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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 pre-amp 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 P-P / 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


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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,

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 pre-amp 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 P-P / 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


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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,

Thank you for the data!

How were you measuring and calculating the peak power?

73,


Mark
_._,_._,

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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 p-p:
10 volts p-p /2  =  5 Vpeak,  so  5 x 5 = 25, then  25 / 50 = 0.5 watts
See:  https://www.allaboutcircuits.com/tools/ohms-law-calculator/

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,

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

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