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The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl Measurements >> RF Measurements >> About power measurement https://designers-guide.org/forum/YaBB.pl?num=1212573477 Message started by didac on Jun 4th, 2008, 2:57am |
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Title: About power measurement Post by didac on Jun 4th, 2008, 2:57am Hi, I have a silly(I think) question: in RF one way to measure the power delivered to a load it's using a directional coupler and a "power detector"-like a schottky diode-, as I understand this method provides a measurement of the real(active) power since the input of the "power detector" it's matched to the output of the circuit(in theory). But recently I've seen efforts of integration of power detectors on-chip, they assume a high input impedance to avoid loading and are based in rms voltage so I think that they are measuring "apparent" power, is it right? Thanks for the help, |
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Title: Re: About power measurement Post by pancho_hideboo on Jun 28th, 2008, 12:47am didac wrote on Jun 4th, 2008, 2:57am:
I don't understand a meaning of "apparent" power. Do you mean reactive power ? Assume a situation where output of DUT is connected to didac wrote on Jun 4th, 2008, 2:57am:
I don't understand a mean of "apparent" power. Do you mean reactive power ? Assume a situation where output of DUT is connected to load impedance(=z) by very short line(volatge is uniform over line length). You can calculate true power delivered to load by knowing voltage of line and load impedance(=z). Here voltage is measured by infinite input impedance probe. This is a same method as a definition of dbm() function in Agilent ADS. Syntax : y = dbm(voltage, z) Examples : y = dbm(100) returns 50 y = dbm(8-6*j) returns 30 Here z is omitted and it is assumed as 50Ohm. |
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Title: Re: About power measurement Post by didac on Jun 28th, 2008, 2:06am Hi, First of all thanks for your reply pancho. The situation that I was thinking was a little bit different: I'm considering an unknown load with a frequency variation(time variation it's also possible), transmission line effects at this moment are neglected but probably they can be included. In the end I'm thinking in a situation like electrical line measurements were power is defined as: S=P+jQ Where S is the apparent power (complex power), P the active(real) and Q the reactive. So I followed the path of Power Factor measurements to made my mind clear. Thanks, |
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Title: Re: About power measurement Post by pancho_hideboo on Jun 28th, 2008, 3:23am didac wrote on Jun 28th, 2008, 2:06am:
How do you define and calculate power without knowing current or impedance ? |
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Title: Re: About power measurement Post by didac on Jun 28th, 2008, 3:28am Hi, You pointed it I need to know current and voltage and take into account modulus and phase... |
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Title: Re: About power measurement Post by pancho_hideboo on Jun 28th, 2008, 3:54am didac wrote on Jun 28th, 2008, 3:28am:
What do you mean by the above statement ? Complex power is defined as V*conj(I)=abs(V)**2/conj(Z)=conj(Y)*abs(V)**2 conj(V)*I=conj(Z)*abs(I)**2 |
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Title: Re: About power measurement Post by didac on Jun 28th, 2008, 4:07am Hi, I mean(maybe I didn't express myself clear I apologize) that in the moment you apply conj(V),conj(I) you are implicitly taking into account the phase and the modulus of a waveform. Thanks, |
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Title: Re: About power measurement Post by pancho_hideboo on Jun 28th, 2008, 8:20am didac wrote on Jun 28th, 2008, 4:07am:
What do you mean by "the modulus of a waveform" ? Considering energy conservative system, you need two quantities, "Potential" and "Flow". In electrical system, Voltage is "Potential" and Current is "Flow". So even though you know V(t)=Real[(Vi(t)+j*Vq(t))*exp(j*omega*time)], you can't define power without knowing current or impedance. But as normalized power, you can use 1 Ohm. The following is a normalized power in assuming load=1.0Ohm. RMS[V(t)]**2=0.5*{RMS[Vi(t)]**2 + RMS[Vq(t)]**2} So you don't have to know Vi(t) and Vq(t), that is, phase. You only have to know envelope level. |
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Title: Re: About power measurement Post by didac on Jun 28th, 2008, 8:30am Hi, I have to think a little bit but suppose that you go to two limits of load open circuit and short circuit, in the first case if you consider something like a voltage divider between source and load you will have maximum RMS voltage but power will not be maximum(assuming sensing of the voltage at the load), in the other case you will have minimum voltage maximum current and power will not be maximum either... It all started when I considered rms detectors over unknown loads that I think give wrong results in terms of power because as I understand how they work they give a monotonic increase of the lecture as magnitude of the load increases... Thanks, PD:I will think about your last post more in depth |
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Title: Re: About power measurement Post by pancho_hideboo on Jun 28th, 2008, 8:48am didac wrote on Jun 28th, 2008, 8:30am:
What do you want to do ? Do you want to define Tevenin equivalent source circuit ? |
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Title: Re: About power measurement Post by didac on Jun 28th, 2008, 8:57am Hi, Pfff... at this point I must be a little bit fuzzy since it's for research... and I don't even know at this moment if it will end up beeing something that I will have to thrown to the sewer. The general context was in terms of BIST for RF circuits so the source and load are defined and was more a question that I asked myself while returning home in the train(things like what to measure?where to measure?how to measure?...) and I ended up with this little thing. Thanks, PD:thanks for your time but don't worry, I think that I've managed to turn one or two possible solutions that I'm checking this days, but I'm not very convinced about how practical they are. |
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