Part I Diodes Purpose PHYS 3152 Methods of Experimental Physics I E2. In this experiment, you will investigate the current-voltage characteristic of a semiconductor diode and examine the applications of the diode for DC power supplies. Equipment and components Two DMMs, signal generator, oscilloscope, adjustable DC power supply. Diode [1N4001 (Si, rectifier) x4 ], ten-turn 1 kω potentiometer, 1 kω resistors (x3), 470Ω resistor, capacitors (22µF, 220 µf). ackground A simplified model for the I-V characteristics of an ideal semiconductor diode relates the current and voltage by qva kt I = I 0(e -1) (1) where I o is the reverse saturation current, V A is the applied voltage across the diode and is positive for the forward bias condition, T is the temperature (in Kelvin), k is oltzmann s 19 constant, and q = 1.602 10 C is the charge for an electron (magnitude only). I ideal diode I = I o qv A kt ( e 1) I o V A Figure 1 The current characteristics of an ideal diode Figure 1 shows the characteristics of an ideal diode under both forward and reverse bias. In forward bias the current increases exponentially, and in reverse bias the current is relatively constant and is equal to I 0. At room temperature (T 300 K), we have V A 0.0259 qva VA = e >> 1 k T 0.0259 V (2) for positive applied voltages (forward bias) much larger than 25.9mV. E2-1/10
Consequently, Equation (1) can be approximated by I qva kt Ie (3) 0 In reverse bias (V A <0) the current is negative and, at room temperature (~300 K), modest values of V A result in the exponential term being much smaller than unity. For V A = -0.1 V the exponential term has the value 1/48 at room temperature and the current is within 2% of I o for the ideal diode. Consequently, Equation (2) becomes I = -I reverse bias V < 0 (4) o A At larger negative values of V A, the approximation of Equation (4) is even better. Hence, under reverse bias conditions the current does not change much with the applied voltage; it is said to saturate and hence I o is termed the reverse saturation current. Less than ideal diodes For real diodes, the current versus applied voltage curve deviates from the ideal case. This deviation is dependent on the applied voltage. To quantify the behavior of real diodes, the ideal diode equation (Eq. 1) is modified to include an ideality factor in the exponent. For a real diode, qva n kt I = I 0(e -1) (6) where n is the ideality factor and is equal to one for an ideal diode. Again, under sufficient forward bias, this expression can be written I I e in forward bias V >> 0 (7) qva n kt 0 A Taking the natural log of this expression yields, q ln I = V A + ln I0 n k T Equation (8) shows that under sufficient forward bias a semi-log plot of current versus applied voltage for real diodes is a straight line. The ideality factor n is not constant for all forward bias voltages but changes depending on the forward bias voltage and the characteristics of real diodes. For many diodes, at low forward currents n 1, and at high currents n 2. Also apparent in many diodes is a region at high currents where the resistance of the bulk diode material decreases with the current in such a way that a specific value of n cannot be determined. Figures 2 and 3 show the semi-log plots of the I-V curve for real diodes with the listed values of n being obtained from the least squares fitting. In many cases, Equation (8) can be further simplified as ln I V /V + ln I (9) A 0 0 where V o is used to characterize the forward voltage drop, above which the diode can be treated as a good approximation to an ideal one-way conductor. (8) E2-2/10
ln I n = 1.69 ln I O n = 1.06 V A Figure 2 Plot of ln I as a function of the forward bias V A for a real diode ln I n=1.06 Figure 3 Plot of ln I as a function of the forward bias V A for another real diode V A E2-3/10
Procedure Exercise 1 I-V curve of a diode Construct the circuit shown in Fig. 4 using the ten-turn 1 kω potentiometer, the 1 kω resistors and the DC power supply. Attach one multimeter so that it measures the voltage drop across the diode (1N4001) and attach another multimeter across R2 so that the current through the diode can be monitored (you need to measure the exact resistance of R2). Calculate the maximum applied voltage V max of the power supply. NOTE: The power dissipated in the resistor R1 (P = V 2 / R1), do not let the power dissipated in the resistor exceed its power rating. Adjust the voltage across the diode by turning the ten-turn 1 kω potentiometer and figure out what is the maximum forward and reverse voltages obtainable from the circuit shown in Fig. 4 for the maximum applied voltage V max. Measure the voltage across the diode and the current through the diode in both the forward and reverse bias conditions. Plot the current (y-axis) versus the bias voltage on linear scales. Plot the natural logarithm of the current (ln I) versus the forward bias voltage. Determine I 0, V 0 and the ideality factor. What is the purpose of R2 in the circuit? Figure 4 Circuit used for measuring the I-V characteristic of a diode Exercise 2 The half-wave rectifier Construct a half-wave rectifier using a 1N4001 diode as shown in Figure 5 with R = 470 Ω. In this circuit the oscilloscope monitors the input (ch1) and output (ch2) of the diode. Record the input and output waveforms with 1) no capacitor, ( Measure the peak voltage. Why is it not equal to Vpk-pk/2? ) 2) a 22 µf capacitor, and 3) a 220 µf capacitor in the circuit. Reverse the diode and see if there is any change. Figure 5 The half-wave rectifier circuit E2-4/10
Exercise 3 The full-wave rectifier Now convert the circuit into a full-wave rectifier using four 1N4001 diodes as shown in Figure 6 with R = 470 Ω. A signal generator with a floating output must be used in the experiment. What will happen if the one end of the signal generator is connected to ground by mistake? NOTE: Do NOT attempt to connect the output of the signal generator with an oscilloscope when the signal generator is connected to the circuit. Record the output waveform with 1) no capacitor, (Why is the peak voltage less than the previous circuit? Adjust the setting of the oscilloscope to zoom in the region of the output waveform near zero volt. Explain why there are flat regions.) 2) a 22 µf capacitor, and 3) a 220 µf capacitor in the circuit. Measure the peak-to-peak value of the ripple voltage across the load resistor R. How does the value of RC affect the output ripple? Why? Figure 6 The full-wave rectifier circuit Questions Explain the shapes of the output waveforms in the half-wave and full-wave rectifier circuits. Exercise 4 The diode limiter Construct the diode limiter as shown in Fig. 7. At the input, apply sine, triangle and square waves at different amplitudes. Measure the output with an oscilloscope. Describe what the diode limiter does and how it works. Figure 7 The diode limiter E2-5/10
Part II Transistors 1 Purpose In this experiment, the basic working principle of a bipolar transistor will be investigated and a basic transistor circuit emitter follower will be introduced. Equipment and components Adjustable dual channels DC power supply, ± 12 V power supply (in ACT-1 circuit box), three DMMs, oscilloscope, signal generator. Transistor (2222A), resistors (100 Ω, 3.3 kω, 47 kω and 100 kω) and a 4.7µF capacitor. ackground ipolar Transistor A bipolar transistor has three pieces of semiconductor materials connected in series. It can be thought as two pn junctions connected back to back. There are two major classifications of the transistors: npn and pnp. The npn transistor consists of two layers of n-type silicon sandwiching one p-type layer (see Figure 8a). The pnp is similar to npn, but with two p-type layers sandwiching an n-type layer (see Figure 8b). 8a 8b Figure 8 A diagrammatic representation of npn and pnp transistors The layers and thus the wires connecting them to the 'outside world' are called, respectively, the emitter (E), the base (), and the collector (C). The symbols for npn and pnp transistors are shown in Figure 9 below. The only difference between the two types of transistors is the direction of the arrow for the emitter, which indicates the direction of the emitter current when the transistors are under operation. Figure 9 The standard graphical symbol for npn and pnp transistors E2-6/10
Emitter follower Figure 10 shows an example of an emitter follower. It is called that because the output terminal is the emitter, which follows the input (the base), less by one diode voltage drop: V = V - Voltage drop across diode (typically ~ 0.6V) E At first glance, this circuit may appear useless because the output is almost identical to the input. The important point is that the input impedance is much larger than the output impedance. This means that the circuit draws less current from the signal source to drive a given load (R E ) than would be the case if the signal source were to drive the load directly. Figure 10 An npn emitter follower Input and output impedances of emitter followers To calculate the input and output impedances of the emitter follower: Consider R E to be the load and make a voltage change V at the base; the corresponding change at the emitter is VE = V. Then the change in emitter current is IE = V /RE IE V so I = = using IE = IC + I and IC = β I 1+ β R (1 + β) E The input resistance is V / I, therefore R IN = (1 + β ) R E. The current gain β is typically larger than 100, so a low-impedance load looks like a much higher impedance at the base. Although resistances are used in this derivation, it can be generalized to complex impedances by allowing V, I, etc., to become complex numbers. It can be found that the same transformation rule applies for impedances: Z IN = (1 + β ) Zload A similar calculation can be done to find that the output impedance Z out of an emitter follower (the impedance looking into the emitter) driven from a source of internal impedance Z source is given by Z OUT Zsource = 1 + β (10) E2-7/10
Procedure Exercise 1 Determining npn and pnp transistors You are given a transistor, 2222A. Its physical picture and pin-outs are shown in Fig. 11 below. Figure 11 Pin-outs of a 2222A transistor Use the diode check position on the digital multi-meter provided to determine whether the transistor is an npn or a pnp transistor. Note that the 'VΩ' terminal is a high potential terminal, whereas the 'COM' terminal is a low potential terminal. If the junction under test is forward-biased, the readout shall be the forward voltage of the junction. If the readout is zero that means the junction is short. If the readout is 1 that means the junction is open or reverse-biased. Record your results in the notebook. Is the transistor an npn or a pnp transistor? Exercise 2 Current gain of a transistor Connect the circuit as shown in Fig. 12. Note that the two voltage sources are referenced to ground, as shown in the two boxes with dashed lines. (For the rest of the course, this simplified notation for voltage sources will be used. You need to remember that the voltages are referenced to ground.) The circuit is designed to use the base () current I or voltage V E to control the current flowing into the collector (C). Fix V CE to be 3V and take E as a common ground. The voltage of C is then +3V. To understand how this transistor circuit works, the first step is to figure out what happens to and E. We will ignore C for now and pretend that the connection to C does not exist. In the first approximation, -E works like a diode. We have seen in earlier parts of this lab that for a diode, current increases exponentially with the voltage across it. To prevent the current from becoming too large, we need to put a 100 kω resistor to limit the current. For current to flow, the applied voltage (voltage source 1) must be positive. Now we are ready to consider the third terminal, the collector C. One important property of a transistor is that the current I C is almost proportional to the current I. The proportionality constant, called the current gain β (different for each device), is typically larger than 100. In other words, I C is controlled by what happens in another part of the circuit, namely the base, even when V CE is almost constant. E2-8/10
Figure 12 The circuit control the conductivity between C and E with the base current Vary the output of the voltage source 1 so that I changes from 0 to 0.1mA and record I C, I, and V E. Plot log (I ) & log (I C ) versus V E on the same graph. You can see that the current flowing into C is now controlled by voltage elsewhere, namely V E. ecause both I C and I vary exponentially with V E, one can also control I C by adjusting I. Using the data collected, plot I C versus I. Note that the relationship is almost linear. This linear relationship implies that for a fixed V CE, one can use I to control I C and the response function is a simple linear function. Determine the current gain β. Use a multimeter with the measurement function of transistor h FE (which has the same definition as β) to measure β and compare the value with your previous result. Note that the multimeter measures the h FE value under the test conditions I E = 10µA and V CE = 2.8V. Exercise 3 Emitter follower Connect the emitter follower circuit as shown in Fig. 13. At the input, connect a 1 khz sine wave with amplitude 2V and no dc offset. Use channel 1 of the oscilloscope to measure the input voltage at and channel 2 to measure the output voltage at E (with oscilloscope probes). Record the output waveform and explain what you see. Next, connect V EE (the low voltage end of the resistor R E ) to -12V instead of ground. Compare the amplitude of the input and output sine waves. Figure 13 The emitter follower with V EE at ground E2-9/10
From what you just measured, you might be tempted to conclude that the transistor does nothing to the input signal. Next we are going to see that emitter followers are actually very useful. Figure 14 Circuit for measuring input and output impedance of emitter follower In Fig. 14, the 100 Ω resistor at the base in the previous circuit is replaced by a 47 kω resistor (R source ). The function generator together with the 47 kω resistor represents an ac voltage source of moderately high output impedance, i.e. low current capability. Measure Z OUT, the output impedance of the emitter follower. Measure the V OUT and then connect a 1 kω load through a blocking capacitor to the output (as shown in the dashed box of Fig. 14) and measure V OUT again. From the drop in the output signal amplitude, calculate the output impedance Z OUT of the emitter follower. NOTE: The voltage drop is likely to be small. Measure Z IN, the input impedance of the emitter follower. Remove the 1kΩ load and the blocking capacitor. Now treat the resistor R E as the load. Use the following steps to determine the input impedance: a. Measure the voltages at the two ends of the 47 kω resistor (that is, V IN and V ) with an oscilloscope. Use the MATH function of the oscilloscope, display a function of the voltage difference (V IN - V ) across the 47 kω resistor (R source ) and measure the voltage change v (peak-to-peak value). Deduce the base current change i = v/rsource. b. Measure the change of base voltage V, v (peak-to-peak value). c. Calculate the input impedance Z IN = v / i. Questions Why do you need to put in the blocking capacitor when measuring Z OUT? Work out Equation (10). With values of Z OUT and Z IN, calculate β respectively and compare them to the measured value from the multimeter. E2-10/10