Legal Limit on 160?

[KA3EKH]

Ok, now that I can develop three to four hundred watts of carrier with no issues from my old RCA transmitter the question is what should my carrier power be? People have been telling me something about it having to be no greater than 325 watts of carrier because when you figure in both sidebands the total will be one kW. Sounds fishy to me, I am new to the AM 160 "Ham" world but spent around thirty years working in both radio and TV broadcasting and in the commercial world have seen were the old analog television stuff was rated in peak envelope but every AM and FM radio station I have ever dealt with is rated in carrier power. A 1 kW AM station is 1 kW CW power and I have always assumed 100 percent peak positive to be 2 kW although most broadcasters run above 100 % positive and 100 % negative is 0 kW. Is an armature station responsible for its peak envelope power or its carrier power? And last but not least if I limit my positive peaks to 90% my PEP at 0.5 kW will still be below 1.0 kW so where does 0.325 kW come from?  I have gone thru several field inspections for some of the 5.0 kW AM station I take care of and never have had any issues deterring power with the FCC, Its always plate E and I times efficiency but this whole ham thing is a mystery.
Ray Fantini

[KD6VXI]

On a 100 percent modulated carrier, your PEP value will be 4 times carrier.

IE, you run 350 watts of carrier, you have 1400 watts PEP, modulated 100 percent in the pos and neg directions.

Welcome to the 'armature' world..


--Shane
KD6VXI

[WA1GFZ]

1KW carrier with two 375 watt side bands = 100% modulation 1500 watts PEP

[Steve - WB3HUZ]

It's all there in Part 97, specifically Subpart A, 97.3(b)(6) and Subpart D, 97.313.

[K1JJ]

For 160M, to figure legal peak power using a shaded dipole, use:


watts= f(x) = f(-x)
f(x) = 2/PI  cos xy dy f(t) cos yt dt
if f(-x) = -f(x) then
f(x) = 2/PI  sin xy dy sin yt dt



If using BIRD watts with a class E rig then:

watts = g(x) = (2/PI)f(t) cos xt dt
xk+1 = (xk + y / (xk)n-1) / 2
g(x) = (2/PI)f(t) sin xt dt




If using CPI watts on 27 mhz then use:
 

Watts = f(-x) = f(x) then
(f(x)) ) = f(x)
If f(-x) = -f(x) then
(f(x)) ) = f



**** (BTW, the above formulas are bogus, just kidding...    - don't ya hate it when someone answers a simple question with complex formulas?)



The real answer:  375 watts of AM carrier will equate to about 1500w pep for 100% modulation. (X4)

Many AMers like to have extra headroom available for asymmetrical audio and use 300 watts of AM carrier and 1500 w pep for ~120%+ modulation. (X5)

T

[WD8BIL]

Peak reading wattmeter makes it easy. Just crank the power till it reads legal limit!

[k4kyv]

1000 watts carrier power + 500 watts peak sideband power on voice crests @ 100% modulation = 1500 watts.

[Bill, KD0HG]

Huh? Whaz this 1500 watts PEP stuff?

Through the history of ham radio, 1000 watts DC input x 90% class C efficiency = 900 carrier watts out.
4x900 watts = 3600 watts PEP into the antenna load w/ 100% modulation.

That too simple for some people. Eh, Don? Hook up an AC wattmeter and read the load.
1 KW - blowers and filaments.

What about us open-line feed guys?

;-)

[K1JJ]

Clark Kent = 375w carrier

Superman = 1KW carrier

Batman = 1500w carrier

The Thing = 5KW carrier

Spiderman = 10KW carrier

The Shadow = 50KW carrier


Become a superhero - take your pick.


T

[KM1H]

You dont bring a knife to a gunfight so dont expect to bring a peanut whistle to a brawl.

Ive decided to grandfather myself in as Im using the same power I ran in the late 50's which was 1000W input to 250TH's Probably a realistic 700-750W out back then.

Pretty soon I can double that with a single switch and go from DSB to SSB AM ....well almost anyway, just need bigger bottles for linear service.

[K5IIA]

man wa1hlr went into detail on this on 75m the other night. was very entertaining.

i asked a ssb station on 3.882 and he told me a.m. legal limit was 20w carrier with 50 percent modulation.

hihi 73

[k4kyv]

Through the history of ham radio, 1000 watts DC input x 90% class C efficiency = 900 carrier watts out.
4x900 watts = 3600 watts PEP into the antenna load w/ 100% modulation.

I'd bet the typical plate modulated amateur transmitter is lucky to get much more than 50% efficiency to the antenna, once you consider  losses in the tank circuit, feedline, antenna tuner, etc.  That 70-75% figure listed in the tube manuals is the ideal plate efficiency of the tube itself, based on the assumption  that the tube is perfectly good and that all the parameters are correct. Pretty much hypothetical in the real world.  90% reportedly is achievable with some of the class D and class E rigs, as well as class C tube rigs with the 3rd harmonic traps installed.  The latter would probably complicate QSYing too much for use in a ham rig, unless you operate like some of those slopbucket groups who never budge a kilocycle off "their" frequency.

[ke7trp]

My new rig is 72 percent as measured.its all sold state except for output and modulators. Good audio processing is the key to over come the power limitation. More then half the stations out there have low and weak audio. The dap processor really helped my station.

[W5COA]

I did the calculation years ago, and came up with the 4X carrier power for PEP on a 100% plate modulated class C carrier.

But, I wonder about AM transmitters that only send out a carrier and one sideband. Doesn't the Collins KWS-1 do that? So what would the PEP vs carrier formula be for that mode?

Jim (Who used to run a pair of link coupled 813's modulated by a pair of 810's, but now only has 375 Watts output available)

[Opcom]

A calibrated scope can be used if you know the impedance of the load and it is resistive. It will work for any signal, AM or carrier plus one sideband. The GRC-106 makes carrier + USB and this works fine (but that set is only 400W pep)

Observe peak to peak RF voltage at the crest of the modulation cycle at a convenient point after the transmitter.
1.) p-p volts / 2.828 = RMS volts
2.) RMS volts / load resistance = load current
3.) RMS volts * load current = watts.

If you have an old scope to dedicate to this, you can mark a line on the screen where 1500W is, assuming the load resistance is the same.

or just use this method which is not as accurate but comes close:

1.) peek twice at the bolts, roughly divide by twice the relative SQRT of the sum of the two right sides, then see how many of the bolts are a mess.
2.) divide the messy bolts by the futile resistance to find the borgs' currants.
3.) bolts that are a mess divided by the value of the currants is what.

[flintstone mop]

1.) peek twice at the bolts, roughly divide by twice the relative SQRT of the sum of the two right sides, then see how many of the bolts are a mess.
2.) divide the messy bolts by the futile resistance to find the borgs' currants.
3.) bolts that are a mess divided by the value of the currants is what.

Some very valuable info here...
The non-calibratede way I do it is:
SCOPE connect to I.F. out on Receiver
Time base set 2ms/div
Adjust for 2 divisions of dead carrier BIG thick GREEN trace

When you modulate you will see the pos and negs. The positves may go two or more divisons UP THats good

FRED

[W1AEX]

As Steve HUZ mentioned, it's all in the regs, however, it's subject to interesting interpretations. I have several crappy-amateur-consumer-grade "peak power meters" of varying quality and they all seem to agree on my power output within a plus or minus 500 watt range (one says 500, another says 1000, another says 1500 watts PEP). Which one is correct? Well, they all report output that is within the legal limit, so who cares? My observations are that it appears to fall under a "don't ask don't tell" policy. Run what you brung, and if someone asks you over the air how much power you are running, just say, "I'm running a substantial amount of power, enough to maintain contact with you."

[WQ9E]

Like Patrick, the only measurement I really trust is using my Tektronix scope and dummy load.  The scope calibration is very easy to check/verify (as is the dummy load impedance) and peak output level is clearly visible.  I haven't bothered yet but I really should write a simple program for my Tektronix 7854 scope/waveform analyzer and let it calculate and display power level.

[KA1ZGC]

Nothing beats a thermal RF current meter. Here's an easy way to build one:

1. Kill a deer.
2. Gut said deer. Save intestines for sausage casings, other organs for dinner.
3. At one end of deer, attach appropriate RF connector as the input.
4. At other end of deer, attach appropriate RF connector as the output.
5. Attach amplifier to deer input, antenna to deer output.
6. Hang deer.
7. Apply RF. Dependent on frequency, the RF will cause the deer to heat and expand in the RF path only, causing the deer to swing in one direction.

The amount of deer deflection will indicate the amount of heating caused by RF. Once calibrated, this instrument will be very accurate, based on the well-known physical law which states that the Angle of the Dangle is directly proportional to the Heat of the Meat.

I think Bill Orr did a writeup on this, or maybe it was Buddly.

[k4kyv]

I have several crappy-amateur-consumer-grade "peak power meters" of varying quality and they all seem to agree on my power output within a plus or minus 500 watt range (one says 500, another says 1000, another says 1500 watts PEP). Which one is correct? Well, they all report output that is within the legal limit, so who cares? My observations are that it appears to fall under a "don't ask don't tell" policy. Run what you brung, and if someone asks you over the air how much power you are running, just say, "I'm running a substantial amount of power, enough to maintain contact with you."

I have a Mirage amateur radio grade wattmeter that was calibrated against a Bird 43.  The SWR function works HI HI FB OM, but the output power indication shows me running more power output than I am running DC input to the final.  I think maybe I have accidentally stumbled upon a solution to the world's energy crisis.  Now, in addition to international fame, I can forget about the money I was about to receive from the former bank CEO turned political refugee who needs my help getting his fortune out of Nigeria... chicken feed.

As for RF power measurement from my ham transmitter, I'll continue to depend on the trusty old thermocouple rf ammeter that has served me well for the past 45-plus years.

[KA3EKH]

Wow, all these formulas make my head hurt! If I get a dead deer from the side of the road will that work just as well as a hunted one? Did read 97.313 and that’s about useless, all it says is thou shall not exceed 1.5 kW peak envelope, so does that mean I can run 1.2 kW of carrier and 10 % modulation? A word about how I am deterring power now. I have the transmitter connected to a Bird 1.5 kW load that has a -40 Db sample port that I use to feed a spectrum analyzer and also have a Bird 43 with a 1 kW slug in front of the load. Have been using the Bird to determine actual power and figuring efficiency backwards by the difference between input power to the PA and power to the load. Can always read the level on the analyzer in Dbm, add 40 to that and convert that back to watts but have not done that yet. Have one of those new fancy HP digital power meters that I can borrow from the TV station but that will only confuse things further. So for now just going to stay at the "Clark Kent" level of 375 watts of carrier but if it warrants it can always stuff the second tube back in and be Superman. How many "Supermen, Batmans and Things" are their out on the bands?
Ray Fantini KA3EKH

[K1JJ]

Ray,

Yes, on AM you CAN legally set your output carrier to 1KW and then run 1500w pep output - and some guys do. The problem is, yes, it makes a quiet frequency to blot out the static, but the audio is way down. Perhaps the "perceived" s/n ratio is the same as 375w at 1500w pep, I dunno.  

Another advantage of lower audio is that many of the older tube receiver detectors cannot handle high levels of modulation before they distort. So running 700w carrier at 1500 pep for local, quiet contacts works for some guys.

But in general, I would set the ratio to at least X4 most of the time. (and even X5 or X6 for times when cornditions are rough)

Good questions, OM. Keep axing away.


T

[k4kyv]

A clear demonstration that the FeeCee shot themselves in the food with that p.e.p. bullsh!t.  Even experienced knowledgeable hams either don't agree exactly on what the power limit is, or don't fully comprehend what the regulations say.  That being the case, what could one expect from Joe Bloe Hammy Hambone with the typical level of technical expertise we see from to-day's question-pool licensees?

With 1000 watts DC input, the rule was unambiguous, very simple to understand and simple to measure. Full compliance capability was realistically expected even from brand new holders of the old 5 wpm, 75-watt Novice Class.   How do you attach a Bird 43 to an open wire fed antenna from a parallel or series tuned circuit and no 50 or 75Ω coax link between the transmitter's tank circuit and the ATU?  What about an antenna fed directly off the output tank circuit with no feed line at all?

[Steve - WB3HUZ]

97.3, (b)(6) PEP (peak envelope power). The average power supplied to the antenna transmission line by a transmitter during one RF cycle at the crest of the modulation envelope taken under normal operating conditions.

What is ambiguous about this?

[KF1Z]

:-)