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NMT EE 321L - EE321 – Lab 11 Bipolar Junction Transistors, Part II

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EE 321 Lab 11 Fall 2004EE321 – Lab 11Bipolar Junction Transistors, Part IIIn this lab we will investigate two bipolar junction transistor (BJT) amplifiers.The Common Emitter AmplifierThe amplifier circuit of last week’s lab is impractical because the bias or operating pointdepends on the β of the transistor. The following bias technique controls the voltage across REand hence the emitter current IE. It is similar to Example 5.10 of Sedra and Smith.1. Construct the circuit in Figure 1 and measure the DC bias voltages at the base, emitter andcollector. From your measurements, determine VBEand VBC. Is the transistor biased in itsactive mode? What are the DC (Q ‘quiescent’) values of ICand IE?2. Apply a triangle wave to the input and measure the voltage gain. Is the gain close to thatfound in the prelab?3. The voltage gain can be increased by decreasing RE. Reduce REto zero for signal frequenciesby ‘bypassing’ it with a large capacitor (100 µF) in parallel with RE. (Be sure to observe thecorrect polarity of the electrolytic capacitor.) Reduce the input amplitude and look at theoutput waveform.Further reduce the input amplitude until the output is approximately linear and measure thegain. The gain has been increased to the large values, but at what expense?VinO2N39041uFVCCVC182K10K7.5K1KQ1RERC= 15VFigure 1.VinOV2N3904REQ2VEE= −15VVCC= +15V3.3K270RSFigure 2.The Emitter Follower Amplifier4. When the input signal is applied to the base and the output is taken at the emitter (Figure 2)the amplifier is called an emitter follower. This is because the emitter voltage ‘follows’ thebase voltage. Emitter follower amplifiers are useful because they have a high input impedanceand a low output impedance. They amplify the signal power by increasing its current, andare therefore sometimes called a ‘current booster’.Construct the circuit in Figure 2.1EE 321 Lab 11 Fall 2004• With zero (ground input) or small input signal, measure and note the DC bias voltages.• With an input, measure the voltage gain (magnitude and sign).• The dynamic range of the output is the largest that it can be without saturating orclipping. Measure the dynamic range of the output voltage (decrease the supply voltagesto ±12 and measure both voutlimits). What causes the output to limit in each case (whenis the transistor saturated, active , cutoff)?• What would happen to voutif the emitter resistor were connected to ground instead of-15V? Try it. What does this mean?5. Measure Zinat the base of the transistor in the follow ing way.• Increase the source resistance RSuntil voutis reduced by about 1/2 (or by some othervalue).• Sketch the equivalent circuit (use an equivalent circuit of an amplifier from Chapter 1,not the transistor model) and compute Zinfrom your meas urement.• From the theoretical result that Zin= (β + 1)[re+ RE] and re= α/gm, estimate the βof your transistor.6. Simulate an input source with 10 kΩ source resistance by setting RS= 10 kΩ (see Figure 3).Measure the output impedance Zoutby loading the output with about 220 Ω resistance. Use aDC blocking capacitor as shown to prevent the load from changing the bias voltages. Ensurethat the capacitive impedance is small compared to 220 Ω by working at a signal frequencyof about 10 kHz. Decrease the input until the output is not clipped.Sketch an equivalent circuit for the amplifier output and compute Zoutfrom your measure-ment. Compare with the theoretical value Zout= [re+(RS/(β +1))] || RE. How much outputvoltage would you have gotten by connecting RLdirectly to the source? (RSis part of thesource.) What does this show?VinOVRE2N3904VEE= −15VVCC= +15VRS10KQ33.3KRLC21uFFigure 3.2N3904BCE2N3904Q4Q5Figure 4.7. Extra Credit. In the emitter follower (Figure 3) replace the transistor by two transistorsconnected in a Darlington configuration, (Figure 4): What is the effective beta of this ‘superbeta’ transistor? Measure Zinand Zoutof the follower now.2EE 321 Lab 11 Fall 2004Pre-Lab1. Find the gain for the circuit in Figure 1. The gain is approximately equal to the ratio RC/RE.This is because vincontrols ve, and ve= ieREand because ic≈ ie.2. Consider the emitter follower in Figure 2.• With VB= 0 V, find the bias currents. Us e thes e to find refor the transistor model.• Find the input and output resistance of this circuit. The formulas are given in Parts 5and 6. Assume a β of 200. (See Pages 478 to 483 of Sedra and Smith for the derivations.)3. Consider a source, an amplifier, and a load (as was done in Chapter 1). The source is a sinewave with a 1 V amplitude and a source resistance of 10 kΩ. The amplifier has a voltage gainof 1, an input resistance and an output res istance as found above. The load is 220 Ω.(a) What is the output voltage if the load is connected directly to the source?(b) What is the output voltage if the source is connected to the input of the amplifier andthe load to its


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NMT EE 321L - EE321 – Lab 11 Bipolar Junction Transistors, Part II

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