Here's an easy conversion:
1 watt DC = 1 watt RF. Add to that 1 mechanical watt = 1 electrical watt = 1 chemical watt.Or in other words a watt's a watt, no matter what.
If your power meter responds differently to RF than it does to DC then you simply can't use a watt at DC to calibrate it for RF power measurements.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Hi,
Is-it possible to calibrate a RF Power Meter using DC current?
I suppose a direct relation can be compute between RF Watt and DC Watt. I have a RF power meter that I need to check and recalibrate. I have no way to do it using RF signal, I have no source, nethier than any reference. So I assume that since the principle is only the measure of a voltage pickup from a 50 ohms load, it can be done using DC. The meter did react to DC, but return value around half of what it suppose to be, correct or not, I am not sure.
What I did; since the load is 50 ohms, I applied 316ma to it using 15.8vdc source. P=R(I*I) so 50 * (316ma*316ma) = 5 watts. The meter indicate different value on different scale, so It must be out of alignment, and I never get 5 watts on any scale.
What I don't exaclty now, is the relation between RF watt and DC watt, since RF signal is AC signal and should probably be view as RMS watt.
Any explanation or correction will be welcome.
Bye Jacques
1.53 million files (or is that flies?) can't be wrong.... can they?
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Watts are _never_ rms. Assuming volts and amps are in phase, _average_ watts are rms volts times rms amps.
Ted
You can only do a DC calibration if the meter is rated down to DC and it is a thermal type. Watts are always RMS regardless of the frequency unless noted otherwise.
Rene
Refers to another thread. 1.53E6 is the approximate number of files (html and pdf mostly) found in a Google search for "watts RMS" (actually it's 1.76E6 this morning, so maybe the term is catching on).
In the case of 'flies', it's an old joke about following the crowd...
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
*IF* the RF wattmeter measures power "directly" by measuring the heat generated, then any waveform (including DC, either polarity) will do for calibration. However, if it uses a diode for (log type or square law type) conversion, then forgeddit; gotza use AC. And if there are any capacitors involved, then ther is a lowest frequency involved.
??? "1.53 million files" ??? Non-sequitor.
Jacques St-Pierre wrote: ...
...
If we assume that the diode drop is negligible and that the capacitor results in peak-rectification, then the voltage across the cap (that the meter is measuring) will be sqrt(2) times the same DC voltage when using AC of a given RMS value.
The meter scale calibration will take this into account and give about half the power indication using DC as you get with AC. I think this is what you stated in your original post. So it looks as if the meter calibration is approximately correct.
kevin
And watts and science part company when marketing of consumer audio products is the name of the game.
Yes a diode is involve, in fact it's a quite simple schematics:
RF power is applied to a set of 100 ohms resistors in paralelle, making it
50 ohms to ground. On one of the 100 ohms resistor, a tap is insert at about 10 ohms from ground. From that tap, a diode couple the energie to a small cap, creating a DC voltage. That DC voltage is then applied trought different resistors (upond range selection) to a DC amp meter of 100ua.The meter label tell me it's usable from 2mhz to 500mhz. The problem is that I don't have any powerfull source to calibrate the unit, only a generator that can output up to 19dbi of signal, with is not enought to do anything usefull on the lowest range of 5 watts of the meter. Maybe I can use a old CB transmitter, measure Peak to Peak output with my Scope and go from there to get a crude alignment.
Pushing DC to it did raise the scale, since it goes trought the diode, but how to compute the relation between DC and RF energie? Is it possible to get a crude alignment from that?
Jacques
you would be better off using a 60 Hz AC signal instead of a DC signal for calibration.
Maybe you have a low voltge transformer you can use for a 60 Hz AC source.
The rectifer responds to the peak but is calibrated to RMS voltage which is average watts.
Mark
I guess that many files CAN be wrong, then.
If you multiply RMS volts by RMS current, you get the AVERAGE wattage. The proof is in the math, not in how many Google hits you get.
Mark
Take Note:
I understand about the lowest frequency, the capacitor is too small to get a good rectification from the AC signal under 2mhz. But if I applied a DC voltage, the cap is irrelevent, so a relation between the value indicate on the meter with DC should have a relation with the value indicate from an RF AC Signal if we take in count the fact that the AC signal is rectified from only one diode.
Bye Jacques
The setup you described may give a square-law type of response, and the meter a crowded "scale" at its low end. Without knowing the loading at the diode to ground, one cannot determine the DC relationship.
I tough of that, but using 60hz will result in pulse current going trough the meter. The capacitor is way too small for 60hz, I will have to replace it temporarily.
Bye
Jacques
Or one of the Muppet TV shows where Kermit was working on the schedule for a TV network. The shows included the X-Flies...
-- ? Michael A. Terrell Central Florida
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