Wilkinson Power Divider Calculator — Equal & Unequal Split

Design a two-way Wilkinson power divider for any power ratio: the quarter-wave arm impedances, the isolation resistor, the output terminations and their matching transformers, and the physical quarter-wave length, with every line checked against the impedance range your process can make.

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Formula

K^2 = \frac{P_3}{P_2},\quad Z_{03} = Z_0\sqrt{\frac{1+K^2}{K^3}},\quad Z_{02} = K^2 Z_{03},\quad R = Z_0\left(K + \frac{1}{K}\right),\quad R_2 = Z_0 K,\quad R_3 = \frac{Z_0}{K}

Reference: Pozar, Microwave Engineering, 4th ed. (2012), §7.3, Eqs. (7.37a)–(7.37c)

Z_0— System impedance (Ω)

K^2— Power ratio P₃/P₂

Z_{02}, Z_{03}— Quarter-wave arm impedances to ports 2 and 3 (Ω)

R— Isolation resistor between the arm ends (Ω)

R_2, R_3— Impedances the arms are terminated in (Ω)

Z_{T2}, Z_{T3}— Output transformers back to Z₀: √(Z₀R₂), √(Z₀R₃) (Ω)

How It Works

A Wilkinson power divider splits a signal between two outputs while keeping every port matched and the two outputs isolated from each other. It is two quarter-wave lines leaving a common input, with a resistor across their far ends. In the equal split each line is $\sqrt{2}\,Z_0$ and the resistor is $2Z_0$ : 70.71 Ω and 100 Ω in a 50 Ω system. Each line transforms its 50 Ω output to 100 Ω at the input, and the two 100 Ω in parallel present 50 Ω. A wave arriving at one output reaches the other by two paths, through the resistor and back through both lines, and the two cancel, so no power crosses between the outputs.

Unequal split

For a power ratio $K^2 = P_3/P_2$ the design equations are (Pozar, Microwave Engineering, 4th ed., equations 7.37a to 7.37c):

Z_{03} = Z_0\sqrt{\frac{1+K^2}{K^3}}, \qquad Z_{02} = K^2 Z_{03}, \qquad R = Z_0\left(K + \frac{1}{K}\right)

The arms are then terminated not in $Z_0$ but in $R_2 = Z_0 K$ and $R_3 = Z_0/K$ , and each output is brought back to $Z_0$ by a further quarter-wave transformer of $\sqrt{Z_0 R_2} = Z_0\sqrt{K}$ and $\sqrt{Z_0 R_3} = Z_0/\sqrt{K}$ . At $K = 1$ everything reduces to the equal split. Replacing the ratio by its reciprocal swaps the two arms and the two terminations, and the resistor stays the same.

Line length

Every line is a quarter wavelength at the centre frequency:

\ell = \frac{\lambda}{4} = \frac{c}{4 f_0\sqrt{\varepsilon_{eff}}}

On microstrip each impedance has its own width and so its own effective permittivity. The calculator uses the one value you enter for every line, so find each width's $\varepsilon_{eff}$ with the microstrip calculator when you lay the lines out.

Validity

The equations assume ideal TEM lines that are exactly a quarter wave long and a resistor with no parasitics at the arm ends. A single section is matched and isolated only at the centre frequency. The practical limit is the line impedances themselves: an unequal split pushes one arm up and the other down, and a line narrower or wider than your process can make cannot be built. The calculator checks the two arms and the two output transformers against the range you enter, 20 to 150 Ω by default, and flags any line outside it while still showing its value. Pozar's Example 8.6 takes 20 Ω and 120 Ω as the practical extremes of a microstrip process.

Worked Example

Problem: A 2.4 GHz feed network on FR-4 microstrip must send twice as much power to one antenna as to the other, P₃/P₂ = 1/2, in a 50 Ω system. The board can make lines from 20 to 120 Ω, and the effective permittivity is about 3.3.

Step 1 - Ratio: K² = 0.5, K = 0.7071

Step 2 - Arm impedances (Pozar equations 7.37a and 7.37b): Z₀₃ = 50 × √((1 + 0.5) / 0.7071³) = 102.99 Ω Z₀₂ = 0.5 × 102.99 = 51.49 Ω

Step 3 - Isolation resistor: R = 50 × (0.7071 + 1/0.7071) = 106.07 Ω

Step 4 - Arm terminations: R₂ = 50 × 0.7071 = 35.36 Ω R₃ = 50 / 0.7071 = 70.71 Ω

Step 5 - Output transformers back to 50 Ω: Z_T2 = √(50 × 35.36) = 42.04 Ω Z_T3 = √(50 × 70.71) = 59.46 Ω

Step 6 - Power at each output: Port 2: −10 log₁₀(1 + 0.5) = −1.76 dB (two thirds of the input) Port 3: 10 log₁₀(0.5 / 1.5) = −4.77 dB (one third)

Step 7 - Line length: λ/4 = 299792458 / (4 × 2.4e9 × √3.3) = 17.19 mm

All four lines fall between 20 and 120 Ω, so the divider can be built. The nearest E96 resistor to 106.07 Ω is 107 Ω. Asked for P₃/P₂ = 1/8 instead, the port-3 arm would need 252.27 Ω, and the calculator flags it as outside the 120 Ω the board can make.

Practical Tips

✓Use a small thin-film resistor, 0402 or 0603, placed straight across the arm ends; at gigahertz frequencies its pad spacing matters more than its tolerance.
✓As a divider, the resistor dissipates only power reflected back from the outputs; as a combiner, it absorbs whatever part of the inputs is unequal or out of phase, so rate it for the worst case.
✓Check the arm impedances against your process before committing to a split ratio: unequal splits push one arm up quickly, to 252 Ω at P₃/P₂ = 1/8 in a 50 Ω system.
✓Turn each impedance into a width with the microstrip calculator, then recompute each line's quarter-wave length from that width's own effective permittivity.

Common Mistakes

✗Terminating an unequal divider's arms directly in 50 Ω. The arms are designed for R₂ = Z₀K and R₃ = Z₀/K; without the output transformers both outputs are mismatched and the split drifts from the ratio asked.
✗Giving every quarter-wave line the same physical length. Each impedance has its own width and effective permittivity, so each line has its own length for a quarter wave.
✗Placing the isolation resistor at the end of long tracks. The isolation depends on the resistor sitting across the arm ends; extra track between them adds phase and degrades isolation at higher frequencies.
✗Using a single-section divider far from its centre frequency. Match and isolation are exact only at the design frequency; a wider band needs several stepped sections, each with its own resistor.

Frequently Asked Questions

What does the isolation resistor in a Wilkinson divider do?

It isolates the two outputs from each other. Power travelling from one output to the other arrives by two paths, through the resistor and through both quarter-wave lines, and the two cancel at the far output. With matched loads no current flows in the resistor and the divider is lossless; it only dissipates power reflected from a mismatched output.

Why are the lines 70.7 Ω in a 50 Ω Wilkinson divider?

Each quarter-wave line transforms its 50 Ω output to Z² / 50 at the input. For the two arms in parallel to present 50 Ω, each must present 100 Ω, so Z² = 50 × 100 and Z = 70.7 Ω, which is √2 times 50 Ω.

How do I design an unequal Wilkinson divider?

Set K² to the power ratio P₃/P₂. The arm to port 3 is Z₀√((1 + K²)/K³), the arm to port 2 is K² times that, and the resistor is Z₀(K + 1/K). The arms end in Z₀K and Z₀/K, so each output needs a quarter-wave transformer back to Z₀. The calculator gives all of them.

Can a Wilkinson divider be used as a combiner?

Yes. Driven from the two outputs with equal, in-phase signals, it combines them into the common port without loss. Any difference between the inputs, in amplitude or phase, is dissipated in the isolation resistor, which is why the resistor's power rating matters in combiner use.

What limits the power ratio a Wilkinson divider can achieve?

The line impedances. As the ratio moves away from 1, one arm rises and the other falls: at P₃/P₂ = 1/8 in 50 Ω the arms are 252 Ω and 32 Ω. A microstrip process can only make lines over a limited range, so very unequal splits are built another way, for example with a coupled-line coupler.

Methodology & References

References

Microwave Engineering, 4th ed. — David M. Pozar (2012), §7.3, pp. 328–333 — the Wilkinson divider, Example 7.2, and the unequal-split design Eqs. (7.37a)–(7.37c)
An N-Way Hybrid Power Divider — E. J. Wilkinson, IRE Transactions on Microwave Theory and Techniques, vol. MTT-8, pp. 116–118, January 1960

The equal split reproduces Pozar's Example 7.2 (70.7 Ω and 100 Ω); the unequal equations reduce to it at K = 1, and exchanging P₃/P₂ for its reciprocal swaps the two arms and the two terminations while leaving R unchanged.

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