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1.0 kn. RE-390 Ω, r-15 Ω. and ßac-75. 5. For a common-emitter amplifier, Rc Assuming that Rg is completely bypassed

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1.0 kn. RE-390 Ω, r-15 Ω. and ßac-75. 5. For a common-emitter amplifier, Rc Assuming that Rg is completely bypassed at the op
1.0 kn. RE-390 Ω, r-15 Ω. and ßac-75. 5. For a common-emitter amplifier, Rc Assuming that Rg is completely bypassed at the operating frequency, the voltage gain is (a) 66.7 (d) 75 (b) 2.56 (c) 2.47 6. In the circuit of Question 5, if the frequency is reduced to the point where Xctbypass) RE, the voltage gain (a) remains the same (b) is less (c) is greater 7. In a common-emitter amplifier with voltage-divider bias, Rimlbase) 68 k2, Ri 33 k2, and R2 15 k2. The total ac input resistance is (a) 68 kΩ (b) 8.95 kΩ (c) 22.2 kΩ (d) 12.3 kΩ 2.2 k Ω and ,-10 Ω, the voltage gain is 8, A CE amplifier is driving a 10 k Ω load. IfRc approximately (a) 220 For a common-collector amplifier, RE = (b) 1000 (d) 180 (c) 10 9. 100 Ω, = 10 Ω, and ßac-150. The ac input resistance at the base is (a) 1500 Ω 15kΩ (b) (c) 1102 (d) 16.5 k2 10. If a 10 mV signal is applied to the base of the emitter-follower circuit in Question 9, the output signal is approximately (a) 100 mV (b) 150 mV() 1.5V (d) 10 mV 11. In a certain emitter-follower circuit, the current gain is 50. The power gain is approximately 12. In a Darlington pair configuration, each transistor has an ac beta of 1 25. If RE is 560 Ω, the (c) 8.75 ΜΩ (a) 50Ay (b) 50 (c) 1 (d) answers (a) and (b) input resistance is (a) 560 Ω (b) 70k 140 kΩ (d)

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