Many hams have seen the classic classroom example:
“A transmitter with a 50-ohm output impedance feeding a 300-ohm antenna will deliver only 49 W instead of 100 W. That must mean SWR causes huge efficiency loss.”
Mathematically, that example is correct — but only within a model that does not represent how HF transmitters actually work. In this article we will look at why that model is misleading, why reflected power does not equal lost power, and why an antenna with SWR 2:1 or even 6:1 can still radiate extremely well.
The Textbook Model: Great for Exams, Terrible for Real Radios
The classic example assumes a very simple source model:
- A perfect voltage source
- Followed by a fixed 50-ohm series resistor (Rout)
- Connected directly to the antenna
If the antenna is 50 Ω, the “transmitter” delivers 100 W. If the antenna is 300 Ω, it becomes a simple voltage divider and the power falls to about 49 W. That is exactly what the maximum power transfer theorem describes — for that model.
But no RF power amplifier is designed like that. Real PA stages are not built to burn half their power in a 50-ohm resistor. The textbook circuit exists because it is easy to draw on a whiteboard, not because it resembles an HF transceiver.
This schoolbook model is useful to pass an exam, but it is so oversimplified that it often confuses hams more than it helps.
What a Real HF Power Amplifier Actually Does
A modern HF PA behaves very differently from the “voltage source + 50-ohm series resistor” picture. In practice, a PA:
- Is designed as an almost ideal low-impedance RF source
- Uses an output matching and low-pass filter network to present 50 Ω toward the outside world
- Regulates drive, voltage, and current to maintain roughly the rated output power
- Reduces power only when the SWR protection threshold is exceeded
In other words, “50-ohm output impedance” is a design target for the output network, not a literal resistor where half the power is supposed to be wasted. The 49 W result you compute in the textbook model is the behavior of a fictional transmitter — the kind RF engineers work hard not to build.
Why Reflected Power Is Not the Same as Lost Power
Consider an SWR of 6:1. The magnitude of the reflection coefficient is:
For SWR = 6:1,
|Γ| = (SWR − 1) / (SWR + 1) = 5 / 7 ≈ 0.714
Reflected power fraction = |Γ|² ≈ 0.51 → about 51% of the power is reflected at the load.
That number looks scary, but it does not mean 51% of your transmitter power is lost as heat.
What actually happens is this:
- The reflected power returns toward the transmitter.
- At the transmitter end, part of that wave is re-reflected back into the line.
- The power bounces back and forth (“ping-pong”) between load and source.
On each trip, small amounts of energy are removed by:
- Coax losses
- Tuner losses
- Balun and ferrite losses
- Conductor and dielectric losses in the antenna system
The end result is that the extra loss caused by SWR is typically modest. For a realistic HF installation with a 20 m feedline and SWR around 6:1, you are usually looking at a few dB of additional loss — not 50% of your power disappearing in smoke.
Reflected ≠ dissipated. That is the single most misunderstood part of SWR discussions in amateur radio.
Why a Tuner Can Fix the PA’s View of the Load
If you stay inside the textbook model, you are forced to conclude:
“A tuner cannot correct this, because the loss happens in the source resistor.”
That conclusion is only valid in the artificial world where a literal 50-ohm resistor sits inside the transmitter as the output impedance.
In a real station:
- The tuner sits near the radio.
- It transforms the complex impedance of the coax + antenna back to 50 Ω at the PA ports.
- The PA “sees” a good match again and can deliver its rated forward power (within its protection limits).
- The unavoidable extra loss happens in the coax and in the tuner’s components — not in a fictitious internal series resistor.
So the tuner absolutely can fix what matters most for the transmitter: the loading condition at the PA. What the tuner cannot do is magically improve a poor radiation pattern or a very low radiation resistance — that is a separate antenna design problem.
SWR and Field Strength: Why a 3 dB Loss Is Not the End of the World
Even if you did permanently lose 50% of your RF power (and in a sane HF station you usually do not), the effect on field strength is smaller than many hams imagine:
- 100 W → 50 W is a 3 dB drop in power.
- Field strength at the receiver changes only by √2 ≈ 1.41.
- That is less than one S-unit in most real-world receivers.
On HF, normal fading (QSB) is often 10 dB or more. Band conditions and noise floor will dominate long before a 1–3 dB SWR-related loss does. Chasing a perfect 1.0:1 SWR at all costs rarely buys you the performance you think it does.
Why EFHW vs EFOC Has Almost Nothing to Do with SWR Alone
It is very common to compare EFHW and EFOC antennas based purely on their SWR curves. That is misleading, because far-field strength depends mainly on:
- Current distribution along the wire
- Transformer and matching losses (for example, 49:1 vs 4:1)
- Common-mode currents on the feedline
- Height above ground and geometry over real earth
In practice, you can easily end up in a situation where:
- A well-designed 4:1 EFOC with SWR around 2.5:1 and low transformer loss
- Outperforms a “perfect-match” EFHW with SWR 1.0:1 but a hotter transformer, more common-mode current, and a less favorable current distribution.
SWR is just one small symptom in a much larger system. Chasing the lowest number on the meter without looking at losses, pattern, and common-mode behavior can send you in the wrong direction.
The Real Role of the Reactive Component
It is absolutely correct that only the real part of the impedance dissipates power. However, the reactive part still matters because it determines the ratio of voltage to current at various points in the system. Extreme reactance can create:
- High RF voltages → dielectric losses, arcing, corona, capacitor stress
- High RF currents → conductor losses, ferrite heating, hot spots in coils and baluns
So while the reactive component X does not directly “burn” power, it indirectly increases losses in the surrounding components. Again, the system matters more than any single number the meter gives you.
One-Sentence Summary
A 6:1 SWR does not cause a 50% radiation loss; that only happens in a schoolbook model with a fictional 50-ohm series resistor. Real HF transmitters, tuners, feedlines, and antennas behave completely differently — and the true extra loss is usually only a few dB.
A Concrete Example: 6:1 SWR on 80 m with 20 m Coax and a Tuner
Let’s put some realistic numbers on it. Suppose you are on the 80 m band with:
- 100 W transmitter into an ATU in the shack
- The tuner output connected to 20 m of decent 50 Ω coax (think RG-213 / LMR-400 class)
- An antenna that presents about 6:1 SWR at the end of the coax
The matcher in the shack transforms the complex impedance of the line+antenna back to 50 Ω at the rig. From the PA’s point of view, life is good again: it sees a proper match and delivers close to 100 W forward power into the tuner.
On 80 m, 20 m of good coax typically has a matched loss well below 0.5 dB. With 6:1 SWR, the effective loss of that same run increases — but it is still usually on the order of about 1 dB or so, not 6 or 10 dB.
In round numbers, that means:
- 100 W out of the tuner
- Something like 70–80 W reaches the antenna feedpoint after coax loss under mismatch
- A bit more is lost inside the tuner’s inductors and capacitors (maybe another fraction of a dB)
You might end up with, say, 60–75 W actually available at the feedpoint — compared to perhaps 80–90 W in a perfectly matched 1:1 SWR case with the same hardware. The difference on the air is only a couple of dB, far less than normal QSB and band variations.
The key takeaway from this example: even with 6:1 SWR on 80 m and 20 m of decent coax, a well-matched system with a tuner still delivers the majority of your transmitter power to the antenna. The “huge loss” that the textbook model predicts simply does not appear in a realistic station.
Closing Thought
The ideal-source-plus-series-resistor model is useful for teaching basic circuit theory, but in HF practice it is so oversimplified that it often does more harm than good. Real antennas deserve real system thinking: look at current distribution, losses, common-mode control, and pattern first — and treat SWR as just one diagnostic tool among many.
Mini-FAQ: SWR, Loss, and Real Stations
- Does a 3:1 SWR mean my antenna is inefficient? — No. A 3:1 SWR only tells you there is a mismatch. The real efficiency depends on feedline loss, matching loss, transformer loss, and how much power actually reaches the radiator.
- Is all reflected power lost as heat in the transmitter? — No. Reflected power bounces between source and load. Only a small fraction is dissipated on each trip in coax, tuner, and matching components.
- Can a tuner “fix” SWR without fixing the antenna? — A tuner can present a safe 50-ohm load to your PA, but it does not improve a poor radiation pattern or a very small radiation resistance. It protects the rig and moves the loss to the tuner and feedline.
- Why do many hams obsess over 1.0:1 SWR? — Because meters make SWR highly visible and schoolbook models over-emphasize it. In real HF work, a slightly higher SWR with lower system loss can easily outperform a “perfect-match” but lossy setup.
Interested in more technical content? Subscribe to our updates for deep-dive RF articles and lab notes via our RF.Guru newsletter signup page.
Questions or experiences to share? Feel free to contact RF.Guru via our RF and antenna support contact page.