ECE 5317-6351 Microwave Engineering Fall 2011 Prof. David R. Jackson Dept. of ECE Notes 18 Multistage Transformers 1 Single-stage Transformer The transformer length is arbitrary in this analysis. Step = 1 Z0 Z1 line 0 S21 Z1 ZL 0 S22 S110 , Z in Z in - Z 0 Z in Z 0 L e- j1 Load S120 Z L is real e- j1 L Z L - Z1 Z L Z1 From previous notes: Step Impedance change 0 S110 - S 22 0 S 21 S120 Z1 - Z 0 Z1 Z 0 2 Z 0 Z1 Z0 1 S110 Z1 Z1 Z 0 2 Single-stage Transformer (cont.) From the self-loop formula, we have (as derived in previous notes) - j 21 0 0 S S e S110 21 012 L - j 21 1 - S22 L e For the numerator: 2 2 Z 0 Z1 4 Z 0 Z1 0 0 S21S12 Z1 Z 0 Z Z 2 1 0 Next, consider this calculation: 1 S110 2 2 2 Z1 Z 0 Z1 - Z 0 Z12 Z 02 2 Z 0 Z1 Z1 Z 0 Z1 Z 0 2 Z 0 Z1 1 1 1 2 2 2 Z Z Z1 Z 0 Z1 Z 0 Z1 Z 0 0 1 2 2 2 Z 2 1 Z 02 2 Z 0 Z1 Z12 Z 02 2 Z 0 Z1 2 Z1 Z 0 2 4 Z 0 Z1 Z1 Z 0 2 Hence S S 1- S11 0 21 0 12 2 0 3 Single-stage Transformer (cont.) We then have S110 1 S 02 11 L e- j 21 0 1 - S22 L e- j 21 Putting both terms over a common denominator, we have S110 S110 2 L e- j 21 1 S110 2 L e- j 21 1 S110 L e- j 21 or S110 L e- j 21 1 S110 L e- j 21 4 Single-stage Transformer (cont.) S110 L e- j 21 1 S110 L e- j 21 Assuming small reflections S 0 11 L 1 Note: It is also true that 0 S110 S21 S120 L e- j 21 But S210 S120 1- S110 2 1 S110 L e- j 21 e j1 0 Denote 0 S11 , 1 L 0 1 e- j 21 Z -Z Z -Z 0 1 0 ; 1 L 1 Z1 Z 0 Z L Z1 0 S110 L e j1 5 Multistage Transformer i i 1 Z0 Z1 3 2 Z2 i N -1 . . . Z3 Z N -1 N ZN ZL Assume 1 2 3 N Assuming small reflections: e j 0 e j e j 1 e j e j 2 e j e j 3 N -2 e j e j N -1 N L e j 6 Multistage Transformer (cont.) Hence 0 1 e- j 2 2 e- j 4 3e- j 6 ..... N e- j 2 N Z n1 - Z n n Z n1 Z n Note that this is a polynomial in powers of z = exp(-j2). i i 1 Z0 Z1 2 Z2 3 Z3 . . . i N -1 N Z N -1 ZN ZL Assume 1 2 3 N 7 Multistage Transformer (cont.) 0 1 e- j 2 2 e- j 4 3e- j 6 ..... N e- j 2 N If we assume symmetric reflections of the sections (not a symmetric layout of line impedances), we have 0 N , 1 N -1 , 2 N -2 , . . . e- jN 0 e jN e- jN 1 e j ( N -2) e- j ( N -2) . . . N odd last term N -1 e e j - j Last term 2 N even last term N 2 8 Multistage Transformer (cont.) Hence, for symmetric reflections we then have - jN 2e 2e- jN 1 cos N cos N 2 ... cos N 2 n ... 1 n N ; N even 0 2 2 cos N cos N 2 ... cos N 2 n ... cos 1 n N -1 0 ; N odd 2 Note that this is a finite Fourier cosine series. 9 Multistage Transformer (cont.) Design philosophy: If we choose a response for ( ) that is in the form of a polynomial (in powers of z = exp(-j2 )) or a Fourier cosine series, we can obtain the needed values of n and hence complete the design. 10 Binomial (Butterworth*) Multistage Transformer Consider: A 1 e- j 2 N Ae- jN e j e- j A e- jN 2 N cos N N A 2 N cos N Choose all lines to be a quarter wavelength at the center frequency so that f f 0 i 2 0 2 (We have a perfect match at the center frequency.) dn Also, 0 for n 1, 2, ..., N -1 d n 2 1st N - 1 derivatives are zero maximally flat *The name comes from the British physicist/engineer Stephen Butterworth, who described the design of filters using the binomial principle in 1930. 11 Binomial Multistage Transformer (cont.) Use the binomial expansion so we can express the Butterworth response in terms of a polynomial series: N 1 z N CnN z n where CnN n 0 N! N - n ! n ! A binomial type of response is obtained if we thus choose A 1 e - j 2 N N A CnN e- j 2 n n 0 We want to use a multistage transformer to realize this type of response. 0 1 e- j 2 2 e- j 4 3 e- j 6 ...... N e- j 2 N A 1 e - j 2 N N A CnN e- j 2 n n 0 Set equal (Both are now in the form of polynomials.) 12 Binomial Multistage Transformer (cont.) Note that as f 0 0 zero length transmisison lines Z L - Z0 A 2N Z L Z0 Hence Z L - Z0 A2 Z Z 0 L -N Note: A could be positive or negative. Equating responses for each term in the polynomial series gives us: n ACnN , n 1,2,......., N Hence Z -Z Z n 1 - Z n 2- N L 0 CnN Z n 1 Z n Z L Z0 This gives us a solution for the line impedances. Z0 Z0 , Z N 1 Z L 13 Binomial Multistage Transformer (cont.) Note on reflection coefficients n ACnN , n 1,2,......., N CnN N! N - n ! n ! Note that CNNn N! N! CnN N - ( N n) ! ( N n)! n! ( N n)! Hence n N n Although we did not assume that the reflection coefficients were symmetric in the design process, they actually come out that way. 14 Binomial Multistage Transformer (cont.) Note: The table only shows data for ZL > Z0 since the design can be reversed (Ioad and source switched) for ZL < Z0 . 15 Binomial Multistage Transformer (cont.) Example showing a microstrip line g1 / 4 50 line Z1 g 2 / 4 Z2 g 3 / 4 100 line Z3 ZL Z0 A three-stage transformer is shown. 16 Binomial Multistage Transformer (cont.) Note: Increasing the number of lines increases the bandwidth. Figure 5.15 (p. 250) Reflection coefficient magnitude versus frequency for multisection binomial matching transformers of Example 5.6. ZL = 50Ω and Z0 = 100Ω. 17 Binomial Multistage Transformer (cont.) Use a series approximation for the ln function: X -1 1 ln X ; X 1 2 X 1 Recall Z n 1 - Z n - N Z L - Z0 N 2 Cn Z n 1 Z n Z L Z0 1 ln Z n1 2- N CnN 1 ln Z L 2 Zn 2 Z0 Hence ve recursi ship n o i t a l e r ZL ln Z n 1 2 C ln ln Z n Z0 -N N n 18 Binomial Multistage Transformer (cont.) Bandwidth m 2 N A cos N m 1 N -1 1 m m cos 2 A Maximum acceptable reflection m m The bandwidth is then: f f0 f -f 2 0 m f0 f / 2 fm /2 f / 2 f0 fm m 4 m 4 1 2-2 2-2 2 2 - cos -1 m f0 /2 2 A Hence 1 N f 4 1 2 - cos-1 m 2 A f0 - m 2 f0 - fm 1 N 19 Binomial Multistage Transformer (cont.) f f0 2 Summary of Design Formulas A 1 e - j 2 N N A CnN e- j 2 n Reflection coefficient response n 0 CnN Z L - Z0 A2 Z Z 0 L -N A coefficient ZL ln Z n 1 2 C ln ln Z n Z0 -N N n 1 N f 4 1 2 - cos-1 m 2 A f0 N! N - n ! n ! Design of line impedances Bandwidth 20 Example Example: three-stage binomial transformer Given: Z L 50 [] 100 N 3 Z1 Z2 Z3 50 50 -100 A 2-3 -0.0417 50 100 1 3 f 4 -1 1 0.05 BW 2 - cos 2 0.0417 f0 0.713 Z 0 100 [] m 0.05 dB m -26.0 [dB] BW 71.3% 21 Example (cont.) ZL ln Z n1 ln Z n 2 C ln Z0 -N Z1 : N n 50 ln Z1 ln Z 0 2-3 C03 ln 4.519 100 Z1 91.7 [] Z2 : 50 ln Z 2 ln Z1 2-3 C13 ln 4.259 100 CnN N! N - n ! n ! C 30 = 1 C13 = 3 C 32 = 3 C 33 = 1 Z 2 70.7 [] Z3 : 50 ln Z 3 ln Z 2 2-3 C23 ln 3.999 100 Z 3 54.5[] 22 Example (cont.) Using the table in Pozar we have: ZL / Z0 2 : Z1, Z2 , Z3 / Z0 1.0907, 1.4142,1.8337 (The above normalized load impedance is the reciprocal of what we actually have.) Hence, switching the load and the source ends, we have Z1, Z2 , Z3 / Z0 1.8337, 1.4142, 1.0907; Z0 50[] Therefore Z1 91.685 [] Z 2 70.711 [] Z 3 54.585 [] 23 Example (cont.) g1 / 4 50 line Z1 g 2 / 4 Z2 g 3 / 4 Z3 100 line ZL Z0 S11 dB 20 log10 f -26 3.29 GHz f0 5.0GHz 6.74 GHz BW 69.0% Response from Ansoft Designer 24 Chebyshev Multistage Matching Transformer Chebyshev polynomials of the first kind: cos n cos 1 x , x 1 Tn x 1 cosh n cosh x, x 1 T1 x x T2 x 2 x 2 1 For -1 x 1: Tn x 1 T3 x 4 x 3 - 3 x For x 1: Tn x 1 Tn x 2 xTn -1 x - Tn -2 x We choose the response to be in the form of a Chebyshev polynomial. 25 Chebyshev Multistage Transformer (cont.) Figure 5.16 (p. 251) The first four Chebyshev polynomials Tn(x). 26 Chebyshev Multistage Transformer (cont.) A Chebyshev response will have equal ripple within the bandwidth. Ae- jN TN secm cos This can be put into a form involving the terms cos (n ) (i.e., a finite Fourier cosine series). Tn x m 1 B A A B n= 2 m A n= 1 2 n ni n= 2 n= 3 g sin a e cr x -1 1 f0 - f 2 m f0 /2 f0 f 2 f - m Note: As frequency decreases, x increases. 27 Chebyshev Multistage Transformer (cont.) We have that, after some algebra, T1 sec m cos sec m cos T2 sec m cos sec 2 m 1 cos 2 -1 T3 sec m cos sec3 m cos 3 3cos - 3sec m cos Tn sec m cos 2 sec m cos Tn-1 sec m cos - Tn-2 sec m cos T1 x x T2 x 2 x 2 1 Hence, the term TN (sec, cos) can be cast into a finite cosine Fourier series expansion. T3 x 4 x 3 - 3 x Tn x 2 xTn -1 x - Tn -2 x 28 Chebyshev Multistage Transformer (cont.) Transformer design Ae- jN TN sec m cos 2 e- jN 0 cos N 1 cos N - 2 .... n cos N - 2n ..... From the above formula we can extract the coefficients n (no general formula is given here). As f 0 0 0 ATN sec m Z L - Z0 Z L Z0 Z -Z 1 A L 0 Z L Z 0 TN sec m 29 Chebyshev Multistage Transformer (cont.) At m m m A TN sec m cos m A TN 1 A A m At 0 : 0 ATN sec m TN sec m 0 Z L - Z0 Z L Z0 sec m 1 A has the same sign as Z L - Z 0 Hence A sgn ZL - Z0 m 30 Chebyshev Multistage Transformer (cont.) Note: The table only shows data for ZL > Z0 since the design can be reversed (Ioad and source switched) for ZL < Z0 . 31 Chebyshev Multistage Transformer (cont.) Bandwidth BW 4 f 2- m f0 At f 0 : ATN sec m Z L - Z0 1 Z - Z 1 Z L - Z0 TN sec m L 0 Z L Z0 A Z L Z0 m Z L Z0 cosh N cosh -1 sec m 1 1 Z L - Z0 -1 sec m cosh cosh m Z L Z0 N Hence X -1 1 ln X ; X 1 2 X 1 1 1 Z L -1 sec m cosh cosh ln N 2 m Z 0 32 Chebyshev Multistage Transformer (cont.) f f0 2 Summary of Design Formulas Ae- jN TN secm cos Reflection coefficient response 1 1 Z L -1 secm cosh cosh ln N 2 m Z0 A sgn ZL - Z0 m A coefficient No formula given for the line impedances. Use the Table from Pozar or generate (“by hand”) the solution by expanding ( ) into a polynomial with terms cos (n ). f 4 2 - m f0 m term Design of line impedances Bandwidth 33 Example Example: three-stage Chebyshev transformer Given Z L 100[] 50[] Z1 Z2 Z3 100[] Z0 50[] m 0.05 Assumed symmetry : Γ3 = Γ0 , Γ 2 = Γ1 A sgn Z L - Z 0 m A 0.05 N 3 A e- j 3 T3 sec m cos Ae- j 3 sec3 m cos 3 3cos - 3sec m cos 2 e- j 3 0 cos 3 1 cos (finite Fourier cosine series form) Equate 34 Example: 3-Section Chebyshev Transformer Equating coefficients from the previous equation on the last slide, we have 2 0 A sec3 m 2 1 3 A sec3 m - 3 A sec m 1 1 100 ln 2 0.05 50 Also, sec m cosh cosh -1 3 1.408 m 44.7 o 0.780[rad] BW 1.007 100.7% BW 0 1 1 3 0.051.408 2 f 4 2 - m f0 0 3 0.0698 1 3 3 0.05 1.408 - 3 0.05 1.408 2 1 2 0.1037 35 Example: 3-Section Chebyshev Transformer Next, use Z n 1 - Z n n Z n 1 Z n Z n 1 Z n 1 n 1- n 1 0.0698 57.5 1- 0.0698 1 0.1037 70.8 Z 2 57.5 1- 0.1037 1 0.1037 87.2 Z 3 70.8 10.1037 Z1 50 Z1 57.5 Z 2 70.8 Z 3 87.2 Checking consistency : Z 4 Z L 87.2 1 0.0698 100.3 1- 0.0698 36 Example: 3-Section Chebyshev Transformer Alternative method: Z n 1 - Z n 1 Z n 1 n ln ln Z n 1 ln Z n 2 n Z n 1 Z n 2 Z n ln Z1 ln Z 0 2 0 ln 50 2 0.0698 4.051 Z1 57.49 ln Z 2 ln Z1 21 4.259 Z 2 70.74 ln Z 3 ln Z 2 2 2 4.466 Z 3 87.05 37 Example: 3-Section Chebyshev Transformer From Table m 0.05, N 3, Z L / Z 0 2, Z 0 50 Z1 1.1475 50 57.4 Z 2 1.4142 50 70.7 Z 3 1.7429 50 87.1 38 Example: 3-Section Chebyshev Transformer g1 / 4 50 line Z1 g 2 / 4 Z2 g 3 / 4 Z3 100 line ZL Z0 S11 dB 20 log10 f -26 2.51 GHz 7.5 GHz f0 5.0GHz BW 99.8% Response from Ansoft Designer 39 Example: 3-Section Chebyshev Transformer Comparison of Binomial (Butterworth) and Chebyshev The Chebyshev design has a higher bandwidth (100% vs. 69%). The increased bandwidth comes with a price: ripple in the passband. Note: It can be shown that the Chebyshev design gives the highest possible bandwidth for a given N and m. 40 Tapered Transformer The Pozar book also talks about using continuously tapered lines to match between an input line Z0 and an output load ZL. (pp. 255-261). Please read this. 41