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Re: [Amps] Amp question

To: "Will Matney" <craxd@engineer.com>, <amps@contesting.com>
Subject: Re: [Amps] Amp question
From: Karl-Arne Markström <sm0aom@telia.com>
Reply-to: Karl-Arne Markström <sm0aom@telia.com>
Date: Tue, 26 Jul 2005 16:46:23 +0200
List-post: <mailto:amps@contesting.com>
A "conventional combiner" is more often than not a "zero-degree combining 
network".

For a through discussion of power dividers and combiners take a look at:
http://www.minicircuits.com/appnote/an10006.pdf, and also pp 498 - 510
in "Single Sideband Systems & Circuits" 2nd edition.

It is the designer's choice if the combining system should be zero-degree or 
180-degree,
but as it simplifies construction and testing somewhat the zero-degree 
approach appears to be usually taken.

A special case is when you use 90-degree hybrids to absorb load reflections, but
they seem to be not so very common for multi-octave combiner networks, as those 
encountered in HF
amplifiers.

73/

Karl-Arne
SM0AOM


----- Original Message ----- 
From: "Will Matney" <craxd@engineer.com>
To: <amps@contesting.com>
Sent: Tuesday, July 26, 2005 12:23 PM
Subject: Re: [Amps] Amp question


> Karl,
> 
> Please see below;
> 
> > 
> > The reason why this works is that the output combining network in 
> > most solid-state PA's are
> > "zero-degree hybrids", which means that they expect the voltages 
> > (or currents) driving them to be
> > in-phase to provide a combined output. Output transformers in p-p 
> > stages can on the other hand
> > be classified as  "180-degree hybrids", which require 
> > opposite-phase driving voltages to operate properly.
> 
> 
> By looking over the schematic of this amp, everything looks the same as one 
> for 180 degrees phase difference except it being minus a splitter in the 
> front end. To me if it were to be zero degrees you wouldn't need a splitter 
> or combiner, just merely paralell the two modules the same as paralelling 
> tubes to add up the power output. They have two conventional RF output and 
> input transformers per module. These work in push-pull to provide a linear 
> signal being it's class AB1. Then each output from the RF transformer goes to 
> either end of the combiner transformer just like all the others being 180 
> degrees out of phase. However, on the schematic, they dont show any phasing 
> dots on the combiner. Still though, I would think it would still have to be 
> out of phase if it's wound the same as a conventional combiner. The only way 
> it could be different is the two coils be in phase instead out of phase? It 
> has to be one of the two by the schematic they show. I still just can't see 
> how it
 's
>   doing it by having them both on at the same time and not buck each other at 
> the end. I'll have to do some research on this as I've not seen that 
> arrangement in the Motorola app notes, nor in any book I have on the subject.
> 
> > 
> > As the phase shifts in the two amplifier halves can be assumed to be equal,
> > it remains to drive the halves with in-phase voltages.
> > 
> > The RC-network on the input side just acts as a resistive power 
> > dividing network with its outputs in-phase,
> > and as the amplifier power gains decrease with frequency, capactive 
> > shunt elements
> > can compensate for this decrease, as the insertion loss of the 
> > network also will decrease with frequency.
> 
> 
> Agreed.
> 
> > 
> > Finally, the reason for using 200 ohms for the balancing resistor 
> > is simply that the designer has chosen a 100 ohm
> > input design impedance for the combining network, which requires 
> > the amplifier outputs to be 100 instead of 50 ohms.
> > This is "perfectly legal" in an environment where you are at 
> > liberty to choose the design impedances freely, and do not have to 
> > take long runs of interconnecting RF cables into account. But the 
> > stray and winding capacitances present can be more difficult to 
> > handle in a system with a higher design impedance.
> 
> 
> Exactly. I figured it was because of it going into the low pass filters the 
> way they had it designed, and allowing them to finish stepping it down to 50 
> ohms. But, like you said, using 100 ohms increases the windings on every 
> transformer by X2.
> 
> > 
> > In professional circles this is not very common, as most 
> > solid-state professional PA's are built from modules
> > having a design impedance of 50 ohms.
> 
> Agreed. That just didn't make a lot of sense to me.
> 
> Best,
> 
> Will
> 
> > 
> > 
> > 73/
> > 
> > Karl-Arne
> > SM0AOM
> > 
> > ----- Original Message -----
> > From: "Will Matney" <craxd@engineer.com>
> > To: <amps@contesting.com>
> > Sent: Tuesday, July 26, 2005 6:57 AM
> > Subject: [Amps] Amp question
> > 
> > 
> > > All,
> > >
> > > Ok, most know how splitter/combiner transformers work in solid 
> > > state amps. Here's the question, how can an amp work using two 
> > > sets of push-pull amps, having a conventional combiner, but not 
> > > have a splitter in the front end? What this amp has is a divider 
> > > network made up of a RC network using several resistors in 
> > > paralell with a capacitor on either side of the RF input. In 
> > > other words, when the RF comes into this RC divider, it applies 
> > > RF equally to both sets of push-pull amps at one time. There is 
> > > no splitter transformer there. How can this work by having both 
> > > push-pull sets on where they should be 180 degrees out of phase, 
> > > and one being off while the other is on? To use a combiner 
> > > transformer, the RF from one set makes 1/2 wave and so from the 
> > > other set. The combiner then adds these two halves together 
> > > making a full sine wave. So, how can this work having RF going to 
> > > both sets at once instead of using a splitter which would put 
> > > them 180 Deg out of phase? This absolutely has  me stumped. By 
> > > the way, it's for an Ameritron ALS-500M.
> > >
> > > One last question, most combiners/splitters use a 100 ohm 
> > > resistor in paralell with it's inputs for a 50 ohm load. How can 
> > > this work if the resistance is 200 ohms? That would be for a 100 
> > > ohm load wouldn't it?
> > >
> > > Best,
> > >
> > > Will
> > >
> > > -- ___________________________________________________________
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> 
> 
> 
> 
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