R 1 re if 1 r2 3 r r e so far we have assumed

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Unformatted text preview: sistance required to set R to any desired value: R2 (1 + β ) Rf ￿ . rπ + (1 + β )Re If β ￿ 1: β R2 . (3) rπ + β R e So far, we have assumed that β ￿ 1 and R ￿ Re + rπ + gm rπ Re . In addition, if we assume that Rf ￿ R, the voltage gain becomes: −β R Av ￿ . rπ + β R e When the amplifier is terminated with load resistance RL = R, the input impedance will be equal to R. In this special case, the power gain is equal to the square of the voltage gain: Rf ￿ G￿ β 2 R2 . (rπ + β Re )2 1 This result is a factor of 4 smaller than the result for the lab amplifier. Thus, the 4:1 impedance transformer employed in the lab amplifier provides 6 dB more power gain under the conditions required for a simultaneous match. 7-13 Solution The ABCD parameters are defined as follows: V1 = AV2 − BI2 (4) I1 = CV2 − DI2 (5) The input impedance of a 2-port that is terminated in load impedance ZL is: ZIN = A V2 − B V1 AV2 − BI2 AZL + B I = = V2 = . 2 I1 CV2 − DI2 CZL + D C I2 − D The input...
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This homework help was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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