本文参考Kian Sen Ang and I. D. Robertson, "Analysis and design of impedance-transforming planar Marchand baluns, " in IEEE Transactions on Microwave Theory and Techniques, vol. 49, no. 2, pp. 402-406, Feb. 2001, doi: 10.1109/22.903108.
keywords: {Impedance matching;Couplers;Coupling circuits;Diodes;Microstrip;Frequency;Microwave integrated circuits;MMICs;Monolithic integrated circuits;Wideband},同时参考 洁仔爱吃冰淇淋的博客
仅供学习使用
图 1. 作为两个相同耦合器的对称Marchand巴伦框图。
巴伦的框图如图1所示。它为一个源阻抗为 Z o Z_o Zo的非平衡输入提供平衡输出，连接到负载阻抗 Z 1 Z_1 Z1。通常，阻抗 Z 1 Z_1 Z1和 Z o Z_o Zo是不同的。例如，在平衡二极管混频器应用中，平衡信号需要馈入一对二极管，其阻抗可能与50Ω源阻抗不同。因此，除了提供平衡输出外，巴伦还需要在源阻抗和负载阻抗之间执行阻抗变换。
图1.1：端口顺序定义，1输入，2耦合，3直通（这里2和3与常见定向耦合器不同，想按常见定向耦合器定义将结果交换2与3顺序即可），4隔离
如图1所示，平面Marchand巴伦由两个耦合段组成，每个段在工作中心频率处为四分之一波长。对于对称巴伦，巴伦的散射矩阵可以从两个相同耦合器的散射矩阵推导出来。我们首先考虑源阻抗和负载阻抗都等于 Z o Z_o Zo的情况。根据图1中定义的电压波，具有无限方向性和耦合系数 C C C的理想耦合器（端口顺序如图1.1所示）的散射矩阵由下式给出：
S c o u p l e r = 0 C − j 1 − C 2 0 C 0 0 − j 1 − C 2 − j 1 − C 2 0 0 C 0 − j 1 − C 2 C 0 . ( 1 ) \begin{align*}S_{coupler}=&\begin{bmatrix}0&C&-j\sqrt{1-C^{2}}&0\\ C&0&0&-j\sqrt{1-C^{2}}\\-j\sqrt{1-C^{2}}&0&0&C\\ 0&-j\sqrt{1-C^{2}}&C&0\end{bmatrix}.\end{align*}\qquad(1) Scoupler= 0C−j1−C2 0C00−j1−C2 −j1−C2 00C0−j1−C2 C0 .(1)
然后，利用图1中指示的电压波关系（其形式如下）可以获得巴伦的S参数：
对于图1中第一段耦合线
由耦合器的散射参数关系，可定义归一化电压波：
{ b 1 = C a 2 − j 1 − C 2 a 3 b 2 = C a 1 − j 1 − C 2 a 4 b 3 = − j 1 − C 2 a 1 + C a 4 b 4 = − j 1 − C 2 a 2 + C a 3 \left\{ \begin{array}{l} b_{1} = C a_{2} -j\sqrt{1-C^{2}}a_{3} \\ b_{2} = Ca_{1} -j\sqrt{1-C^{2}}a_{4} \\ b_{3} = - j\sqrt{1-C^{2}}a_{1}+ Ca_{4} \\ b_{4} =- j\sqrt{1-C^{2}} a_{2} + Ca_{3} \end{array} \right. ⎩ ⎨ ⎧b1=Ca2−j1−C2 a3b2=Ca1−j1−C2 a4b3=−j1−C2 a1+Ca4b4=−j1−C2 a2+Ca3
其中 b i b_i bi为反射波 a i a_i ai为入射波
为简化表达式，令：
A = C , β = − j 1 − C 2 A =C, \quad \beta = -j\sqrt{1-C^{2}} A=C,β=−j1−C2
则上式可写为：
{ b 1 = A a 2 + β a 3 b 2 = A a 1 + β a 4 b 3 = β a 1 + A a 4 b 4 = β a 2 + A a 3 \left\{ \begin{array}{l} b_{1} =A a_{2} + \beta a_{3} \\ b_{2} = A a_{1} + \beta a_{4} \\ b_{3} = \beta a_{1} + A a_{4} \\ b_{4} = \beta a_{2} +A a_{3} \end{array} \right. ⎩ ⎨ ⎧b1=Aa2+βa3b2=Aa1+βa4b3=βa1+Aa4b4=βa2+Aa3
考虑巴伦的连接条件（端口1短路， Γ = Z L − Z 0 Z L + Z 0 = − 1 \Gamma=\dfrac{Z_L-Z_0}{Z_L+Z_0}=-1 Γ=ZL+Z0ZL−Z0=−1, b 1 = Γ a 1 b_1=\Gamma a_1 b1=Γa1）：
{ b 1 = A a 2 + β a 3 = − a 1 b 2 = A a 1 + β a 4 b 3 = β a 1 + A a 4 b 4 = β a 2 + A a 3 \left\{ \begin{array}{l} b_{1} =A a_{2} + \beta a_{3}=-a_1 \\ b_{2} = A a_{1} + \beta a_{4} \\ b_{3} = \beta a_{1} +A a_{4} \\ b_{4} = \beta a_{2} +A a_{3} \end{array} \right. ⎩ ⎨ ⎧b1=Aa2+βa3=−a1b2=Aa1+βa4b3=βa1+Aa4b4=βa2+Aa3
消去 a 1 a_1 a1
{ b 2 = − A 2 a 2 − A β a 3 + β a 4 b 3 = − A β a 2 − β 2 a 3 + A a 4 b 4 = β a 2 + A a 3 ( 1.1 ) \left\{ \begin{array}{l} b_{2} = -A^2 a_{2}-A\beta a_3 + \beta a_{4} \\ b_{3} =-A\beta a_{2} -\beta^2 a_{3}+Aa_4 \\ b_{4} = \beta a_{2} +A a_{3} \end{array} \right.\quad (1.1) ⎩ ⎨ ⎧b2=−A2a2−Aβa3+βa4b3=−Aβa2−β2a3+Aa4b4=βa2+Aa3(1.1)
图1中第二段耦合线
由耦合线方程，带入边界条件，端口 1 ′ 1^{\prime} 1′开路，端口 2 ′ 2^{\prime} 2′短路：
{ b 1 ′ = A a 2 ′ + β a 3 ′ = a 1 ′ b 2 ′ = A a 1 ′ + β a 4 ′ = − a 2 ′ b 3 ′ = β a 1 ′ + A a 4 ′ b 4 ′ = β a 2 ′ + A a 3 ′ \left\{ \begin{array}{l} b_{1}^{\prime} =A a_{2}^{\prime} + \beta a_{3}^{\prime}=a_{1}^{\prime}\\ b_{2}^{\prime} = A a_{1}^{\prime} + \beta a_{4}^{\prime}=-a_{2}^{\prime} \\ b_{3}^{\prime} = \beta a_{1}^{\prime} + A a_{4}^{\prime} \\ b_{4}^{\prime} = \beta a_{2}^{\prime} +A a_{3}^{\prime} \end{array} \right. ⎩ ⎨ ⎧b1′=Aa2′+βa3′=a1′b2′=Aa1′+βa4′=−a2′b3′=βa1′+Aa4′b4′=βa2′+Aa3′
可解得 a 1 ′ a_1^{\prime} a1′ 和 a 2 ′ a_2^{\prime} a2′关于 a 3 ′ a_3^{\prime} a3′ a 4 ′ a_4^{\prime} a4′的表达式
将解代回 b 3 ′ b_3^{\prime} b3′ 和 b 4 ′ b_4^{\prime} b4′ 的表达式：
{ b 3 ′ = β a 1 ′ + A a 4 ′ = ( β 2 − β 2 A 2 1 + A 2 ) a 3 ′ + ( A − A β 2 1 + A 2 ) a 4 ′ b 4 ′ = β a 2 ′ + A a 3 ′ = ( A − A β 2 1 + A 2 ) a 3 ′ + ( − β 2 1 + A 2 ) ( 1.2 ) \left\{ \begin{aligned} b_{3}^{\prime} = \beta a_{1}^{\prime} + A a_{4}^{\prime}=(\beta^2-\dfrac{\beta^2A^2}{1+A^2})a_3^{\prime}+(A-\dfrac{A\beta^2}{1+A^2})a_4^{\prime} \\b_{4}^{\prime} = \beta a_{2}^{\prime} +A a_{3}^{\prime}=(A-\dfrac{A\beta^2}{1+A^2})a_3^{\prime}+(-\dfrac{\beta^2}{1+A^2}) \end{aligned} \right.\quad (1.2) ⎩ ⎨ ⎧b3′=βa1′+Aa4′=(β2−1+A2β2A2)a3′+(A−1+A2Aβ2)a4′b4′=βa2′+Aa3′=(A−1+A2Aβ2)a3′+(−1+A2β2)(1.2)
结合两者链接关系（第一段线四端口与第二段线三端口连在一起）我们有
{ b 4 = a 3 ′ b 3 ′ = a 4 ( 1.3 ) \left\{ \begin{array}{l} b_{4} = a_{3}^{\prime}\\ b_{3}^{\prime} = a_4 \end{array} \right.\quad (1.3) {b4=a3′b3′=a4(1.3)
将 a 4 a_4 a4、 b 4 b_4 b4、 a 3 ′ a_3^{\prime} a3′、 b 3 ′ b_3^{\prime} b3′ 全部消去后，得到仅关于 a 2 a_2 a2、 a 3 a_3 a3、 a 4 ′ a_4^{\prime} a4′ 及对应 b b b 的方程组：
{ b 2 = ( β 4 − A 2 β 2 1 + A 2 − A 2 ) a 2 + ( A β 3 − A 3 β 3 1 + A 2 − A β ) a 3 + ( A β − A β 3 1 + A 2 ) a 4 ′ b 3 = ( A β 3 − A 3 β 3 1 + A 2 − A β ) a 2 + ( A 2 β 2 − A 4 β 2 1 + A 2 − β 2 ) a 3 + ( A 2 − A 2 β 2 1 + A 2 ) a 4 ′ b 4 ′ = ( A β − A β 3 1 + A 2 ) a 2 + ( A 2 − A 2 β 2 1 + A 2 ) a 3 − ( β 2 1 + A 2 ) a 4 ′ ( 1.4 ) \left\{ \begin{array}{l} \begin{aligned} b_{2} &=(\beta^4-\dfrac{A^2\beta^2}{1+A^2}-A^2)a_2+(A\beta^3-\dfrac{A^3\beta^3}{1+A^2}-A\beta)a_3+(A\beta-\dfrac{A\beta^3}{1+A^2})a_4^\prime \end{aligned} \\15pt \begin{aligned} b_{3} &= \left( {A\beta^{3}} -\dfrac{A^3\beta^3}{1+A^2}- A \beta \right) a_{2} + \left(A^2\beta^{2}- \frac{A^{4} \beta^{2}}{1+A^{2}} -\beta^{2} \right) a_{3} + (A^2-\frac{A^2\beta^2}{1+A^2})a_{4}^{\prime} \end{aligned} \\15pt \begin{aligned} b_{4}^{\prime} &=(A\beta-\frac{A\beta^3}{1+A^2})a_{2} +(A^2-\frac{A^2\beta^2}{1+A^2})a_{3} - \left( \frac{\beta^{2}}{1+A^{2}} \right) a_{4}^{\prime} \end{aligned} \end{array} \right.\quad(1.4) ⎩ ⎨ ⎧b2=(β4−1+A2A2β2−A2)a2+(Aβ3−1+A2A3β3−Aβ)a3+(Aβ−1+A2Aβ3)a4′b3=(Aβ3−1+A2A3β3−Aβ)a2+(A2β2−1+A2A4β2−β2)a3+(A2−1+A2A2β2)a4′b4′=(Aβ−1+A2Aβ3)a2+(A2−1+A2A2β2)a3−(1+A2β2)a4′(1.4)
A = C , β = − j 1 − C 2 A =C, \quad \beta = -j\sqrt{1-C^{2}} A=C,β=−j1−C2
将我们一开始的条件带入即可得到下式：
S balun = 1 − 3 C 2 1 + C 2 j 2 C 1 − C 2 1 + C 2 − j 2 C 1 − C 2 1 + C 2 j 2 C 1 − C 2 1 + C 2 1 − C 2 1 + C 2 2 C 2 1 + C 2 − j 2 C 1 − C 2 1 + C 2 2 C 2 1 + C 2 1 − C 2 1 + C 2 . ( 2 ) S_{\text{balun}} = \begin{bmatrix} \frac{1 - 3C^2}{1 + C^2} & j \frac{2C \sqrt{1 - C^2}}{1 + C^2} & -j \frac{2C \sqrt{1 - C^2}}{1 + C^2} \\ j \frac{2C \sqrt{1 - C^2}}{1 + C^2} & \frac{1 - C^2}{1 + C^2} & \frac{2C^2}{1 + C^2} \\ -j \frac{2C \sqrt{1 - C^2}}{1 + C^2} & \frac{2C^2}{1 + C^2} & \frac{1 - C^2}{1 + C^2} \end{bmatrix} . \qquad(2) Sbalun= 1+C21−3C2j1+C22C1−C2 −j1+C22C1−C2 j1+C22C1−C2 1+C21−C21+C22C2−j1+C22C1−C2 1+C22C21+C21−C2 .(2)