Question regarding measuring power output.
Bernard A. Poskus
I keyed my QCX into a dummy load and measure peaktopeak voltage with an oscilloscope across the dummy load. It was 19.2 volts, peaktopeak. 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%2Frmsvoltagecalculator&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 peaktopeak 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 peaktopeak voltage with an oscilloscope across the dummy load. It was 19.2 volts, peaktopeak. 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:




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. DC meters don't measure PP (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:




Hans Summers
Hi Paul There are errors in this... you are mixing peak voltage, and peaktopeak voltage.
No... Vp / sqrt (2). The difference is Vpp (Vpeakpeak) and Vp (Vpeak). Vpp = 2 x Vp.
No... again, this is all correct for Vp. But not Vpp which is "peaktopeak" and is twice "peak". P = Vp^2 / 100 and P = Vpp^2 / 400
Hmm... Anyway... FYI, I wrote an article about power measurement here: http://qrplabs.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 voltmeterSorry, 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
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:




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:




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:




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 halfwave
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:


