Hi Gary,
B = 20Log(1+10^(-A/20))
Let's say you measured the input IMD3 at -44 dB and the output IMD3 at
-32 dB. The difference is 12 dB and that is A. Divide A by 20 which
equals 0.6. Negate that and raise 10 to that power which is 0.251. Add
1 to that which is 1.251 and take the Log of that which is 0.973.
Multiply by 20 which is 1.945 which is ~2 dB. This is B which is the
contribution to the output IMD3 due to the input IMD3. Therefore the
output IMD3 is really -34 dB if there was no contribution from the input.
I hope that explains it. If not get back to me and I'll give it another
try. If I confused you, I am sure I confused someone else.
73, Tom W0IVJ
On 5/6/2012 1:49 PM, Gary Schafer wrote:
> 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
>>>>>>>> ========================
>>>>>>>> ______________________________**_________________
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