Does 3 W vs 5 W TX power make a huge difference in QRP CW operation?

Amateur Radio Asked on January 13, 2021

I’ve bought the QCX+ 40 m band radio. I will be powering this off a 12 V battery. I read on the manual the RF power at 12 V is just 3 W whereas on 15 V it is 5 W. Should I buy a boost converter for the battery to up the voltage to 15 V?

RF power output on 40 m

5 Answers

If you haven't done much QRP operating, QRP is a funny thing. Sometimes propagation is great, and you wonder why anyone would bother with 100 W or more. I'll never forget my first DX QSO, from Oregon to Estonia, running 11 W to a vertical, by gray-line propagation at the bottom of the sunspot cycle in 2006. Other times it seems like a complete waste of time, calling station after station that boom into your headphones but can't hear you at all.

The difference between 3 W and 5 W is 2.2 dB, which is about a third of an S-unit. (An S-unit is 6 dB.) The question is, how much does that mean to you? That depends on what kind of operating you do, or want to do, with your QRP rig.

  • If you're a QRP contester, then 2 dB can make quite a difference, because every decibel matters to contesters. Contesters generally have very good ears, and serious ones are willing to work hard to decode signals just above the noise floor to get the contact. Losing 2 dB means potentially losing a few contacts in your next major contest.

  • If you're a casual QRPer who makes contacts on the QRP calling frequency, then you'll potentially lose a fraction of potential contacts. QRPers are also generally willing to work at pulling in your weak signal, so every decibel helps. You can always spot yourself on, which encourages other operators to listen for you.

  • If you're a Morse ragchewer, then lots of hams either won't be able to hear you at all, or won't tune slowly enough to hear your CQ, or won't be willing to work at pulling in a very weak signal on a noisy band, so you're generally limited to stations that are exactly one skip away. To those stations you might be S-5 on their meters. For those stations, there usually isn't much difference between S-5 and S-4.6. I'd think you wouldn't notice much difference between 3 W and 5 W when ragchewing with random hams.

My advice would be to try the radio at 3 W, and see how you like QRP. If you like it enough, it might be worth buying that battery booster. The battery booster will drain your battery faster though, because the radio would consume more power and the booster itself consumes power, since it's not 100% efficient.

Correct answer by rclocher3 on January 13, 2021

You can test the value of this difference. Emit a WSPR beacon, vary every other one in transmit power by 2.2 dB, and gather statistics over multiple cycles of the number of spots at each power level. Then turn down the WSPR power util you get about the same number or distance of WSPR spots as contacts you expect to make in you desired mode, and rerun the A/B trial to see how much difference 2.2 dB makes in actual contact potential from your location and antenna.

Answered by hotpaw2 on January 13, 2021

Generally QRP is considered to be 10 watts or less. A small RF amp could get your 3 watts up to 9 which would be a more powerful signal but this is something you would have to homebrew. You can get the power needed to drive this transceiver and amp by wiring the batteries to increase the voltage to the level required, (series vs. parallel wiring).

This all depends on how hardy the radio circuitry is. If it can take the power - no problems - if it can't, then you have a short duration heating unit.

I understand the thinking of less is more when it comes to using CW and QRP, and a different radio would be an expense you may not want to go for, but all of this depends on how you intend to approach the hobby.

I am not into QRP as a discipline of the practice of amateur radio operations, however I can understand the appeal. If it were me, on the other hand, I would get a 10 watt radio that I could adjust the output power level on for times when I wanted to transmit at a reduced power level.

Of course, that might mean more of an investment in equipment which may or may not be what you are after in the amateur radio hobby.


Answered by SFC on January 13, 2021

It makes a difference of:

$$ 10 times log_{10}left(3 over 5right) = 2.2 :mathrm{dB} $$

How big is 2.2 dB? See How big is a decibel?

It's also relevant to note: if the 12V battery is lead acid, the charged voltage can be as high as 13.7.

Furthermore, while a boost converter might further increase your transmit power, it's likely to make noise which will degrade your receive performance.

Answered by Phil Frost - W8II on January 13, 2021

For any transmitter, if you double the power, this might seem a lot, but at the other end at a receiver the signal will increase by only 3 dB which is half an S point.

So an increase of 3W to 5W is 2.2 dB which is a bit less than half an S point.

This change is negligible and you won't notice much difference.

However if the level of your transmitted signal is down in the noise at a receiver, it could make the difference between being able to hear it or not.

To answer your question, it is probably not worth the effort to try and increase the supply because no one will notice. Having said that every bit counts and increasing the supply will help a little bit.

Personally i would only make the supply 15 V if it was easy to do.

Answered by Andrew on January 13, 2021

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