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| TomC:
--- Quote from: G0HZU on April 10, 2016, 03:02:01 pm ---Dunno if this helps but I have a HP 8473C detector here that works up to 26GHz. It looks very similar to the one you have and it uses a simple Schottky diode. However, it is a very expensive device because the diode is packaged to work well up to 26GHz. But apart from this I think the technology is similar to the detector you are using. http://cp.literature.agilent.com/litweb/pdf/5952-8299.pdf If you look at the spec for input VSWR and frequency response this detector is very flat indeed up to many GHz. I also have a couple of Agilent ESGD 4433 (4000) sig gens and the typical flatness spec to 4GHz is very impressive. See below for the datasheet spec for a typical generator of this type. If I connect the 8473C to the sig gen at 100MHz and set the level down at -20dBm this gets me close to square law operation. If I do this the 8473C detector output is about 6.6mVDC. If I turn up the level 1dB it goes to about 8.1mV and if I turn it down 1dB it goes down to 5.25mV DC. If I then sweep across the full 4GHz range of the sig gen I see very little change in the detector voltage it stays at 6.6mV and probably only changes by +/- 0.1mV. So the detector flatness is very good indeed. However, if I cool or heat the 8473C then the detector voltage changes quite a bit. i.e. it's only flat and consistent at a fixed temperature when used down at these tiny signal levels. So it's great for checking flatness but not so great at measuring absolute power levels. Your detector should be similar although the performance in terms of flatness and input VSWR etc will be dictated by the diode package and also by how well it is fitted inside the body of the detector in terms of stray inductance etc. If this is done with a diode package that has significant stray inductance then you might only get good flatness performance up to 500MHz or maybe 1GHz. --- End quote --- Thanks for the input and for taking the time to test the 8473C! One big difference between the 8473C and the CRYSTEK detector is that the Agilent datasheet for their series of LBSD detectors is thorough and answers all pertinent questions, in contrast, the CRYSTEk datasheet leaves way too much room open for speculation. http://www.crystek.com/microwave/spec-sheets/rfdetector/CPDETLS-4000.pdf For example, for the 8473C the Agilent datasheet gives an operating temperature of -20°C to +85°C, then at the end of the table there is a note that states: "Above specifications are at 25°C and <= -20 dBm unless otherwise specified.". In contrast, the CRYSTEK datasheet gives an operating temperature of -20°C to 70°C, but it remains mute as to any temperature restrictions for the table of calibration point values given on page 2. Maybe this table, or part of it, is only valid at 25°C or a slightly lower temperature. If that's the case, since the experiment described on my previous post was done at around 28°C, the dismal results at power levels below 0dBm should have been expected! Although the 8473C and the CPDETLS-4000 both use LBSD detectors, looking at the data sheets it seems to me that there aren't very many additional similarities. As you stated, the 8473C has impressive broadband flatness, frequency response and low VSWR, all of which makes it great for checking flatness. As far as measuring absolute power, I think it can probably also do a good job, but the user must have the appropriate test equipment to calibrate it. In contrast, the CPDETLS-4000, as evidenced by the graph and table on page 2 of the datasheet, has lousy flatness, and as far as VSWR is concerned, there is no spec. However, the fact that during my experiment it seemed to exhibit > 60 ohms impedance at 10MHz gives some insight as to how lousy its VSWR specs may be. The one redeeming factor, in my opinion, is that the datasheet suggests that the CPDETLS-4000 can be used to measure input power in the range of -10dBm to 10dBm by merely referencing the calibration points table given on page 2. But even this one thing is a riddle, since the datasheet doesn't provide any guidance as to the degree of accuracy that can be expected. Judging from my experiment results, it seems that the window of usable accuracy is limited to input power in the range of 0dBm to 10dBm, at least this seems to be true at 10MHz. So, unless appropriate test equipment is available to calibrate it, it seems to me that readings in the square law region will have too much uncertainty and therefore will not be very useful. Since I don't own the appropriate test equipment to calibrate this thing, it seems that I'm left with a device that can only provide usable readings while operating in the linear region. As you pointed out earlier, this should work fine if the input signal is a clean sine wave, but where harmonics are present the uncertainty can potentially become large enough to make readings useless. So I want to do a little more research and conduct some experiments regarding how harmonics affect the behavior of LBSD detectors under different circumstances. First I'd like to simulate a harmonic that varies in phase in respect to the fundamental, for example, as the harmonics associated with AM modulation could be expected to behave. Then I'd like to experiment with square wave harmonics, whose phase is fixed with respect to the fundamental. I suspect that in the case of square waves, although the reported dBm level will be incorrect, the corresponding Vpp value will be correct or close to it. |
| G0HZU:
--- Quote ---So I want to do a little more research and conduct some experiments regarding how harmonics affect the behavior of LBSD detectors under different circumstances. --- End quote --- I've done this a few times in the past. I've also tested popular logamps like the AD8307 for similar issues. The AD8307 is a popular device to use as a power meter up to a few hundred MHz. However, it does suffer significant uncertainty from harmonics if the phase is rotated. It's 15 years or more since I played with the AD8307 at my place of work like this but I recall that it is odd order harmonics that upset it the most. Diode detectors like my 8473C are affected by both even and odd harmonics when used in the linear region although not as much as theory suggests for a perfectly linear diode. The second harmonic causes more issues with it than the third but both are problematical if you are after low uncertainty. If it helps, have a read of this classic old app note from HP. It's the one I used many years ago. I think you will find the whole document to be interesting and it may give you a few ideas. But have a look at page 23 onwards and this shows how to model a diode detector wrt the second harmonic as it is used across the square law and linear regions. Up to +/- 20% (or approx +/- 0.8dB) uncertainty in reported power level for a -20dBc harmonic! I don't think your detector will be quite this bad unless you use it way up near its upper limits of range in the linear region. You may find the uncertainty window becomes very asymmetric (wrt phase angle) at high harmonic levels as well. So your research results may show a one sided 'pointy/dippy' effect in the uncertainty window rather than a regular +/- ripple effect as you rotate the phase through 360 degrees. http://www.hparchive.com/seminar_notes/Pratt_Diode_detectors.pdf Hope it helps... |
| TomC:
--- Quote from: G0HZU on April 12, 2016, 11:14:06 pm ---I've done this a few times in the past. I've also tested popular logamps like the AD8307 for similar issues. The AD8307 is a popular device to use as a power meter up to a few hundred MHz. However, it does suffer significant uncertainty from harmonics if the phase is rotated. It's 15 years or more since I played with the AD8307 at my place of work like this but I recall that it is odd order harmonics that upset it the most. Diode detectors like my 8473C are affected by both even and odd harmonics when used in the linear region although not as much as theory suggests for a perfectly linear diode. The second harmonic causes more issues with it than the third but both are problematical if you are after low uncertainty. If it helps, have a read of this classic old app note from HP. It's the one I used many years ago. I think you will find the whole document to be interesting and it may give you a few ideas. But have a look at page 23 onwards and this shows how to model a diode detector wrt the second harmonic as it is used across the square law and linear regions. Up to +/- 20% (or approx +/- 0.8dB) uncertainty in reported power level for a -20dBc harmonic! I don't think your detector will be quite this bad unless you use it way up near its upper limits of range in the linear region. You may find the uncertainty window becomes very asymmetric (wrt phase angle) at high harmonic levels as well. So your research results may show a one sided 'pointy/dippy' effect in the uncertainty window rather than a regular +/- ripple effect as you rotate the phase through 360 degrees. http://www.hparchive.com/seminar_notes/Pratt_Diode_detectors.pdf Hope it helps... --- End quote --- Thanks for the reply and link! As it turns out I have already read this article and found it quite interesting, particularly the model you refer to on page 26. In fact, the experiment that I'm working on at the moment is based on this model. I've seen the associated graph on a couple other articles, I believe one was from Agilent, but apparently they had altered the formulas and I couldn't make them work. However, the formulas on this original vintage article by Pratt worked just fine for me! The only thing is the percentage error that he quotes, ±20%, I see that there is about 20% between 0.83dB and -9.92dB if you relate this back to power input levels, but then I come up with about ±10%. For example, if you are expecting 0dBm (632.5mVpp) but you could end up with as much as 0.83dBm (695.9mVpp) or as little as -0.92dBm (568.9mVpp), then I think you would calculate the percentage error as follows: (695.9 - 632.5) / 632.5 x 100 = 10.02% (568.9 - 632.5) / 632.5 x 100 = -10.06% So as I'm working on my experiment I'm still wondering if I'm missing something silly or if there is actually a typo on the article and the correct figure is ±10%. |
| G0HZU:
In the linear region the detected voltage will be the true peak of the overall waveform if the diode was perfect and we ignore any Vdrop in the diode for the sake of simplicity. So if we look at a simple theoretical case for a very large signal being fed to a suitable linear detector diode and the Vpk of the carrier was 10Vpk and the -20dBc harmonic was 1Vpk then the worst case uncertainty one way would be 11V peak (additive) and the other would be 9V peak (subtractive). power = (Vpk*Vpk)/100 for a 50R system if you assume a perfect sine wave. (but we know this ISN'T a perfect sine wave) So the uncertainty window would be a falsely indicated power level of 81/100W to 121/100W or 0.81 to 1.21W as the phase is changed. This is roughly a (reported) power error of +/- 20% or -0.92dB to +0.83dB with the -20dBc harmonic present. Obviously, the detector you are using can't cope with voltages this big but I've used big numbers to demonstrate linear behaviour so the numbers are easy to deal with. |
| TomC:
--- Quote from: G0HZU on April 13, 2016, 05:16:29 pm ---In the linear region the detected voltage will be the true peak of the overall waveform if the diode was perfect and we ignore any Vdrop in the diode for the sake of simplicity. So if we look at a simple theoretical case for a very large signal being fed to a suitable linear detector diode and the Vpk of the carrier was 10Vpk and the -20dBc harmonic was 1Vpk then the worst case uncertainty one way would be 11V peak (additive) and the other would be 9V peak (subtractive). power = (Vpk*Vpk)/100 for a 50R system if you assume a perfect sine wave. (but we know this ISN'T a perfect sine wave) So the uncertainty window would be a falsely indicated power level of 81/100W to 121/100W or 0.81 to 1.21W as the phase is changed. This is roughly a (reported) power error of +/- 20% or -0.92dB to +0.83dB with the -20dBc harmonic present. Obviously, the detector you are using can't cope with voltages this big but I've used big numbers to demonstrate linear behaviour so the numbers are easy to deal with. --- End quote --- Thanks, that cleared it up! What I was overlooking in my calculation is that when you convert from dBm to Vpp you also have to consider the impedance when talking about power levels. So I correctly found the percentage error for the voltage levels, not the power levels, and although I went over it several times I still wasn't seeing it. Once I saw your example it hit me like a flash of lighting, well, at least I figured that I was probably missing something silly! So I think that my example would be correct If I add the impedance: If you are expecting 0dBm (632.5mVpp @ 50 ohms = 1mW) but you could end up with as much as 0.83dBm (695.9mVpp @ 50 ohms = 1.21mW) or as little as -0.92dBm (568.9mVpp @ 50 ohms = 0.81mW), then you can calculate the voltage percentage error as follows: (695.9 - 632.5) / 632.5 x 100 = 10.02% (568.9 - 632.5) / 632.5 x 100 = -10.06% and the power percentage error as follows: (1.21 - 1) / 1 x 100 = 21% (0.81 - 1) / 1 x 100 = -19% So Thanks again G0HZU, that really helped! |
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