Question regarding measuring power output.


Bernard A. Poskus
 

I keyed my QCX into a dummy load and measure peak-to-peak voltage with an oscilloscope across the dummy load.  It was 19.2 volts, peak-to-peak.  I think this means my QCX is putting out a little under 3.7 watts into the dummy load.  Would you agree?


Jerry Wolczanski
 

More like .92 Watts.
Can you show your math?

Jerry
KI4IO


Evan Hand
 

Here is a link to a site with the formulas to convert from pea to peak to RMS
https://www.google.com/url?sa=i&url=https%3A%2F%2Fcircuitdigest.com%2Fcalculators%2Frms-voltage-calculator&psig=AOvVaw0uIsM6_fPSgim15j2357Uc&ust=1585857950592000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCMDUs7CDyOgCFQAAAAAdAAAAABAD

For your situation it would be
Vrms = 0.35355 *Vpp.  or
19.2 volts * 0.35355 = 6.79 Vrms

Power is E^2/R or 6.79^2/50 = 46.1/50 = 0.922 watts or 922 mWatts

Hope this helps.
73
Evan
AC9TU


Richard G4TGJ
 

19.2V peak would give you 3.7W. But if it was peak-to-peak then you need to divide the power by 4 giving 0.92W.
--
73
Richard
G4TGJ


James Daldry W4JED
 

Hi, Bernard

Lets see - 19.2 volts divided by 2 pi (convert to RMS) equals 6.79 volts RMS. Power is E squared over R, so it's 46.1041 divided by 50, or a skosh less than a watt. Of course that assumes that the load is 50 ohms at RF, that the scope is good for the frequency and properly calibrated ( in other words, not a $20 ebay special).

Somebody (probably Allison) posted a little chart of RMS, P to P, and watts in 50 ohms on the board a while back. It seems that 5 watts is 44.7 volts P to P in 50 ohms.

73

JIm W4JED

On 4/1/20 4:03 PM, Bernard A. Poskus wrote:
I keyed my QCX into a dummy load and measure peak-to-peak voltage with an oscilloscope across the dummy load.  It was 19.2 volts, peak-to-peak.  I think this means my QCX is putting out a little under 3.7 watts into the dummy load.  Would you agree?


Bernard A. Poskus
 

Well, my Oak Hills Research QRP wattmeter shows 2.5 watts out, though that's when it's transmitting into an antenna (that's a good match).  I see that all of you come up with just under a watt. 

I did my math by squaring 19.2, and then dividing the result by 100, to arrive at 3.68 watts.

I have measured the dummy load and it comes out right around 50 ohms.


Jim Allyn - N7JA
 

On Wed, Apr 1, 2020 at 01:24 PM, James Daldry W4JED wrote:
Lets see - 19.2 volts divided by 2 pi (convert to RMS)
     No.  Divide the peak value by 2 to get the peak value, then multiply that by .7071 to get the RMS value.  No idea how you came up with the right answer by using the wrong formula.


Jim Allyn - N7JA
 

On Wed, Apr 1, 2020 at 03:31 PM, Bernard A. Poskus wrote:
I did my math by squaring 19.2, and then dividing the result by 100, to arrive at 3.68 watts.

You measured 19.2 volts peak to peak.  Divide that by 2 to get the peak value of 9.6 volts, then multiply by .7071 to get the RMS voltage of 6.788.  Power = E^2/R = 6.788^2/50 = .92154 watts.


Paul AI4EE
 

No, Bernard is correct. The voltage he is measuring is made with a DC meter.


On 4/1/2020 6:47 PM, Jim Allyn - N7JA wrote:
On Wed, Apr 1, 2020 at 03:31 PM, Bernard A. Poskus wrote:
I did my math by squaring 19.2, and then dividing the result by 100, to arrive at 3.68 watts.

You measured 19.2 volts peak to peak.  Divide that by 2 to get the peak value of 9.6 volts, then multiply by .7071 to get the RMS voltage of 6.788.  Power = E^2/R = 6.788^2/50 = .92154 watts.


Jim Allyn - N7JA
 

On Wed, Apr 1, 2020 at 08:59 PM, Paul AI4EE wrote:
No, Bernard is correct. The voltage he is measuring is made with a DC meter.

Quoting Bernard in his first post:

I keyed my QCX into a dummy load and measure peak-to-peak voltage with an oscilloscope across the dummy load.

DC meters don't measure P-P (AC) voltages.


Paul AI4EE
 

Vrms = Vpp/sqrt(2)

Vrms^2 = Vpp^2/2

P =Vrms^2/R = Vpp^2/(2*R)

For a R = 50 ohm load

P= Vpp^2/100

Same as you get if you use a DC voltmeter


On 4/2/2020 2:02 AM, Jim Allyn - N7JA wrote:
On Wed, Apr 1, 2020 at 08:59 PM, Paul AI4EE wrote:
No, Bernard is correct. The voltage he is measuring is made with a DC meter.

Quoting Bernard in his first post:

I keyed my QCX into a dummy load and measure peak-to-peak voltage with an oscilloscope across the dummy load.

DC meters don't measure P-P (AC) voltages.


Hans Summers
 

Hi Paul

There are errors in this... you are mixing peak voltage, and peak-to-peak voltage. 

Vrms = Vpp/sqrt(2)

No... Vp / sqrt (2). 

The difference is Vpp (Vpeak-peak) and Vp (Vpeak). Vpp = 2 x Vp. 

Vrms^2 = Vpp^2/2

P =Vrms^2/R = Vpp^2/(2*R)

For a R = 50 ohm load

P= Vpp^2/100

No... again, this is all correct for Vp. But not Vpp which is "peak-to-peak" and is twice "peak". 

P = Vp^2 / 100
and
P = Vpp^2 / 400

Same as you get if you use a DC voltmeter

Hmm...

Anyway... FYI, I wrote an article about power measurement here: http://qrp-labs.com/qcx/rfpower.html

73 Hans G0UPL


Jim Allyn - N7JA
 

On Wed, Apr 1, 2020 at 11:39 PM, Paul AI4EE wrote:
Same as you get if you use a DC voltmeter
Sorry, but DC voltmeters still don't measure AC/RF signals.  They measure DC, which is why they are called DC voltmeters.


Andrew Lenton
 

Hi,  I am finding it hard to believe all the trouble people are have with some basic Maths (well sums really), using a scope is a good method, however, you must be aware of the bandwidth of you scope, if a scope has a bandwidth of 100 MHz, remember, most scope specs then related to 100MHz bandwidth are typically 3dB down at this frequency, so your results may seem a bit pessimistic.

 

Why not build a simple Watt meter, the germanium point contact diode has a very low drop voltage 0.15 to 0.2 of a volt

 

 

 


 

Use the table below to read off Watts

 

 

W

V

I

R

Check

Linear Display %

 

0.01

0.707107

0.014142

50

0.01

3.162277628

 

0.02

1

0.02

50

0.02

4.47213591

 

0.03

1.224745

0.024495

50

0.03

5.47722552

 

0.04

1.414214

0.028284

50

0.04

6.324555257

 

0.05

1.581139

0.031623

50

0.05

7.071067741

 

0.06

1.732051

0.034641

50

0.06

7.745966614

 

0.07

1.870829

0.037417

50

0.07

8.366600181

 

0.08

2

0.04

50

0.08

8.94427182

 

0.09

2.12132

0.042426

50

0.09

9.486832885

 

0.1

2.236068

0.044721

50

0.1

9.999999899

 

0.25

3.535534

0.070711

50

0.25

15.81138814

 

0.5

5

0.1

50

0.5

22.36067955

 

0.75

6.123724

0.122474

50

0.75

27.3861276

 

1

7.071068

0.141421

50

1

31.62277628

 

2

10

0.2

50

2

44.7213591

 

3

12.24745

0.244949

50

3

54.7722552

 

4

14.14214

0.282843

50

4

63.24555257

 

5

15.81139

0.316228

50

5

70.71067741

 

6

17.32051

0.34641

50

6

77.45966614

 

7

18.70829

0.374166

50

7

83.66600181

 

8

20

0.4

50

8

89.4427182

 

9

21.2132

0.424264

50

9

94.86832885

 

10

22.36068

0.447214

50

10

99.99999899

 

Also bear in mind the PIV of the diode you are using, as it will break,  with too much power, down as 5 Watts is nearly 16 Volts !

 

73

 

Andrew G8UUG


George Korper
 

Very handy Andrew. I will print out chart.
Thank God April Fools is over!

On Thu, Apr 2, 2020 at 8:45 AM Andrew Lenton <a@...> wrote:

Hi,  I am finding it hard to believe all the trouble people are have with some basic Maths (well sums really), using a scope is a good method, however, you must be aware of the bandwidth of you scope, if a scope has a bandwidth of 100 MHz, remember, most scope specs then related to 100MHz bandwidth are typically 3dB down at this frequency, so your results may seem a bit pessimistic.

 

Why not build a simple Watt meter, the germanium point contact diode has a very low drop voltage 0.15 to 0.2 of a volt

 

 

 


 

Use the table below to read off Watts

 

 

W

V

I

R

Check

Linear Display %

 

0.01

0.707107

0.014142

50

0.01

3.162277628

 

0.02

1

0.02

50

0.02

4.47213591

 

0.03

1.224745

0.024495

50

0.03

5.47722552

 

0.04

1.414214

0.028284

50

0.04

6.324555257

 

0.05

1.581139

0.031623

50

0.05

7.071067741

 

0.06

1.732051

0.034641

50

0.06

7.745966614

 

0.07

1.870829

0.037417

50

0.07

8.366600181

 

0.08

2

0.04

50

0.08

8.94427182

 

0.09

2.12132

0.042426

50

0.09

9.486832885

 

0.1

2.236068

0.044721

50

0.1

9.999999899

 

0.25

3.535534

0.070711

50

0.25

15.81138814

 

0.5

5

0.1

50

0.5

22.36067955

 

0.75

6.123724

0.122474

50

0.75

27.3861276

 

1

7.071068

0.141421

50

1

31.62277628

 

2

10

0.2

50

2

44.7213591

 

3

12.24745

0.244949

50

3

54.7722552

 

4

14.14214

0.282843

50

4

63.24555257

 

5

15.81139

0.316228

50

5

70.71067741

 

6

17.32051

0.34641

50

6

77.45966614

 

7

18.70829

0.374166

50

7

83.66600181

 

8

20

0.4

50

8

89.4427182

 

9

21.2132

0.424264

50

9

94.86832885

 

10

22.36068

0.447214

50

10

99.99999899

 

Also bear in mind the PIV of the diode you are using, as it will break,  with too much power, down as 5 Watts is nearly 16 Volts !

 

73

 

Andrew G8UUG


Paul AI4EE
 

After I wrote this, I got to thinking (something I should have done first). Hans is correct. My apologies to those who think more methodically.

Paul - AI4EE 😢


On 4/2/2020 3:06 AM, Hans Summers wrote:
Hi Paul

There are errors in this... you are mixing peak voltage, and peak-to-peak voltage. 

Vrms = Vpp/sqrt(2)

No... Vp / sqrt (2). 

The difference is Vpp (Vpeak-peak) and Vp (Vpeak). Vpp = 2 x Vp. 

Vrms^2 = Vpp^2/2

P =Vrms^2/R = Vpp^2/(2*R)

For a R = 50 ohm load

P= Vpp^2/100

No... again, this is all correct for Vp. But not Vpp which is "peak-to-peak" and is twice "peak". 

P = Vp^2 / 100
and
P = Vpp^2 / 400

Same as you get if you use a DC voltmeter

Hmm...

Anyway... FYI, I wrote an article about power measurement here: http://qrp-labs.com/qcx/rfpower.html

73 Hans G0UPL


Paul AI4EE
 

Am I correct in saying that since the output of this circuit is rectified ac, the peak voltage can be measured with a DC multimeter?

If so, the peak reading squared divided by 100 does indeed give the correct answer, which is what the QRPGuys tell you.

Paul, AI4EE


On 4/2/2020 8:44 AM, Andrew Lenton wrote:

Hi,  I am finding it hard to believe all the trouble people are have with some basic Maths (well sums really), using a scope is a good method, however, you must be aware of the bandwidth of you scope, if a scope has a bandwidth of 100 MHz, remember, most scope specs then related to 100MHz bandwidth are typically 3dB down at this frequency, so your results may seem a bit pessimistic.

 

Why not build a simple Watt meter, the germanium point contact diode has a very low drop voltage 0.15 to 0.2 of a volt

 

 

 


 

Use the table below to read off Watts

 

 

W

V

I

R

Check

Linear Display %

 

0.01

0.707107

0.014142

50

0.01

3.162277628

 

0.02

1

0.02

50

0.02

4.47213591

 

0.03

1.224745

0.024495

50

0.03

5.47722552

 

0.04

1.414214

0.028284

50

0.04

6.324555257

 

0.05

1.581139

0.031623

50

0.05

7.071067741

 

0.06

1.732051

0.034641

50

0.06

7.745966614

 

0.07

1.870829

0.037417

50

0.07

8.366600181

 

0.08

2

0.04

50

0.08

8.94427182

 

0.09

2.12132

0.042426

50

0.09

9.486832885

 

0.1

2.236068

0.044721

50

0.1

9.999999899

 

0.25

3.535534

0.070711

50

0.25

15.81138814

 

0.5

5

0.1

50

0.5

22.36067955

 

0.75

6.123724

0.122474

50

0.75

27.3861276

 

1

7.071068

0.141421

50

1

31.62277628

 

2

10

0.2

50

2

44.7213591

 

3

12.24745

0.244949

50

3

54.7722552

 

4

14.14214

0.282843

50

4

63.24555257

 

5

15.81139

0.316228

50

5

70.71067741

 

6

17.32051

0.34641

50

6

77.45966614

 

7

18.70829

0.374166

50

7

83.66600181

 

8

20

0.4

50

8

89.4427182

 

9

21.2132

0.424264

50

9

94.86832885

 

10

22.36068

0.447214

50

10

99.99999899

 

Also bear in mind the PIV of the diode you are using, as it will break,  with too much power, down as 5 Watts is nearly 16 Volts !

 

73

 

Andrew G8UUG


Jim Allyn - N7JA
 

On 4/2/20 7:54 AM, Paul AI4EE wrote:
Am I correct in saying that since the output of this circuit is rectified ac, the peak voltage can be measured with a DC multimeter?

Yes, a peak detector combined with a DC voltmeter can be used as an RF power indicator.


Ed Kwik
 

Take a look at section 1.10 RF Power Measurements

http://www.arrl.org/files/file/Product%20Notes/chapter_1.pdf

Ed

AB8DF

 

Sent from Mail for Windows 10

 

From: Jim Allyn - N7JA
Sent: Thursday, April 2, 2020 1:40 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Question regarding measuring power output.

 

On 4/2/20 7:54 AM, Paul AI4EE wrote:

> Am I correct in saying that since the output of this circuit is

> rectified ac, the peak voltage can be measured with a DC multimeter?

 

 

Yes, a peak detector combined with a DC voltmeter can be used as an RF

power indicator.

 

 

 

 


James Daldry W4JED
 

Hi, Jim

I guess I pulled it out of my navel. What I divided the 19.2 by was 2.828, which is 4 times .707. Don't know where the 2 pi came from. Using .7071 requires the first step of dividing P to P by 2. The only times I use peak voltage is when figuring out half-wave rectified AC. Anyway, dividing by 2.828 does it in one step. That's the way I did it when working on Pioneer cube amplifiers and Yamaha M80's.

73

Jim W4JED

On 4/1/20 6:39 PM, Jim Allyn - N7JA wrote:
On Wed, Apr 1, 2020 at 01:24 PM, James Daldry W4JED wrote:
Lets see - 19.2 volts divided by 2 pi (convert to RMS)
     No.  Divide the peak value by 2 to get the peak value, then multiply that by .7071 to get the RMS value.  No idea how you came up with the right answer by using the wrong formula.