EE 321 Analog Electronics, Fall 2013 Homework #8 solution


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1 EE 321 Analog Electronics, Fall 2013 Homework #8 solution The following table summarizes some of the basic attributes of a number of BJTs of different types, operating as amplifiers under various conditions. Provide the missing entries. I will just provide the explicit equations for the first column, (a): β = α 1 α I E = I C α I B = I C β g m = I C V T r e = α g m = β g m Transistor a b c d e f g α β I C (ma) I E (ma) I B (ma) g m (ma/v) r e (Ω) (Ω) K A biased BJT operates as a groundedemitter amplifier between a signal source, with a source resistance of 10kΩ, connected to the base and a 10kΩ 1
2 load connected as a collector resistance R C. In the corresponding model, g m is 40mA/V and is 2.5kΩ. Draw the complete amplifier model using the hybridπ BJT equivalent circuit. Calculate the overall voltage gain v c /v s. What is the value of BJT β implied by the values of the model parameters? To what value must β be increased to double the overall voltage gain? vs Rs vc Rp Rc v o = g m R C v i = g m R C v s = R s = 80 The value of β can be found from or = β g m β = g m = = 100 To double the gain we would want to double the factor It is currently equal to R s = 0.2 To double it we would need to change : R s + = 0.4 = R s = 0.67R s = 6.7kΩ The factor increase in β is the same as the factor increase in : new 6.7 β new = β old = 100 old 2.5 = For the circuit shown in Fig. P5.115, draw a complete smallsignal equivalent circuit utilizing an appropriate T model for the BJT (use α = 0.99). Your 2
3 circuit should show the values of all components, including the model parameters. What is the input resistance R in? Calculate the overall gain v o /v sig. Here is the smallsignal equivalent Rc RL re Rsig where The overall voltage can be found from v o = (R C R L )i c i c = αi e so i e = v sig R sig +r e 3
4 v o 1 =(R C R L )α v sig R sig +r e 1 =(10 10) = A commonemitter amplifier of the type shown in Fig 5.60(a) is biased to operate at I C = 0.2mA and has a collector resistance R C = 24kΩ. The transistor has β = 100 and a large V A. The signal source is directly coupled to the base and C c1 and R B are eliminated. Find R in, the voltage gain A vo, and R o. Use these results to determine the overall voltage gain when a 10 kω resistor is connected to the collector and the source resistance R sig = 10kΩ. This is a commonemitter amplifier with R B =, so the input resistance is R in = = βv T = = 12.5kΩ I C 0.2 A vo = g m R C = I C V T R C = = 80 4
5 The overall voltage gain, v o /v sig is then R O = R C = 10kΩ R L G v = v o = A vo v sig R sig + R L +R O = ( 80) = For the commonemitter amplifier shown in Figure P5.130, let V CC = 9V, = 27kΩ, = 15kΩ, R E = 1.2kΩ, and R C = 2.2kΩ. The transistor has β = 100, and V A = 100V. Calculate the dc bias current I E. If the amplifier operates between a source for which R sig = 10kΩ and a load of 2kΩ, replace the transistor with its bybridπ model, and find the values of R in, the voltage gain v o /v sig, and the current gain i o /i i. We have V B = (i 1 i B ) V B = V BE +(β +1)R E i B V B = V CC i 1 and we can eliminate i D and i B from the first equation. ( VCC V B V B = V B [ 1+ + (β +1)R E V B = V CC R +V 2 BE (β+1)r E 1+ + ] (β+1)r E = 9 V B V BE (β +1)R E ) = V CC +V BE (β +1)R E = 3.03V 5
6 Note that for (β +1)R E and V CC V BE the expression for V B reduces to the voltage divider expression, R V B = V 1 CC = V 1+ CC = 9 15 R = 3.21V 1 I will proceed with the result from the fully correct expression. Next, the emitter current is I E = V E = V B V BE R E R E The smallsignal model looks like this = = 1.94mA Rs ii vo io vs R1 R2 Rp ro Rc RL The input resistance is where such that R in = = β = βv T = βv T = (β +1)V T = = 1302Ω g m I C αi E I E 1.94 The voltage gain is R in = = = 1.15kΩ G v = R in A v = R in g m (R C r o R L ) = R in βi E (R C r o R L ) R in +R s R in +R s R in +R s (β +1)V T where and then r o = V A I C = V A αi E = (β +1)V A βi E = = 52.1kΩ
7 G v = R in R in +R s = = 8.07 βi E (R C r o R L ) (β +1)V T ( ) Using the topology of Fig. P5.130, design an amplifier to operate between a 10kΩ source and a 2kΩ load with a gain v o /v sig of 8. The power supply available is 9V. Use an emitter current of approximately 2mA and a current of about onetenth of that in the voltage divider that feeds tbe base, with the dc voltage at the base about one third of the supply. The transistor available has β = 100 and V A = 100V. Use standard 5% resistors (See Appendix G) This is the same circuit as before, except choose the nearest 5% resistors. In that case choose = 27kΩ, = 15kΩ, R E = 1.2kΩ, R C = 2.2kΩ. Well, it turns out the problem is identical to P That was easy. 7
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