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Power output tests at 24v PA
Chris [N7CCX]
Howdy all, I did last night some key down test (keyed the radio with a straight key) into a dummy load and did some power measurements..
I am seeing a bit higher than expected power outputs on 12V and then also 24! I took a single measurement on areas on the bands and then keyed down for a couple seconds and took the average VMM reading, I then took two formulas. One was QRPGuys Dummy load formula (which my dummy load is based on  Except the diode is swapped out for a BAV21 instead of a 72v diode). Other is K4EAA and even Elecrafts Calculations.. My results: The graph is with the QRPGuys calculations which the formula is shown above.. My question is are these even close or accurate? The forumlas used above base on the exact builds they did. K4EAA uses 20 resistors to achieve a dummy load, meanwhile QRPGuys uses 4 resistors then they originally have a smaller voltage diode... Does someone have a better formula I should use? Signal is dumping into 4 3W 200OHM Resistors, cooled by a fan (cause they def are not rated for 20+ W :) then they feed into a BAV21 Diode with a .01UF capacitor on there to help smooth it out.. Either way the IRF510s didnt seem too upset from the extra power, but I am running a fan on one of the cases Sunil sells and its held up great (minus some fan noise i need to fix). I also need to run a probe on the 510s, since they are still on default heatsinks :) Any help on confirming a better formula or even if I am accurate with my measurements helps! Thanks


Mike Kilpatrick <mike.kilpatrick@...>
I could be way wrong but I measure DC volts, add 0.4V then divided by 1.414.
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Multiply the result by itself (square it) and divide that result by the 50 ohms load. Result is power in watts.
On 10 Nov 2019, at 04:56, Chris [N7CCX] <chriscoel@...> wrote:


Jerry Gaffke
That's correct, assuming a forward voltage of 0.4 Volts across the diode.
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According to fig 7 of the datasheet, that happens at a current of 20 uA when the temp is 25 C. https://media.digikey.com/pdf/Data%20Sheets/Fairchild%20PDFs/BAV21.pdf Dividing the peak voltage (what you measure with a diode detector) by 1.414 converts it to an RMS voltage, assuming we are dealing with a sine wave. We must use that RMS voltage when calculating power. Jerry
On Sat, Nov 9, 2019 at 12:42 PM, Mike Kilpatrick wrote:
I could be way wrong but I measure DC volts, add 0.4V then divided by 1.414.


Chris [N7CCX]
So my pemdas order of operations is wrong!


Miguel Angelo Bartie
Hi Chris
I usually measure RF power with a diode and metal film resistors.
The diode barrier voltage is important for low power (below 2W).
The graph you get shows problems at 1.6, 7 and 10 MHz, in my view is that you are having problems with the low pass filter that has a lower than the expected frequency cut, thus reducing the power in these bands.
If possible change the value of the inductors downwards (by taking a loop, each inductor, for example). After this measure the power again
73 from py2ohh miguel


Hi Chris,
In your first column of power calculations, the formula is correct but you must have made a mistake in the spreadsheet calculation. Either that or you didn't take into account the the probe gives Vpeak not Vrms so it must be converted. This is shown correctly in the second column formula: your reading squared divided by 100. In converting Vpeak to Vrms you divide by the square root of 2. However, since this is subsequently being squared, you can just put that in the divisor with the 50 and that makes it 100. Of course both voltage measurements should have the diode voltage drop correction added back in before and any calculations are done. Be aware that dummy loads/detectors can be built in different configurations so the formulas may vary (e.g. Elecraft's DL1 where the probe is measuring across only half the load). 73, Mark PS: What is a VMM?


ajparent1/KB1GMX <kb1gmx@...>
A note on 24V operating...
If you do that there will only be a bit more power is the result. Since the gain of the finals does not increase with greater applied voltage you still need more drive to take advantage of it. It only enables it to produce more power. That goes back the the initial problem, not enough drive on the higher bands from the decaying gain from the 3904s. As to dummy loads and detectors, Diodes are terrible load and when connected to dummy load they change the effective impedance presented back to the radio. They also have a hard to predict offset (threshold) that at best is approximated and untested. Above a few watts its not a big deal unless some one can show a significant difference for 15V vs 15.5V (peak) across a 50 ohm load. Then we get into peak vs pp and RMS power. Note most diode detectors are usually PEAK so the actual is .707(give or take considering diode offset) times the peak voltage (correct only for sine wave). Also detected waveform makes a difference and assuming a sine wave is not always valid. Its those details that impact read power and why many people get sometimes wildly different numbers. the simplest test is pit a fresh 1.5V battery across hte dummy load. That should cause 30mA to flow and the diode detector if perfect (and DC connected) will read 1.56V (fresh alkaline cell). Likely it will be a diode drop lower and that depends on the diode and meter load presented to the diode. Typical for 1n34 is about 1.45V, 1n5711 Schottky maybe 1.35V, and 1n4148 silicon about .96. FYI the resistor in the load if 50 ohms is dissipating .045W (45 milliwatts) DC. If it was 1.56V AC PP its a lot less (about 6mW). Allison


Chris [N7CCX]
@mIguel, it is not designed to do 160, this was just a test.
7/10 ill check.. @mark variable multi meter

