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 grounded-emitter 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 small-signal 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 small-signal 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 common-emitter 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 common-emitter 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 common-emitter 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 bybrid-π 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 small-signal 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 one-tenth 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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