Hi from Adrien F4IJA
Woody
Hi Adrien,
First, I will say my uBITX is set for "true" QRP, only 5 watts out. It is a Version 4 board with Axicom relays. I measured at two supply voltages, 12.0v and 13.6v. I have additional circuits (Rx preamp and RF attenuator) which will increase the current from completely "stock" uBITX (about 100 ma?). Power out was measured using my homebrew QRP watt meter and my 200 MHz oscilloscope at 50 ohms (oscilloscope is more accurate). Measured at 14.2 MHz Tone applied to microphone input sufficient to reach maximum output.... 12 Volts:  Rx = 419 ma. Tx = 1.4 amps. Pout: Watt meter = 3 watts. Scope: 30v Peak to peak / 15 v peak. Calculated Pout: 2.25 watts RMS, 9 watts peak. 13.6 Volts:  Rx= 452 ma. Tx = 1.73 amps Pout: Watt meter = 5 watts. Scope: 35v PP / 17.5v peak. Calculated Pout: 3 watts RMS, 12.3 watts peak.  Documentation of my build is available on my web site as a ZIP file: www.albe24.com/kz4akubitx.zip  Hope this data will help! 73, Woody  KZ4AK 


Adrien F4IJA
Awesome, thanks! I'm on my phone right now but I'll see that deeply tonight. 73 Adrien F4IJA
Le mar. 27 août 2019 à 17:43, woody <woody@...> a écrit : Hi Adrien,


Adrien F4IJA
Now I've my uBitx, I'm making some tests with a lab monitored power supply and you can find here my first results upon power drain consumption : http://www.tobeca.fr/f4ija/doku.php?id=ubitx:measurements
So, about 29.5W during WSPR transmit and 1.92W when idle. Interesting fact during the idle phase is that if I push to the maximum the output volume on the external speaker, there is the same curent drawn.  73's Adrien F4IJA https://www.qrz.com/db/F4IJA


Hi Woody,
Thank you for the data! How were you measuring and calculating the peak power? 73, Mark


Woody
Ran the output into a 50 ohm dummy load and measured the voltage with a 200 MHz 'scope.
Woody On 9/23/2019 23:18, Mark  N7EKU wrote: Hi Woody, 


Hi Woody,
What I don't get is your calculation for peak power? I guess I'm not to familiar with the calculation and the theory for that. 73, Mark.


Woody
"What I don't get is your calculation for peak
power? "
Hi Adrien,
The oscilloscope displays a full sine wave with positive and negative peaks about zero  across the 50 ohm dummy load. This is peak  to  peak voltage into the 50 ohm dummy load. One half of the wave amplitude is the peak voltage (for example, the positive peak voltage above zero). I = current in amps, E = voltage, R = resistance (50 ohms), P= power in watts Combining ohms law (I=E/R) and the formula for power (P= I x E) we get P = E squared / R. For example with a measured value of 10 volts pp: 10 volts pp /2 = 5 Vpeak, so 5 x 5 = 25, then 25 / 50 = 0.5 watts See: https://www.allaboutcircuits.com/tools/ohmslawcalculator/ To obtain RMS power ("average") do the same  but first divide the peak voltage by the square root of 2. Woody  KZ4AK


Hi Woody,
You've got double the peak power in your original post, which made me wonder how you came up with it. 73, Mark.


Woody
On 9/27/2019 14:49, Mark  N7EKU wrote:
Hi Woody,Without digging out past posts, don't remember what error I made where.... Any issues with my math with my latest reply to Adrien about how to calculate? Woody 

