Hi Tom,
I can't seem to make your math work for finding input IMD contribution?
Where does the divide by 20 come from in finding B?
73
Gary K4FMX
> -----Original Message-----
> From: Tom Thompson [mailto:tlthompson@qwest.net]
> Sent: Saturday, May 05, 2012 8:54 PM
> To: garyschafer@comcast.net
> Cc: 'Roger (K8RI)'; amps@contesting.com
> Subject: Re: [Amps] LDMOS Solid State Amplifiers
>
> Roger is correct. The two exciters, which were about 70 kHz apart in
> frequency, were combined with a 6-dB combiner and the input IMD was
> monitored with a through sampler as it went into the amp. The amp under
> test was fed into a dummy load and monitored with another sampler into a
> spectrum analyzer. Both samplers replicate the sampled signal 40 dB
> down. A single tone from the amp was referenced and then the two
> exciters were adjusted to give two carriers at a level 6-dB below the
> reference.
>
> The input IMD3 was -44 dBt and the output was -32 dBt. Using the
> following equation:
>
> B = 20Log(1+10^(-A/20))
>
> A = 44 - 32 = 12. Therefore B = 2 dB which is the contribution to the
> output IMD3 from the input IMD3. It then follows that the output IMD3
> would be 2 dB better, thus -34 dBt if the input had no IMD3. This only
> applied to IMD3 in my case because IMD5 and higher were not measurable
> on the input.
>
> I hope this is clear.
>
> 73, Tom W0IVJ
>
>
>
>
>
>
>
>
> On 5/5/2012 1:15 PM, Gary Schafer wrote:
> > Roger,
> > as I understood it he used the two independent exciters thru a
> combiner to
> > create the two tone signal for the IMD test. They would be separated
> in
> > frequency by whatever amount (a few KHz or so could be used). Equal
> levels
> > would have to be use but phase is irrelevant.
> >
> >
> > 73
> > Gary K4FMX
> >
> >> -----Original Message-----
> >> From: amps-bounces@contesting.com [mailto:amps-
> bounces@contesting.com]
> >> On Behalf Of Roger (K8RI)
> >> Sent: Saturday, May 05, 2012 2:36 PM
> >> To: amps@contesting.com
> >> Subject: Re: [Amps] LDMOS Solid State Amplifiers
> >>
> >> On 5/5/2012 11:30 AM, Tom Thompson wrote:
> >>> Bob,
> >>>
> >>> Vdd was fed in at the U point on the brass tube, single turn, point.
> >> If
> >>> I switch back to that transformer, I'll try the bifilar choke feed
> and
> >>> report. I did some IMD measurements this morning. I fed the amp
> with
> >>> two Norcal 40 QRP transceivers on 40 m.
> >> If I have followed this correctly:
> >> As this is a single, PP amp fed with two independent exciters how do
> you
> >> maintain proper phasing which is critical? Even if the exciters are
> >> synchronized a tiny difference in path length can make a difference.
> >>
> >> 73
> >>
> >> Roger (K8RI)
> >>
> >>> The input IMD due to the
> >>> combiner isolation was -44 dBt where dBt means below one of the two
> >> tone
> >>> peaks instead of the carrier which if present would be 6 dB higher.
> >> The
> >>> contribution from the input IMD is given by B = 20Log(1+10^(-A/20))
> >>> where A is difference in dB between the input IMD and the output
> IMD.
> >> I
> >>> measured an output IMD3 of -32 dBt at 100 watts on the amp under
> test.
> >>> That makes A = 12dB therefore B = 2dB which makes IMD3 = -34dBt.
> IMD5
> >>> was -60dBt and IMD7 was -52dBt.
> >>>
> >>> 73 Tom W0IVJ
> >>>
> >>> On 5/4/2012 11:59 PM, Bob Henderson wrote:
> >>>> Tom
> >>>>
> >>>> Interesting. Thanks.
> >>>>
> >>>> With your brass tube transformer, how was Vdd fed to the drains?
> >>>>
> >>>> The 10db reduction in H3& H5 is a significant improvement but I
> am
> >>>> wondering how much is due to the bifilar choke feed of Vdd and how
> >> much due
> >>>> to transition to a TLT? It would have been interesting to see what
> >> change
> >>>> resulted from adding the bifilar choke feed to your brass tube
> >> transformer
> >>>> set up.
> >>>>
> >>>> H3& H5 at -22dBc or better is easily good enough. After that,
> it's
> >> all
> >>>> about IMD performance.
> >>>>
> >>>> 73 Bob, 5B4AGN
> >>>>
> >>>> On 4 May 2012 23:43, Tom Thompson<tlthompson@qwest.net> wrote:
> >>>>
> >>>>> Bob,
> >>>>>
> >>>>> I am experimenting with the 300 W Freescale part. Using the brass
> >> tube
> >>>>> output transformer with a single turn on the primary and 2 turns
> on
> >> the
> >>>>> secondary without a harmonic filter I measured the following:
> >>>>> Vdd = 50 V
> >>>>> Id = 10.5 A
> >>>>> Po = 200 W
> >>>>> 3H = -11.8 dBc
> >>>>> 5H = -20.5 dBc
> >>>>> I then followed Manfred's suggestions and used a 4:1 transmission
> >> line
> >>>>> transformer wound with 30 ohm coax, a bifilar wound power combiner
> >> to
> >>>>> supply drain voltage, and a choke balun on the output of the
> >> transmission
> >>>>> line transformer. I then measured the following with no harmonic
> >> filter:
> >>>>> Vdd = 50 V
> >>>>> Id = 7.5 A
> >>>>> Po = 200 W
> >>>>> 3H = -22dBc
> >>>>> 5H = -30 dBc
> >>>>> When I reduced the power output to 100 W, 3H went to -30 dBc. In
> >> all
> >>>>> cases the total Idq was 1.5 A.
> >>>>> I hope this helps.
> >>>>>
> >>>>> 73, Tom W0IVJ
> >>>>>
> >>>>>
> >>>>>
> >>>>> On 5/3/2012 8:21 PM, Manfred Mornhinweg wrote:
> >>>>>
> >>>>>> Bob,
> >>>>>>
> >>>>>> My problem area was the extent of harmonics generated within
> the
> >>>>>>> device. H3 was within a dB or two of fundamental energy levels
> >> and
> >>>>>>> H5 only marginally better.
> >>>>>>>
> >>>>>> That typically happens when your output network isn't correctly
> >> done.
> >>>>>> There is an incredible amount of equipment, including HF ham
> >>>>>> transceivers comemrcially made today, that have incorrectly
> >> implemented
> >>>>>> power amplifiers, due to their designers not understanding of the
> >> basic
> >>>>>> principles under which transformers operate.
> >>>>>>
> >>>>>> A serious problem. My output arrangement focused largely
> upon a
> >> 1:9
> >>>>>>> coax wound RF2000 from RF Parts as used in the Granberg designs
> at
> >>>>>>> the 1kW level.
> >>>>>>>
> >>>>>> Granberg apparently was the one who "invented", or at least
> >> popularized,
> >>>>>> the wrong output network. Several of his papers contain the
> >> mistake,
> >>>>>> but others do not. It seems to me that he really didn't
> understand
> >> this
> >>>>>> issue, at least not when he published those old papers.
> >>>>>>
> >>>>>> How are you feeding the drains? If you are using a bifiliar
> choke,
> >>>>>> designed in such a way that it can act as a balancing
> >> autotransformer,
> >>>>>> then that should be fine, and you have to look elsewhere for the
> >> reason
> >>>>>> of the high harmonics. But if you are using two individual
> chokes,
> >> then
> >>>>>> that's wrong, and if you are feeding the drains through some sort
> >> of
> >>>>>> center point on the transformer, then there is a pretty good
> chance
> >> that
> >>>>>> it's wrong too!
> >>>>>>
> >>>>>> Typical symptoms of the incorrect output configuration are:
> >> Extremely
> >>>>>> high distortion (harmonics, IMD), horrible waveform at the
> drains,
> >> that
> >>>>>> includes peaks well above twice Vdd, low efficiency, low gain,
> and
> >> a
> >>>>>> sort of gain breakpoint: Up to a certain power the amp is easy to
> >> drive,
> >>>>>> and from that point up it gets suddenly very hard to drive
> further.
> >>>>>>
> >>>>>> Harmonics were not a consequence of transformer saturation
> >>>>>> That could hardly ever happen at HF. Before you saturate a
> ferrite
> >> core
> >>>>>> at HF, you will melt it down with the losses!
> >>>>>>
> >>>>>> But DC saturation can happen, in very tricky situations,
> specially
> >> if
> >>>>>> you have hugely more inductance than needed.
> >>>>>>
> >>>>>> No problem in a single frequency amp but I am way short of
> >> clever
> >>>>>>> enough to figure out a scheme which will handle that over 5
> >> octaves.
> >>>>>> Use either an output transformer that has a true center point, or
> a
> >>>>>> bifiliar choke to supply power. Note that the typical RF power
> >>>>>> transformers made from two ferrite tubes, with a single-turn
> >> primary, DO
> >>>>>> NOT HAVE A CENTER POINT. The junction of the two metal tubes is
> NOT
> >> a
> >>>>>> center point! Using this junction as a makeshift center point
> >> causes
> >>>>>> endless trouble, and many amplifiers, based on some of Granberg's
> >>>>>> designs, contain exactly this mistake.
> >>>>>> With transmission line transformers, a center point is usually
> also
> >>>>>> unavailable, but some transmission line configurations can have
> >> one.
> >>>>>> The basic point is this: Class B or class AB push-pull amps MUST,
> I
> >>>>>> repeat _MUST_ have something that provides balance around a true
> >> center
> >>>>>> point. It cannot work in pure differential mode, because each FET
> >>>>>> conducts for half of each cycle, and is in high impedance during
> >> the
> >>>>>> other half cycle. You cannot draw current between one transistor
> >> that is
> >>>>>> on and another that is off! That's why balun or balbal type
> output
> >>>>>> transformers only work correctly in conjunction with a bifiliar
> >> feed
> >>>>>> choke that provides the center point.
> >>>>>>
> >>>>>> Class A push pull amps do not have this restriction, and can work
> >> well
> >>>>>> in pure balanced mode.
> >>>>>>
> >>>>>> So, check your feed arrangement, maybe that's where your problem
> >> is!
> >>>>>> Manfred
> >>>>>>
> >>>>>> ========================
> >>>>>> Visit my hobby homepage!
> >>>>>> http://ludens.cl
> >>>>>> ========================
> >>>>>> ______________________________**_________________
> >>>>>> Amps mailing list
> >>>>>> Amps@contesting.com
> >>>>>>
> >>
> http://lists.contesting.com/**mailman/listinfo/amps<http://lists.contest
> >> ing.com/mailman/listinfo/amps>
> >>>>>>
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