Decibel Budget on a 630 m Experimental Station
An advanced licensee is setting up a low-power station on the 630 m band (472–479 kHz). The exciter delivers 200 mW into a 50 Ω load. A band-pass filter follows with 2.1 dB insertion loss. Then a broadband amplifier provides 26 dB of gain with a measured output compression point of +30 dBm. A directional coupler with 0.4 dB through-loss and 20 dB coupling feeds a power meter. After the coupler, 18 m of RG-58 type coax (0.55 dB per 10 m at 475 kHz) runs to a short vertical with a measured feedpoint impedance of 12 − j140 Ω. The antenna's radiation resistance is estimated at 0.9 Ω, and the loading coil has a measured Q of 180. The operator wants to know the ERP and whether the amplifier is being overdriven.
1. Calculate the drive power arriving at the amplifier input and the amplifier output power in dBm and watts. Is the amplifier being overdriven?
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Drive chain: Exciter 200 mW = 10·log10(200) = +23.0 dBm. Filter loss 2.1 dB, coupler through-loss 0.4 dB → total 2.5 dB. Drive at amplifier input = 23.0 − 2.5 = +20.5 dBm (about 112 mW).
Amplifier output: 20.5 + 26 = +46.5 dBm. That is 10^(46.5/10) mW ≈ 44.7 W. The P1dB/compression point is +30 dBm, so the amplifier is driven 16.5 dB into compression - grossly overdriven. Output will be clipped, splattered, and far below the 44.7 W linear prediction. The operator must add at least ~16.5 dB of attenuation (or reduce exciter drive) before the amplifier to keep peak envelope power below the compression point. Note that with the amplifier compressed, the actual output is unpredictable and likely rich in harmonics - a compliance problem on a band with strict spurious limits.
2. Estimate the power delivered to the antenna terminals and the ERP, given the mismatch loss at the feedpoint.
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At the antenna terminals: Amplifier output (assume linear, 44.7 W = +46.5 dBm) minus coax loss. Coax: 18 m × 0.55 dB/10 m = 0.99 dB. Power at feedpoint = 46.5 − 0.99 = +45.5 dBm ≈ 35.5 W.
Mismatch loss: Reflection coefficient |Γ| = |(12 − j140 − 50)/(12 − j140 + 50)|. Numerator = −38 − j140, |num| = √(1444 + 19600) = 145.1. Denominator = 62 − j140, |den| = √(3844 + 19600) = 153.1. |Γ| = 0.948. Mismatch loss = −10·log10(1 − 0.948²) = −10·log10(0.1013) = 9.95 dB. So only about 3.6 W is accepted by the antenna system.
Coil loss: The loading coil cancels +j140; its loss resistance is R_loss = X/Q = 140/180 = 0.78 Ω. Total antenna circuit resistance = 0.9 + 0.78 + small ground loss (assume ground loss 2 Ω for a short vertical) ≈ 3.7 Ω. Radiation efficiency = 0.9/3.7 ≈ 24%. So radiated power ≈ 3.6 × 0.24 ≈ 0.87 W. ERP referenced to a short dipole ≈ 0.87 × 1.76 (2.45 dB) ≈ 1.5 W EIRP-equivalent, or roughly 2.1 W ERP over a half-wave dipole. This is well within typical 630 m power limits (often 5 W EIRP in many jurisdictions) but the operator should verify the national limit and the spurious emissions from the overdriven amplifier.
3. What single change gives the largest improvement in radiated power, and why?
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The 9.95 dB mismatch loss dominates. Matching the antenna (an L-network or tapped coil transforming 12 − j140 Ω to 50 Ω) recovers nearly 10 dB - a factor of ~10 in power at the antenna. By contrast, eliminating the 2.5 dB pre-amplifier loss or the 1 dB coax loss would gain only a few dB. Improving ground loss from 2 Ω to 1 Ω would raise efficiency from 24% to ~34% (about 1.5 dB). So the correct engineering judgement is: fix the match first, then reduce ground loss, then worry about feedline. Also, the operator must first back off the drive to keep the amplifier linear; otherwise the recovered power is squandered as splatter and harmonic energy.