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Explain terms for Elecraft DL1 Power Calculation

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Chip Stratton

Explain terms for Elecraft DL1 Power Calculation

The DL1 50 ohm dummy load is basically composed of two 25 ohm resistors in
series. For power calculations, RF is taken off between the two resistors,
rectified via a diode with a 0.3v drop. The DC is then smoothed with a .01
uf capacitor. before reaching the test point.

The power calculation is then given as P(watts)=((Voltsx1.414) + 0.15)^2 /
50

I'm wanting to understand where the terms in this equation come from? I
presume 0.15 is somehow related to the 0.3 Volt drop across the diode, but
how is 0.15 arrived at and why not 0.3? And where does 1.414, obviously the
sqrt of 2 come from - the divided voltage from the dummy load?

Thanks for any enlightenment!
Chip
AE5KA
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Kevin Stover

Re: Explain terms for Elecraft DL1 Power Calculation

Volts X 1.414 gives you peak volts.
I'm not quite sure why only half the diode voltage drop (.15) is added.
I've only ever seen it all added.
All that squared and divided by the impedance (50Ohms) gives you power.

On 1/22/2012 12:05 PM, Chip Stratton wrote:

> The DL1 50 ohm dummy load is basically composed of two 25 ohm resistors in
> series. For power calculations, RF is taken off between the two resistors,
> rectified via a diode with a 0.3v drop. The DC is then smoothed with a .01
> uf capacitor. before reaching the test point.
>
> The power calculation is then given as P(watts)=((Voltsx1.414) + 0.15)^2 /
> 50
>
> I'm wanting to understand where the terms in this equation come from? I
> presume 0.15 is somehow related to the 0.3 Volt drop across the diode, but
> how is 0.15 arrived at and why not 0.3? And where does 1.414, obviously the
> sqrt of 2 come from - the divided voltage from the dummy load?
>
> Thanks for any enlightenment!
> Chip
> AE5KA
> ______________________________________________________________
> Elecraft mailing list
> Home: http://mailman.qth.net/mailman/listinfo/elecraft
> Help: http://mailman.qth.net/mmfaq.htm
> Post: mailto:[hidden email]
>
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> Please help support this email list: http://www.qsl.net/donate.html
>

--
R. Kevin Stover
AC0H
ARRL
FISTS #11993
SKCC #215
NAQCC #3441

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Chip Stratton

Re: Explain terms for Elecraft DL1 Power Calculation

OK, I get the 1.414 term now. The RF is sampled at the voltage divider, and
the diode/capacitor circuit rests at the peak voltage it sees - which is
half what the transmitter is delivering. To get peak voltage of the RF
waveform at the input to the DL1, we multiply by two. To get the RMS we
then divide by 1.414. But 2/1.414 is 1.414, so we simply multiply the
voltage by 1.414.

Now for the 0.15 term....?

Chip
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Alan Bloom

Re: Explain terms for Elecraft DL1 Power Calculation

In reply to this post by Chip Stratton

I think it can be made clearer by re-casting the equation from:

P(watts)=((Voltsx1.414) + 0.15)^2 / 50

to

P(watts) = [2 x 0.707 x (Volts + 0.106)]^2 / 50

A diode rectifier is a peak detector. So you multiply by 0.707 to
convert from peak to RMS and then multiply by 2 to correct for the 2:1
voltage divider. I'm not sure why only 0.1V is added to compensate for
the diode voltage drop. It's true that the voltage drop is less with a
high load impedance, but 0.1V seems too small.

Alan N1AL

On Sun, 2012-01-22 at 18:05 +0000, Chip Stratton wrote:

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Jack Smith-6

Re: Explain terms for Elecraft DL1 Power Calculation

Alan:

A couple years ago, I measured the DC output for a 1N5711 Schottky diode
at 2, 10 and 21 MHz with RF input from essentially 0 to 1.8V peak
(half-wave rectification). The diode operated into a high Z load (an HP
digital multi-meter) and the RF was from an HP signal generator.

A plot of this is at
http://www.cliftonlaboratories.com/diodes_for_rf_probes.htm

After reading your question about the diode drop today, I ran a linear
regression fit of the average of the 2, 10 and 21 MHz data, over the
range starting a bit above the knee, at 0.3V peak up to the 1.8V peak.
Below this point, the diode is in square law range and a simple fixed
voltage drop assumption fails.

Over this linear range, the Y intercept is -0.11V, with an R^2 > 0.999
and standard error of 0.00275V.

Into a high Z load, therefore, 0.11V is a suitable diode offset. The
actual diode offset, leaving aside unit-to-unit variations, will be a
function of temperature and load impedance amongst other things. Given
the other inaccuracies in the process, 0.1V is a good estimate.

Jack K8ZOA

On 1/22/2012 1:57 PM, Alan Bloom wrote:

> I think it can be made clearer by re-casting the equation from:
>
> P(watts)=((Voltsx1.414) + 0.15)^2 / 50
>
> to
>
> P(watts) = [2 x 0.707 x (Volts + 0.106)]^2 / 50
>
> A diode rectifier is a peak detector. So you multiply by 0.707 to
> convert from peak to RMS and then multiply by 2 to correct for the 2:1
> voltage divider. I'm not sure why only 0.1V is added to compensate for
> the diode voltage drop. It's true that the voltage drop is less with a
> high load impedance, but 0.1V seems too small.
>
> Alan N1AL
>
> On Sun, 2012-01-22 at 18:05 +0000, Chip Stratton wrote:
>> The DL1 50 ohm dummy load is basically composed of two 25 ohm resistors in
>> series. For power calculations, RF is taken off between the two resistors,
>> rectified via a diode with a 0.3v drop. The DC is then smoothed with a .01
>> uf capacitor. before reaching the test point.
>>
>> The power calculation is then given as P(watts)=((Voltsx1.414) + 0.15)^2 /
>> 50
>>
>> I'm wanting to understand where the terms in this equation come from? I
>> presume 0.15 is somehow related to the 0.3 Volt drop across the diode, but
>> how is 0.15 arrived at and why not 0.3? And where does 1.414, obviously the
>> sqrt of 2 come from - the divided voltage from the dummy load?
>>
>> Thanks for any enlightenment!
>> Chip
>> AE5KA
>> ______________________________________________________________
>> Elecraft mailing list
>> Home: http://mailman.qth.net/mailman/listinfo/elecraft
>> Help: http://mailman.qth.net/mmfaq.htm
>> Post: mailto:[hidden email]
>>
>> This list hosted by: http://www.qsl.net
>> Please help support this email list: http://www.qsl.net/donate.html
>>
>
> ______________________________________________________________
> Elecraft mailing list
> Home: http://mailman.qth.net/mailman/listinfo/elecraft
> Help: http://mailman.qth.net/mmfaq.htm
> Post: mailto:[hidden email]
>
> This list hosted by: http://www.qsl.net
> Please help support this email list: http://www.qsl.net/donate.html
>

Don Wilhelm-4

Re: Explain terms for Elecraft DL1 Power Calculation

In reply to this post by Alan Bloom

Alan,

Actually the correct formula is even easier. If you note, you are
squaring the 2 x .707 term, and the overall result of that is 2. So the
formula becomes:

P (watts) = 2(Volts + diode drop)^2/50

or even easier

P (watts) = (Volts + diode drop)^2/25

Substitute whatever you think is correct for the diode drop - see Jack
Smith's recent post for details.

The formula above was supposed to make it into the DL2 manual at one
time, but I see it has not been inserted.

73,
Don W3FPR

On 1/22/2012 1:57 PM, Alan Bloom wrote:

Chip Stratton

Re: Explain terms for Elecraft DL1 Power Calculation

OK, thanks everyone for the education. This came up because I wanted to
build an RF detector to use with a multimeter as an RF field strength meter
to compare different ground/counterpoise schemes for HF on a boat.

I ended up copying the circuit for the MFJ-802 field strength meter, which
puts a load of 1 meg across a 1n270 diode bridge. Using a .3 volt drop
assumption for those diodes gives a very linear correlation of voltage
versus transmitted power in the near field. So the .15 volt assumption in
the dummy load calc was a puzzle to me.

AE5KA
Chip
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Alan Bloom

Re: Explain terms for Elecraft DL1 Power Calculation

In reply to this post by Jack Smith-6

Jack,

Interesting. I didn't realize it was a Schottky diode. But it makes
sense. Using the rule that the current increases by e (2.718) for every
26 mV change in forward voltage: If you assume that the forward voltage
drop is 0.3V at (let's say) 1 mA forward current, then at 1 uA (1V into
a 1M load), the voltage should drop by a factor of 0.026V x ln(1000) =
0.18V. 0.3 - 0.18 = 0.12V, pretty close.

With a silicon diode the answer would be something like 0.6 - 0.18V =
0.42V. That's why Schottky-diode detectors have better linearity.

Alan N1AL

On Sun, 2012-01-22 at 14:21 -0500, Jack Smith wrote:

> Alan:
>
> A couple years ago, I measured the DC output for a 1N5711 Schottky diode
> at 2, 10 and 21 MHz with RF input from essentially 0 to 1.8V peak
> (half-wave rectification). The diode operated into a high Z load (an HP
> digital multi-meter) and the RF was from an HP signal generator.
>
> A plot of this is at
> http://www.cliftonlaboratories.com/diodes_for_rf_probes.htm
>
> After reading your question about the diode drop today, I ran a linear
> regression fit of the average of the 2, 10 and 21 MHz data, over the
> range starting a bit above the knee, at 0.3V peak up to the 1.8V peak.
> Below this point, the diode is in square law range and a simple fixed
> voltage drop assumption fails.
>
> Over this linear range, the Y intercept is -0.11V, with an R^2 > 0.999
> and standard error of 0.00275V.
>
> Into a high Z load, therefore, 0.11V is a suitable diode offset. The
> actual diode offset, leaving aside unit-to-unit variations, will be a
> function of temperature and load impedance amongst other things. Given
> the other inaccuracies in the process, 0.1V is a good estimate.
>
> Jack K8ZOA
>
>
> On 1/22/2012 1:57 PM, Alan Bloom wrote:
> > I think it can be made clearer by re-casting the equation from:
> >
> > P(watts)=((Voltsx1.414) + 0.15)^2 / 50
> >
> > to
> >
> > P(watts) = [2 x 0.707 x (Volts + 0.106)]^2 / 50
> >
> > A diode rectifier is a peak detector. So you multiply by 0.707 to
> > convert from peak to RMS and then multiply by 2 to correct for the 2:1
> > voltage divider. I'm not sure why only 0.1V is added to compensate for
> > the diode voltage drop. It's true that the voltage drop is less with a
> > high load impedance, but 0.1V seems too small.
> >
> > Alan N1AL
> >
> > On Sun, 2012-01-22 at 18:05 +0000, Chip Stratton wrote:
> >> The DL1 50 ohm dummy load is basically composed of two 25 ohm resistors in
> >> series. For power calculations, RF is taken off between the two resistors,
> >> rectified via a diode with a 0.3v drop. The DC is then smoothed with a .01
> >> uf capacitor. before reaching the test point.
> >>
> >> The power calculation is then given as P(watts)=((Voltsx1.414) + 0.15)^2 /
> >> 50
> >>
> >> I'm wanting to understand where the terms in this equation come from? I
> >> presume 0.15 is somehow related to the 0.3 Volt drop across the diode, but
> >> how is 0.15 arrived at and why not 0.3? And where does 1.414, obviously the
> >> sqrt of 2 come from - the divided voltage from the dummy load?
> >>
> >> Thanks for any enlightenment!
> >> Chip
> >> AE5KA
> >> ______________________________________________________________
> >> Elecraft mailing list
> >> Home: http://mailman.qth.net/mailman/listinfo/elecraft
> >> Help: http://mailman.qth.net/mmfaq.htm
> >> Post: mailto:[hidden email]
> >>
> >> This list hosted by: http://www.qsl.net
> >> Please help support this email list: http://www.qsl.net/donate.html
> >>
> >
> > ______________________________________________________________
> > Elecraft mailing list
> > Home: http://mailman.qth.net/mailman/listinfo/elecraft
> > Help: http://mailman.qth.net/mmfaq.htm
> > Post: mailto:[hidden email]
> >
> > This list hosted by: http://www.qsl.net
> > Please help support this email list: http://www.qsl.net/donate.html
> >
> ______________________________________________________________
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>
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>