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W4RFM
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« on: May 01, 2013, 12:16:43 PM »

Simple question, how much audio power do you need to modulate a 250 watt AM transmitter?  Answer 125 watts. Yes, no.  I am finding several differing thoughts on that subject.  Some say, if you put 125 watts of audio into your calss C amplifier, you must also de-rate that amount by the efficiency of the class C amplifier, others have said other things.  I would like for you folks to discuss this.
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« Reply #1 on: May 01, 2013, 12:52:41 PM »

Bob,

Here is my understanding of the topic. This applies to high level class C plate modulation only.

For 100% modulation with sine wave audio tone, modulator audio OUTPUT power must be 50% of class c rf INPUT power. For asymmetric voice audio, average output power will be somewhat less than 50% for 100% modulation. In either case, spare audio power capability is a good idea.

For example, if you build a transmitter with 500W input power, say 2000VDC @ 250mA, you would probably get about 350-380 watts of output power from the RF stage. To modulate 100% with sine wave tone, you would need 250 watts audio output. To modulate 100% with voice audio, you would probably need somewhere between 200 to 250 watts of average audio power. I would try to design a modulator capable of between 270-300 watts output to have some spare audio capability.

Regarding the modulation transformer, if you run the RF plate current through the secondary of the mod transformer, then you would need a 250W unit to fully modulate a 500W input RF section. If you use a mod reactor so that the RF plate current does not traverse the modulation transformer secondary, then you could probably get by with a slightly smaller unit.

In my recent build, I am running the 822 final at 2000VDC and 250mA and am getting 375 watts output. The mod transformer is a UTC S-22 rated at 250 watts. I am using a mod reactor and there is no DC current through the S-22 secondary. My modulators are 811A's at 1350VDC and are capable of about 320W audio output. So you can see that I have "overbuilt" my modulation system.

This is my crude way of looking at class C plate modulation, but it has worked for me. I'm sure the EE's on the board can give you a more technical explanation.

Ron
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kb3ouk
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« Reply #2 on: May 01, 2013, 01:16:23 PM »

A 250 watt transmitter is around 335 watts input power, assuming average class C efficiency of 75%. So your 250 watt transmitter has a modulator that would at minimum put out 170 watts. However, you have to take transformer losses into consideration. The best advice I've heard is to just simply build a modulator that is capable of the same power output as the RF final. So if you have a 250 watt final, build a 250 watt modulator.
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« Reply #3 on: May 01, 2013, 04:39:05 PM »

Just do it the way it was always done, use Input power for the Class C calculation and dont fuss over efficiency if its only 60% on 10M or less on 6M, and 75% on 75/160.

Build a modulator that can provide about a 125% minimum (this assumes no processing otherwise go for 150%) of the 50% audio Output needed since that is based upon tube manual audio output and not what comes out of the transformer. This overhead also allows for tube aging since hams are notorious for buying used tubes of unknown remaining life......as long as they are cheap or free Roll Eyes Grin

A pair of 811's is fine for a single 813 as an example and very little of their 340W theoretical capability will be wasted.

There is nothing that irks me more than listening to a strapping carrier and pipsqueak audio.

Carl
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« Reply #4 on: May 01, 2013, 04:53:52 PM »

The plate dissipation rule of thumb...

Well, the books I've read all say that the power required to plate modulate a class C stage is 50% of its input power. The efficiency of the modulated stage is irrelevant and does not enter into the calculation once you're thinking input power. So the answer would be 50% of the input power of the modulated stage must be available at the output of the modulation transformer. However, some allowances should be made. The human voice is asymmetrical, not a sine wave, so add some power for that. A modulator running at its limit has more distortion than if it has some room to spare so add some more power for that. Something on the order of 100% would have a good safe design margin.

A common setup for AM broadcast transmitters is to use a pair of tubes modulated by a pair of the same tubes. So a comfortable rule of thumb would be to have as much plate dissipation in the modulator tubes as there is in the RF final tube(s). A classic AM ham setup (as Carl mentions) is an 813 modulated by a pair of 811As. 125 W Pd for an 813 and 65x2=130W Pd for the 811As. A good match. Anything beyond that is just for bragging rights Wink

Sure, less may work ok, but if you're going to build a rig - make it a good one!

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K1JJ
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« Reply #5 on: May 01, 2013, 05:13:20 PM »

A common setup for AM broadcast transmitters is to use a pair of tubes modulated by a pair of the same tubes. So a comfortable rule of thumb would be to have as much plate dissipation in the modulator tubes as there is in the RF final tube(s). -Don


This was my first reaction.  Pick the tubes.   Over the years I have often run a single RF final modulated by a pair of the same. This has produced excellent headroom and cleanliness.  I've run a single 813 by a pair -  a single 4X1 by a pair...   a single 833A by a pair... and many more.  All were able to do 150% positive easily and cleanly.  This is well suited for a 2:1 step down mod xfmr.

The the tetrode / triode equivalent might be a 4X1  X  3X1's.   Or a 4-400A X 3-500Z's.   Or as already suggested, an 813 X 811A's.  (Though, with air, an 813 is probably more of a 250 watt tube)


The other choice is a pair modulated by a pair. A 1:1 mod xfmr ratio is suited here.  I now like 2X2 the best. Over the last five years I have leaned towards running audio around 125% positive and have no need for bigger headroom.  But it MUST be clean.

After running THD tests on many AM rigs, I saw how quickly distortion gets bad when we get at or very close to modulator saturation. We want the modulators to loaf up to the peak power we require - or we must pay the price in the form of a wider signal.  It takes a good ear to hear the audio quality difference ON frequency - but splatter OFF freq can be heard by everyone.

Also, the old Tron rule -  the mod transformer needs 10 pounds per 100 watts of audio power. This is a  good idea, especially if extreme clean lows are desired.  The classic Gates, RCA-type  BC mod iron rated at 1KW, etc., can't be beat. (used with a Heising reactor)

T

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« Reply #6 on: May 01, 2013, 07:50:50 PM »

There is nothing that irks me more than listening to a strapping carrier and pipsqueak audio.

It's the opposite that irritates me: those that insist on running a tiny carrier but big audio. Sure it may sound good on a sync detector, but not everyone has one. When done right, running less than 100% modulation (but not less than maybe 75%) can have its advantages. One thing I noticed is weaker modulation (less than 100%) is a little easier to copy in high static than heavy modulation (over 100%).
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« Reply #7 on: May 01, 2013, 11:29:06 PM »

I have to agree. Most BA receiver detectors don't really like mod % over 100 (or 90). Sync or not doesn't matter. It's one thing to listen to 125% mod on a fine signal ground wave, and impressive. Beautiful. But here in TX, an awful lot of the 'east coast sound' stuff we hear is unintelligible mush, regardless of % modulation. More mod % is just louder mush.

In my humble opinion, the best (distortion free, loud, always intelligible ... and always intelligent!) ham AM signal on the air anywhere is K4KYV, period.  Whatever Don is  doing, he's doing it right.
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K1JJ
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« Reply #8 on: May 02, 2013, 12:42:53 AM »

Other than severe distortion, which is rare amongst the AMers, the transmitted EQ curve may be a big part of why it  may be hard to copy some.  Boosting too much low end or having a muddy middle because of no EQing at all can cause intelligibility problems.  We all have different voices and must find the best curve to give the desired result.

Running a high % of positive modulation will cause intelligibility problems when there is selective fading. But, yes, many older receivers can't handle over 100% positive from diode detector overload. Pity, cuz most male voices, when limited to -95% negative, naturally reach up to 125% positive or higher. I know mine does.  So if I limit my positive peaks to  + 100% positive, my negative peaks (pinched carrier) are modulated only about 70% or so.  Not an optimum signal to noise ratio.


There's not much that can be done on the receive end for poor transmit EQing, but by adding a simple SoftRock that has a sync detector, the higher modulation levels above 100% could be handled better.  It pays to have two AM receivers going in the shack for this purpose.

**  BTW, I noticed today for the first time, a spell checker is working as I type a reply. Fantastic!!!!  Tnx to Steve / HX or whoever got it working.  Lords knows I need it.  Now  a grammar / phrasing checker. Just imagine.  Wink

T

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« Reply #9 on: May 02, 2013, 08:23:40 AM »

**  BTW, I noticed today for the first time, a spell checker is working as I type a reply. Fantastic!!!!  Tnx to Steve / HX or whoever got it working.  Lords knows I need it.  Now  a grammar / phrasing checker. Just imagine.  Wink

Wat spel cheker? I bet it's the spell checker built into your browser, because if it's on this site, it's obviously not working.


Anyway, one thing I have noticed that makes high positive modulation easier to copy on some receivers is to slow the AGC down if possible, or just use the manual RF gain.
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« Reply #10 on: May 02, 2013, 02:16:36 PM »


**  BTW, I noticed today for the first time, a spell checker is working as I type a reply. Fantastic!!!!  Tnx to Steve / HX or whoever got it working.  Lords knows I need it.  Now  a grammar / phrasing checker. Just imagine.  Wink

T

It's in your browser. Grammar/phrasing checkers are generally in most word processors. If you have a long and wordy technical or otherwise dissertation, you can type it into a word processor, check it for grammar, punctuation, and phrasing (i.e. no dangling participles, etc.), and when satisfied, copy and paste it into a new post. Most likely, it can add an additional level of clarity to your post(s).
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« Reply #11 on: May 02, 2013, 03:28:09 PM »

My current plan (subject to change based on input from you folks) was to run a pair of 811a's using a TRIAD 12AL 125 watt transformer, to modulate a single 4-125a running at 1400vdc @ 255 mA input.  357 watts in, maybe 250 out.  I can build the modulator power supply to do whatever, as I will also have about 60 HY of modulation reactor in the line.  Both designs are straight out of the 1956 (14th edition) west coast handbook. I have talked with several of you about this rig over the past years and I appreciate all of your input.
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« Reply #12 on: May 02, 2013, 06:31:38 PM »

Quote
My current plan (subject to change based on input from you folks) is to run a pair of 811a's using a TRIAD 12AL 125 watt transformer, to modulate a single 4-125a running at 1400vdc @ 255 mA input.  357 watts in, maybe 250 out.

Sounds like a plan for piss weak audio; did you read and understand what I and others said?.
Half or 357 is 178.5 which is what you need for audio at 100% in a perfect world of theory. On a real earth at least 250W is needed and better yet at 300W for some reserve and not pushing the modulator hard.

Carl
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« Reply #13 on: May 02, 2013, 07:26:12 PM »

Quote
My current plan (subject to change based on input from you folks) is to run a pair of 811a's using a TRIAD 12AL 125 watt transformer, to modulate a single 4-125a running at 1400vdc @ 255 mA input.  357 watts in, maybe 250 out.

Sounds like a plan for piss weak audio; did you read and understand what I and others said?.
Half or 357 is 178.5 which is what you need for audio at 100% in a perfect world of theory. On a real earth at least 250W is needed and better yet at 300W for some reserve and not pushing the modulator hard.

Carl

Agreed, Carl.  In my original correspondence with Bob, I had recommended the following alternate choices: Thordarson T11M77, UTC CVM4 or its daddy the VM4 or the Stancor A-3898.  The problem is that old friend voltage across a given class C impedance load.  If we can assume approximate 125% positive peaks on modulation, we have to design in more than the classic ½ audio power for a given class C power. 

Interesting about HLR’s rule of thumb of 10 lbs per 100 watt af power rule.  Turns out that those modulation transformers I recommended weigh out about 20+ lbs.  Now – if one wishes to design the classic “yellowey” audio parameters of 300 to 3000 cycles than we can take out a little from the mod iron requirements but you still cannot fool the typical requirements of the male asymmetrical voice characteristics.  My positive peaks come out to about 150% positive peaks with 100% negative points.

Interesting dialogue

Al
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« Reply #14 on: May 02, 2013, 07:57:11 PM »

... but you still cannot fool the typical requirements of the male asymmetrical voice characteristics.  My positive peaks come out to about 150% positive peaks with 100% negative points.

Interesting dialogue

Al

The thang that really makes me take notice about modulator overhead is how much power it takes to modulate a carrier at various percentage levels.

Someone correct me if I'm wrong, but 100% modulation requires X4 peak power.   150% requires X6 power.   200% requires X8 peak power.   (corrected numbers)

So even if we are looking at a natural 125% positive peak modulation level, we need X5 peak power, no?  
So much for the X4 requirement.  

I think many of the older boat anchor rigs were designed with X4 in mind and show it as they flat top around 110%. Or they even start to show poor linearity when approaching 90% positive.   But when we build our own, we can do away with these limitations.

The problem sometimes is we are limited by our modulation iron. The tubes may handle it but we may crap out our mod iron in the process. The worst risk is voltage.   To produce more peak power often means running a higher voltage.  The mod iron may be able to take the power, but even using arc gaps, the risk of internal flashovers is real when pushed.

That old heavy-duty BC mod iron is "almost" worth its weight in gold ... Wink

T

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« Reply #15 on: May 02, 2013, 08:07:20 PM »

Bob,  

Those 811A's can easily modulate a single 4-125 at 350 or so watts input well over 100%. At 1000VDC on the plates, they can produce about 250 watts of audio, and at 1500VDC, they can go up to 340 watts of audio. Your design has provided for plenty of spare audio capability.

The modulation transformer however, is a bit small for this design, but if you use a modulation reactor setup, you might be OK. I agree that it would be desirable to get a larger mod transformer if possible, or alternatively, run the final at a lower power input, say 275-300 watts input.

If you want to run 350 watts input, I have a Merit 175 watt mod transformer that might fit the bill nicely. Let me know if you'd like to have it and I'll send you the details.

Ron
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« Reply #16 on: May 02, 2013, 08:40:56 PM »

Quote
My current plan (subject to change based on input from you folks) is to run a pair of 811a's using a TRIAD 12AL 125 watt transformer, to modulate a single 4-125a running at 1400vdc @ 255 mA input.  357 watts in, maybe 250 out.

Sounds like a plan for piss weak audio; did you read and understand what I and others said?.
Half or 357 is 178.5 which is what you need for audio at 100% in a perfect world of theory. On a real earth at least 250W is needed and better yet at 300W for some reserve and not pushing the modulator hard.

Carl

Agreed, Carl.  In my original correspondence with Bob, I had recommended the following alternate choices: Thordarson T11M77, UTC CVM4 or its daddy the VM4 or the Stancor A-3898.  The problem is that old friend voltage across a given class C impedance load.  If we can assume approximate 125% positive peaks on modulation, we have to design in more than the classic ½ audio power for a given class C power. 

Interesting about HLR’s rule of thumb of 10 lbs per 100 watt af power rule.  Turns out that those modulation transformers I recommended weigh out about 20+ lbs.  Now – if one wishes to design the classic “yellowey” audio parameters of 300 to 3000 cycles than we can take out a little from the mod iron requirements but you still cannot fool the typical requirements of the male asymmetrical voice characteristics.  My positive peaks come out to about 150% positive peaks with 100% negative points.

Interesting dialogue

Al


If the bandwidth is rolled off below 300 hz the natural asymmetry disappears. I spelled all those words incorrectly.
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« Reply #17 on: May 02, 2013, 10:15:22 PM »

Yes Carl, I understand, that is why I said: my current plan subject to change etc. and  I can build the modulator power supply to do whatever, I can get a bigger transformer and do what you and the other prescribe, or I will de rate the RF output of this transmitter, (all of which is 85% finished), and build another that is larger in all respects. And thanks for the offer Ron, I have sent you an email.
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« Reply #18 on: May 02, 2013, 10:52:14 PM »


Someone correct me if I'm wrong, but 100% modulation requires X4 peak power.   150% requires X8 power.   200% requires X16 peak power.  

So even if we are looking at a natural 125% positive peak modulation level, we need X6 peak power, no?  
So much for the X4 requirement.  


not sure if you are referring to the N times carrier = PEP (the PEP bs) but at 150% positive, your peak power out should be six times dead carrier power.
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« Reply #19 on: May 02, 2013, 11:40:50 PM »

Thanks for the info, Rob.

So then it's X5 carrier power for 125% modulation -   X6 for 150%  - X8 for 200% and X16 for 300%.


T





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« Reply #20 on: May 03, 2013, 09:19:47 AM »

300 percent is 10x I think.

Maybe we need a chart!  Haha
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« Reply #21 on: May 03, 2013, 10:48:18 AM »

On modulation peaks, the electrical plate power input to the transmitter's rf output stage is (1+k)(1+k) x the electrical plate power input at carrier... where k is the peak modulation index (as a percentage)/100%.

For example, if the positive peak modulation index is 150%, then k= 150%/100% = 1.5, and (1+k)(1+k) = (2.5)(2.5) = 6.25

Therefore, if the peak positive modulation index is 150%, the peak electrical plate power input (on audio peaks) to the transmitter's output stage will be 6.25 x the electrical power input to the transmitter's output stage at carrier.

HOWEVER, there is a very non-intuitive and possibly confusing aspect to this.

While (for the particular example of 150% positive peak modulation index), the peak plate power input to the transmitter's output stage is 6.25 x the plate power input at carrier, the peak audio power that the modulator is providing is only: k/(1+k) of this peak power... i.e. 6.25 x 1.5/2.5 x the plate power input at carrier= 3.75 x the electrical plate power input at carrier.  

3.75 + 1 does not equal 6.25 (!). So where is the extra peak power coming from???

The extra peak power is coming from the stored energy in the capacitor at the output of the DC power supply that provides the unmodulated B+ to the RF output stage. [I.e., on 150% positive modulation peaks, the current being drawn from that capacitor is 2.5 x as large as the current being drawn from that capacitor at carrier... while the voltage across that capacitor is (essentially) the same as it is at carrier]

Therefore the DC supply for the unmodulated B+ is providing (from its stored energy) 2.5 x as much power on modulation peaks as it does at carrier.

3.75 + 2.5 = 6.25

As a second non-intuitive and possibly confusing aspect of this... the peak power being provided by the modulator's audio amplifier (i.e. not including power being drawn from the stored magnetic energy of the modulation transformer) is k/(1+k) x the total peak power being provided by the modulator (including power being drawn on positive modulation peaks from the stored magnetic energy in the modulation transformer).

Therefore, for the example we are considering (150% positive modulation peaks), the peak electrical power being supplied to the plate of the rf stage by the audio amplifier of the modulator is 3.75 x 1.5/2.5 x the electrical plate power being supplied to the rf stage at carrier = 2.25 x the electrical power being supplied to the plate of the rf stage at carrier.  

The total peak power budget is:

(k x k) x the electrical power being supplied to the plate of the rf stage at carrier from the audio amplifier

+

k x the electrical power being supplied to the plate of the rf stage at carrier from the magnetic energy stored in the modulation transformer

+

(1+k) x the electrical power being supplied to the plate of the rf stage at carrier from the energy stored in the capacitor of the unmodulated B+ supply

The bottom line is:

At a minimum, the modulator's audio amplifier must provide a peak audio power output of k x k x the electrical plate power input to the rf output stage at carrier.

Stu
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« Reply #22 on: May 03, 2013, 05:10:21 PM »

The nice thing about using 811As is that you can swap to 572Bs in the future if you want more headroom. Leave some extra room for them. A pair of 572Bs will modulate your 4-125A right out of its socket. Shocked

-Don
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« Reply #23 on: May 03, 2013, 10:43:26 PM »

Stu,

Thanks for the detail explanation on this subject of PEP.  I'll have read a few more times to understand everything.

Fred
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« Reply #24 on: May 16, 2013, 10:03:29 PM »

Simple question, how much audio power do you need to modulate a 250 watt AM transmitter?  Answer 125 watts. Yes, no.  I am finding several differing thoughts on that subject.  Some say, if you put 125 watts of audio into your calss C amplifier, you must also de-rate that amount by the efficiency of the class C amplifier, others have said other things.  I would like for you folks to discuss this.

Another view of modulation power is to use the AM carrier power and modulation index formula:

Pt = Pc X (1+m^2/2)

where Pt is the total output power consisting of carrier plus modulation power, Pc is carrier power, and m is the index of modulation.

Here is a table for determining the Total ouput power at percentage modulation and the minimum modulation power required:

For example 25% modulation is a modulation index of 0.25. m^2 = 0.0625, m^2 divided by 2 = 0.03125, 1 + 0.01325 = 1.03125, so total power output Pt for 25% modulation for a 1kW carrier is 1,031 Watts. Modulation power required would thus be 31 Watts.

Modulation index          multiplier coefficient (1+m^2/2)


0.25                           1.031

0.50                           1.125

0.90                           1.405

1.00                           1.500

1.10                           1.605

1.25                           1.780

1.5                             2.125


Taking for example the maximum power output for an AM broadcast station with a carrier power of 1kW and a modulation percentage of 125%;

Total Power Output at 125% Modulation is Pt = Pc X  1.780 = 1,780 Watts
Pm is Modulation Power is = Pt - Pc = 1,780 Watts - 1,000 Watts = 780 Watts.

Usual design considerations for this example would be to have an 800 Watt modulator circuit for a little headroom.

If one were to design a Class B P-P modulator circuit for a plate modulated 1kW Class C final, and assuming an efficiency of 70% for the 800 Watt P-P modulator,

the DC input power to the Class B P-P circuit would have to be 1,143 Watts.


Phil - AC0OB
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