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[Technical article, rfforum Editorial Team]
Matching a low-impedance device (a power transistor, a patch edge, a parallel array) to 50 ohm is one of the most common jobs in RF design. A quarter-wave transformer is the simplest answer. This note works through one example and shows how much bandwidth extra sections buy.
1. Single section
The transformer impedance is the geometric mean: Z1 = √(Z0·ZL) = √(50 × 12.5) = 25 ohm. Its length is λg/4 at the centre frequency. On a substrate with εeff ≈ 3.3 at 2.4 GHz, λ0 = 125 mm, so λg/4 = 125/(4·√3.3) ≈ 17.2 mm.
2. How much bandwidth?
For a maximum acceptable reflection Γm, the fractional bandwidth of a single section is
Δf/f0 = 2 − (4/π)·arccos[ (Γm/√(1−Γm²)) · 2√(Z0·ZL)/|ZL − Z0| ]
With Γm = 0.1 (return loss 20 dB), the term in brackets is 0.1005 × 50/37.5 ≈ 0.134, and the result is Δf/f0 ≈ 17 %. At 2.4 GHz that is about 410 MHz with RL better than 20 dB. A larger impedance ratio narrows this quickly.
3. Two-section binomial (maximally flat)
Using the usual approximation ln(Z(n+1)/Zn) = 2^(−N)·C(N,n)·ln(ZL/Z0):
Z1 = 50·exp(−¼·ln 4) ≈ 35.4 ohm
Z2 = 35.4·exp(−½·ln 4) ≈ 17.7 ohm
The bandwidth for the same Γm = 0.1 is
Δf/f0 = 2 − (4/π)·arccos[ ½·(Γm/A)^(1/N) ], with A = 2^(−N)·|ZL − Z0|/(ZL + Z0) = 0.25 × 0.6 = 0.15
which gives about 54 %, three times the single-section value, at the cost of one more line section (twice the length).
4. Chebyshev (equal ripple)
A Chebyshev transformer lets the reflection ripple up to Γm across the band instead of being maximally flat at the centre. For the same number of sections it gives a wider band than binomial, or for the same band, fewer sections. Its impedances come from tables (for example in Pozar's Microwave Engineering, chapter 5) or a short script. For two sections the difference from binomial is modest; it grows with three or more sections.
5. Practical points
- The impedances are physical line widths: on a thin substrate 17.7 ohm can mean a very wide line. Check that the width stays well below λg/4 or transverse modes appear.
- Junction discontinuities between sections of very different width add a shunt capacitance. EM-simulate the steps, then shorten the lines slightly to compensate.
- If the load is complex (a transistor), first add a line or a reactive element to make it real at the centre frequency, then transform.
- Copper loss in the low-impedance section matters at high power: currents are larger there.
Discussion: in your designs, at what impedance ratio do you switch from a quarter-wave transformer to a tapered line (Klopfenstein or exponential)?
— rfforum 编辑部 / Editorial Team
Matching a low-impedance device (a power transistor, a patch edge, a parallel array) to 50 ohm is one of the most common jobs in RF design. A quarter-wave transformer is the simplest answer. This note works through one example and shows how much bandwidth extra sections buy.
1. Single section
The transformer impedance is the geometric mean: Z1 = √(Z0·ZL) = √(50 × 12.5) = 25 ohm. Its length is λg/4 at the centre frequency. On a substrate with εeff ≈ 3.3 at 2.4 GHz, λ0 = 125 mm, so λg/4 = 125/(4·√3.3) ≈ 17.2 mm.
2. How much bandwidth?
For a maximum acceptable reflection Γm, the fractional bandwidth of a single section is
Δf/f0 = 2 − (4/π)·arccos[ (Γm/√(1−Γm²)) · 2√(Z0·ZL)/|ZL − Z0| ]
With Γm = 0.1 (return loss 20 dB), the term in brackets is 0.1005 × 50/37.5 ≈ 0.134, and the result is Δf/f0 ≈ 17 %. At 2.4 GHz that is about 410 MHz with RL better than 20 dB. A larger impedance ratio narrows this quickly.
3. Two-section binomial (maximally flat)
Using the usual approximation ln(Z(n+1)/Zn) = 2^(−N)·C(N,n)·ln(ZL/Z0):
Z1 = 50·exp(−¼·ln 4) ≈ 35.4 ohm
Z2 = 35.4·exp(−½·ln 4) ≈ 17.7 ohm
The bandwidth for the same Γm = 0.1 is
Δf/f0 = 2 − (4/π)·arccos[ ½·(Γm/A)^(1/N) ], with A = 2^(−N)·|ZL − Z0|/(ZL + Z0) = 0.25 × 0.6 = 0.15
which gives about 54 %, three times the single-section value, at the cost of one more line section (twice the length).
4. Chebyshev (equal ripple)
A Chebyshev transformer lets the reflection ripple up to Γm across the band instead of being maximally flat at the centre. For the same number of sections it gives a wider band than binomial, or for the same band, fewer sections. Its impedances come from tables (for example in Pozar's Microwave Engineering, chapter 5) or a short script. For two sections the difference from binomial is modest; it grows with three or more sections.
5. Practical points
- The impedances are physical line widths: on a thin substrate 17.7 ohm can mean a very wide line. Check that the width stays well below λg/4 or transverse modes appear.
- Junction discontinuities between sections of very different width add a shunt capacitance. EM-simulate the steps, then shorten the lines slightly to compensate.
- If the load is complex (a transistor), first add a line or a reactive element to make it real at the centre frequency, then transform.
- Copper loss in the low-impedance section matters at high power: currents are larger there.
Discussion: in your designs, at what impedance ratio do you switch from a quarter-wave transformer to a tapered line (Klopfenstein or exponential)?
— rfforum 编辑部 / Editorial Team
