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- 1. Microwave Engineering and Antennas Constant-gain circles Domine Leenaerts, Professor Department of Electrical Engineering Center for Wireless Technologies Eindhoven
- 2. Constant-gain circles Objective of this lecture β’ Discuss available gain circle β’ Discuss unilateral constant-gain circles β’ Provide an example
- 3. Circles in the Π-plane β’ Take as example the expression for πΊππ£ as a function of reflection coefficients and scatter parameters: β’ A certain πΊππ£ can be realized with many combinations of Ξπ and Ξππ’π‘ β’ The locus of such points in the Π-plane is typically a circle of the form β’ with C the center and r the radius of the circle and dependent on the realized gain πΊππ£ π€ π, π = πππ£,ππ’π‘ππ’π‘ πππ£,π = 1 β π€ π 2 1 β π11π€ π 2 π21 2 1 1 β π€ ππ’π‘ 2 Ξ β πΆ = π
- 4. Available gain circles β’ An amplifier has the following s-parameter values: β’ S11 = 0.61β 165o, S21 = 3.72β 59o, S12 = 0.05β 42o, S22 = 0.45β β48ΒΊ β’ We can plot the Ξπ , Ξππ’π‘ circles for Gav = 13, 14 and 15dB β’ Note that Gmax is equal to 16.2dB β’ Note that for Gav, the output is matched, not the input! 13dB 14dB 15dB Available power gain circles Ξπ + - Output Network ML π 11 π 12 π 21 π 22 Z0 Z0 Input Network MS Active Device G0 Ξππ’π‘ ΞπΏ = Ξππ’π‘ β Ξπ Ξππ Ξππ’π‘ for Gav=15dB Ξππ’π‘ β
- 5. Unilateral transducer Gain β’ In the unilateral case, the reverse transmission coefficient is zero (π 12 = 0), yielding for the unilateral transducer (power) gain πΊπ,π’ β’ Maximum gain improvement is observed when one port is conjugately matched. For instance, if the input port is matched, we have ππ = 1 β π€ π 2 1 β π11π€ π 2 yields α ππ,πππ₯ Ξπ =π 11 β = 1 1 β π11 2 πΊπ,π’ π = 1β π€π 2 1βπ11π€π 2 π21 2 1β π€πΏ 2 1βπ22π€πΏ 2 = ππ π21 2ππΏ
- 6. Unilateral transducer Gain β’ The βgainβ π21 2 is defined based on 50β¦ termination. As π 11 and π 22 are mostly not 50β¦, power loss occurs. Hence the difference between πΊπ,π’ π€ π, π and π21 2 . By optimizing MS and ML we do not improve gain, we reduce the losses! + - Output Network ML οΊ ο» οΉ οͺ ο« ο© 22 21 11 s s 0 s Z0 Z0 Input Network MS Active Device G0 GL GS s22 s11 πΊπ,π’ π€ π, π = 1β π€π 2 1βπ11π€π 2 π21 2 1β π€πΏ 2 1βπ22π€πΏ 2 = ππ π21 2 ππΏ = πππΊπππΏ
- 7. Unilateral constant-gain circle, source β’ Equating ππ to a constant, we can solve the unilateral transducer gain for terminations Ξπ, yielding equations of βconstant-gain circlesβ for source terminations: ππ = 1 β π€ π 2 1 β π11π€ π 2 ππ = ππ 1 β π 11 2 πΆπ = πππ 11 β 1 β π 11 2 1 β ππ ππ = 1 β π 11 2 1 β ππ 1 β π 11 2 1 β ππ Source circles s11 s* 11 s22 ππ,πππ₯ (center) (radius) (normalization) π < ππ,πππ₯
- 8. Unilateral constant-gain circle, source Source circles s11 s* 11 s22 ππ,πππ₯ π < ππ,πππ₯ β’ Some observations: 1. the centers of source plane constant gain circles always lie on a line drawn between the point π 11 β and the origin of the source plane. 2. The radius of constant gain circles decreases with increasing gain or reduced losses. The maximum gain is reached for just a single point
- 9. Unilateral constant-gain circle β’ We can apply the same strategy for the load terminations, simple change ππ β ππΏ, ππ β ππΏ, π 11 β π 22, Ξπ β ΞπΏ Source circles Load circles s11 s* 11 s* 22 s22
- 10. Example: maximize gain while minimizing NF β’ A unilateral amplifier has the following s-parameter values: S11 = 0.8β 120o, S21 = 4β 60o, S12 = 0, S22 = 0.2β β30ΒΊ β’ For the amplifier also the noise parameters are given: ππΉπππ = 1.6ππ΅, ππ = 0.16, Ξπππ‘ = 0.26β β152π
- 11. Example: investigate the unilateral gain β’ Using β’ We find π21 2 = 12.04ππ΅ , ππ,πππ₯ = 4.44ππ΅ , ππΏ,πππ₯ = 0.18ππ΅ 1. The maximum unilateral gain is therefore 4.44+12.04+0.18 = 16.66 dB 2. There is not much loss in the output. In other words, the ouput impedance is close to 50β¦. 3. We would like to keep the matching losses at the input to a minimum, but also have a good noise figure. πΊπ,π’ π€ π, π = ππ π21 2ππΏ
- 12. Example: investigate the unilateral gain Ms=3dB Ms=2dB Ms=1dB π 11 β π 22 β Ms_max=4.44dB ML=0.1dB ML_max=0.18dB πΊπ,π’ π€ π, π = ππ π21 2 ππΏ
- 13. Example: plot the noise circles Ξπππ‘ ππΉ = 1.7ππ΅ ππΉ = 1.8ππ΅ ππΉ = 1.9ππ΅ ππΉ = 2.0ππ΅ ππΉπππ = 1.6ππ΅ ππΉπππ = 1.6ππ΅, ππ = 0.16, Ξπππ‘ = 0.26β β152π
- 14. Example: choose Ms π 11 β π 22 β Ξπ Ms=3dB ππΉ = 1.9ππ΅ πΊπ,π’ π = 3 + 12.04 + 0.18 = 15.22ππ΅ Plotting the noise circles and the gain circles in the same Smith chart reveals that ππ = 3ππ΅ is a good trade off between maximizing the gain and minimizing the noise figure.
- 15. Example: input matching network β’ With ππ = 3ππ΅, we have ΞS = 0.468β β120o β’ This leads to the source impedance β’ Which we then need to transform to 50β¦ π 11 β π 22 β 3dB ZS = 23.15 β j24 Ξ©, Ξπ + - Output Network ML οΊ ο» οΉ οͺ ο« ο© 22 21 11 s s 0 s Z0 Z0 Input Network MS Active Device G0 GL GS s22 s11
- 16. What about Gav at optimum noise match? β’ The Ms = 1.3dB intersects with Ξπππ‘. β’ For πΊππ£ the output is conjugately matched, π 22 β Ξπππ‘ ππΉ = 1.6ππ΅ Ms=1.3dB ML_max=0.18dB πΊπ,π’ π = πΊππ£,πππ‘ = 1.3 + 12.04 + 0.18 = 13.52ππ΅
- 17. Summary β’ Discussed available gain circle β’ Discussed unilateral constant-gain circles β’ Example to optimize between unilateral gain and noise figure