Frǝd's Ham Radio Scratchpad - W6BSD¶
SWR along the line animation: vswr_wave_0.html
What about SWR¶
Mathematically we calculate SWR by determining the reflection coeficient gamma ($\Gamma$) which is a function of the load impedance ($Z_L$) and the source impedance ($Z_0$). These 2 impedances are complex number.
$$ \Gamma = \frac{Z_L - Z_0}{Z_L + Z_0} = \frac{V_{rev}}{V_{fwd}} = \left| \frac{VSWR-1}{VSWR+1} \right| = \sqrt{\frac{P_{rev}}{P_{fwd}}} = 10^{\frac{-\text{Return Loss}}{20}} $$
VSWR is them calculated using $\Gamma$ in the following equation:
$$ VSWR = \frac{1+|\Gamma|}{1-|\Gamma|} $$
In our example above we have the forward voltage is 20V, and the reverse voltage is 15V:
$$ \begin{align} \Gamma &= \frac{10}{20} && = 0.5 \\ \nonumber VSWR &= \frac{1+0.5}{1-0.5} && = 3 \end{align} $$
We can calculate the SWR from the Return Loss: $$ VSWR = \frac{10^{\frac{-RL}{20}}+1}{10^{\frac{-RL}{20}}-1} $$
The following equation is especially useful when using a directional power meter. It allows you to calculate the VSWR from the forward and reverse power.
$$ VSWR = \frac{1+\sqrt{\frac{P_{ref}}{P_{fwd}}}}{1-\sqrt{\frac{P_{ref}}{P_{fwd}}}} $$
The return loss ($R_L$) expressed in $dB$ is normally calculated as follows: $$ \text{RL}_{dB} = -10 \log_{10}{\left(\frac{P_{ref}}{P_{fwd}}\right)} $$
It is them possible to calculate the Return loss ($R_L$) from the VSWR using the equation: $$ \text{RL}_{dB} = -20 \log_{10}{\left(\frac{VSWR-1}{VSWR+1}\right)} $$
Relected power¶
Reflected power
$$ \text{Reflected Power}_{\%} = |\Gamma|^2 \cdot 100 $$
Return loss
$$ \text{Return Loss}_{dB} = -20 \log_{10} \left({|\Gamma|} \right) $$
Mismatch loss
$$ \text{Mismatch Loss}_{dB} = -10 \log_{10} \left(1 - {|\Gamma|}^{2} \right) $$
Do not use this formula to calculate the losses. See the section above.
Misc¶
$$ \begin{align} |Z| &= \sqrt{R^2 + jX^2} \\ \nonumber \theta &= \arctan{\frac{X}{R}} \end{align} $$
Q Factor of an inductor¶
$$ Q=\frac{X_L}{R} $$
or
$$ \begin{align} Q &= \frac{2{\pi}fL}{R}\\ \nonumber &= \frac{2{\pi} \times 14.15 \times {10^6} \times 220 \times {10^{-9}}}{.05} \\ \nonumber &= 391 \end{align} $$
LC Q factor equations¶
Series LC¶
$$ Q = \frac{1}{R} \times \sqrt{\frac{L}{C}} $$
Parallel LC¶
$$ Q = R\times\sqrt{\frac{C}{L}} $$
Power: dBm -> Watt¶
$$ \begin{align} P_{mW} &= 10^{\left(\frac{P_{dBm}}{10}\right)} \\ \nonumber P_{W} &= 10^{-3} \times 10^{\left(\frac{P_{dBm}}{10}\right)} \\ \end{align} $$
Quick reference
0 dBm = 1 mW
10 dBm = 10 mW
20 dBm = 100 mW
30 dBm = 1 W
40 dBm = 10 W
50 dBm = 100 W
Power: Watt -> dBm¶
$$ \begin{align} P_{\mathrm{dBm}} &= 10 \times \log_{10}(P_{\mathrm{W}} \times 10^{3}) \\ \nonumber &= 10 \times \log_{10}(P_{\mathrm{W}}) + 30 \end{align} $$
Volt -> dBm¶
$$ \begin{align} P_{\mathrm{dBm}} &= 10 \log_{10}\left(\frac{V_{\mathrm{rms}}^2 \cdot 10^3}{Z}\right) \end{align} $$
dBm -> Voltage¶
$$ \begin{align} V_{\mathrm{rms}} &= \sqrt{R \cdot \frac{10^{\frac{P_{\mathrm{dBm}}}{10}}}{1000}} \\ &= \sqrt{R \cdot 10^{\frac{P_{\mathrm{dBm}} - 30}{10}}} \end{align} $$
Antenna Tuning¶
To determine the target length of the antenna elements:
- Measure the current length and the resonant frequency
- Apply the following formula to determine the new length
$$ L_{target} = L_{measured} \times \frac{f_{measured}}{f_{target}} $$
Example¶
- Length of an element 19.37m
- Frequency measured = 3.6MHz
- Target Frequency = 3.75MHZ
$$ \begin{align} L_{target} &= 19.3 \times \frac{3.6}{3.75} \\ \nonumber \\ \nonumber &= 19.3 \times 0.96 \\ \nonumber &= 18.595 \\ \nonumber \end{align} $$
- The new length for each element is 18.6 meter
LC Circuits¶
The lower case omega - $ \omega $ - is angular frequency - $ \omega = 2 \pi f $
The impedance of L and C, for ideal components its equal to the imaginary reactance.
$$ \begin{aligned} X_{L} &= \omega L \\ X_{C} &= \frac{1}{\omega C} \\ \end{aligned} $$
Resonance occurs when: $$ \begin{aligned} X_{L} &= X_{C} \\ \omega L &= \frac{1}{\omega C} \\ \omega^2 &= \frac{1}{LC} \\ \omega &= \frac{1}{\sqrt{LC}} \\ \end{aligned} $$
Series RLC Circuit¶
Impedance¶
$$ Z = R + j\omega L - \frac{j}{\omega C} $$
Magnitude of impedance:
$$ |Z| = \sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2} $$
Differential Equation (for voltage source $V(t)$):¶
$$ V(t) = R i(t) + L \frac{di(t)}{dt} + \frac{1}{C} \int i(t) \, dt $$
Taking the derivative:
$$ L \frac{d^2i(t)}{dt^2} + R \frac{di(t)}{dt} + \frac{1}{C} i(t) = \frac{dV(t)}{dt} $$
Parallel RLC Circuit¶
Admittance¶
$$ Y = \frac{1}{R} + j\omega C - \frac{j}{\omega L} $$
Magnitude of admittance:
$$ |Y| = \sqrt{ \left( \frac{1}{R} \right)^2 + \left( \omega C - \frac{1}{\omega L} \right)^2 } $$
Resonant Frequency¶
The resonant frequency $f_0$ (same for both series and parallel RLC circuits) is:
$$ f_0 = \frac{1}{2 \pi \sqrt{LC}} $$ $$ C = \frac{1}{4 \pi^2 f^2 L} $$ $$ L = \frac{1}{4 \pi^2 f^2 C} $$
Transmission line losses¶
Coax Cable Loss and the Impact of Bad SWR
$$ -10 \log_{10} \left( \frac{ \alpha^2 - |\Gamma|^2}{\alpha \cdot (1 - |\Gamma|^2)} \right) $$
Where $\alpha$ (Alpha) is the matched-line loss ratio:
$$ \alpha = 10^\frac{\text{-cable loss in decibels}}{10} $$
And the reflection coefficient $\Gamma$ (Gamma) is related to the SWR by:
$$ \Gamma = \bigg| \frac{\text{SWR}−1}{\text{SWR}+1} \bigg| $$
Dielectric Constants¶
For wire insulation, the dielectric constant is a key factor that affects signal speed and capacitance. While the ideal value depends on the application, fluoropolymer-based materials like PTFE (Teflon) are widely regarded as the gold standard, offering one of the lowest dielectric constants at around 2.1.
Here are the dielectric constant values for common wire insulation materials, organized by material type:
High-Performance Fluoropolymers
| Material Type | Dielectric Constant (εᵣ) | Key Characteristics / Applications |
|---|---|---|
| PTFE (Teflon) | 2.1 | Excellent thermal and chemical resistance; very low loss; ideal for high-speed data, aerospace, and downhole equipment. |
| Fluoropolymer Composite | 2.85 | Composite film used for aerospace wire; balances mechanical durability with good electrical properties. (e.g. PTFE + glass or ceramic filler), this value varies widely by formulation. It's not a universally standardized value. |
High-Temperature Resins
| Material Type | Dielectric Constant (εᵣ) | Key Characteristics / Applications |
|---|---|---|
| Polyimide | ≈ 2.8 - 3.6 | Exceptional heat resistance and mechanical strength; used in motors, transformers, and aerospace applications. Widely depends on the formulation. |
| PPE (Polyphenylene Ether) | ≈ 3.3 - 3.5 | Low signal transmission loss; suitable for high-frequency and LSI package applications. |
Standard & Power Cables
| Material Type | Dielectric Constant (εᵣ) | Key Characteristics / Applications |
|---|---|---|
| XLPE (Crosslinked Polyethylene) | ≈ 2.3 | Standard insulation for medium and high-voltage power cables (e.g., 36-136 kV). |
| Pure Polyethylene (PE) | ≈ 2.25 - 2.30 | Baseline for non-polar, pure polymer; very stable and low. |
| Foamed Polyethylene | ≈ 1.3 - 1.8 | Air pockets are introduced (εᵣ ≈ 1), significantly lowering the overall value. |
| Pure PVC (Rigid) | ≈ 3.0 - 4.0 | Baseline for polar PVC; higher than PE due to its molecular structure. |
| PVC with Plasticizers | 5.0 to 30+ | Plasticizers (used to make wire flexible) can dramatically increase the value. |
| PVDF (polyvinylidene fluoride) | ≈ 8 - 12 | Extreme cases where specific additives create very high permittivity for niche applications. |
→ Moisture / Frequency¶
Effect of Moisture Polar plastics like PVC absorb moisture, which raises the dielectric constant and reduces insulation quality.
Effect of Frequency As signal frequency increases, the dielectric constant typically drops from the static value.
→ Understanding the Trade-offs¶
- PTFE has excellent electrical properties (very low dielectric constant and loss) and high heat resistance, but it is more expensive and can be more difficult to process than other materials .
- Polyimide offers a good balance of high heat resistance, mechanical toughness, and reliable electrical performance, making it a top choice for demanding environments like aerospace and high-performance motors .
- Standard materials like XLPE and PVC are cost-effective and have adequate performance for general power and control applications .
→ How to Use This Data¶
When selecting a material, keep in mind that the dielectric constant can vary slightly based on the material's specific formulation, the frequency of the signal, and the test temperature . For critical high-frequency designs, always refer to the manufacturer's official datasheet.
I hope this detailed breakdown is helpful. If you can share more about your specific application (e.g., high-speed data, high-voltage power, or flexible robotics), I may be able to provide more targeted information.
Single wire velocity factor¶
The Velocity factor can be calculated from the dielectric constant of the material with the following formula.
$$ V = \frac{1}{ \sqrt{εᵣ} } $$
Where:
V = velocity in the material
c = speed of light in vacuum (≈ 3 × 10⁸ m/s)
εᵣ = relative dielectric constant
For example:
The velocity factor of a conductor insulated with pure polyethylene (PE)
$$ \begin{aligned} V &= \frac{1}{ \sqrt{2.25} } \\ &= 0.66 \end{aligned} $$
Parallel wire feed line¶
Characteristic impedance (Z₀) is the ratio of voltage to current for a wave traveling down the transmission line. It's determined purely by the geometry and materials of the line, not by what's connected to the ends.
$$ Z_0 = \frac{120}{\sqrt{\varepsilon_r}} \ln\left(\frac{D}{d} + \sqrt{\left(\frac{D}{d}\right)^2 - 1}\right) $$
Z₀ = characteristic impedance in ohms
D = center-to-center spacing between wires
d = wire diameter
εᵣ = relative permittivity of the dielectric medium
Or the simplifyed version:
$$ Z_0 ≈ \frac{120}{εr} \ln\left(\frac{2d}{D}\right) $$
Cross sectional area¶
The cross-sectional area of a (round) wire is just the area of a circle. If you know the radius
If you know the diameter (The diameter d = 2 r):
$$ A = \pi \cdot r^2 $$
Calculation of the cross section A.
$$ A = \frac{\pi \cdot d^2}{4} $$
Wire diameter = 1.5 $$ \begin{align} A &= \frac{\pi \cdot 1.5^2}{4} \\ &= \frac{7.0685}{4} \\ &= 1.76 \end{align} $$