Experiment 8&9 BJT AMPLIFIER 1
BJT AS AMPLIFIER 1. Objectiv e: 1- To demonstrate the operation and characteristics of small signals common emitter amplifiers. 2- What do we mean by a linear amplifier and why we need it linear 3- Effect of Q point on the shape of output signal. 4- Analyze the dc operation. 5- Stability of Q point 6- See the effect of coupling capacitors 7- Analyze the ac operation. 8- Determine the input resistance. 9- Determine the output resistance. 1- Determining the voltage gain. 11- Understand the model of a transistor 12- Explain the effects of an emitter-bypass capacitor. 13- Demonstrating the effect of a load resistor on the voltage gain. 14- Demonstrating the phase shift. 2. Introduction: In this experiment, we emphasize the use of the bipolar transistor in linear amplifier applications. Linear amplifiers imply that, for the most part, we are dealing with analog signals. A linear amplifier then means that the output signal is equal to the input signal multiplied by a constant, where the magnitude of the constant of proportionality is, in general, greater than unity. A linear amplifier provides amplification of a signal without any distortion so that the out-put signal is an exact amplified replica of the input signal. We want the output signal to be linearly proportional to the input signal so that the output of the speakers is an exact (as much as possible) reproduction of the signal generated from the compact disc. Therefore, we want the amplifier to be a linear amplifier. The transistor is the heart of an amplifier. Bipolar transistors have traditionally been used in linear amplifier circuits because of their relatively high gain. 2
Figure shows the circuit where BB is a dc voltage to bias the transistor at a particular Q-point and vs is the ac signal that is to be amplified. To use the circuit as an amplifier, the transistor needs to be biased with a dc voltage at a quiescent point (Q-point), as shown in the figure, such that the transistor is biased in the forward-active region. If a time-varying (e.g., sinusoidal)signal is superimposed on the dc input voltage, BB, the output voltage will change along the transfer curve producing a time-varying output voltage. If the time varying output voltage is directly proportional to and larger than the time-varying input voltage, then the circuit is a linear amplifier. From this figure, we see that if the transistor is not biased in the active region (biased in either cutoff or saturation), the output voltage does not change with a change in the input voltage. Thus, we no longer have an amplifier. The time-varying signals are assumed to be small signals, which means that the amplitudes of the ac signals are small enough to yield linear relations. 3. Effect of biasing on the Q point 3
Bias establishes the dc operating point (Q-point) for proper linear operation of an amplifier. If an amplifier is not biased with correct dc voltages on t he input and output, it can go into saturation or cutoff when an input signal is applied. Figure below shows the effects of proper and improper dc biasing of an inverting amplifier. In part (a), the output signal is an amplified replica of the input signal except that it is inverted, which means that it is out of phase with the input. The output signal swings equally above and below the dc bias level of the output,dc(out). Improper biasing can cause distortion in the output signal, as illustrated in parts (b) and (c). Part (b) illustrates limiting of the positive portion of the out-put voltage as a result of a Q point (dc operating point) being too close to cutoff. Part (c) shows limiting of the negative portion of the output voltage as a result of a dc operating point being too close to saturation. 4
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4. Effect of Q point on the output signal. 1-connect the below circuit 2-In order to make the input signal, S, small enough to prevent distortion of the amplifier output, use the voltage divider circuit shown in Figure below, where S is the voltage across the 1Ω resistor. The voltage divider circuit shown in the Figure below divides input voltage from FGEN by 1:1.Thus, to get S =.2P-P, set FGEN to provide 2.P-P as input to the voltage divider. You may find, it is not possible to see the correct amplitude of S on the oscilloscope, but if you are applying 2.P-P, 1kHz, at the input of the voltage divider shown in Figure below, you can be assured that S =.2P-P, 1kHz. 7
11% 83% R2 (kω) O Shape of the waveform (e.g.sinusoidal or distorted) 1 5 1 15 2 25 3 s O Shape of the waveform (e.g.sinusoidal or distorted) 1m 2m 5m 15m 2m 25m 5m R2(1) R2 1k R2 68k A C1(1) C1 1u Q1 2N2222 +88.8 AC olts B C D R1 1k R1 1k C2 22u 8
5. Common Emitter Amplifier BJT transistor amplifiers are frequently used in the common-emitter configuration (CE), since this design gives both a high current gain (AI) and a high voltage gain (A). This experiment explores the dc and ac characteristics of the common-emitter circuit and how changing the Q-point affects circuit performance. The collector current (IC ) in a BJT circuit depends on the β of the transistor as well as the transistor s temperature and the other circuit elements. Good amplifier design requires choosing biasing resistors such that the quiescent (DC) collector current remains constant, regardless of whether a transistor with a different β is being used. Selecting the proper biasing network and resistors keeps the collector current relatively constant. A key element that stabilizes the circuit to changes in the transistor β is the emitter resistor, RE. Using a bias network with an emitter resistor is a good way to keep IC relatively constant if the β or temperature changes. Any increase in IC will cause the feedback voltage drop, FB, across RE, to increase, thus lowering the base-emitter voltage, BE, and causing IB to decrease. This is a negative feedback effect. Therefore, any changes in β will not cause IC to change significantly, since IB will scale with 1/β. However, using an emitter resistor lowers the ac voltage gain of the circuit, since the ac component, ic, must flow through RE to ground. Therefore, RE impedes ic. The negative feedback voltage, FB across RE "kills" the ac gain. To get around this undesired ac response and still maintain the excellent control of ICQ that RE provides, a capacitor is placed across RE in order to short the ac current, ic, around RE to ground. This capacitor, CE, is called an emitter bypass capacitor and is generally large, 1μF or greater. This capacitor allows most of the ac current to flow around RE directly to ground. Note that the dc current, ICQ, still flows through RE, since the capacitor acts as an open circuit to dc, and the stabilizing effect of RE is maintained. If RE is physically removed from the circuit, the negative feedback effect on the gain is taken away The common emitter amplifier is characterized by moderate input impedance and slightly high output impedance. Both the voltage and current gain are high resulting in po wer gain. 9
6. Dc analysis B R R 2 CC E B BE 1 R2 I E E CE CC ICR C IERE R E r e 26m IE Question: 1-For the circuit below find Q-point and find r e mathematically and using Orcad. 2-Change β from 9 to 1 and compute the change in Q point Ic ce 3-change the transistor itself and find the change in Q point Ic ce. Comment the result in your own words Change Is =2e-15 B=9 11
7. Effect of input (coupling ) capacitors The capacitors at the input and output serve to isolate the signal source and load from the voltage source CC it is called coupling capacitors.a coupling capacitor is used to connect two circuits such that only the AC signal from the first circuit can pass through to the next while DC is blocked. This technique helps to isolate the DC bias settings of the two coupled circuits. Use of Coupling Capacitors Coupling capacitors are useful in many types of circuits where AC signals are the desired signals to be output while DC signals are just used for providing power to certain components in the circuit but should not appear in the output. For example, a coupling capacitor normally is used in audio circuits, such as a microphone circuit. DC power is used to give power to parts of the circuit, such as the microphone, which needs DC power to operate. So DC signals must be present in the circuit for powering purposes. However, when a user talks into the microphone, the speech is an AC signal, and this AC signal is the only signal in the end we want passed out. When we pass the AC signals from the microphone onto the output device, say, speakers to be played or a computer to be recorded, we don't want to pass the DC signal; remember, the DC signal was only to power parts of the circuit. We don't want it showing up on the output recording. On the output, we only want the AC speech signal. So to make sure only the AC passes while the DC signal is blocked, we place a coupling capacitor in the circuit. How to Place a Coupling Capacitor in a Cirucit In order to place a capacitor in a circuit for AC coupling, the capacitor is connected in series with the load to be coupled. 11
Question : Connect the circuit and see the output before and after capacitor. C1 OFF = AMPL = 1 FREQ = 1k 1 1u R1 1k 2 2 8. Ac analysis. R R R 1 2 RR 1 2 R R 1 2 Zo R C ro Zo RC ro 1RC Zi R re A v o C o i R r r e Question : Compute :Zo,Zi,Av assume ro=5k mathematically and using orcad compute Av. 22 C1 56k Rc 6.8k 1u Q1 OFF = AMPL = 1m FREQ = 1k 8.2k Q2N2222 1.5k 2u 12
To find from orcad plot Iin and find the peak to peak value then find Zin To find = from Orcad plot Ic and find the peak to peak value of Ic then plot o and find peak to peak value then find Zo 9. Explain the effects of an emitter-bypass capacitor. When we remove the emitter-bypass capacitor the voltage gain decrease. A v R r o C o i re RE Question :Compute the gain without the capacitor and see the effect in Orcad. 1. Demonstrating the effect of a load resistor on the v oltage gain. When we insert a RL the gain decrease significantly, so we need a matching circuit before connecting the load. A v o C L o i R R r r e Question : 1-Give an example of RL in practical 2-Try Rl=5, 1k, 1k, 1Tera 1 22 R2 C1 R1 56k Q1 6.8k C3 1u R5 OFF = AMPL = 1m FREQ = 1k 2 1u Q2N2222 1k R4 8.2k R3 1.5k C2 22u 13
11. AC analysis using Transistor model in Orcad : 1. Insert the dependent current source from the Analog library. 2. Connect the circuit 3. Compute Av using Orcad 22 C1 1u 56k GAIN = 9 Bre F1 1.66k F Rc 6.8k r 5k OFF = AMPL = 1m FREQ = 1k 8.2k 1.5k 2u 12. Compute Zi,Zo output using Orcad Note that we can not measure Zin and Zo directly as usual impedance because they are an AC parameter which mean that the transistor must be in the active mode to measure it. Insert I_test (dc current source equal 1A) and deactivate all the independent sources and short all the capacitor then measure _test. Z _ test I _ test C1 56k Bre GAIN = 9 F1 Rc 6.8k C1 56k Bre GAIN = 9 F1 Rc 6.8k OFF = AMPL = FREQ = 1k 1u I_test 1 8.2k 1.66k F r 5k OFF = AMPL = FREQ = 1k 1u 8.2k 1.66k F r 5k 1 I_test 1.5k 2u 1.5k 2u 13. Compute Zi,Zo practically In practical, we cannot measure Zi and Zo using the previous technique in Orcad but we measure it as follow. Zin=Iin/in and Zo=o/Io so we just want to measure In,Io,in and o but sometimes it is difficult to measure them accurately so we will use this techniques Zi: Measure the input resistance by inserting a variable resistance in series with the signal generator and input coupling capacitor and vary the resistance until o equal half of the o without this resistance, the value of the variable resistance equal to Rin(why?) Zo: The output resistance is measured as follows. a- Measure the output voltage gain at no load. b- Connect a variable resistance as a load and change its value until half of the outputin (a)is obtained c- The value of the variable resistance is equal to Rout (why?) 14
Lab Report 1- Answer all the questions of the experiment (try to justify and explain all the result s )? 2- Summary of your measurement? Parameter Mathematically Using Orcad Q-point Ic ce Ie re Av(NL) Av(unbypassed) Av(RL=5) Av(RL=1T) Zi Zo 3- Bring the following circuit in a breadboard, make sure to choose R2 such that your Q point lay in the middle of the active region (ce=.5 cc) before coming to the lab? C1=C2=1u m Ce=22u Bonus: use variable resistor and find Zo,Zi in Orcad 15