Topics

Measuring power

WA4EFS
 

Several posts have referred to the RF power output of QRP rigs including the Hilltopper, the relation to amplifier class, or to final transistor quality, etc.  I am curious as to how power is being measured or assessed in relation to various reports.

 

Thanks and 73,

-Lloyd

 

Charles W. Powell
 

Lloyd,

I don’t know about anyone else, but I am measuring power on mine, key down, using a 50 ohm dummy load, and connecting my high impedance oscilloscope leads across the load.  The scope is a a Siglent 200 mHz scope, so it is measuring far away from the rolloff.  Using the peak-to-peak voltage, I make my calculation by deriving the RMS voltage, then using RMS-squared/resistance.  So with about 13.4 DC volts input, my peak output voltage on the HT-20 was 52 volts.  ((52 * 0.707/2)^2)/50 = 6.8 watts into 50 ohms.  I get approximately the same readings with my MFJ tuner’s power readout and also my QRP-o-meter, although I don’t expect those to be nearly as accurate.

Caveats:  I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.

72,

Chas - NK8O

On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...> wrote:

Several posts have referred to the RF power output of QRP rigs including the Hilltopper, the relation to amplifier class, or to final transistor quality, etc.  I am curious as to how power is being measured or assessed in relation to various reports.
 
Thanks and 73, 
-Lloyd

KB9BVN - <kb9bvn@...>
 

If I measure my HT20 into a dummy load with my WM2 power meter, I see close to 7 watts.  Perfect situation. 

If I measure my HT20 into my attic dipole thru my MFJ antenna tuner, I see close to 5 watts. Imperfect situation. 

My attic dipole is a halfwave 40m dipole...so tuning it up on 20m is not real efficient anyway. 






On Wed, Apr 17, 2019 at 10:03 AM Charles W. Powell via Groups.Io <doctorcwp=yahoo.com@groups.io> wrote:
Lloyd,

I don’t know about anyone else, but I am measuring power on mine, key down, using a 50 ohm dummy load, and connecting my high impedance oscilloscope leads across the load.  The scope is a a Siglent 200 mHz scope, so it is measuring far away from the rolloff.  Using the peak-to-peak voltage, I make my calculation by deriving the RMS voltage, then using RMS-squared/resistance.  So with about 13.4 DC volts input, my peak output voltage on the HT-20 was 52 volts.  ((52 * 0.707/2)^2)/50 = 6.8 watts into 50 ohms.  I get approximately the same readings with my MFJ tuner’s power readout and also my QRP-o-meter, although I don’t expect those to be nearly as accurate.

Caveats:  I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.

72,

Chas - NK8O

On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...> wrote:

Several posts have referred to the RF power output of QRP rigs including the Hilltopper, the relation to amplifier class, or to final transistor quality, etc.  I am curious as to how power is being measured or assessed in relation to various reports.
 
Thanks and 73, 
-Lloyd



--
73 de KB9BVN
Brian Murrey

chuck adams <chuck.adams.k7qo@...>
 

You can save a few steps on the calculator by using  P = V_pk^2/100 for a 50 ohm load.
Always saves my bacon on a calculator.  Formula works only for 50 ohm load.

You can get the above formula from yours below by replacing  0.707/2 by  1/sqrt(2),
thus (0.707/2)^2 = 1/2 and (1/2)/50 = 1/100.  For those that missed that day of
class.  :-)

FYI

chuck, k7qo


On 4/17/19 7:03 AM, Charles W. Powell via Groups.Io wrote:
Lloyd,

I don’t know about anyone else, but I am measuring power on mine, key down, using a 50 ohm dummy load, and connecting my high impedance oscilloscope leads across the load.  The scope is a a Siglent 200 mHz scope, so it is measuring far away from the rolloff.  Using the peak-to-peak voltage, I make my calculation by deriving the RMS voltage, then using RMS-squared/resistance.  So with about 13.4 DC volts input, my peak output voltage on the HT-20 was 52 volts.  ((52 * 0.707/2)^2)/50 = 6.8 watts into 50 ohms.  I get approximately the same readings with my MFJ tuner’s power readout and also my QRP-o-meter, although I don’t expect those to be nearly as accurate.

Caveats:  I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.

72,

Chas - NK8O

On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...> wrote:

Several posts have referred to the RF power output of QRP rigs including the Hilltopper, the relation to amplifier class, or to final transistor quality, etc.  I am curious as to how power is being measured or assessed in relation to various reports.
 
Thanks and 73, 
-Lloyd


WA0ITP
 

My personal favorite is (Vpp/2.8)^2/feedline R   This makes sense to me and is easy to remember.  Same as yours Chas. 
72 WAØITP
I love this radio stuff.
www.wa0itp.com
www.4sqrp.com
On 4/17/2019 9:03 AM, Charles W. Powell via Groups.Io wrote:

Lloyd,

I don’t know about anyone else, but I am measuring power on mine, key down, using a 50 ohm dummy load, and connecting my high impedance oscilloscope leads across the load.  The scope is a a Siglent 200 mHz scope, so it is measuring far away from the rolloff.  Using the peak-to-peak voltage, I make my calculation by deriving the RMS voltage, then using RMS-squared/resistance.  So with about 13.4 DC volts input, my peak output voltage on the HT-20 was 52 volts.  ((52 * 0.707/2)^2)/50 = 6.8 watts into 50 ohms.  I get approximately the same readings with my MFJ tuner’s power readout and also my QRP-o-meter, although I don’t expect those to be nearly as accurate.

Caveats:  I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.

72,

Chas - NK8O

On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...> wrote:

Several posts have referred to the RF power output of QRP rigs including the Hilltopper, the relation to amplifier class, or to final transistor quality, etc.  I am curious as to how power is being measured or assessed in relation to various reports.
 
Thanks and 73, 
-Lloyd


WA4EFS
 

Thanks for the several informative replies re. RF power measurement.  With a fully charged 12 volt battery as power source and a 50 ohm resistive load my scope indicates 40 volts P-P from the Hilltopper 40, equating to 4 watts, which seems about right.  13.8 volts might produce an additional watt or two out.

 

73, -Lloyd

 

From: HilltopperKit@4SQRP.groups.io [mailto:HilltopperKit@4SQRP.groups.io] On Behalf Of WA0ITP
Sent: Wednesday, April 17, 2019 2:03 PM
To: HilltopperKit@4SQRP.groups.io
Subject: Re: [HilltopperKit] Measuring power

 

My personal favorite is (Vpp/2.8)^2/feedline R   This makes sense to me and is easy to remember.  Same as yours Chas. 

72 WAØITP
I love this radio stuff.
www.wa0itp.com
www.4sqrp.com

On 4/17/2019 9:03 AM, Charles W. Powell via Groups.Io wrote:

Lloyd,

 

I don’t know about anyone else, but I am measuring power on mine, key down, using a 50 ohm dummy load, and connecting my high impedance oscilloscope leads across the load.  The scope is a a Siglent 200 mHz scope, so it is measuring far away from the rolloff.  Using the peak-to-peak voltage, I make my calculation by deriving the RMS voltage, then using RMS-squared/resistance.  So with about 13.4 DC volts input, my peak output voltage on the HT-20 was 52 volts.  ((52 * 0.707/2)^2)/50 = 6.8 watts into 50 ohms.  I get approximately the same readings with my MFJ tuner’s power readout and also my QRP-o-meter, although I don’t expect those to be nearly as accurate.

 

Caveats:  I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.

 

72,

 

Chas - NK8O



On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...> wrote:

 

Several posts have referred to the RF power output of QRP rigs including the Hilltopper, the relation to amplifier class, or to final transistor quality, etc.  I am curious as to how power is being measured or assessed in relation to various reports.

 

Thanks and 73, 

-Lloyd