SWR Vs. Gas Discharge Tubes and MOV's

Why your gas discharge tube is crunchy. For a 100-watt output into a 200-ohm load, the RMS voltage is approximately 141.4 volts. This is calculated using the formula for RMS voltage: V RMS ​ = P×R ​ V RMS ​ = 100W×200Ω ​ = 20,000 ​≈141.4V If you are concerned with peak voltage (relevant for component ratings like switches or gaps), the value is approximately 200 volts. Peak voltage is calculated as: V peak ​ =V RMS ​ × 2 ​ ≈141.4×1.414≈200V Note that these calculations assume a perfect 1:1 SWR (standing wave ratio). If the impedance mismatch is significant, voltages can spike much higher, particularly at points of high reflection or in antenna tuners. The peak-to-peak voltage ($V_{pp}$) for 100 watts into 200 ohms is approximately 282.8 volts. This is derived from the RMS voltage ($141.4 , \text{V}$) or the peak voltage ($200 , \text{V}$) calculated previously: V pp ​ =V peak ​ ×2=200V×2=400V Correction: Let's re-calculate precisely. $V_{RMS} = \sqrt{20,000} \approx 141.42 , \text{V}$ $V_{peak} = 141.42 \times \sqrt{2} \approx 200.0 , \text{V}$ $V_{pp} = 200.0 \times 2 = \mathbf{400.0 , \text{V}}$ Wait, $\sqrt{20000} = 141.421356$. $141.421356 \times 1.41421356 = 200$. So $V_{peak}$ is exactly 200V. Therefore, $V_{pp} = 400 , \text{V}$. This value is critical for determining the breakdown voltage requirements of components like capacitors in antenna tuners or the spacing of open-wire feed lines. For 100 watts into a 100-ohm load: RMS Voltage: 100 V V RMS ​ = 100W×100Ω ​ = 10,000 ​ =100V Peak Voltage: ~141.4 V V peak ​=100V× 2 ​ ≈141.4V Peak-to-Peak Voltage: ~282.8 V V pp ​ =141.4V×2≈282.8V This setup results in significantly lower voltage stress compared to the 200-ohm scenario, making it easier on component ratings in tuners and feed lines. 1. The Critical Role of Peak Voltage GDTs are rated by their DC spark-over voltage, but in an AC or RF circuit, the voltage constantly swings. The GDT will fire if the instantaneous voltage at any point in the cycle exceeds its breakdown threshold. Normal Operation: The GDT's minimum DC breakdown rating must be higher than the circuit's maximum peak voltage ($V_{peak}$), not just the RMS voltage. The Risk: If you select a GDT based only on RMS voltage, the tube will likely short-circuit the signal during the positive and negative peaks of every cycle, causing continuous conduction, overheating, and failure. For your specific HF scenarios: 100W into 200Ω: The peak voltage is 200 V ($400 \text{ V}_{pp}$). You would need a GDT with a minimum DC breakdown voltage significantly higher than 200 V (e.g., a 230V or 250V rated tube, accounting for tolerance). 100W into 100Ω: The peak voltage is ~141 V ($283 \text{ V}_{pp}$). A GDT rated for roughly 150V–180V minimum would be required. 2. RF Specific Considerations In RF applications (like HF transmitters), there is an additional safety margin often recommended beyond just the peak voltage. The 60% Rule: Some manufacturers suggest that the continuous RF peak voltage should not exceed 60% of the GDT's rated DC breakdown voltage. This accounts for: Tolerance: GDT breakdown voltages often have a $\pm 20%$ tolerance. A "200V" tube could fire as low as 160V. SWR Spikes: High Standing Wave Ratio (SWR) can cause voltage peaks to multiply significantly along the feed line, potentially exceeding the calculated 200V or 141V. Impulse Response: Fast rise-time transients can cause the spark-over voltage to "overshoot" the DC rating, but relying on this is risky for continuous wave (CW) or SSB peaks. 3. Selection Guideline To choose the correct GDT: Calculate $V_{peak}$ ($V_{RMS} \times 1.414$). Add a safety margin for SWR and tolerance (often doubling the peak or using the 60% rule). Select a GDT where the minimum DC breakdown voltage (rating minus tolerance) is higher than this value. For the 200-ohm case ($200 \text{ V}_{peak}$), a standard 230V GDT might be too close if tolerances and SWR are considered; a 300V or 350V GDT would be safer to ensure it only fires during actual surges, not during full-power transmission. What MOV will protect 240 volt line To protect a 240-volt AC line, you should select a Metal Oxide Varistor (MOV) with a Maximum Continuous Operating Voltage (MCOV) rating of at least 300 VAC (RMS). 1. Recommended Voltage Ratings The industry standard (IEC 60950-1 and IEC 62368-1) requires the MOV's rated voltage to be at least 125% of the nominal line voltage to prevent leakage and thermal runaway during normal operation. Nominal Line: 240 VAC RMS Peak Line Voltage: $240 \times 1.414 \approx 340 \text{ V}_{peak}$ Minimum MCOV Required: $240 \times 1.25 = 300 \text{ VAC}$ Common Part Selection: A 300 VAC or 320 VAC rated MOV is standard. In part numbers (which often denote DC voltage), this corresponds to roughly 470V DC (e.g., a 7D471K or 10D471K). Warning: Do not use a MOV rated for 240V DC or ~275V DC (like a 10D271K) on a 240V AC line. The AC peak voltage (340V) exceeds their rating, causing them to conduct continuously, overheat, and fail. 2. Clamping Voltage While the MOV must withstand 340V peaks normally, it should clamp dangerous surges to a safe level. A 300 VAC MOV typically begins conducting significantly around 470V–500V and clamps high-current surges to approximately 775V–800V. Ensure your protected equipment can withstand this clamping voltage. 3. Critical Safety Considerations Thermal Protection: MOVs degrade over time and can fail short-circuit, leading to thermal runaway and fire. For 240V mains, it is highly recommended to use a Thermally Protected MOV (TMOV) or place the MOV in series with a Gas Discharge Tube (GDT) and a dedicated fuse. Follow Current: Unlike GDTs, MOVs can struggle to stop conducting if the line voltage remains high after a surge. Proper fusing is mandatory to disconnect the MOV if it fails.