1. QRPLabs@groups.io

Re: 5W Power Amp Peak Voltage

J68HZ
 

You are mixing Peak Power with RMS power…

 

Peak power = 16 watts as measured by the scope (80 volts peak-to-peak).

 

Average (RMS) power = 16 watts / ( 2 * 2^1/2) = 5.65 watts.

 

Power input is related to power output this way:  Power out = Power input * efficiency (a function of class, components, etc).

 

If the input voltage is 12V (is it 12 or 5v?) and the current is 1.17 amps on average, the INPUT power is 12V* 1.17 A = 14 watts average power.

 

The efficiency is therefore 100% * 5.65 watts average power output / 14 watts average input power = 40.3%

 

 

 

 

 

 

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From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of James Anderson
Sent: Saturday, August 19, 2017 3:50 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] 5W Power Amp Peak Voltage

 

Hmmmm. I'm still a little confused. Sorry.

After reading the second half of this site by Rex Greenwell http://www.eham.net/ehamforum/smf/index.php?topic=13255.0;wap2

If as represented on my scopes display shows an 80 volt peak to peak then according to the above sites equation my output power at the theoretical 50ohm antenna has an output power of 16 watts...

Thats the output power from the 5w power amp kit... I know Hans has explained that greater power can be squeezed from the kit but that assumes higher supply voltages than I'm currently using.

Input power from the U3S is just a mere 62mw output feeding into the 5w power amp kit. Peak to Peak oscilloscope output from the U3S kit at TX is 5 volts. Given the equation from the said site


So P=(5*.354) =1.77 squared =3.1329 divided by 50 = .062658 watts.

How have you got the result of 5.6 watts and I have got it to 16 watts.

The current loading of the 5 watt power amp is 1.17 amps if that helps anymore. 

Kind regards, James.

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