Lab Report 2 - Simple Transistor Amplifiers

# Lab Report 2 - Simple Transistor Amplifiers - Scott Smith...

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Scott Smith ENEE306-0104 12 October 2005 Lab Report Lab 2: Simple Transistor Amplifiers 2.1 – Introduction In this lab, transistors are explored as common emitter (CE) and emitter follower (EF) amplifiers. The advantages and disadvantages are explored. The combination of said circuits can offset the disadvantages of each circuit. Starting with the general theory behind all BJTs, we then move on to the CE amp and its implications, then on to the EF amp, then finally the combination of the CE amp cascaded with the EF amp. 2.2 BJT Forward Active Operation, Equivalent Circuit and β 2.2.1 – Theory The NPN transistor, used in the beginning of this lab, is a p-doped region sandwiched by two n-doped regions. The p-doped region is the base and the two n-doped regions are the collector and emitter as shown in figure 1. Figure 2 shows what the 2N3904 and 2N3906 physically look like. N + P N - Base Collector Emitter Figure 1 Figure 2 For the BJT to function as an amplifier, it must be biased into the forward active region. To accomplish this, the base-emitter junction must be forward biased. The equivalent circuit is shown in figure 3.

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Smith, 0104 I B I E I C 2N3904 I B I C I E β I B + V BE - Figure 3 The currents displayed in figure 2 are DC values. The DC currents are related by simple equations. By using Kirchhoff’s Current Law, B C E I I I + = (1) As seen from the collector branch in the equivalent circuit in figure 2, B C I I β = (2) Substituting equation (2) into equation (1), ( ) + = 1 B E I I (3) 2.2.2 – DC Levels and Loop Equations V C V B V E R C R E R B V CC Figure 4 Using Kirchhoff’s Voltage Law on the circuit figure 4, B B CC B R I V V = (4) E E B BE B E R I V V V V = = 7 . 0 (5) 0 7 . 0 = E E B B CC R I R I V (6) 2
Smith, 0104 Since C E I I >> , 1 β , then the following approximation can be made: C E CC C C CC C R I V R I V V = (7) 2.2.3 – Experiment: Determining BJT Current Gain β and Verifying the Equivalent Circuit Step 1: Set up the circuit in figure 5. Let Ω = Ω = Ω = k R R R B C E 200 , 500 , 500 . By measuring the voltages V C , V B , and V B E , and solving the appropriate loop equations, determine I C , I B B , and I E , respectively. Verify that . From the ratio E C I I E C I I , calculate β . Circuit Setup, Measured Values, and Equations: R C R E R B V CC 179 9 . 194 500 039 . 10 = Ω = Ω = = k R R V V B C CC (11) (10) (9) (8) B C B C B E E E C C CC C I I I I I R V I R V V I = = = = Figure 5 Step 2: Change R E to three different values. Make a table of R E , I B , I B C , I E , V BE and β for each value of R E . The data should verify that β is fairly constant and V BE is approximately 0.7 V (one diode drop). R E ( ) I B ( mA ) I C ( mA ) I E ( mA ) V BE ( V ) β V B ( V ) V C ( V ) V E ( V ) 503 0.02334 9.09800 9.35586 0.784 390 5.490 5.490 4.706 456 0.03219 6.58800 6.87939 0.628 205 3.765 6.745 3.137 227 0.03702 8.15600 8.29075 0.942 220 2.824 5.961 1.882 Table 1 The data shows an average V BE of approximately 0.7 V . The calculation of β is off by quite a bit. This is probably due to the accuracy of the oscilloscope used to measure the voltages or the multimeter used to measure the values for the resistors.

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## This note was uploaded on 05/04/2008 for the course ENEE 306 taught by Professor Goldsman during the Spring '04 term at Maryland.

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Lab Report 2 - Simple Transistor Amplifiers - Scott Smith...

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