Topics

QCX 30 meter power output measurements

Stephen Farthing G0XAR JO92ON97
 

Evening all,

After a exhausting trip to Friedrischaven where I met up with Hans and my VNWA pals I decided to have a relaxing afternoon building Hans’s Dummy load kit and then measure the output power of my 30 meter QCX. 

At 12 volts. :-

The QCX tells me it’s 3.75 Watts 

The dummy load with diode test point tells me it’s 2.49 Watts

The scope tells me it’s 2.2 Watts - this is the most accurate method. I have a 100 MHz scope and followed the method in EMRED. 

The graph on page 129 tells me I should expect about 3 Watts. I may fiddle with the turns on the LPF to see if I can up it a bit but I’m happy enough. If it’s of interest I can duplicate the tests at other input voltages. 

Hope this is of interest,

Regards,

Steve G0XAR

Julian Forsey
 

Sorry to be so ignorant, but what is EMRED?
I have a 'scope and would like to make some comparative measurements.
Julian
G4ETS

Roger Hill
 

Hi Julian.


It's a book,   Experimental methods in RF Design, by Wes Hayward, Rick Campbell, and Bob Larkin.


73

Roger

G3YTN

---
***************************
Roger Hill
***************************


On 2018-06-09 09:13, Julian Forsey wrote:

Sorry to be so ignorant, but what is EMRED?
I have a 'scope and would like to make some comparative measurements.
Julian
G4ETS

zfreak@...
 

Just in case you were wondering, reprints are now available from the ARRL. $49 and includes a CD with copies of a few older (but very good) books.
Get a coupon and it should cost you $40 plus shipping. I just bought mine a few weeks ago.

Ebay prices for the original version went through the roof, so the ARRL wisely reprinted it for us slackers that didn't buy it in 2004. It is an outstanding book and very detailed.
Regards,
Mike M.
KU4QO

On Sat, Jun 9, 2018 at 4:16 AM, Roger Hill <rhill@...> wrote:

Hi Julian.


It's a book,   Experimental methods in RF Design, by Wes Hayward, Rick Campbell, and Bob Larkin.


73

Roger

G3YTN

---
***************************
Roger Hill
***************************


On 2018-06-09 09:13, Julian Forsey wrote:

Sorry to be so ignorant, but what is EMRED?
I have a 'scope and would like to make some comparative measurements.
Julian
G4ETS


Stephen Farthing G0XAR JO92ON97
 

Hi guys,

Apologies for not having explained EMRED in my original post. 

Basically the method is to use an oscilloscope to measure the peak to peak voltage of the output of the QCX across a 50 ohm load. 
Once you have done this you can calculate the output power by substituting this value in the following equation :-

Output power in Watts = (peak to peak voltage * peak to peak voltage) / (load resistance * 8) 

So , if you measure a peak to peak voltage of 30 volts across a 50 ohm load, substituting the measurements in the equation we get :- 

Output power in Watts = (30*30)/(50*8)   = 900/400 = 2.25 

I don’t have a more accurate method of measuring output power in my shack. 

Hope this helps,

Steve 

Alan Richmond
 

Know your oscilloscope. The bandwidth of an oscilloscope is the the frequency at which its response is 3 dB down. That is, it will indicate 70.7% of the true voltage.

Ideally, its response will be Gaussian, and will start to roll off noticeably from, say, half the 3 dB point, but some are not, and the response is very irregular, perhaps peaked near the limit, up in order to achieve a higher specification, so indicating a higher voltage just below the limit.

A good test is to apply a good square-wave whose 10-90% rise-time is less than one third of the period of the three dB frequency. A constant-amplitude sine-wave generator would also show any anomalies, as you run the frequency up. If you characterise the scope this way, it can be used to measure sine-wave amplitudes above the three dB point with a fair degree of accuracy. A 100 MHz scope should be pretty accurate up to 50 Mhz, but a 20 MHz one would be doubtful at anything over 10 MHz.

I worked for Tektronix for over 25 years, initially as a test engineer in a factory (in Guernsey), so have known many oscilloscopes, not just Tek's ones.

Alan Richmond
GU3ONJ
 


E & O E

Joe Cotton
 

Alan
Since you have the credentials, it would be foolish to argue with you.  But isn't a 20 MHz scope good up to that frequency without rolloff?  Isn't that the point of the frequency spec?  I ask because I have a (cheap) 20 MHz scope which is a pleasure to use as I build and design circuits.
thanks
Joe

AA7US
 

Hi Joe,

The simple answer is NO... but it depends on what you mean by "good."

The detailed answer is more complicated as Alan eluded to.  I'd suggest spending some time doing a web search on terms like "understanding oscilloscope bandwidth" or something similar.  Some examples of a good explanations can be found HERE and also in THIS Textronix tech brief.

73,

John
AA7US

Alan
 

Joe

You can only specify a point BELOW the  nominal deflection at lower frequencies so a half power point ( -3dB) is the norm for such measurements. Scopes are designed to give a clean pulse response to a very fast squarewave, that demands that the harmonics have  a phase shift which smoothly increases with frequency. If the phase response is not linear then all kinds of aberrations  are visible on the corners  ( or after) as the   variously delayed harmonics  combine to produce irregular shapes. The phase linear response produces a smooth decline in sensitivity with frequency but the exact shape of the curve and how quickly it falls off at higher frequencies is not simple, there are also nulls at certain frequencies with digital and analogue scopes.

 

So a 20MHz scope will produce a half power indication at 20MHz. However it’s very much worse than that if you use a probe in front of the scope. A very slight mis adjustment of the compensation capacitor in the probe will greatly influence the indicated power. The impedance at 20MHz or higher is very much lower than most users think, it’s K ohms not M ohms while the inductance of the earth lead (6-12”) can cause the impedance to be very low around low VHF frequencies.  An x1 passive probe normally has an extremely poor HF response and a very low tip impedance. Tek do publish impedance curves for many of their probes which reveals the pitfalls of passive probes.

 

An active probe can overcome many of these issues but is very expensive.  A way around all of these issues is to use a simple diode voltmeter with the dummy load. You  can roughly calibrate the meter using  AF frequencies ( that includes 50/60Hz) where a digital multimeter can be used to check the V and A.  As a check you can also pass DC through the load and check the temperature rise of the resistors. This thermal transfer method was very widely used as a method of measuring RF power for a very long time, it still remains a fundamental way of comparing  power.

 

 

73

Alan G8LCO

 

Sent from Mail for Windows 10

 

Alan Richmond
 

Scopes were always specified at the 3 dB point, and as far as I know, still are.

You should be able to find a detailed specification somewhere for your scope, and it should state how the upper frequency is measured. I think they are spec'd this way because they are often used just to look at a waveform, without measuring it, and are still useful for this even though their response is attenuated. Especially true with digital circuitry.

If you have a good transmitter covering all the HF bands, you could tune it up into a 50 Ohm load, and measure the peak-to-peak voltage with your scope on 14,21 and 28 MHz. That should tell you a lot about its response. If the power is the same on all bands, the voltages seen should be the same, unless the 'scope response is decreasing with frequency.

Going even further, you could record the response at intermediate frequencies, to make a simple response curve by interpolating between the test points, and this would greatly increase the accuracy of your measurements.

As I said at the beginning, know your oscilloscope.

And try to use 10X probes as much as possible, as they will load the measured circuit with on tenth of the capacitance, if they are properly adjusted (frequency compensated, using the square wave that is usually provided on the 'scope. Use probes rated at least as high a frequency as the scope. Even more work, characterise the 'scope with both 1X and 10X probes. They could be quite different.

Happy testing,

Alan
GU3ONJ

E & O E

Bob Macklin <macklinbob@...>
 

Measuring voltage or frequency with a scope is only an approximation.

What's of interest is the shape of the waveform.

You need to know the limits of what a scope is good for.

Bob Macklin
K5MYJ
Seattle, Wa.
"Real Radios Glow In The Dark"

----- Original Message -----
From: "Alan Richmond" <@AlanOnATriumph>
To: <QRPLabs@groups.io>
Sent: Sunday, June 10, 2018 10:19 AM
Subject: Re: [QRPLabs] QCX 30 meter power output measurements


Scopes were always specified at the 3 dB point, and as far as I know, still
are.

You should be able to find a detailed specification somewhere for your
scope, and it should state how the upper frequency is measured. I think they
are spec'd this way because they are often used just to look at a waveform,
without measuring it, and are still useful for this even though their
response is attenuated. Especially true with digital circuitry.

If you have a good transmitter covering all the HF bands, you could tune it
up into a 50 Ohm load, and measure the peak-to-peak voltage with your scope
on 14,21 and 28 MHz. That should tell you a lot about its response. If the
power is the same on all bands, the voltages seen should be the same, unless
the 'scope response is decreasing with frequency.

Going even further, you could record the response at intermediate
frequencies, to make a simple response curve by interpolating between the
test points, and this would greatly increase the accuracy of your
measurements.

As I said at the beginning, know your oscilloscope.

And try to use 10X probes as much as possible, as they will load the
measured circuit with on tenth of the capacitance, if they are properly
adjusted (frequency compensated, using the square wave that is usually
provided on the 'scope. Use probes rated at least as high a frequency as the
scope. Even more work, characterise the 'scope with both 1X and 10X probes.
They could be quite different.

Happy testing,

Alan
GU3ONJ

E & O E

Adam Goler
 

Hi Steve,

Apologies, I don't have a copy of EMRED. Why the factor of 8 in the denominator? I figure you're getting some 1/sqrt(2)'s from the conversion of Vpp to Vrms, but in this case wouldn't it be

Output power = (Vpp/sqrt (2))^2 / (load) = (Vpp^2)/ (2* load)

I would like to do a similar measurement

Thanks!
--
73, Adam KM6PHD NAQCC#9838

geoff M0ORE
 

I can remember seeing the denominator as 8 times the load. i.e.400 for a 50 ohm load. I can't remember where I saw it and I can't see any mention in my copy of EMRFD 1st edition. ( Pity about all the errors,takes so long to check the errata file to make sure what you are reading hasn't been corrected)
The RadCom Handbook says it's just the V(p-p) ^2/R but that means  my 20 Watt rig is pushing out 98 Watts which is ridiculous.

ajparent1/KB1GMX
 

The correct math is V*V/R where V is RMS volts.  So if you use read P-P volts then its .707V(squared)/R

The amount of mystery is because everywhere you go its specified differently.

Allison

James Daldry W4JED
 

Hi, Folks

I think the main source of confusion is peak voltage and peak-to-peak voltage. For some reason most formulas I've seen use peak voltage, even though the only time you actually use peak voltage is to compute the output voltage of a half-wave rectifier. That's about 160 volts under load from a 120 volt power line.

One volt RMS into a one ohm resistance is 1 watt. One volt RMS = 2.828 volts peak to peak. 2.828 volts squared divided by 1 ohm is about 8 watts instantaneous peak power. That's where the "magic 8" comes from. Stereo equipment used to be rated either in IPP or RMS. Divide IPP by 8 to get RMS. Or you could multiply the 8 ohm load by 8. The answer would be the same.

Working backwards, if you take the square root of 50 ohms, you find that 1 watt in 50 ohms is about 7 volts RMS. Since it's a square, 14 volts RMS becomes 4 watts. Since my scope isn't calibrated in RMS, that comes to around 39 volts p-p.

Back when I was in the TV shop (here he goes again), I used to have to rebuild the output circuits of Yamaha M-80 receivers. Aptly named. They had regulated + and - 120 volts going to the output circuits. Once repaired they would roast my 8 ohm 200 watt load resistors for a few minutes before tripping the 4 amp breaker on my bench Powerite. When one of these was connected to a shorted speaker, fire would come up out of the ventilation screen, and it would take a lunch bag of parts to make it work again.

73

Jim W4JED

On 5/20/19 1:44 PM, ajparent1/KB1GMX wrote:
The correct math is V*V/R where V is RMS volts.  So if you use read P-P volts then its .707V(squared)/R

The amount of mystery is because everywhere you go its specified differently.

Allison

Adam Goler
 

Allison, Jim, thanks!

I suspect that further above, the calculation with V_peak, rather than V_peaktopeak should have that factor of 8, since Vp = Vpp/2 for a sine wave; therefore we have

P = V_RMS^2 / R = (Vpp/sqrt(2))^2 / R = (Vp/2*sqrt(2))^2 / R = Vp^2 / (8 R)

Curious of EMRED makes this distinction...

In any event, I'm trying not to blow up my oscilloscope, so with a nominal 50 Ohm load, and output power of 5W, I expect a Vpp of ~23V.

--
73, Adam KM6PHD NAQCC#9838

geoff M0ORE
 

I've been looking thro my copy of EMRFD to find the method and formula but can't find it. Can you point me to a chapter and section or a page number please. My EMRFD is first edition, second printing.

Geoff

On 21/05/2019 16:44, Adam Goler wrote:

Allison, Jim, thanks!

I suspect that further above, the calculation with V_peak, rather than V_peaktopeak should have that factor of 8, since Vp = Vpp/2 for a sine wave; therefore we have

P = V_RMS^2 / R = (Vpp/sqrt(2))^2 / R = (Vp/2*sqrt(2))^2 / R = Vp^2 / (8 R)

Curious of EMRED makes this distinction...

In any event, I'm trying not to blow up my oscilloscope, so with a nominal 50 Ohm load, and output power of 5W, I expect a Vpp of ~23V.

--
73, Adam KM6PHD NAQCC#9838

Ronald Taylor
 

Geoff, I have the “Revised First Edition” of EMRFD. Don’t know if that’s the same as yours or not. In mine, that info is in Fig. 7.18 on page 7.9

73 …. Ron

On Tue, May 21, 2019 at 9:20 AM geoff M0ORE via Groups.Io <m0ore=tiscali.co.uk@groups.io> wrote:

I've been looking thro my copy of EMRFD to find the method and formula but can't find it. Can you point me to a chapter and section or a page number please. My EMRFD is first edition, second printing.

Geoff

On 21/05/2019 16:44, Adam Goler wrote:

Allison, Jim, thanks!

I suspect that further above, the calculation with V_peak, rather than V_peaktopeak should have that factor of 8, since Vp = Vpp/2 for a sine wave; therefore we have

P = V_RMS^2 / R = (Vpp/sqrt(2))^2 / R = (Vp/2*sqrt(2))^2 / R = Vp^2 / (8 R)

Curious of EMRED makes this distinction...

In any event, I'm trying not to blow up my oscilloscope, so with a nominal 50 Ohm load, and output power of 5W, I expect a Vpp of ~23V.

--
73, Adam KM6PHD NAQCC#9838

geoff M0ORE
 

Thank you Ron, I have found it.

Looking at the waveforms in the diagrams, it would suggest that is for a single carrier and not a two tone test waveform.

A two tone oscillator is shown in fig 7.26 on page 7.14 but no mention of how to use it for checking linearity.

Geoff

On 21/05/2019 20:28, Ronald Taylor wrote:
Geoff, I have the “Revised First Edition” of EMRFD. Don’t know if that’s the same as yours or not. In mine, that info is in Fig. 7.18 on page 7.9

73 …. Ron

On Tue, May 21, 2019 at 9:20 AM geoff M0ORE via Groups.Io <m0ore=tiscali.co.uk@groups.io> wrote:

I've been looking thro my copy of EMRFD to find the method and formula but can't find it. Can you point me to a chapter and section or a page number please. My EMRFD is first edition, second printing.

Geoff

On 21/05/2019 16:44, Adam Goler wrote:

Allison, Jim, thanks!

I suspect that further above, the calculation with V_peak, rather than V_peaktopeak should have that factor of 8, since Vp = Vpp/2 for a sine wave; therefore we have

P = V_RMS^2 / R = (Vpp/sqrt(2))^2 / R = (Vp/2*sqrt(2))^2 / R = Vp^2 / (8 R)

Curious of EMRED makes this distinction...

In any event, I'm trying not to blow up my oscilloscope, so with a nominal 50 Ohm load, and output power of 5W, I expect a Vpp of ~23V.

--
73, Adam KM6PHD NAQCC#9838

Ronald Taylor
 

Yes, this is just a method of calculating the power represented by the measurement of peak to peak voltage using a scope and a known impedance load and assuming a pure sine wave. 

The two tone test is for checking the intermodulation distortion of a SSB transmitter. You can look for flat topping in a scope display or distortion products on a spectrum analyzer while running the two tones through your mic input (with properly set levels). There is a bit of a discussion of the topic on p. 6.81 as well.

On Tue, May 21, 2019 at 12:54 geoff M0ORE via Groups.Io <m0ore=tiscali.co.uk@groups.io> wrote:

Thank you Ron, I have found it.

Looking at the waveforms in the diagrams, it would suggest that is for a single carrier and not a two tone test waveform.

A two tone oscillator is shown in fig 7.26 on page 7.14 but no mention of how to use it for checking linearity.

Geoff

On 21/05/2019 20:28, Ronald Taylor wrote:
Geoff, I have the “Revised First Edition” of EMRFD. Don’t know if that’s the same as yours or not. In mine, that info is in Fig. 7.18 on page 7.9

73 …. Ron

On Tue, May 21, 2019 at 9:20 AM geoff M0ORE via Groups.Io <m0ore=tiscali.co.uk@groups.io> wrote:

I've been looking thro my copy of EMRFD to find the method and formula but can't find it. Can you point me to a chapter and section or a page number please. My EMRFD is first edition, second printing.

Geoff

On 21/05/2019 16:44, Adam Goler wrote:

Allison, Jim, thanks!

I suspect that further above, the calculation with V_peak, rather than V_peaktopeak should have that factor of 8, since Vp = Vpp/2 for a sine wave; therefore we have

P = V_RMS^2 / R = (Vpp/sqrt(2))^2 / R = (Vp/2*sqrt(2))^2 / R = Vp^2 / (8 R)

Curious of EMRED makes this distinction...

In any event, I'm trying not to blow up my oscilloscope, so with a nominal 50 Ohm load, and output power of 5W, I expect a Vpp of ~23V.

--
73, Adam KM6PHD NAQCC#9838