Introduction PNP C NPN C

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1 Introduction JT Transistors: A JT (or any transistor) can be used either as a switch with positions of on or off, or an amplifier that controls its output at all levels in between the extreme on or off positions. And for either a switching or an amplification application, either a PNP or NPN JT can be used. This lab will focus primarily on the NPN JT. Never Points IN Points In Proudly I V NPN I V I V PNP I V Figure 1: JT Schematic Symbols V I V I Figure 2: JT Voltages and urrents Whether used as a switch or in an amplifying circuit, the main idea of the transistor is to control the transistor s relatively large output with a small amount of input. More specifically, the JT allows us to control the large amount of current flowing through the collector-emitter (I) junction with a small amount of base current (I). Often, it is useful to model an important part of these transistors behavior by imagining a diode between the base-emitter () junction in the direction of the schematic symbols arrows. The forward-biasing of this junction is the first thing that needs to happen in order to turn the transistor on. Specifically, this means that for a NPN, V = +0.7V and for a PNP, V = -0.7V. Figure 3 shows an example of a JT circuit. V V R R V I I R V junction Figure 3 I = I + I I base loop collector loop 1

2 Since the two circuit loops share the emitter terminal, this is an example of a specific common emitter circuit, and so the loops are named after the terminals that are unique to each loop. I (very small) is being used to control the current flowing in the collector loop (I I). The base terminal is essentially a valve controlling the current in the collector loop. Its only role is to control I via its input, the much smaller current I. JT as an Amplifier: When a JT is used as an amplifier, for every amp that goes into the base terminal (I), there will be βd times that current flowing out of the collector terminal. βd (or simply β) is known as the D current gain of the JT. Unfortunately, β is difficult to precisely control and varies from JT to JT, within those with the same part number and even the same production lot. It is typically between 100 and 300. This unpredictable current gain can be troublesome for an amplifier, but can be mitigated with methods discussed later in this course. Also, the voltage drop experienced as the collector current flows through the junction (V) is greater than 0.3V. When these things are happening, the JT is said to be in the active region. Active JT (amplifier): I I D V 0.3V JT as a Switch: As you know, a switch has two positions: off and on. To turn the transistor off, all that needs to be done on the base terminal is to cut its current off, then the junction stops conducting. Turning the transistor to the on position means letting nearly all current possible flow through the junction. A JT is a passive device. This means that its output, I, can only be as large as the circuitry in the collector loop allows. Again, the base terminal is a like a valve controlling current through the junction. The only control the transistor has over this current comes from varying how well its junction conducts current (or, thus, varying the junction s resistance). See Figure 4 for an analogous circuit representing the collector loop in Figure 3. I I R V R Figure 4 Junction Resistance collector loop 2

3 To turn this device on, we want to allow the most current possible through the junction. All the transistor can do is to turn its resistance between the and terminals down as low as it can go, until there is less than 0.3V across the junction. Looking at the circuit above, the maximum possible emitter current is going to be determined entirely by V and R, things external to the transistor. I max V R To make this fully on position happen, the transistor s input (base) must have enough current flowing into it so the transistor is demanding more current than the collector can supply. When this happens, the transistor is operating in the saturated region. Saturated JT (switch, on): I I D 0V V 0.3V If there was 1μA flowing into the base terminal, and β was 200, then this would demand that 200μA flow in the collector loop. If V was 5V, and R was 10kΩ, then Imax would be 50μA. The transistor would be saturated, letting nearly all of the 50 possible μa flow through its junction. When it is desired that a transistor be put in saturation under all possible varying conditions, this is called hard saturation. There needs to be enough I so that I Imax for all possible values of β and all possible values of V while saturated (0-0.3V). Junction onduction: In order for the junction to conduct at all, the diode must be forward biased. That is, for an NPN JT, V must be +0.7V. The conductivity of the junction is controlled by I as described earlier. To control I in the circuit in Figure 3, we can manipulate both V and R. I can be found by applying Ohm s law to R, since, assuming the transistor is conducting, the voltage on either side of R is known. 3

4 ollector urves: ollector curves give us visual of how much collector current (y-axis) we can pass through a diode as the diode s collector-emitter voltage varies (x-axis). Figure 5 I (ma) saturation ollector urves (I vs V ) active I = 40μA I = 30μA I = 20μA breakdown 1 1V I = 10μA cutoff breakdown (~40V) V ach curve is for a set input, or base current (I). As we can see from the plot, base current has a definite effect on the transistor s output (I). In order to more directly see the effect the transistor (I) has on its output (I), we will set up a base biased circuit, and then change its input (I). We will record values for the output (I) and then plot I versus I, or output versus input. We will then be able to see the cutoff, active, and saturated regions in a different way. These will not be collector curves, but, simply, a plot of output current versus input current. A base-biased circuit is one that establishes a fixed base current. ecause the current is fixed, we can change the circuit to have controllable values of I. Since D current gain varies from transistor to transistor, we will first measure βd for three different transistors. In the last part of the lab, we will take our highest and lowest values of βd and plot their I versus their I and see how D current gain affects the circuit s operating regions. 4

5 Pre-Lab Unless your instructor says otherwise, it is recommended that you do this lab in a spreadsheet. There are repetitive calculations required and plots will be made from the data. 1) Devise a way to obtain a measured value for D current gain using for the circuit in Figure 6 using only your voltmeter and ohmmeter (no ammeter). State clearly what you intend to measure directly, and what calculation(s) you will do to obtain a measured value for D current gain. V 15V Figure 6 R 1kΩ 2 10µF V 5V 1 R 10µF The capacitors are only there as a precaution against noise. 2) reate xcel tables for your Part A,, and measurements and calculations. 3) reate xcel tables for your Part predictions for I (in µa) and I (in ma) for each of the base resistor R1 through R12. Here, you assume that βd is 200. If you use xcel properly, you can do this by copying and pasting your work from the previous step, and making a few small changes. Remember, I has an upper limit that it reaches when saturation occurs. You can have xcel check for saturation using the IF function. 4) Plot your results for I versus I from steps 3 and 4 on the same plot. Required quipment: Resistors (Ω): R = 1k, R1 = 1M, R2 = 560k, R3 = 180k, R4 = 100k, R5 = 82k, R6 = 68k, R7 = 56k, R8 = 47k, R9 = 39k, R10 = 33k, R11 = 18k, R12 = 10k apacitors (F): 1 = 2 = 10µ (or larger) three 2N3904 transistors one 2N3904 transistor (bad, get from instructor) one 2N3906 transistor DMM Power supply 5

6 Part A: Transistor Measurements 1) Look up the datasheet for your 2N3904 and 2N3906 transistors and sketch their pinouts in your lab book. 2) Use your DMM s diode test mode to obtain the barrier voltages for all possible junctions (pairs of transistor pins) for the 2N3904 and 2N3906 transistors. Record your DMM s results in a table like the one in Figure 7. DMM Measurements of V barrier w/ Diode Tester Diode Good NPN ad NPN Good PNP Figure 7 3) Obtain a bad NPN transistor from your instructor and complete the same tests for this transistor, recording your results in the same table. 4) Using your results for the good NPN transistor, sketch a rough model of what is going on inside the transistor. Label the pin names on your transistor model. Part : Measuring D urrent Gain 1) Assemble the circuit in Figure 6 using R4 and one of your working 2N3904 transistors. 2) Use the method you developed in the Prelab to obtain a measured value for D current gain. 3) Repeat steps 1 and 2 for your other two good 2N3904 transistors. e sure to keep track of which physical transistor had what value of gain. Part : Plotting I vs. I 1) Select the transistor with the lowest value for D current gain from Part. 2) You should still have your circuit built from Figure 6 with R1. Directly measure V. Perform the measurements necessary to obtain measured values I and I without using an ammeter. 6

7 3) Repeat step 2 using each of the remaining base resistors (R1 through R12) as your base resistor. 4) Select the transistor with the highest value for D current gain from Part and repeat Part steps ) reate a plot of I versus I for both of these transistors. Plot both of these data sets on the same graph as your prelab s graph (right click the graph and click Select Data to get started). Indicate which plots are the prelab prediction, and which are your higher and lower beta-valued transistors. 6) Where is each transistor in its active and saturated regions? From your plot, determine the precise location (coordinates) of this border between the two regions. 7) Which of the two transistors saturates more easily? Why? 8) If you needed a higher maximum collector current out of this circuit, what things could you change to make that happen? 7