KPA-1500 Efficiency As A Function of Input Power

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KPA-1500 Efficiency As A Function of Input Power

john@kk9a.com
Interesting data, Charles.  I assumed that running the KPA1500 (or KPA500)
at lower power would keep the fan from running as much. I knew that
efficiency changed with output power but I did not expect that it was this
significant. If your data is correct there is 1000++ watts of heat to
dissipate no matter what power level you use.

John KK9A


From: charles k5ua
Date: Fri Aug 17 01:17:49 EDT 2018

Upon further testing, I have found that the efficiency of the KPA-1500
is varies greatly with driving power.  The following table illustrates
the relationships between exciter power, voltage, current,
power-in(voltage x amps), power-out, efficiency, and dissipated power.
The following test was made with the KPA-1500 into a dummy load and
readings from the KPA-1500 utility software.

(Exciter)(Amps) (Voltage) (Power-In) (Power-Out) (Efficiency Pout/Pin)
(Dissipated Pwr)
   10w     29a     52.7v     1528w       375W        375/1528 = 24.5%
1205w
   15w     37a     52.7v     1950w       620w        620/1950 = 31.7%
1335w
   20w     43a     52.6v     2262w       861w        861/2262 = 38.0%
1413w
   25w     47a     52.6v     2472w      1060w       1060/2472 = 42.8%
1451w
   30w     51a     52.6v     2682w      1227w       1227/2682 = 45.7%
1420w
   35w     53a     52.5v     2782w      1380w       1380/2782 = 49.5%
1412w
   40w     55a     52.5v     2887w      1497w       1497/2887 = 51.9%
1382w
   45w     56a     52.5v     2940w      1600w       1600/2940 = 54.4%
1313w
   50w     57a     52.5v     2992w      1703w       1703/2992 = 56.9%
1315w
   53w     59a     52.5v     3097w      1825w       1825/3097 = 58.9%
1306w

The remarkable thing is that DISSIPATED POWER appears to be relatively
constant from 375w output through 1825w output.  This implies the
KPA-1500 will need to dissipate nearly the same amount of heat at low
power output as at high power output, if I am interpreting the data
correctly. I would like to know from Elecraft if this is normal
behavior.

Charles  K5UA

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KPA-1500 Efficiency As A Function of Input Power

ANDY DURBIN
"There was a similar post about the KPA500 in the last year or so. Elecraft responded that the amp was designed for 500 Watts running less was less efficient and running amp at low power was doing it no favors. "

Here is an example of measured PA dissipation for a KPA500:

https://www.dropbox.com/s/bq0v2740t3iztnj/KPA500%20PA%20DISS%20for%2040%20meters.pdf?dl=0

73,
Andy k3wyc
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Re: KPA-1500 Efficiency As A Function of Input Power

Charlie T, K3ICH
In reply to this post by john@kk9a.com
I really do NOT understand why this data is so surprising.
It all seems to me to be perfectly normal with the amplifier's highest
efficiency occurring at near max output.
Which curiously, I would assume,  is the way the amp was designed.

Think zero output with zero drive = zero efficiency.
Apply some drive, read some output and the efficiency goes up from there.

73, Charlie k3ICH




-----Original Message-----
From: [hidden email] <[hidden email]> On
Behalf Of [hidden email]
Sent: Friday, August 17, 2018 8:57 AM
To: [hidden email]
Cc: [hidden email]
Subject: [Elecraft] KPA-1500 Efficiency As A Function of Input Power

Interesting data, Charles.  I assumed that running the KPA1500 (or KPA500)
at lower power would keep the fan from running as much. I knew that
efficiency changed with output power but I did not expect that it was this
significant. If your data is correct there is 1000++ watts of heat to
dissipate no matter what power level you use.

John KK9A


From: charles k5ua
Date: Fri Aug 17 01:17:49 EDT 2018

Upon further testing, I have found that the efficiency of the KPA-1500 is
varies greatly with driving power.  The following table illustrates the
relationships between exciter power, voltage, current, power-in(voltage x
amps), power-out, efficiency, and dissipated power.
The following test was made with the KPA-1500 into a dummy load and readings
from the KPA-1500 utility software.

(Exciter)(Amps) (Voltage) (Power-In) (Power-Out) (Efficiency Pout/Pin)
(Dissipated Pwr)
   10w     29a     52.7v     1528w       375W        375/1528 = 24.5%
1205w
   15w     37a     52.7v     1950w       620w        620/1950 = 31.7%
1335w
   20w     43a     52.6v     2262w       861w        861/2262 = 38.0%
1413w
   25w     47a     52.6v     2472w      1060w       1060/2472 = 42.8%
1451w
   30w     51a     52.6v     2682w      1227w       1227/2682 = 45.7%
1420w
   35w     53a     52.5v     2782w      1380w       1380/2782 = 49.5%
1412w
   40w     55a     52.5v     2887w      1497w       1497/2887 = 51.9%
1382w
   45w     56a     52.5v     2940w      1600w       1600/2940 = 54.4%
1313w
   50w     57a     52.5v     2992w      1703w       1703/2992 = 56.9%
1315w
   53w     59a     52.5v     3097w      1825w       1825/3097 = 58.9%
1306w

The remarkable thing is that DISSIPATED POWER appears to be relatively
constant from 375w output through 1825w output.  This implies the
KPA-1500 will need to dissipate nearly the same amount of heat at low power
output as at high power output, if I am interpreting the data correctly. I
would like to know from Elecraft if this is normal behavior.

Charles  K5UA

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Re: KPA-1500 Efficiency As A Function of Input Power

Bill Frantz
In reply to this post by ANDY DURBIN
For those curious about the reasons for high power draw at low
power out, it might be useful to review the handbook discussion
of amplifiers, and the characteristics of class A, B, and C designs.

73 Bill AE6JV

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408-356-8506       | Fourth Amendment that it operates only by
www.pwpconsult.com | accident.  -  William Hugh Murray

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Re: KPA-1500 Efficiency As A Function of Input Power

Bob McGraw - K4TAX
In reply to this post by ANDY DURBIN
I've said this and measured this for years with tube type amps, where as
I hear hams say they are running reduced power to "save the tubes".  
That's HOGWASH.  Determine the efficiency at reduced power vs. rated
power.    The excessive heat at reduced power has to go somewhere.

I view running an amp, tube or solid state types, at reduced power is
much like driving with your right foot on the accelerator and the left
foot on the brake  {US style} and wondering why your gas mileage suffers
and your brakes wear out.

Tune it up, load it up, and run that sucker at full power.  After all,
........that's why you have an amp.

73

Bob, K4TAX



On 8/17/2018 8:11 AM, ANDY DURBIN wrote:

> "There was a similar post about the KPA500 in the last year or so. Elecraft responded that the amp was designed for 500 Watts running less was less efficient and running amp at low power was doing it no favors."
>
> Here is an example of measured PA dissipation for a KPA500:
>
> https://www.dropbox.com/s/bq0v2740t3iztnj/KPA500%20PA%20DISS%20for%2040%20meters.pdf?dl=0
>
> 73,
> Andy k3wyc
> ______________________________________________________________
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> Help: http://mailman.qth.net/mmfaq.htm
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> Message delivered to [hidden email]
>


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Re: KPA-1500 Efficiency As A Function of Input Power

Bob McGraw - K4TAX
In reply to this post by Charlie T, K3ICH
Yes yes and yes.    Amps are designed with the fixed output matching
network chosen for a value of impedance and energy transfer to be at
"rated power".  Running any amp at reduced power then does not confirm
to the design of the output matching network and the efficiency
decreases.   You aren't doing any favor to yourself, the amp, or others
to run the amp otherwise.

73

Bob, K4TAX


On 8/17/2018 8:24 AM, Charlie T wrote:

> I really do NOT understand why this data is so surprising.
> It all seems to me to be perfectly normal with the amplifier's highest
> efficiency occurring at near max output.
> Which curiously, I would assume,  is the way the amp was designed.
>
> Think zero output with zero drive = zero efficiency.
> Apply some drive, read some output and the efficiency goes up from there.
>
> 73, Charlie k3ICH
>
>
>
>
> -----Original Message-----
> From: [hidden email] <[hidden email]> On
> Behalf Of [hidden email]
> Sent: Friday, August 17, 2018 8:57 AM
> To: [hidden email]
> Cc: [hidden email]
> Subject: [Elecraft] KPA-1500 Efficiency As A Function of Input Power
>
> Interesting data, Charles.  I assumed that running the KPA1500 (or KPA500)
> at lower power would keep the fan from running as much. I knew that
> efficiency changed with output power but I did not expect that it was this
> significant. If your data is correct there is 1000++ watts of heat to
> dissipate no matter what power level you use.
>
> John KK9A
>
>
> From: charles k5ua
> Date: Fri Aug 17 01:17:49 EDT 2018
>
> Upon further testing, I have found that the efficiency of the KPA-1500 is
> varies greatly with driving power.  The following table illustrates the
> relationships between exciter power, voltage, current, power-in(voltage x
> amps), power-out, efficiency, and dissipated power.
> The following test was made with the KPA-1500 into a dummy load and readings
> from the KPA-1500 utility software.
>
> (Exciter)(Amps) (Voltage) (Power-In) (Power-Out) (Efficiency Pout/Pin)
> (Dissipated Pwr)
>     10w     29a     52.7v     1528w       375W        375/1528 = 24.5%
> 1205w
>     15w     37a     52.7v     1950w       620w        620/1950 = 31.7%
> 1335w
>     20w     43a     52.6v     2262w       861w        861/2262 = 38.0%
> 1413w
>     25w     47a     52.6v     2472w      1060w       1060/2472 = 42.8%
> 1451w
>     30w     51a     52.6v     2682w      1227w       1227/2682 = 45.7%
> 1420w
>     35w     53a     52.5v     2782w      1380w       1380/2782 = 49.5%
> 1412w
>     40w     55a     52.5v     2887w      1497w       1497/2887 = 51.9%
> 1382w
>     45w     56a     52.5v     2940w      1600w       1600/2940 = 54.4%
> 1313w
>     50w     57a     52.5v     2992w      1703w       1703/2992 = 56.9%
> 1315w
>     53w     59a     52.5v     3097w      1825w       1825/3097 = 58.9%
> 1306w
>
> The remarkable thing is that DISSIPATED POWER appears to be relatively
> constant from 375w output through 1825w output.  This implies the
> KPA-1500 will need to dissipate nearly the same amount of heat at low power
> output as at high power output, if I am interpreting the data correctly. I
> would like to know from Elecraft if this is normal behavior.
>
> Charles  K5UA
>
> ______________________________________________________________
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> Home: http://mailman.qth.net/mailman/listinfo/elecraft
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> delivered to [hidden email]
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Re: KPA-1500 Efficiency As A Function of Input Power

k6dgw
In reply to this post by Charlie T, K3ICH
Well, only if you've figured out a way to divide by zero. [:-)

The term "amplifier efficiency" must have changed dramatically since I
worked in broadcast nearly a lifetime ago.  Then, it was the ratio
[expressed as a percentage] of the RF power delivered to the 3 1/8"
hardline divided by the DC power supplied to the amplifier plate
circuit.  Power to the filaments was generally excluded by manufacturers
seeking to have higher efficiency numbers.  One 10 KW FM transmitter I
helped build used four 4-1000A's in parallel and the filament power was
600 watts. Efficiency, including filament power was just over 70%.

To measure the efficiency of a KPA1500 [or 500, or KXPA100, or KPA2] one
would measure the key down RF output and divide it by the key down DC
input power.  If it was water-cooled, I suppose one should include the
input power to the pump(s), but no one ever did.  SS amplifiers like the
KPA(1)500 probably run class AB<mumble> or its equivalent and have
substantial DC input power even at zero RF input.

RF output power vs RF input power defines the amplifer's transfer
function, not its efficiency.

73,

Fred ["Skip"] K6DGW
Sparks NV DM09dn
Washoe County

PS:  I really don't recommend trying to get four parallel 4-1000A's to
work anything close to "well" at 90.7 MHz. Apparently, it seemed like a
good idea at the time. [:-)

On 8/17/2018 6:24 AM, Charlie T wrote:
> Think zero output with zero drive = zero efficiency.
> Apply some drive, read some output and the efficiency goes up from there.
>
> 73, Charlie k3ICH
>

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Re: KPA-1500 Efficiency As A Function of Input Power

Bill Frantz
And I would guess it was running in class C, which is not
linear, so only useful for certain modes (including FM). Class C
can convert most of the input power to RF power.

73 Bill AE6JV

On 8/17/18 at 11:48 AM, [hidden email] (Fred Jensen) wrote:

>One 10 KW FM transmitter I helped build used four 4-1000A's in
>parallel and the filament power was 600 watts. Efficiency,
>including filament power was just over 70%.
-----------------------------------------------------------------------
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(408)356-8506      | contact sport.               | 16345
Englewood Ave
www.pwpconsult.com |  - Ken Widelitz K6LA / VY2TT | Los Gatos,
CA 95032

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Re: KPA-1500 Efficiency As A Function of Input Power

W2xj
Actually a class C amplifier can be made to perform as a linear. The technology is over 60 years old but I don’t think it has ever been tried in amateur service.

Sent from my iPad

> On Aug 17, 2018, at 3:06 PM, Bill Frantz <[hidden email]> wrote:
>
> And I would guess it was running in class C, which is not linear, so only useful for certain modes (including FM). Class C can convert most of the input power to RF power.
>
> 73 Bill AE6JV
>
>> On 8/17/18 at 11:48 AM, [hidden email] (Fred Jensen) wrote:
>>
>> One 10 KW FM transmitter I helped build used four 4-1000A's in parallel and the filament power was 600 watts. Efficiency, including filament power was just over 70%.
> -----------------------------------------------------------------------
> Bill Frantz        | Ham radio contesting is a    | Periwinkle
> (408)356-8506      | contact sport.               | 16345 Englewood Ave
> www.pwpconsult.com |  - Ken Widelitz K6LA / VY2TT | Los Gatos, CA 95032
>
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Re: KPA-1500 Efficiency As A Function of Input Power

W2xj
In reply to this post by k6dgw
Today in broadcast we generally rate transmitters by AC in to RF out. For AM transmitters that figure is in the low 90 percent range and the current FMs run at about 74% AC in to RF out.

Sent from my iPad

> On Aug 17, 2018, at 11:48 AM, Fred Jensen <[hidden email]> wrote:
>
> Well, only if you've figured out a way to divide by zero. [:-)
>
> The term "amplifier efficiency" must have changed dramatically since I worked in broadcast nearly a lifetime ago.  Then, it was the ratio [expressed as a percentage] of the RF power delivered to the 3 1/8" hardline divided by the DC power supplied to the amplifier plate circuit.  Power to the filaments was generally excluded by manufacturers seeking to have higher efficiency numbers.  One 10 KW FM transmitter I helped build used four 4-1000A's in parallel and the filament power was 600 watts. Efficiency, including filament power was just over 70%.
>
> To measure the efficiency of a KPA1500 [or 500, or KXPA100, or KPA2] one would measure the key down RF output and divide it by the key down DC input power.  If it was water-cooled, I suppose one should include the input power to the pump(s), but no one ever did.  SS amplifiers like the KPA(1)500 probably run class AB<mumble> or its equivalent and have substantial DC input power even at zero RF input.
>
> RF output power vs RF input power defines the amplifer's transfer function, not its efficiency.
>
> 73,
>
> Fred ["Skip"] K6DGW
> Sparks NV DM09dn
> Washoe County
>
> PS:  I really don't recommend trying to get four parallel 4-1000A's to work anything close to "well" at 90.7 MHz. Apparently, it seemed like a good idea at the time. [:-)
>
>> On 8/17/2018 6:24 AM, Charlie T wrote:
>> Think zero output with zero drive = zero efficiency.
>> Apply some drive, read some output and the efficiency goes up from there.
>>
>> 73, Charlie k3ICH
>>
>
> ______________________________________________________________
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Re: KPA-1500 Efficiency As A Function of Input Power

k6dgw
In reply to this post by Bill Frantz
Yes, it was Class C ... FM is constant envelope modulation.  The ancient
50's/60's rules of thumb were:
   Class A: 25% - really linear
   Class B: 50% - linear with crossover distortion
   Class C: 75% - exceedingly non-linear
   Class AB1: between A & B, no grid current - mitigates crossover
distortion
   Class AB2:  AB1 but draws grid current - also mitigates crossover dist.

For audio, Classes B and AB<mumble> needed two tubes and operated push-pull.

We were interested in efficiency of the PA because higher efficiency
meant lower dissipation for a fixed output [10 KW].  As it was, at 90.7
MHz, the anodes were already nearly white.  Again ... I do not recommend
using four parallel 4-1000A's anywhere near 90.7 MHz. [:-))

The RCA Ampliphase [aka "Amplifuzz"] series of AM transmitters ran Class
C through the entire RF chain.  They were a bear to align ... you never
wanted to start a PoP after sign-off if you'd been awake all day.  The
Ampliphase history is moderately interesting however, especially if
you're from or in Sacramento CA.

This thread however has been discussing "efficiency" of a KPA1500 in
terms of output power vs drive power which isn't even remotely close to
measuring efficiency.  As amateurs, I can't come up with a single reason
why we'd be concerned about efficiency as long as the amplifier design
is solid and robust.  Our A/C uses far more energy over a summer than my
K3/KPA500 did all year, and I leave it on a lot.  If this is all about
fans and their noise, the thread needs a new subject.

73,

Fred ["Skip"] K6DGW
Sparks NV DM09dn
Washoe County

On 8/17/2018 3:06 PM, Bill Frantz wrote:

> And I would guess it was running in class C, which is not linear, so
> only useful for certain modes (including FM). Class C can convert most
> of the input power to RF power.
>
> 73 Bill AE6JV
>
> On 8/17/18 at 11:48 AM, [hidden email] (Fred Jensen) wrote:
>
>> One 10 KW FM transmitter I helped build used four 4-1000A's in
>> parallel and the filament power was 600 watts. Efficiency, including
>> filament power was just over 70%.
> -----------------------------------------------------------------------
> Bill Frantz        | Ham radio contesting is a    | Periwinkle
> (408)356-8506      | contact sport.               | 16345 Englewood Ave
> www.pwpconsult.com |  - Ken Widelitz K6LA / VY2TT | Los Gatos, CA 95032
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Re: KPA-1500 Efficiency As A Function of Input Power

k6dgw
In reply to this post by W2xj
Yes.  Electricity cost to generate the RF that earns the revenue is a
major part of the broadcast station budget.  Not so much for amateurs. 
Since FM is constant envelope modulation, it makes sense it would have a
lower overall efficiency.  Even Rush Limbaugh has to take a breath
occasionally on AM ... it makes no difference on FM. [:-))

73,

Fred ["Skip"] K6DGW
Sparks NV DM09dn
Washoe County

On 8/17/2018 3:45 PM, W2xj wrote:
> Today in broadcast we generally rate transmitters by AC in to RF out. For AM transmitters that figure is in the low 90 percent range and the current FMs run at about 74% AC in to RF out.
>
>

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Re: KPA-1500 Efficiency As A Function of Input Power

ke9uw
Seems like this discussion would be functional if it were ways to quiet the fans without compromising the efficiency of the cooling.

Chuck Jack
KE9UW

Sent from my iPhone, cjack

> On Aug 17, 2018, at 7:13 PM, Fred Jensen <[hidden email]> wrote:
>
> Yes.  Electricity cost to generate the RF that earns the revenue is a major part of the broadcast station budget.  Not so much for amateurs.  Since FM is constant envelope modulation, it makes sense it would have a lower overall efficiency.  Even Rush Limbaugh has to take a breath occasionally on AM ... it makes no difference on FM. [:-))
>
> 73,
>
> Fred ["Skip"] K6DGW
> Sparks NV DM09dn
> Washoe County
>
>> On 8/17/2018 3:45 PM, W2xj wrote:
>> Today in broadcast we generally rate transmitters by AC in to RF out. For AM transmitters that figure is in the low 90 percent range and the current FMs run at about 74% AC in to RF out.
>>
>>
>
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Chuck, KE9UW
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Re: KPA-1500 Efficiency As A Function of Input Power

Eric Swartz - WA6HHQ
Administrator
In reply to this post by k6dgw
Also on this related thread:

We're way over the posting limit on this one. Let's close this thread for now.

(And, in the future, please voluntarily close long threads well before I happen
to stop by :-)

73,
Eric
/elecraft.com/

On 8/17/2018 5:13 PM, Fred Jensen wrote:
> Yes. Electricity cost to generate the RF that earns the revenue is a major
> part of the broadcast station budget.  Not so much for amateurs.  Since FM is
> constant envelope modulation, it makes sense it would have a lower overall
> efficiency.  Even Rush Limbaugh has to take a breath occasionally on AM ... it
> makes no difference on FM. [:-))
>
> 73,

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