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K2: HRD Power Control Anomaly

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wrmoore

K2: HRD Power Control Anomaly

I'm operating my K2/100 (KPA100+KAT100 in EC2) from Ham Radio Deluxe,
which works just great. But just now I was playing around with the HRD
power control slider and discovered some anomalies.

If I set the power on the K2 to say 50w, then move the HRD power slider
down to < 11 w, the LCD power readout on the K2 follows the HRD slider
down appropriately, but the KAT100 PWR RANGE LED stays in HIGH. If I
then transmit, the power is actually what I set it to (<11w). So does
that mean that I'm driving the KPA100 at QRP levels instead of using the
K2 PA and bypassing the KPA100??? Moving the slider to other power
levels (without moving the K2 power control) results in the same thing -
power out follows the HRD slider, but the PWR RANGE LED stays at HIGH no
matter if the power is set to <11w.

Now if I start with the K2 power control at <11w, first of all, HRD
displays 10x the actual power setting (i.e., 100 for 10w). That's easy
to get used to. But if I crank the HRD slider up beyond 10w (e.g., HRD
slider to 150 or 15w), the actual power output, and the K2 LCD power
display, go to 15w, but the KAT100 PWR RANGE LED stays in LOW. Again
does that mean that I'm driving the K2 PA to 15w and not going through
the KPA100???

Is this a HRD "issue" or a K2 CAT command issue???

Kinda strange!

73,
Randy, KS4L
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Don Wilhelm-4

Re: K2: HRD Power Control Anomaly

Randy,

If your K2 firmware is less than version 2.02e then the problem is with
your K2 firmware.
If the firmware is 2.02e or greater, then HRD is the problem. What
version of HRD are you using? If it is an older one, that may be your
problem.

See the K2 Programmer's Reference. The Extended form of the PC command
must be used to change the power from the low range to the high range.

I don't know whether Simon has implemented the Extended version of the
PC command in his software. You may want to ask that question on the
HRD forum.

I cannot check it out here, I have only the QRP K2 and no KAT100.

73,
Don W3FPR

Randy Moore wrote:

> I'm operating my K2/100 (KPA100+KAT100 in EC2) from Ham Radio Deluxe,
> which works just great. But just now I was playing around with the HRD
> power control slider and discovered some anomalies.
>
> If I set the power on the K2 to say 50w, then move the HRD power slider
> down to < 11 w, the LCD power readout on the K2 follows the HRD slider
> down appropriately, but the KAT100 PWR RANGE LED stays in HIGH. If I
> then transmit, the power is actually what I set it to (<11w). So does
> that mean that I'm driving the KPA100 at QRP levels instead of using the
> K2 PA and bypassing the KPA100??? Moving the slider to other power
> levels (without moving the K2 power control) results in the same thing -
> power out follows the HRD slider, but the PWR RANGE LED stays at HIGH no
> matter if the power is set to <11w.
>
> Now if I start with the K2 power control at <11w, first of all, HRD
> displays 10x the actual power setting (i.e., 100 for 10w). That's easy
> to get used to. But if I crank the HRD slider up beyond 10w (e.g., HRD
> slider to 150 or 15w), the actual power output, and the K2 LCD power
> display, go to 15w, but the KAT100 PWR RANGE LED stays in LOW. Again
> does that mean that I'm driving the K2 PA to 15w and not going through
> the KPA100???
>
> Is this a HRD "issue" or a K2 CAT command issue???
>
> Kinda strange!
>
> 73,
> Randy, KS4L
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wrmoore

Re: K2: HRD Power Control Anomaly

Don,

My K2 is running 2.04r and I have HRD 4.1 Build 2055 Beta, which I think
is the most recent.

My real question is what is happening in the KPA100 and the K2 when this
happens? Practically speaking, it isn't much of an operational problem.

73,
Randy, KS4L

Don Wilhelm wrote:

> Randy,
>
> If your K2 firmware is less than version 2.02e then the problem is
> with your K2 firmware.
> If the firmware is 2.02e or greater, then HRD is the problem. What
> version of HRD are you using? If it is an older one, that may be your
> problem.
>
> See the K2 Programmer's Reference. The Extended form of the PC
> command must be used to change the power from the low range to the
> high range.
>
> I don't know whether Simon has implemented the Extended version of the
> PC command in his software. You may want to ask that question on the
> HRD forum.
>
> I cannot check it out here, I have only the QRP K2 and no KAT100.
>
> 73,
> Don W3FPR

Don Wilhelm-4

Re: K2: HRD Power Control Anomaly

Randy,

That tells me that HRD is not using the extended PC command format.

The consequences of going from high power to low is that the scaling
resistors in the KAT100 will be active (they would normally switch out
of the circuit going from high to low power), and the power detection at
low power (<11 watts) will be incorrect - there will not be as much
voltage on the VRFDET line as the K2 is expecting and the K2 power
control circuits will try to increase the power to bring up the VRFDET
voltage. The result will be either high power output (i.e 100 watts
instead of 10 watts) *if* the KPA100 is also in high power state like
the KAT100 or if the KPA100 is in low power state (but not the KAT100,
the base K2 will put out maximum power and may give HI CUR messages.

Alternately if the KAT100 stays in low power mode, the KPA100 will only
have 1/10 the expected power.

Since HRD apparently does not do the low power/hi power switching, I
would recommend that you operate only within either the low power range
or the high power range when changing the power through HRD. To switch
ranges, use the K2 power knob and then proceed with HRD as normal.

73,
Don W3FPR

Randy Moore wrote:

> Don,
>
> My K2 is running 2.04r and I have HRD 4.1 Build 2055 Beta, which I think
> is the most recent.
>
> My real question is what is happening in the KPA100 and the K2 when this
> happens? Practically speaking, it isn't much of an operational problem.
>
> 73,
> Randy, KS4L
>
> Don Wilhelm wrote:
>
>> Randy,
>>
>> If your K2 firmware is less than version 2.02e then the problem is
>> with your K2 firmware.
>> If the firmware is 2.02e or greater, then HRD is the problem. What
>> version of HRD are you using? If it is an older one, that may be your
>> problem.
>>
>> See the K2 Programmer's Reference. The Extended form of the PC
>> command must be used to change the power from the low range to the
>> high range.
>>
>> I don't know whether Simon has implemented the Extended version of the
>> PC command in his software. You may want to ask that question on the
>> HRD forum.
>>
>> I cannot check it out here, I have only the QRP K2 and no KAT100.
>>
>> 73,
>> Don W3FPR
>>
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Stephen W. Kercel

: Resonance

In reply to this post by Don Wilhelm-4

Don:

Regarding the complex impedance of an antenna at the end point, you
raise some interesting questions.

I was always of the impression that the definition of resonance of a
half wave radiator is the condition in which the current at the center
is a maximum and the current at the ends is at zero. The current
distribution in a half wave antenna is analogous to the displacement of
a violin string, which when vibrating at resonance has zero displacement
at the ends and maximum displacement at the center. (Such a resonance is
easily detected with a grid dip meter, even if no feedline at all is
connected to the radiator. Admittedly, the feedpoint would need to be
shorted. Sweeping the grid dip meter through a range of frequencies is
analogous to the broadband energy in the "pluck" on the violin string.
In either case (assuming very high Q), only the energy at the resonant
frequency actually gets coupled into the device.)

In the case of center fed half wave element, a zero value of imaginary
component of impedance at a center feed point is a coincidental
indication of antenna resonance rather than the definition of antenna
resonance; it is used by amateurs because it is easy to measure, whereas
the current distribution is almost impossible to measure directly.

If you run the "BY dipole" simulation on EZNEC at a frequency and
radiator length for which the center feedpoint impedance is pure
resistance, and then move the feedpoint around, the radiation pattern
comes out just the same (within the limits of computational error)
regardless of the feedpoint location. Also, the magnitude of the current
distribution remains about the same, big in the center and approaching
zero at the ends. Thus, my sense is that by the analogy to the violin
string, the antenna is resonant at that length for that frequency
irrespective of the feedpoint location, or the fact that the feedpoint
impedance is complex.

Admittedly, that much of the discussion is literally academic, depending
on how we define resonance.

However, you raise another question that is more practical than
academic. You make the perfectly reasonable point that if I play around
with the radiator length, I should find a length that has an end
feedpoint impedance of some big value of R plus J0. I am sure you can do
that, but my question to you is, what is the advantage to doing so?
(Note: This is not intended as a smart aleck comment. If there is some
advantage easily obtained by tweaking the radiator length, I'd really
like to know what it is.) I do not expect that minimizing radiation from
the transmission line is one of those advantages. Changing the radiator
length such that you move away from the length that gives "violin string
resonance" would make the current distribution on the radiator more
asymmetrical and would increase the probability of feed line radiation.

This line of reasoning got me curious about something else. Elecraft
rigs are usually rated as being able to operate normally for any load
that has an SWR of 2 or less compared to a characteristic impedance of
50+J0. Am I correct in assuming that that means that the rig expected to
be able to operate normally into any complex load on or inside the SWR =
2 circle on the Smith Chart?

Why that matters is the following. Using a high impedance quarter wave
(approximately) transformer to Zepp feed a half wave radiator, it is
relatively straightforward to tweak the transformer length such that the
subsequent 50 Ohm coax has an SWR well inside the SWR = 2 circle. If so,
the Elecraft rig should be perfectly happy, even if I seldom if ever
actually find an impedance with a zero imaginary component.

Have I missed something in my thinking? (It would not be the first time.)

TNX & 73,

Steve Kercel
AA4AK

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Guy, K2AV

Re: : Resonance

> I was always of the impression that the definition of resonance of a
> half wave radiator is the condition in which the current at the center
> is a maximum and the current at the ends is at zero.

This would also be true of a 1/4 wave dipole fed in the center, or any
dipole less than a 1/2 wave. The current would be higher in the center of a
less than 1/2 wave dipole, than it would be for the same power fed to the
center of a 1/2 wave dipole. No one to my knowledge considers a 1/4 wave
dipole "resonant".

I'm not aware of any standard reference that does not define as resonant a
1/2 wave dipole having zero reactance at a center feed. The classic Terman's
shows overall circuit current at "resonance" as being entirely resistive.
{p.46, Electronic and Radio Engineering 4th Edition, F E Terman, McGraw
Hill, 1955}

For the dipole this would be the point that the undissipated power from
prior excitation returns exactly in phase with incident excitation. This is
your grid dip meter case of maximum accepted power, hence maximum dip, and
also where a center feed displays zero reactive current.

Perhaps a better definition of a wire resonant at a given frequency would be
*if there exists* a point on the wire where a feed so placed would not
exhibit any reactance. This takes in other cases than center fed 1/2 wave
dipoles.

73, Guy.

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Stephen W. Kercel

Re: : Resonance

Guy:

Interesting points.

73,

Steve
AA4AK

Guy Olinger, K2AV wrote:

>> I was always of the impression that the definition of resonance of a
>> half wave radiator is the condition in which the current at the center
>> is a maximum and the current at the ends is at zero.
>
> This would also be true of a 1/4 wave dipole fed in the center, or any
> dipole less than a 1/2 wave. The current would be higher in the
> center of a less than 1/2 wave dipole, than it would be for the same
> power fed to the center of a 1/2 wave dipole. No one to my knowledge
> considers a 1/4 wave dipole "resonant".
>
> I'm not aware of any standard reference that does not define as
> resonant a 1/2 wave dipole having zero reactance at a center feed. The
> classic Terman's shows overall circuit current at "resonance" as being
> entirely resistive. {p.46, Electronic and Radio Engineering 4th
> Edition, F E Terman, McGraw Hill, 1955}
>
> For the dipole this would be the point that the undissipated power
> from prior excitation returns exactly in phase with incident
> excitation. This is your grid dip meter case of maximum accepted
> power, hence maximum dip, and also where a center feed displays zero
> reactive current.
>
> Perhaps a better definition of a wire resonant at a given frequency
> would be *if there exists* a point on the wire where a feed so placed
> would not exhibit any reactance. This takes in other cases than center
> fed 1/2 wave dipoles.
>
> 73, Guy.
>
>

Don Wilhelm-4

Re: : Resonance

In reply to this post by Guy, K2AV

Guy and Steve,

AFAIK, the condition of zero reactance *defines* resonance whether that
be a dipole or a tuned circuit using lumped components.

On a dipole of any length (whether resonant or not), the current must be
zero at the ends (there is no place for it to go - it is an open
circuit). If that dipole is less than 1/2 wavelength long, the current
will be a maximum at the center - lets restrict the discussion to 1/2
wave or less for simplicity. The thing which changes as the feedpoint
is moved along the antenna is the feedpoint impedance - it can be fed at
any point - the impedance will be lowest in the center and highest at
the ends.
If you plot both the voltage and the current along an antenna, you can
get an *idea* about the feedpoint impedance at any point by dividing the
voltage by the current (there are other factors like the radiation
resistance, so that is not exact) - in the center, the voltage is low
but the current is high, so the impedance (V/I) is low and it becomes
larger as you move toward either end of the dipole.
If the reactance is zero at any feedpoint, it will be zero no matter how
the feedpoint is moved - that fact only occurs if the wire is resonant -
if there is any reactance, the values of resistance and reactance will
move about the constant SWR circle on a Smith chart.

Steve, your analogy of a guitar string is OK, but what you are stating
only applies at resonance - and is thus comparable only to a half wave
dipole. The fact is that a wire of any length can be made to take power
at any frequency by feeding it with the conjugate of its feed impedance
- and a transmission line section can easily provide that at certain
lengths and characteristic impedances (or a lumped element network like
a tuner). I cannot think of an easy analogy to that for a vibrating
string feedpoint. Maybe the MEs in this group can provide that
mechanical analogy.

73,
Don W3FPR

Guy Olinger, K2AV wrote:

>> I was always of the impression that the definition of resonance of a
>> half wave radiator is the condition in which the current at the center
>> is a maximum and the current at the ends is at zero.
>>
>
> This would also be true of a 1/4 wave dipole fed in the center, or any
> dipole less than a 1/2 wave. The current would be higher in the center of a
> less than 1/2 wave dipole, than it would be for the same power fed to the
> center of a 1/2 wave dipole. No one to my knowledge considers a 1/4 wave
> dipole "resonant".
>
> I'm not aware of any standard reference that does not define as resonant a
> 1/2 wave dipole having zero reactance at a center feed. The classic Terman's
> shows overall circuit current at "resonance" as being entirely resistive.
> {p.46, Electronic and Radio Engineering 4th Edition, F E Terman, McGraw
> Hill, 1955}
>
> For the dipole this would be the point that the undissipated power from
> prior excitation returns exactly in phase with incident excitation. This is
> your grid dip meter case of maximum accepted power, hence maximum dip, and
> also where a center feed displays zero reactive current.
>
> Perhaps a better definition of a wire resonant at a given frequency would be
> *if there exists* a point on the wire where a feed so placed would not
> exhibit any reactance. This takes in other cases than center fed 1/2 wave
> dipoles.
>
> 73, Guy.
>
>

Stephen W. Kercel

Re: : Resonance

Don:

Your points are well taken.

TNX & 73,

Steve
AA4AK

Don Wilhelm wrote:

> Guy and Steve,
>
> AFAIK, the condition of zero reactance *defines* resonance whether
> that be a dipole or a tuned circuit using lumped components.
>
> On a dipole of any length (whether resonant or not), the current must
> be zero at the ends (there is no place for it to go - it is an open
> circuit). If that dipole is less than 1/2 wavelength long, the
> current will be a maximum at the center - lets restrict the discussion
> to 1/2 wave or less for simplicity. The thing which changes as the
> feedpoint is moved along the antenna is the feedpoint impedance - it
> can be fed at any point - the impedance will be lowest in the center
> and highest at the ends.
> If you plot both the voltage and the current along an antenna, you can
> get an *idea* about the feedpoint impedance at any point by dividing
> the voltage by the current (there are other factors like the radiation
> resistance, so that is not exact) - in the center, the voltage is low
> but the current is high, so the impedance (V/I) is low and it becomes
> larger as you move toward either end of the dipole.
> If the reactance is zero at any feedpoint, it will be zero no matter
> how the feedpoint is moved - that fact only occurs if the wire is
> resonant - if there is any reactance, the values of resistance and
> reactance will move about the constant SWR circle on a Smith chart.
>
> Steve, your analogy of a guitar string is OK, but what you are stating
> only applies at resonance - and is thus comparable only to a half wave
> dipole. The fact is that a wire of any length can be made to take
> power at any frequency by feeding it with the conjugate of its feed
> impedance - and a transmission line section can easily provide that at
> certain lengths and characteristic impedances (or a lumped element
> network like a tuner). I cannot think of an easy analogy to that for
> a vibrating string feedpoint. Maybe the MEs in this group can provide
> that mechanical analogy.
>
> 73,
> Don W3FPR
>
>
> Guy Olinger, K2AV wrote:
>>> I was always of the impression that the definition of resonance of a
>>> half wave radiator is the condition in which the current at the center
>>> is a maximum and the current at the ends is at zero.
>>>
>>
>> This would also be true of a 1/4 wave dipole fed in the center, or
>> any dipole less than a 1/2 wave. The current would be higher in the
>> center of a less than 1/2 wave dipole, than it would be for the same
>> power fed to the center of a 1/2 wave dipole. No one to my knowledge
>> considers a 1/4 wave dipole "resonant".
>>
>> I'm not aware of any standard reference that does not define as
>> resonant a 1/2 wave dipole having zero reactance at a center feed.
>> The classic Terman's shows overall circuit current at "resonance" as
>> being entirely resistive. {p.46, Electronic and Radio Engineering 4th
>> Edition, F E Terman, McGraw Hill, 1955}
>>
>> For the dipole this would be the point that the undissipated power
>> from prior excitation returns exactly in phase with incident
>> excitation. This is your grid dip meter case of maximum accepted
>> power, hence maximum dip, and also where a center feed displays zero
>> reactive current.
>>
>> Perhaps a better definition of a wire resonant at a given frequency
>> would be *if there exists* a point on the wire where a feed so placed
>> would not exhibit any reactance. This takes in other cases than
>> center fed 1/2 wave dipoles.
>>
>> 73, Guy.
>

David Woolley (E.L)

Re: : Resonance

In reply to this post by Don Wilhelm-4

Don Wilhelm wrote:

> dipole. The fact is that a wire of any length can be made to take power
> at any frequency by feeding it with the conjugate of its feed impedance

Although I agree with most of the preceding, the above statement hints
at a common misunderstanding. In general, you do not want to feed an
antenna from a source impedance that is the the complex conjugate of its
impedance. Doing so guarantees that you cannot exceed 50% efficiency.

Whilst such matching will give you the highest power for any given open
circuit voltage, it will not give you the highest output power for a
given input power.

When one says is a transmitter is matched to 50 ohms, one actually means
that its output network presents the optimum load to the finals when
feeding a 50 ohm load.

--
David Woolley
"The Elecraft list is a forum for the discussion of topics related to
Elecraft products and more general topics related ham radio"
List Guidelines <http://www.elecraft.com/elecraft_list_guidelines.htm>
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Don Wilhelm-4

Re: : Resonance

David,

You are correct, this *is* a common point of misunderstanding. You
referred to the driving "source impedance" in your statement, and I
believe that is the source of the misunderstanding.

What you say is true about the driving generator. Amplifiers are
designed for some efficiency into some load - in our case, usually 50
ohms, and whatever the amplifier designer must do to make that happen is
up to the designer. However, to obtain maximum power transfer *into*
the antenna, the antenna and feedline must have an impedance that
matches the output impedance of the amplifier - we all strive to match
our antennas to a 50 ohm load (or an SWR of 1:1 based on a 50 ohm system).

We typically adjust the parameters of an antenna system to have an input
impedance of 50 +j0 ohms in order to obtain the maximum power into the
antenna (because that the the load the amplifier needs to see for proper
in-spec operation). To accomplish that, we might have a tuner of some
nature between the transmitter and the input of the feedline. If the
antenna feedline has an impedance of (for example) 120 +j30. then the
matching network will have an input impedance of 50 +j0 and an output
impedance of 120 - j30 -- that is as far as it goes. I stand on my
statement (but will not extend it to the internals of amplifier design)
- it is only related to feedline and antenna matching.

Consider that if one builds a matched antenna system at some frequency
as I described above and terminates it at the transmitter end with a 50
ohm pure resistive dummy load. Now split the feedline at any point and
measure the impedance of both the open ends created by the split - you
will measure an R+jX impedance in one direction and an R-jX impedance in
the other direction - that is a conjugate match. If it is something
other than that, there will be significant loss in the antenna system.

Yes, if we put a theoretical (equivalent circuit) generator on that 50
ohm feedpoint, that theoretical generator must have an internal
resistance of 50 +j0 ohms to achieve maximum power transfer between that
theoretical generator and the antenna feed - and the power dissipated by
that theoretical resistance on the theoretical generator will be the
same as that delivered to the load. However, real amplifiers are not
usually built the same way we create equivalent circuits. We do know
that amplifier efficiencies can be much higher than 50%. When we
replace the real amplifier with an equivalent circuit, that will contain
a driving generator of zero loss and a series resistance of 50 ohms, but
we never analyze the internals of an equivalent circuit, it is only the
*representation* of the driving source which makes the system being
analyzed behave as though the real generator (amplifier) were driving it.

73,
Don W3FPR

>
> Although I agree with most of the preceding, the above statement hints
> at a common misunderstanding. In general, you do not want to feed an
> antenna from a source impedance that is the the complex conjugate of its
> impedance. Doing so guarantees that you cannot exceed 50% efficiency.
>
> Whilst such matching will give you the highest power for any given open
> circuit voltage, it will not give you the highest output power for a
> given input power.
>
> When one says is a transmitter is matched to 50 ohms, one actually means
> that its output network presents the optimum load to the finals when
> feeding a 50 ohm load.
>
>