Theory of operation

As seen in Fig. 1, a BCB is formed of two conductors sandwiched by a substrate, with l_P, l_S, w_P, and w_s as lengths and widths of primary (bottom) and secondary (top) conductors, respectively. In addition, h is the substrate thickness. When port 1 is excited, power is transferred from the bottom conductor to the top conductor into ports 2 and 3 [Reference Long18]. BCB, however, suffers from poor isolation between output ports [Reference Ma, Yan, Yeo and Lim19]. Therefore, an IC is placed between ports 2 and 3 to achieve good isolation response.

Figure 1.

Broadside-coupled balun with isolation circuit.

Figure 2 represents the equivalent circuit model of the BCB of Fig. 1 [Reference Jaisson20]. The BCB can be modelled by a transformer comprised of primary inductor, L_P, secondary inductor, L_s, and mutual inductance and capacitance denoted by L_m, and C_m, respectively. In addition, Z ₀: port 1 impedance, Z ₁: ports 2 and 3 impedances, I _P: current of primary inductor, I _S: current of secondary inductor, and I ₀: current at input port. Ohmic and dielectric losses are neglected for simplicity.

Figure 2.

Equivalent circuit model of balun of Figure 1.

Mutual inductance, L_m, can be determined as a function of self-inductances, L_P and L_S, such that ${L_m} = {K_m}\sqrt {{L_p}{L_s}} $, where K_m is the magnetic coupling coefficient. Equation (1) expresses impedance transformation ratio (2Z ₁/Z ₀) in terms of primary and secondary inductors.

(1)\begin{equation}\frac{{2{Z_1}}}{{{Z_0}}} = {K_m}^2\frac{{{L_s}}}{{{L_p}}} = \frac{1}{{{n^2}}} \cong {\left( {\frac{{{N_s}}}{{{N_p}}}} \right)^2}\end{equation}

where N_P and N_s are number of turns for primary and secondary windings, respectively. Also, n is the ratio of primary to secondary turns [Reference Abrie21].

Extracting the S-matrix of isolated balun

Under the lossless condition and neglecting magnetic flux leakage, the S-matrix of the balun of Fig. 1, without the IC, can be given by [Reference Ma, Yan, Yeo and Lim19]:

(2)\begin{equation}\left[ S \right] = \left[ {\begin{array}{*{20}{c}} 0&{j/\sqrt 2 }&{ - j/\sqrt 2 } \\ {j/\sqrt 2 }&{1/2}&{1/2} \\ { - j/\sqrt 2 }&{1/2}&{1/2} \end{array}} \right]\end{equation}

As equation (2) implies, while the input port is matched, the output ports are not matched without the IC. Return losses of output ports (S ₂₂, S ₃₃) and isolation between them (S ₃₂) are 6 dB; emphasizing the need for an IC. The S-matrix of the balun with IC can be obtained if admittance matrix (Y-matrix) of the balun without IC is calculated based on port designations in Fig. 1, as:

(3)\begin{align}{\left[ {\text{Y}} \right]_{{\text{without IC}}}}& = \left( {{y^*}{\text{ - y}}{{\text{k}}^{{\text{ - 1}}}}{\text{S}}{k^*}} \right){\left( {{{\text{k}}^{{\text{ - 1}}}}{\text{S}}{k^*}{\text{ + }}{{\text{U}}_{\text{n}}}} \right)^{{\text{ - 1}}}} \nonumber\\ &= \left[ {\begin{array}{*{20}{c}} {\text{0}}&{\frac{{{\text{ - j}}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}}&{\frac{{\text{j}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}} \\ {\frac{{{\text{ - j}}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}}&{\text{0}}&{\text{0}} \\ {\frac{{\text{j}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}}&{\text{0}}&{\text{0}} \end{array}} \right]\end{align}

where U_n is the identity matrix, and k ⁻¹, k*, y = y* are given by:

(4)\begin{align}{k^{ - 1}} &= \left[ {\begin{array}{*{20}{c}} {\sqrt {{Z_0}} }&0&0 \\ 0&{\sqrt {{Z_1}} }&0 \\ 0&0&{\sqrt {{Z_1}} } \end{array}} \right] \nonumber\\ {k^*} &= \left[ {\begin{array}{*{20}{c}} {{1 \mathord{\left/ {\vphantom {1 {\sqrt {{Z_0}} }}} \right. } {\sqrt {{Z_0}} }}}&0&0 \\ 0&{{1 \mathord{\left/ {\vphantom {1 {\sqrt {{Z_1}} }}} \right. } {\sqrt {{Z_1}} }}}&0 \\ 0&0&{{1 \mathord{\left/ {\vphantom {1 {\sqrt {{Z_1}} }}} \right. } {\sqrt {{Z_1}} }}} \end{array}} \right] \nonumber\\ {y^*} &= \left[ {\begin{array}{*{20}{c}} {{1 \mathord{\left/ {\vphantom {1 {{Z_0}}}} \right. } {{Z_0}}}}&0&0 \\ 0&{{1 \mathord{\left/ {\vphantom {1 {{Z_1}}}} \right. } {{Z_1}}}}&0 \\ 0&0&{{1 \mathord{\left/ {\vphantom {1 {{Z_1}}}} \right. } {{Z_1}}}} \end{array}} \right]\end{align}

Now, the admittance matrix of the IC with input impedance Z _IC can be given as:

(5)\begin{equation}{\left[ Y \right]_{{\text{IC}}}}{\text{ = }}\left[ {\begin{array}{*{20}{c}} {\text{0}}&{\text{0}}&{\text{0}} \\ {\text{0}}&{\frac{{\text{1}}}{{{Z_{{\text{IC}}}}}}}&{\frac{{\text{1}}}{{{Z_{{\text{IC}}}}}}} \\ {\text{0}}&{\frac{{\text{1}}}{{{Z_{{\text{IC}}}}}}}&{\frac{{\text{1}}}{{{Z_{{\text{IC}}}}}}} \end{array}} \right]\end{equation}

Therefore, Y-matrix of the isolated balun can be calculated from equations (3) and (5) as:

(6)\begin{equation}{\left[ {\text{Y}} \right]_{{\text{with IC}}}}{\text{ = }}{\left[ {\text{Y}} \right]_{{\text{without IC}}}}{\text{ + }}{\left[ {\text{Y}} \right]_{{\text{IC}}}}{\text{ = }}\left[ {\begin{array}{*{20}{c}} {\text{0}}&{\frac{{{\text{ - j}}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}}&{\frac{{\text{j}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}} \\ {\frac{{{\text{ - j}}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}}&{\frac{{\text{1}}}{{{{\text{Z}}_{{\text{IC}}}}}}}&{\frac{{\text{1}}}{{{{\text{Z}}_{{\text{IC}}}}}}} \\ {\frac{{\text{j}}}{{\sqrt {{\text{2}}{{\text{Z}}_{\text{0}}}{{\text{Z}}_{\text{1}}}} }}}&{\frac{{\text{1}}}{{{{\text{Z}}_{{\text{IC}}}}}}}&{\frac{{\text{1}}}{{{{\text{Z}}_{{\text{IC}}}}}}} \end{array}} \right]\end{equation}

which in S-parameters reads as:

(7)\begin{align}{\left[ {\text{S}} \right]_{{\text{with IC}}}} &= -{\text{ k}}{\left( {{\text{Y + y}}} \right)^{{\text{ - 1}}}}\left( {{\text{Y - }}{y^*}} \right){{\text{k}}^{{\text{ - 1}}}} \nonumber\\ & = \left[ {\begin{array}{*{20}{c}} {\text{0}}&{\frac{{\text{j}}}{{\sqrt {\text{2}} }}}&{\frac{{{\text{ - j}}}}{{\sqrt {\text{2}} }}} \\ {\frac{{\text{j}}}{{\sqrt {\text{2}} }}}&{\frac{{\text{1}}}{{{\text{2}}{{\text{Z}}_{\text{1}}}/{{\text{Z}}_{{\text{IC}}}}}}{\text{ - 1}}}&{\frac{{\text{1}}}{{{\text{2}}{{\text{Z}}_{\text{1}}}/{{\text{Z}}_{{\text{IC}}}}}}{\text{ - 1}}} \\ {\frac{{{\text{ - j}}}}{{\sqrt {\text{2}} }}}&{\frac{{\text{1}}}{{{\text{2}}{{\text{Z}}_{\text{1}}}/{{\text{Z}}_{{\text{IC}}}}}}{\text{ - 1}}}&{\frac{{\text{1}}}{{{\text{2}}{{\text{Z}}_{\text{1}}}/{{\text{Z}}_{{\text{IC}}}}}}{\text{ - 1}}} \end{array}} \right]\end{align}

Comparison of equations (2) and (5) shows that the IC does not, theoretically, have any effect on the balun insertion losses (S ₂₁, S ₃₁). Moreover, from equation (7) “perfect” output match and isolation can be achieved if 2Z ₁ = Z _IC is satisfied.

Analysis of the wideband IC

Given the 180° phase difference between ports 2 and 3 in the balun, the IC can be realized by a λ/2 TL as depicted in Fig. 3(a). For this phase difference, a single λ/2 TL offers a limited bandwidth (see Fig. 3), and so we try here to provide analysis that results into a wideband IC. Figure 3 shows two different ICs; one with a TL of length l ₀ = λ ₀/2 (Fig. 3(a)), and the other with two TLs of lengths l ₁ = λ ₁/2, and l ₂ = λ ₂/2 (Fig. 3(b)).

Figure 3.

(a) Conventional IC and (b) the proposed IC.

In Fig. 3(a), input impedance Z_in of the IC’s constituent TL with characteristic impedance ${Z_s}$ can be calculated using equation (8):

(8)\begin{align}{Z_{in}}& ={Z_s}\frac{{{Z_{\text{1}}}{\text{ + }}j{Z_s}{\text{tan}}\frac{{{\text{2}}\pi {l_{\text{0}}}}}{\lambda }}}{{{Z_s}{\text{ + }}j{Z_{\text{1}}}{\text{tan}}\frac{{{\text{2}}\pi {l_{\text{0}}}}}{\lambda }}}{\text{ = }}{Z_s}\frac{{{Z_1}{\text{ + }}j{Z_s}{\text{tan}}\frac{{\pi {\lambda _{\text{0}}}}}{\lambda }}}{{{Z_s}{\text{ + }}j{Z_i}{\text{tan}}\frac{{\pi {\lambda _{\text{0}}}}}{\lambda }}} \nonumber\\ & = {Z_s}\frac{{{Z_{\text{1}}}{\text{ + }}j{Z_s}{\text{tan}}\frac{{\pi \omega }}{{{\omega _{\text{0}}}}}}}{{{Z_s}{\text{ + }}j{Z_{\text{1}}}{\text{tan}}\frac{{\pi \omega }}{{{\omega _{\text{0}}}}}}}\end{align}

where ω ₀ is the center frequency, and ω is operating frequency.

The input impedance of the single-TL IC, Z _IC-S, can be given by:

(9)\begin{equation}{Z_{{\text{IC - S}}}}{\text{ = }}{Z_1} + {Z_{in}}\end{equation}

Now for very small bandwidth, Δω ₀, around ω ₀, let ω = ω ₀ + Δω ₀. Thus, bandwidth of the IC of Fig. 3(a) can be derived by calculating Z _IC-S:

(10)\begin{equation}{Z_{{\text{IC - S}}}} \cong \frac{{{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}}{\text{ + }}{Z_{\text{1}}}{Z_s}}}{{{Z_s}}}{\text{ + }}j\left( {\frac{{{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}}}}{{{Z_s}}}} \right)\frac{{\pi \left( {\omega - {\omega _{\text{0}}}} \right)}}{{{\omega _{\text{0}}}}}\end{equation}

Fractional bandwidth (FBW) of IC in Fig. 3(a) can be obtained from equation (10) as:

(11)\begin{equation}\frac{{{\text{2}}\Delta {\omega _{\text{0}}}}}{{{\omega _{\text{0}}}}} = \frac{{\text{2}}}{\pi }\left( {{\text{1 + }}\frac{{{Z_{\text{1}}}{Z_s}}}{{{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}}}}} \right)\end{equation}

Therefore, to obtain maximum bandwidth in a single-TL IC, according to equation (11), one has to choose Z_S close to Z ₁. For the case of a PA, where Z ₁ (and in turn Z_S) is small, realizing correspondingly thick TL proves almost impractical.

Now, assuming ω = ω ₁ + Δω ₁ and ω = ω ₂ + Δω ₂, Z_inm (where m = 1, 2), Z_inm for the double-TL IC can be given by:

(12)\begin{equation}{Z_{in}}_m{\text{ = }}{Z_s}\frac{{{Z_i}{\text{ + }}j{Z_s}{\text{tan}}\frac{{\pi {\lambda _m}}}{\lambda }}}{{{Z_s}{\text{ + }}j{Z_i}{\text{tan}}\frac{{\pi {\lambda _m}}}{\lambda }}} = {Z_s}\frac{{{Z_{\text{1}}}{\text{ + }}j{Z_s}{\text{tan}}\frac{{\pi \omega }}{{{\omega _m}}}}}{{{Z_s}{\text{ + }}j{Z_{\text{1}}}{\text{tan}}\frac{{\pi \omega }}{{{\omega _m}}}}}\end{equation}

(13)\begin{equation}{Z_{in}}_m \cong \frac{{{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}}{\text{ + }}{Z_{\text{1}}}{Z_s}}}{{{Z_s}}}{\text{ + }}j\left( {\frac{{{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}}}}{{{Z_s}}}} \right)\frac{{\pi \left( {\omega - {\omega _m}} \right)}}{{{\omega _m}}}\end{equation}

Therefore, Z _IC-D for Fig. 3(b) can be calculated as:

(14)\begin{equation}{Z_{{\text{IC - D}}}}{\text{ = ((}}{Z_1} + {Z_{in1}}){\text{(}}{Z_1} + {Z_{in2}}))/\left( {2{Z_1} + {Z_{in1}} + {Z_{in2}}} \right)\end{equation}

Substituting equation (13) into (14) gives:

(15)\begin{align}{Z_{{\text{IC - D}}}} &\cong \frac{{{Z_{\text{0}}}}}{{\text{2}}}\left( {\text{1 + }}j\pi \left( {\frac{{\left( {\omega - {\omega _{\text{1}}}} \right)}}{{{\omega _{\text{1}}}}}{\text{ + }}\frac{{\left( {\omega - {\omega _{\text{2}}}} \right)}}{{{\omega _{\text{2}}}}}} \right) \right. \nonumber\\ &\quad \times \left.\left( {\frac{{{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}} - {Z_{\text{1}}}{Z_s}}}{{{\text{2}}\left( {{Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}}{\text{ + }}{Z_{\text{1}}}{Z_s}} \right)}}} \right) \right)\end{align}

Assuming Δω ₁ ≈ Δω ₂, the resonance frequency of the IC of Fig. 3(b) and its FBW can be determined as equation (16):

(16)\begin{align}{ }\frac{{{{2\Delta \omega }}}}{{{{{\omega }}_{\text{r}}}}} &= \frac{{\text{4}}}{{t{\pi }}}\frac{{{{\text{Z}}_{\text{s}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}^{\text{2}}{\text{ + }}{{\text{Z}}_{\text{1}}}{{\text{Z}}_{\text{s}}}}}{{{{\text{Z}}_{\text{s}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}{{\text{Z}}_{\text{s}}}}}{\text{ = }}\frac{{\text{4}}}{{t{\pi }}}\left( {{\text{1 + }}\frac{{{{\text{Z}}_{\text{1}}}{{\text{Z}}_{\text{s}}}}}{{{{\text{Z}}_{\text{s}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}{{\text{Z}}_{\text{s}}}}}} \right){\text{, }} \nonumber\\ {{{\omega }}_{\text{r}}}& = \frac{{{\text{2}}{{{\omega }}_{\text{1}}}{{{\omega }}_{\text{2}}}}}{{{{{\omega }}_{\text{1}}}{\text{ + }}{{{\omega }}_{\text{2}}}}}\end{align}

Now, the bandwidth advantage of the proposed IC of Fig. 3(b) over IC of Fig. 3(a) becomes clear by dividing FBWs such that:

(17)\begin{equation}\frac{{{\text{B}}{{\text{W}}_{{\text{Double - TL }}}}}}{{{\text{B}}{{\text{W}}_{{\text{Single - TL }}}}}}{\text{ = 2}}\left( {{\text{1 + }}\frac{{{{\text{Z}}_{\text{1}}}{{\text{Z}}_{\text{s}}}}}{{{{\text{Z}}_{\text{s}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}^{\text{2}}{\text{ - }}{{\text{Z}}_{\text{1}}}{{\text{Z}}_{\text{s}}}}}} \right)\end{equation}

Let $A = {Z_s}^{\text{2}} - {Z_{\text{1}}}^{\text{2}} - {Z_{\text{1}}}{Z_s}$; provided that A > 0, then, the FBW of double-TL IC of Fig. 3(b) will be at least twice that of the single-TL IC of Fig. 3(a). For A > 0 to be true, Z_S is derived in terms of Z ₁ (see equation (18)) and plotted against it in Fig. 4. The hatched area shows the region that permits positive sign for A. That is, for the proposed double-TL IC and for a given Z ₁, the corresponding Z_S can be obtained in this area, which guarantees at least doubled bandwidth compared to a single-TL IC. Note that the red line indicates the minimum values that Z_S can take for given Z ₁ under this condition.

(18)\begin{equation}{\text{(}}{Z_s} - {\text{(}}\frac{{{\text{1 + }}\sqrt {\text{5}} }}{{\text{2}}}{\text{)}}{Z_{\text{1}}}{\text{)(}}{Z_s} - {\text{(}}\frac{{{\text{1}} - \sqrt {\text{5}} }}{{\text{2}}}{\text{)}}{Z_{\text{1}}}{\text{)0}}\end{equation}

Figure 4.

The hatched region indicates the permitted values of Z_S and Z ₁ for the IC of Figure 3(b).

Let’s for a PA application, assume Z ₁ = 6 Ω; then according to Fig. 4, ${{\text{Z}}_{\text{s}}}{\text{ - (}}\frac{{{\text{1 + }}\sqrt {\text{5}} }}{{\text{2}}}{\text{)}}{{\text{Z}}_{\text{1}}}{\text{0}}$ is forced (above the red line within the hatched area), which for the given Z ₁ translates into Z_S > 9.7 Ω. This means that according to equation (17), in principle, for practical Z_S values (between 10 Ω and 100 Ω), a double-TL IC can offer a BW around 2.1–96.83 times the BW of the single-TL IC.

Here, a practical case of Z_S = 25 Ω is selected for the sake of argument. Equation (17) ensures that the bandwidth of IC of Fig. 3(b) is approximately 2.7 times larger than that of the IC of Fig. 3(a). Now, we can verify this numerically. If we calculate Z _IC-S, and Z _IC-D from equations (9) and (14) and substitute for isolation (S ₃₂ = S ₂₃) in the matrix of equation (7), then we can plot isolation versus frequency for single-TL IC of Fig. 3(a) against double-TL IC of Fig. 3(b). As is evident in Fig. 5, the double-TL IC (red starred solid line) offers 3.2 times (with reference to −20 dB return loss (RL) level) larger bandwidth than single-TL IC (blue circled solid line). To further verify the results, both ICs of Fig. 3 are simulated in AWR Microwave Office software and isolation results are plotted in Fig. 5. Note that the results are in very good agreement. Furthermore, S ₂₃ of the classical BCD balun (i.e. without an IC) is also plotted in Fig. 5 against single-TL and double-TL cases, and as seen below, the S ₂₃ of balun without IC is very poor throughout the band, and in particular around the center frequency (and upward). Adding a single-TL IC improves this response and evidently, the double-TL IC improves the S ₂₃ significantly.

Figure 5.

Isolation of single-TL IC against the proposed double-TL IC.

From a design perspective, the λ/2 length of the IC compared to the compact size of the balun itself, poses a size challenge, and so, it is essential to make the IC compact. This is done by using the equivalent T-network of the IC TLs loaded with shunt capacitors as shown in Fig. 6.

Figure 6.

Equivalent circuit of double-TL ICs formed of shunt loaded capacitors.

Design procedure

Now that different blocks of the isolated BCB are analyzed, design of the proposed balun can be elaborated through a simple straightforward procedure, depicted in the flowchart of Fig. 7.

Figure 7.

Design procedure for the proposed isolated broadside-coupled balun.

Note that the actual design involves much more details to be seen, and that since an all-ports-matched balun suitable for PA applications is intended, the procedure is presented by means of a balun that satisfies the PA requirements.

Case study: isolated balun design for PA applications

Design formulas and considerations for isolated BCB is illustrated through an example based on balun in Fig. 8.

Figure 8.

(a) 3D view of the BCB without isolation circuit: primary (blue), and secondary (red) windings sandwiching the substrate, and (b) associated geometrical parameters of spiral inductor.

For the BCB balun of Fig. 8(a) at f_c, the constitutive parameters of Fig. 2 are derived by:

(19)\begin{equation}{f_c}{\text{ = }}\frac{{\text{1}}}{{{\text{2}}\pi \sqrt {{\text{(}}{C_m}/2)\left( {{L_P}{\text{ + }}n{L_m}} \right) - \left( {1/4} \right){{\left( {{Z_{\text{1}}}{C_m}} \right)}^{\text{2}}}\left( {{\text{1 + }}n} \right)} }}\end{equation}

where $n = \sqrt {{Z_0}/2{Z_1}} $, and L_P and L_S in nH can be found from [Reference Mohan, Del Mar Hershenson, Boyd and Lee22]:

(20)\begin{equation}{L_{{\text{spiral}}}}{\text{ = }}{L_P}{\text{ = }}{L_S} = \frac{{{\text{1}}{\text{.27}}{\mu _{\text{0}}}{N^{\text{2}}}{d_a}}}{{\text{2}}}\left( {{\text{ln}}\left( {\frac{{{\text{2}}{\text{.07}}}}{{{\rho }}}} \right){\text{ + 0}}{\text{.18}}\rho {\text{ + 0}}{\text{.13}}{\rho ^{\text{2}}}} \right)\end{equation}

where ${d_a} = \frac{{{d_o} + {d_i}}}{2}$, $\rho = \frac{{{d_o} - {d_i}}}{{{d_o} + {d_i}}}$, N is the number of turns, and d_o, d_i are the outer and inner diameters of the coil, respectively (see Fig. 8(b)). In addition, for the stacked overlay spiral, C_m which is proportional to the area occupied by the transformer can be found by [Reference Yue and Wong23]:

(21)\begin{equation}{C_m}{\text{ = }}\frac{{\text{1}}}{{\text{2}}}\frac{{{\varepsilon _{\text{0}}}{\varepsilon _r}}}{h}{\text{.}}\left( {{d_o}^{\text{2}} - {d_i}^{\text{2}}} \right)\end{equation}

where h, and ${\varepsilon _r}$ are the substrate thickness, and dielectric constant, respectively.

Also, according to equation (1), L_m can be determined in terms of K_m, L_S, and L_P as ${L_m}{\text{ = }}{K_m}\sqrt {{L_P}{L_S}} $. The value of K_m mostly depends on the geometry of windings and h. While K_m for perfect coupling is unity, it is less than unity in practice. The value of K_m for broadside-coupled structures, which can be calculated from K_m ≈ 0.9 − h/AD for average diameter of AD, is empirically reported to be mostly between 0.7 and 0.9 for strong coupling [Reference Mohan, Yue, Del Mar Hershenson, Wong and Lee24].

In PAs, large step-down impedance transformation ratio is desirable. For one thing, most PAs have low output resistance; in most cases, less than 10 Ω (depending on device voltage and power rating). Here, an isolated balun is designed to have 4:1 impedance transformation, i.e. n = 2. At f_c = 97 MHz and for Z ₀ = 50 Ω and 2Z ₁ = 12.5 Ω, number of turns for primary and secondary windings can be calculated from equation (1) as N_P = 2, N_S = 1. In other words, assuming perfect coupling, K_m = 1, based on equation (2), L_P/L_S = 4. Moreover, assuming d_o = 38 mm, d_i = 10 mm, geometrical and circuit model parameters are calculated using equations (19)–(21) to be L_P = 108.4 nH, L_S = 27.11 nH, C_m = 26.4 pF, w_P = 12 mm, w_S = 5.5 mm, S = 0.7 mm. At this stage, the IC is designed and added to the balun. The optimized IC parameters are obtained using Fig. 4 and equation (18) as Z _S = 80 Ω, C ₁ = 18 pF, C ₂ = 1.3 nF, C ₃ = 68 pF, C ₄ = 1.2 pF, R ₁ = 6.25 Ω. The final schematic of the proposed isolated BCB is depicted in Fig. 9.

Figure 9.

Schematic of the final isolated broadside-coupled balun.

There are also some PHC constraints in choosing design parameters. More specifically, increasing substrate thickness, h, will improve PHC. Moreover, tracks should be wide enough to have suitable current-handling for the transistor. Maximum power rating of the isolation resistors (R ₁–R ₄) should be taken into account when determining the balun output power. For instance, here for a 100 W PA, each resistor should at least withstand 25 W for a robust design. Furthermore, tuning capacitors are used at input and output of the balun to improve phase and amplitude imbalance, insertion loss, and input return loss; though frequency shifting effect of such capacitors cannot be neglected.