Low SWR, low power?

[druid]

I have a ICOM IC-730 and I'm just getting the transmitter "tuned up" before I broadcast for the first time. I put in a SWR meter and sure enough, it's about 1.5-2 on 20 meters (into a 20M center-fed dipole). It's about 8 or so at 40 meters.

Unfortunately, although it puts out about 40 Watts (AM) at 14.2 MHz into a dummy load, it drops to about 7 Watts into the antenna! It puts out 40W at 7.2, though, so effectively I'm radiating about the same power.

Anyone know what's going on - why the transmitter power is so low at 14.2 MHz?

Lloyd
VA7LAS

[John_D]

How long is the coax to your 20 metre dipole? Depending on the length of coax you can actually get VSWR of 1:1 without an antenna on the end! What also raises my suspicions is that it puts out the same power on 40 metres....at 40 metres there should be a big mismatch of impedance and the ICOM is designed to "fold bad" its power when it see that.

[K7KBN]

A "20M center-fed dipole" is going to be about 33 feet long, overall, and feeding it with coax can yield a mismatch of anywhere from 1.5:1 to 2:1, depending on how high the antenna is, and what kind of coax.

Now, trying to use this same 33-foot long antenna, with no changes, on 40 meters really is not going to work well at all. It's now only 1/4 wavelength long rather than 1/2 wavelength, and it will have a very high feedpoint impedance - possibly in the thousands of ohms. If you want to operate well on 40 meters, your half-wavelength dipole needs to be right around 66 feet long overall.

[druid]

Thanks K7KBN - I understand about dipole mismatches. The "issue" was the difference in power output from the transmitter. But I think I have the answer: my little $30 SWR/Power meter doesn't work well at all.

I was transmitting into a dummy load (2 100 Ohm 10W resistors in parallel) and the meter was showing 10 Watts. Then the resistors started smoking... so I hooked up my trusty oscilloscope, and sure enough: 25 Watts at the load. The other thing I noticed was the meter on the radio showed "100" in all tests, which implies to me that it's putting out the same power at the different frequencies and into the dummy load or the antenna (yes, I know at the mismatching frequencies I won't be radiating much power!)

The other test I did was use the field strength feature of the meter, and I saw nearly twice the field strength from the 20M dipole, 30 ft away, then from my 5W VHF on a dipole 6ft away. I'm too lazy to do the math (and I suspect the difference in freq would have a large effect on the reading), but I'd say I'm transmitting more significantly more than 5W at 14 MHz.

So - next step is to <gulp!> plug in the mike and actually start transmitting!

druid - designing his QSL card, just in case...

[AB3NK]

I don't think that someone here knows what they are talking about.

I do not want to come off as looking like the bad guy here.

AM stands for Amplitude Modulation.

Amplitude modulation is where you transmit both the carrier and both the upper and lower side band.

The carrier and the side bands are not unilateral - hence if the radio is rated at 200 watts output, the radio does not put 66 watts into each side band and 66 watts into the carrier.
The carrier is a power hog and absorbs 50 - 60% of the power output in this model of radio.

Hence when operating Amplitude Modulation the total output for the radio is only about 40 watts.

http://www.classicicom.com/manuals/IC-730.pdf

Next time look at the specifications before asking the question and it will save you a lot of agrivation.

[KM3F]

I don't think they do either!
A 100 watt radio on AM will only have a carrier of no more than about 25 watts into a load of less than 2 to 1.
How it comes about is a carrier modulated at 100% = a peak envolope of 4 times the carrier.
Thus 25 watt carrier modulated 100% = 100 watts voice peaks. Can't be very much different.
You can't use a 20m dipole on 40m directly without a tuner, even then the radiation efficiency will be low.
Your 50 ohm load of two 100 ohm resistors! Are they carbon or wire wound and what power dissapation is each resistor and how long did you apply power. It's all important. Can't use wire wound resistors.
All that stuff in the original post is telling nothing worthwhile except lack of knowledge.
Sorry to be harsh but need to get your attention.
Might as well learn correct the first time rather than have to unlearn it later.
An swr meter reads both forward and reflected power. Depending on the meter, a dual needle or seperate meters as to how it's used. The meter can add both directions and make the reading higher or lower than it really is. This is how it will lie to you under a greatly miss matched conditions.
Good luck.

[AB3NK]

When using a transmatch - the transmatch only matches the resistance to make it into something that the radio will accept.

newer radios by nature - will cut back it's transmit power when the match is not good, hence even when you use a transmatch with a radio with a antenna that is the wrong size - electrically - the radio might cut the power back to about 25 watts to protect the finals.

That is what mine does.

[druid]

I don't think they do either!
A 100 watt radio on AM will only have a carrier of no more than about 25 watts into a load of less than 2 to 1.
How it comes about is a carrier modulated at 100% = a peak envolope of 4 times the carrier.
Thus 25 watt carrier modulated 100% = 100 watts voice peaks. Can't be very much different.

I freely admit I know nothing of PEP. I DO know RMS, and I was measuring the carrier-only, which is a true sine wave and therefore RMS (true) power is 1/2 * Vp * Vp /Rload, or at least it was when I took my degree in EE.

You can't use a 20m dipole on 40m directly without a tuner, even then the radiation efficiency will be low.

Of course you can! Just not very effectively. A tuner does not change the radiation characteristics of the antenna - just matches impedances. This gives a better overall efficiency, but the 20M antenna is still a 20M antenna.

Your 50 ohm load of two 100 ohm resistors! Are they carbon or wire wound and what power dissapation is each resistor and how long did you apply power. It's all important. Can't use wire wound resistors.

2 10W carbon resistors. You can "overdrive" carbon resistors to 120% easily, 150% for a few moments, and even 200% for short periods. However, 100W RMS into 2 10W resistors would have them smoking in seconds, and overheating to the point of changing their resistance.

All that stuff in the original post is telling nothing worthwhile except lack of knowledge.

Unfortunately one of my bad habits is to include far too much information when I'm asking a question: my question was about why the power seemed to be so much different at 7MHz than 14 MHz - this had nothing to do with what kind of resisters I was using for my dummy load, or whether I was using an antenna tuner, or the difference between RMS and PEP.

I found that the answer was my cheap-and-nasty meter - once I used a scope I could see the actual situation.

Sorry to be harsh but need to get your attention.
Might as well learn correct the first time rather than have to unlearn it later.

I see SO much mis-information on radio and AC theory on the internet it's hard to separate it from facts. I KNOW the theory, and also am fully aware that theory often is not the same as reality. But myths and legends (like that a 12V lead-acid battery discharges when stored on a cement floor) make it hard to discuss what's really going on. This is why I use a scope - it tells the Real Story instead of looking at misleading meters.

druid
VA7LAS