have the Diode Voltage (Vdiode) on the horizontal (x) axis and the Diode Current (ID) on the vertical (y) axis. 5. What type of V-I characteristic does the diode have? Linear or non-linear? Explain your answer. PART 3. Diode Parameter Extraction By using graphical techniques, the student will extract the parameters for Isat, Vbi and n using EXCEL. 1. The student must use the ID-Vdiode values from Table 1 and create a semilog plot with the data. The student should double-click on the current-axis of the curve and then select the logarithmic scale. The graph should now be almost linear. Now double-click on the current-axis again and change ‘maximum’ in the scale menu to .001. Now the student should extrapolate the most linear part of the current curve to the y-axis in order to find the yintercept. This value is ln(Isat). (See green dotted line in Figure 1). Determine the saturation current, Isat (also indicated as Is on the curve in Figure 1). Record this value (Isat) on Table 3. 2. After extrapolating and finding the Isat value, the student should then record the value that the linear curve crosses the x-axis. Record this value(Vbi) on Table 3. This is the turn-on voltage, Vbi. 3. In order to find n, the ideality factor, take the slope of the most linear part of the semi-logrithmic part of the curve. This value is equal to q/(nkT). Determine the value of n if T is 300K. Record this value(n) on Table 3. 4. Using the ideal diode equation, solve for ID by using the values of VD from Table 1 . Use the values on Table 3 to obtain the modeled ID values and record on Table 4. Graph the expression of ID vs Vdiode_mod. Calculate the root mean square error between the measured data on Table 1 and the modeled ID data from Table 4. (Note: You have just extracted the all the parameters needed to create a diode model.) 5. Use the ‘Curve-Fitting in Excel.pdf’ file to check your extracted values of Is and n. If there are differences between the extraction and the curve-fitting values, explain the potential sources of error. Record your curve-fitting Is and n on Table 3. Supply a copy of your curve-fitted plot along with your measured data. 5. What is a device model? Why are circuit models used? What are the advantages of using a device model versus the measurements? Table 3. Diode Model Parameters Is (A) Vbi (V) n Table 4. Measurement results for simple diode Diode Voltage Vdiode_mod(Volts) Model ID(mA) Note: Remember to use the diode voltage (VD) values from Table 1 to find Model ID. Root mean square error: ____________________ PART 4. Simulating Diode Model in ADS 1. Build the circuit as seen in Figure 2 in ADS. Make sure all the components are retrieved from the ADS library. The Ammeter can be simulated in ADS as a Probe component. The “I_Probe” component is a device that acts as an ammeter, and it can be found in the “Probe Component” library. 2. To produce the I-V curve with our circuit, we must allow the voltage across the diode, Vd (Refer to Lab #1 on how to name a node), to increment giving a corresponding value for the current through the diode, Id. This can be done by sweeping the input voltage, Vdc from 0 to 10V. The DC simulation component will allow you to sweep this voltage. Refer to Lab #3 (DC Simulation) for instructions on how to sweep DC voltages and how to obtain DC source. Use the same step values for the DC sweep as in Part 2. 3. Obtain a silicon 1N4007 from the Diode library

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