Electronic Devices, 9th edition Thomas L. Floyd. Input signal. R 1 and R 2 are selected to establish V B. If the V CE

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1 3/9/011 lectronic Devices Ninth dition Floyd hapter 5: Transistor ias ircuits The D Operating Point ias establishes the operating point (Q-point) of a transistor amplifier; the ac signal (ma) moves above and below Load line this point. For this example, the dc base current is 300 µa. When the input causes the base current to vary between 00 µa and 400 µa, the collector current varies between 0 ma and 40 ma. Q c A Q V Q V ce b 400 µ A 300 µ A Q 00 µ A V (V) The D Operating Point A signal that swings outside the active region will be clipped. For example, the bias has established a low Q- point. As a result, the signal is will be clipped because it is too close to cutoff. Q utoff 0 V ce Q V Q V utoff Q nput signal V A practical way to establish a Q-point is to form a voltagedivider from V. 1 and are selected to establish V. f the 15 V divider is stiff, is small compared to. Then, V 1 Determine the base voltage for the circuit. V 1 1 kω ( 15 V) 4.6 V 7 kω 1 kω 1 7 kω 1 kω What is the emitter voltage, V, and current,? V is one diode drop less than V : V 4.6 V 0.7 V 3.9 V Applying Ohm s law: V 3.9 V 5.76 ma 1 7 kω 4.6 V D V 1 kω The unloaded voltage divider approximation for V gives reasonable results. A more exact solution is to Thevenize the input circuit. V TH V (no load) 4.6 V TH kω The Thevenin input circuit can be drawn TH TH 4.6 V V 8.31 kω 1 7 kω 1 kω 1

2 3/9/011 Now write KVL around the base emitter circuit and solve for. VTH TH V VTH TH D Substituting and solving, 4.6 V 0.7 V k 5.43 ma Ω Ω 00 and V (5.43 ma)(0.68 kω) 3.69 V TH TH 4.6 V V 8.31 kω Multisim allows you to do a quick check of your result. A pnp transistor can be biased from either a positive or negative supply. Notice that (b) and (c) are the same circuit; both with a positive supply V (a) (b) (c) V Determine for the pnp circuit. Assume a stiff voltage divider (no loading effect). 1 V V 1 7 kω ( 15.0 V ) 10.4 V 7 kω 1 kω V V V 10.4 V 0.7 V 11.1 V V 15.0 V 11.1 V 5.74 ma 1 kω 10.4 V 11.1 V 1 7 kω mitter ias mitter bias has excellent stability but requires both a positive and a negative source. For troubleshooting analysis, assume that V for an npn transistor is about 1 V. Assuming that V is 1 V, what is? 1 V 15 V ( 1 V) 7.5 kω 1.87 ma 68 kω 3.9 kω 1 V 7.5 kω V 15 V mitter ias A check with Multisim shows that the assumption for troubleshooting purposes is reasonable. For detailed analysis work, you can include the effect of D. n this case, 1 V D

3 3/9/011 ase ias ase bias is used in switching circuits because of its simplicity, but not widely used in linear applications because the Q-point is dependent. ase current is derived from the collector supply through a large base resistor. What is? V 0.7 V 15 V 0.7 V 5.5 µa 560 kω 560 kω 15 V ase ias ompare V for the case where 100 and 00. For 100: ( 100)( 5.5 µa).55 ma V ( )( ) 15 V.55 ma 10.4 V For 300: ( 300)( 5.5 µa) 7.65 ma V 15 V ( 7.65 ma)( ) 1.3 V 560 kω mitter-feedback ias An emitter resistor changes base bias into emitter-feedback bias, which is more predictable. The emitter resistor is a form of negative feedback. The equation for emitter current is found by writing KVL around the base circuit. The result is: D ollector-feedback ias ollector feedback bias uses another form of negative feedback to increase stability. nstead of returning the base resistor to V, it is returned to the collector. The equation for collector current is found by writing KVL around the base circuit. The result is D Key Terms ollector-feedback ias ompare for the case when 100 with the case when 300. When 100, 15 V 15 V 0.7 V.80 ma 1.8 k 330 kω Ω 100 D 330 kω When 300, V 15 V 0.7 V 1.8 k 330 kω Ω 300 D 4.93 ma Q-point D load line Linear region Stiff voltage A voltage divider for which loading effects divider can be ignored. Feedback The dc operating (bias) point of an amplifier specified by voltage and current values. A straight line plot of and V for a transistor circuit. The region of operation along the load line between saturation and cutoff. The process of returning a portion of a circuit s output back to the input in such a way as to oppose or aid a change in the output. 3

4 3/9/ A signal that swings outside the active area will be a. clamped b. clipped c. unstable d. all of the above. A stiff voltage divider is one in which a. there is no load current b. divider current is small compared to load current c. the load is connected directly to the source voltage d. loading effects can be ignored 3. Assuming a stiff voltage-divider for the circuit shown, the emitter voltage is a. 4.3 V b. 5.7 V c. 6.8 V d. 9.3 V 1 0 kω 4. For the circuit shown, the dc load line will intersect the y-axis at a. 5.0 ma b ma c ma d. none of the above 1 0 kω 5. f you Thevenize the input voltage divider, the Thevenin resistance is a. 5.0 kω b kω c. d. 30 kω 1 0 kω 6. For the circuit shown, the emitter voltage is a. less than the base voltage b. less than the collector voltage c. both of the above d. none of the above 1 kω 1 7 kω 4

5 3/9/ mitter bias a. is not good for linear circuits b. uses a voltage-divider on the input c. requires dual power supplies d. all of the above 8. With the emitter bias shown, a reasonable assumption for troubleshooting work is that the a. base voltage 1 V b. emitter voltage 5 V c. emitter voltage 1 V d. collector voltage V 68 kω 3.9 kω 7.5 kω V 15 V 9. The circuit shown is an example of 10. The circuit shown is an example of a. base bias b. collector-feedback bias c. emitter bias a. base bias b. collector-feedback bias c. emitter bias d. voltage-divider bias d. voltage-divider bias Answers: 1. b 6. d. d 7. c 3. a 8. c 4. a 9. a 5. b 10. b 5

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