Download Advanced Design Techniques for RF Power Amplifiers (Analog by Anna N. Rudiakova, Vladimir Krizhanovski PDF

By Anna N. Rudiakova, Vladimir Krizhanovski

ISBN-10: 1402046383

ISBN-13: 9781402046384

Complex layout recommendations for RF energy Amplifiers presents a deep research of theoretical elements, modelling, and layout techniques of RF high-efficiency energy amplifiers. The ebook can be utilized as a consultant through scientists and engineers facing the topic and as a textual content booklet for graduate and postgraduate scholars. even if basically meant for experienced readers, it offers an outstanding speedy commence for novices.

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This waveform looks like third-harmonic peaking one, that was considered in details in the Chapter 1. Theoretical Analysis of BJT Class-F Power Amplifier 51 Figure 2-8. Functions ε 3 , 5 , MF (ε 5 ) , ε 3 , 5 , ma x ( ε 5 ) , and ε 3 , 5 , ∃ ( ε 5) . 8 2 ω0t Figure 2-9. The waveform of voltage v C , that is appropriate to ε 3 = − 3 and ε 5 = 20 from the region 1. 52 Chapter 2 The function γ (ε 3 , ε 5 ) is the same as function γ 2 (ε 3 , ε 5 ) within the region 1: γ (ε3 , ε5 ) = γ 2 (ε3 , ε5 ) = × (15ε 50 10ε 32ε 5 3 − 3ε 5 + Aε )ε (2-44) 3 1 , for region 1.

Function γ 1 ( ε 3 , ε 5 ) . Figure 2-7. Function γ 2 ( ε 3 , ε 5 ) . Theoretical Analysis of BJT Class-F Power Amplifier 49 ε 3ε 5 ε 3 + ε 5 + ε 3ε 5 = (15ε × 50 10ε 32ε 5 3 − 3ε 5 + Aε )ε (2-41) 3 1 25ε − 30ε 3ε 5 + 20ε ε − 3ε 5 + ( ε 5 − 5ε 3 ) Aε 2 3 2 3 5 . The following expressions for the curves ε 3,5, MF (ε 5 ) and ε 3,5,max (ε 5 ) can be found by solving the Eq. (2-41): ε 3,5, MF (ε 5 ) = − 9ε 5 , ε 5 + 25 ε 3,5,max ( ε 5 ) = − Aε 2 + Aε 3 Aε 12 − 4 Aε 10 2 (2-42) Aε 1 1 − 2 Aε 2 − Aε 3 − , 2 4 Aε 2 + Aε 3 where Aε 1 = − Aε 2 = Aε312 Aε310 Aε212 4 Aε210 3 Aε 3 = + 4 Aε 11 Aε 12 − 2 Aε 11 , 3 Aε 10 Aε210 2 Aε 6 2 Aε 10 3 Aε 4 + 3 3 − 8 Aε 7 , Aε 10 Aε 4 3 2 Aε 10 , Aε 4 = 2 Aε311 + Aε 8 − 9 Aε 7 Aε 11 A12 + 27 Aε 10 Aε27 + + 27 Aε212 Aε 5 − 72 Aε 5 Aε 10 Aε 11 , (2-43) 50 Chapter 2 Aε 5 = ε 52 + ε 53 , Aε 6 = 625 + 2000ε 5 + 2130ε 52 + 1064ε 53 + 256ε 54 , Aε 7 = −6ε 5 − 5ε 52 , Aε 8 = Aε 9 , Aε 9 = 3375000000ε 52 + 24300000000ε 53 + 78306750000ε 54 + 147776400000ε 55 + 179460724500ε 56 + 145015498800ε 57 + 77734274481ε 58 + 26570467584ε 59 + 5239406592ε 510 + 452984832ε 511, Aε 10 = 16ε 5 + 25 , Aε 11 = 25 + 13ε 5 − 8ε 52 , Aε 12 = 75 + 52ε 5 .

The known values are Rin , RL , xL or gin , g L , bL , while the unknown values are x1 , x2 , x3 or b1 , b2 , b3 . One of x values or one of b values should be given in order to calculate the matching network parameters. The remained parameters can be found from the Eqs. (1-35), (1-36) or (1-37), (1-38). 1 Input network example Usually, the input matching network should provide transformation of standard 50 Ohm or 75 Ohm impedance to the much lower transistor input impedance. Example of input network is shown in Fig.

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