EMRFD Message Archive 10885

Message Date From Subject
10885 2015-03-17 20:04:46 farhanbox@gmail.c... Re: Power calculations

there was an article by wes about db, dbm and power. it is no longer public. you might want to read that,  but ...

the relationship between the peak and rms value depends upon the shape of the signal (as seen on a scope) or the spectral purity (as seen on a spectrum analyzer). thr 1:1.414 ratio between the rms and peak is valid only for a pure sinewave. for a square wave it will be 1:1. for a spikey waveform can be much higher. 

i hope this is confusing enough.

- f

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10886 2015-03-17 23:29:33 kerrypwr Re: Power calculations
And different types of power detector behave differently when the RF being measured is not a "pure" sine wave.

Power meters that measure heating in some way or another generally just add-up the powers in the fundamental and any harmonics but the diode sensors (ie voltmeters) that we mostly use in our amateur work behave differently.

The diode charges its capacitor to the total voltage of the fundamental and the harmonics but, because the phase relationship between the fundamental and the harmonics is not known, we don't know to what degree the harmonic(s) add to or subtract from the fundamental and each other.

This means that the voltage on the capacitor might be anywhere within quite a wide range so there is considerable uncertainty in the voltage read on the meter.

The manual for my HP410C voltmeter with the lovely HP 11036A RF probe contains these examples of RF/AC voltage measurement uncertainties;


Incidentally, the final example seems "out-of-kilter" with the others; perhaps it's a misprint or perhaps I'm missing something.  I've never bothered to do the fairly simple maths relating to this.


As if this were not enough to trouble us, remember that we square voltage to calculate power so these uncertainties are squared!

A close-to-true power measurement is difficult for professionals with high-class instruments; it's very much more so for us amateurs.

Kerry VK2TIL.
10887 2015-03-18 00:20:16 Will Re: Power calculations
Hi Kerry,

Slightly off topic but would you have a manual for the HP410C?

Many years ago I bought one at a junk sale but it has problems with
power supply and I wasn't able to find a full schematic online.


10888 2015-03-18 00:40:16 kerrypwr Re: Power calculations
G'day Will.

I have a couple of pdf manuals plus, probably, other stuff that may be of interest; contact me directly at planningpower which is at iprimus stop com stop au.

Kerry VK2TIL.
10889 2015-03-18 09:48:37 K5ESS Re: Power calculations

Try here for manual.







10891 2015-03-18 10:39:48 Dana Myers Re: Power calculations
10892 2015-03-18 10:57:48 Nick Kennedy Re: Power calculations
"What value of the RF voltage is normally used to indicate power output on a transmitter like a QRP rig. The one that is used to compare apples to apples? Peak or RMS?"

You don't usually see power output of a transmitter described in terms of volts.  If you do though, you need to know what units are specified.  Assuming a 50 ohm load, power would be:

P = V^2 / 400 if V is peak to peak
P = V^2 / 100 if V is peak
P = V^2 / 50 if V is RMS

A similar question is, if a magnitude of volts (RF or audio) is indicated on a schematic or circuit description and no specifier is given, what is assumed?  I believe the default is RMS. If it's peak or peak-to-peak, the qualifier should be stated.  Is that the general consensus?


Nick, WA5BDU
10893 2015-03-18 11:01:06 Dana Myers Re: Power calculations
10894 2015-03-19 01:17:32 Eamon Egan Re: Power calculations

Well to start with, if it's a CW rig, it should be pretty straightforward.

If it does CW and SSB I would assume its SSB PEP would correspond to its CW power.

If it's just SSB I would probably assume PEP rating.

As far as voltage is concerned, it's equal to sqrt(P) times the impedance (presumably 50 ohms).

I hope I've understood the question and helped a bit to answer it.


10895 2015-03-19 06:16:59 dx11 Re: Power calculations



You probably mean  ” As far as voltage is concerned, it's equal to sqrt(P times the impedance (presumably 50 ohms)).” ?

Cor PA4Q



10898 2015-03-19 19:59:25 Eamon Egan Re: Power calculations

Yes indeed, thanks for setting me (and others) straight.
This is what I get from trying to do algebra in my head :)