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#311
March 25th 04, 04:03 PM
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#312
March 26th 04, 12:52 AM
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Go to Agilent home page, search for 16334A.
I also have a homebrew design that works fine, resolution to as little as 0.01pF, but haven't had a chance to move it from the breadboard stage to a more finished implementation. It will be tweezers that connect to a readout box through a single coaxial cable (RG-174-type). There can be interchangeable "heads" -- tweezers, spring clips, etc. If/when I get enough round tuits to finish it up, I'll post/publish something on it. Cheers, Tom Rex wrote in message . .. On 23 Mar 2004 23:50:18 -0800, (Tom Bruhns) wrote: with my capacitance tweezers That sounds interesting. I have used regular tweezers to slip sm caps into a fixture I made. This goes pretty fast except when I accidentally twang a part into some far and unknown location. Did you make these youself? Can you give a quick description. I can think how I could make low capacitance tweezers, but not how to flexibly connect them to a capacitance meter and get consistant readings. |
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#313
March 26th 04, 12:52 AM
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Go to Agilent home page, search for 16334A.
I also have a homebrew design that works fine, resolution to as little as 0.01pF, but haven't had a chance to move it from the breadboard stage to a more finished implementation. It will be tweezers that connect to a readout box through a single coaxial cable (RG-174-type). There can be interchangeable "heads" -- tweezers, spring clips, etc. If/when I get enough round tuits to finish it up, I'll post/publish something on it. Cheers, Tom Rex wrote in message . .. On 23 Mar 2004 23:50:18 -0800, (Tom Bruhns) wrote: with my capacitance tweezers That sounds interesting. I have used regular tweezers to slip sm caps into a fixture I made. This goes pretty fast except when I accidentally twang a part into some far and unknown location. Did you make these youself? Can you give a quick description. I can think how I could make low capacitance tweezers, but not how to flexibly connect them to a capacitance meter and get consistant readings. |
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#314
March 27th 04, 12:40 PM
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In
That sounds interesting. I have used regular tweezers to slip sm caps into a fixture I made. This goes pretty fast except when I accidentally twang a part into some far and unknown location. Did you make these youself? Can you give a quick description. I can think how I could make low capacitance tweezers, but not how to flexibly connect them to a capacitance meter and get consistant readings. Wayne Kerr used to make a VHF admittance bridge which serves well for measuring chips, and they can be bought for less than £100 (813? bridge) in UK leaded low value ceramic and silvered mica have their Tc and loss modified by the effect of the encapsulation material. It therefore follows that chips npo chip will be better. -- ddwyer |
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#315
March 27th 04, 12:40 PM
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In
That sounds interesting. I have used regular tweezers to slip sm caps into a fixture I made. This goes pretty fast except when I accidentally twang a part into some far and unknown location. Did you make these youself? Can you give a quick description. I can think how I could make low capacitance tweezers, but not how to flexibly connect them to a capacitance meter and get consistant readings. Wayne Kerr used to make a VHF admittance bridge which serves well for measuring chips, and they can be bought for less than £100 (813? bridge) in UK leaded low value ceramic and silvered mica have their Tc and loss modified by the effect of the encapsulation material. It therefore follows that chips npo chip will be better. -- ddwyer |
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#316
March 27th 04, 06:31 PM
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I read in sci.electronics.design that ddwyer
wrote (in ) about 'Extracting the 5th Harmonic', on Sat, 27 Mar 2004: Wayne Kerr used to make a VHF admittance bridge which serves well for measuring chips, and they can be bought for less than £100 (813? bridge) You might well be able to extend its frequency range downwards, too. IIRC, it needs some very small 1 uF capacitors, which simply weren't available when it was manufactured. There is an associated transistor test set which is a walking disaster. -- Regards, John Woodgate, OOO - Own Opinions Only. The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk |
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#317
March 27th 04, 06:31 PM
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I read in sci.electronics.design that ddwyer
wrote (in ) about 'Extracting the 5th Harmonic', on Sat, 27 Mar 2004: Wayne Kerr used to make a VHF admittance bridge which serves well for measuring chips, and they can be bought for less than £100 (813? bridge) You might well be able to extend its frequency range downwards, too. IIRC, it needs some very small 1 uF capacitors, which simply weren't available when it was manufactured. There is an associated transistor test set which is a walking disaster. -- Regards, John Woodgate, OOO - Own Opinions Only. The good news is that nothing is compulsory. The bad news is that everything is prohibited. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk |
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#318
March 28th 04, 05:37 AM
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Avery Fineman wrote:
In article , Peter John Lawton writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. I just wondered from a theoretical point of view what the program would say about the harmonic content as you decreased the values you put into it for t_r and t_f. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. This is connected with my question. I am pondering why the energy available for higher harmonics is less than for the fundamental and also how your program works out this energy. But, also mentioned before by others is that shortening the rect- angular waveshape DOES increase the 5th harmonic, as evident by the approximate 12 db increase at 40 to 35 percent of the repetition period. Its like pushing the baby on the swing in the park, you only need to give it the occasional push or pull in the right direction. A 5f resonator gets has to go for 2.5 cycles in between refuelling from a square-wave (1:1) of frequency f. With a mark/space ratio of 2:3 it looks to me as though the 'pull' on the 5f resonator as the rectangular wave drops will take out all the energy that was put in on the preceding 'push' - so no 5th harmonic. On the other hand, 1.5:3.5 (30%) should be just as good for 5th as 1:1 (but no better). All that's assuming a zero t_r and t_f. In practice it must be that the energy transfer to a resonator depends on these times. It seems that the energy transfer is not so great to a faster resonator. Pursuing the swing analogy, your arm has to move faster than the swing if you're going to add to the swings energy. An equivalent shortening happens in vacuum tube multipliers through biasing (self, fixed, or both) and that can be adjustable along with the drive level. It's not quite the same with bipolars since the overdrive effects are more saturation than in the self- bias conditions of tubes. It's close, though. From all indications of the Fourier series results, there's a definite reason why so few multipliers went beyond tripling. The amount of energy (relative to fundamental and taking into account the finite rise and fall times) of 4th and higher harmonics just isn't as much as intuition would have everyone believe! What do you mean by intuition here? My intuition suggests to me that as the rise and fall times get shorter, the energy available for the harmonics approaches that for the fundamental. In other words, as a square wave approaches perfection it gains the ability to stimulate all odd harmonic resonators equally - but surely that can't be right. Possibly the energy transfer to a f and (say) 5f resonator approaches the same value but the 5f resonator loses 5 energy at five times the rate of the f one, assuming equal Q. Peter Len Anderson retired (from regular hours) electronic engineer person |
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#319
March 28th 04, 05:37 AM
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Avery Fineman wrote:
In article , Peter John Lawton writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. I just wondered from a theoretical point of view what the program would say about the harmonic content as you decreased the values you put into it for t_r and t_f. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. This is connected with my question. I am pondering why the energy available for higher harmonics is less than for the fundamental and also how your program works out this energy. But, also mentioned before by others is that shortening the rect- angular waveshape DOES increase the 5th harmonic, as evident by the approximate 12 db increase at 40 to 35 percent of the repetition period. Its like pushing the baby on the swing in the park, you only need to give it the occasional push or pull in the right direction. A 5f resonator gets has to go for 2.5 cycles in between refuelling from a square-wave (1:1) of frequency f. With a mark/space ratio of 2:3 it looks to me as though the 'pull' on the 5f resonator as the rectangular wave drops will take out all the energy that was put in on the preceding 'push' - so no 5th harmonic. On the other hand, 1.5:3.5 (30%) should be just as good for 5th as 1:1 (but no better). All that's assuming a zero t_r and t_f. In practice it must be that the energy transfer to a resonator depends on these times. It seems that the energy transfer is not so great to a faster resonator. Pursuing the swing analogy, your arm has to move faster than the swing if you're going to add to the swings energy. An equivalent shortening happens in vacuum tube multipliers through biasing (self, fixed, or both) and that can be adjustable along with the drive level. It's not quite the same with bipolars since the overdrive effects are more saturation than in the self- bias conditions of tubes. It's close, though. From all indications of the Fourier series results, there's a definite reason why so few multipliers went beyond tripling. The amount of energy (relative to fundamental and taking into account the finite rise and fall times) of 4th and higher harmonics just isn't as much as intuition would have everyone believe! What do you mean by intuition here? My intuition suggests to me that as the rise and fall times get shorter, the energy available for the harmonics approaches that for the fundamental. In other words, as a square wave approaches perfection it gains the ability to stimulate all odd harmonic resonators equally - but surely that can't be right. Possibly the energy transfer to a f and (say) 5f resonator approaches the same value but the 5f resonator loses 5 energy at five times the rate of the f one, assuming equal Q. Peter Len Anderson retired (from regular hours) electronic engineer person |
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#320
March 28th 04, 07:51 AM
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In article , Peter John Lawton
writes: Avery Fineman wrote: In article , Peter John Lawton writes: The above will hold true at any fundamental frequency provided the rise and fall times are equal and each equal to 0.02 times the repetition period. Those numbers will change given faster or slower rise/fall times. All db calculated as 20 x Log (voltage). Width is determined at the baseline, not the 50% amplitude point. Len Anderson retired (from regular hours) electronic engineer person I wonder what happens to these numbers as the rise/fall time tends to zero? The harmonic content will increase...but also show dips depending on the percentage width relative to the period. I could present those (takes only minutes to run the program and transcribe the results) but that is academic only. The rise and fall times will NOT be zero due to the repetition frequency being high (repetition time short). Consider that a 3 MHz waveform has a period of 333 1/3 nSec and that Paul is using a TTL family inverter to make the square wave. Even with a Schmitt trigger inverter the t_r and t_f are going to be finite, possibly 15 nSec with a fast device (and some capacitive loading or semi-resonant whatever to mess with on- and off-times). 15 nSec is 4.5% of the repetition period, quite finite...more than I showed on the small table given previously. I'm sure someone out there wants to argue minutae on numbers but what is being discussed is a squarish waveform with a repetition frequency in the low HF range. Periods are valued in nanoSeconds and the on/off times of squaring devices are ALSO in nanoSeconds. There's just NOT going to be any sort of "zero" on/off times with practical logic devices used by hobbyists. I just wondered from a theoretical point of view what the program would say about the harmonic content as you decreased the values you put into it for t_r and t_f. The harmonic values will change, approaching that of an ideal square wave. That's a truism. With zero rise and fall it IS the same as an ideal square wave. There's NO accurate little formula, saying, or myth that will predict any particular harmonic value. That's the reason for using nice, very quick number-crunching computer programs. What is not intuitive to me (and to others) is that harmonic energy of a rectangular waveform drops drastically by the 5th harmonic and is certainly lower than "obvious" numbers bandied about. This is connected with my question. I am pondering why the energy available for higher harmonics is less than for the fundamental and also how your program works out this energy. My program was developed while at RCA Corporation, specifically in the time period of winter 1973-1974 using the core of three ideal waveforms: rectangular, rising triangle, falling triangle. They relate to a singular waveform using a time-delay formula multiplier so that the rising triangle butts up to (in time) to the start of the rectangular waveform and the falling triangle starts at the end of the rectangular waveform. Entry is rise-time (the rising triangle), fall-time (the falling triangle), and 50% amplitude pulse width which is the rectangular waveform length and the length of the rising and falling triangles adjusted for their inputted times. [draw it out to see it better] Each basic waveform generates its own Fourier coefficient set. All sets are simply added algebraically. Mathematically okay to do that. A quick form of proof of that is to use a simple frequency-to-time transform that works at each specified point in time along the repetition period of the waveform. The original was a time-to- frequency transform, mathematically different than the opposite. If a reconstruction of the frequency-to-time results in the original entry specifications, then it is called accurate enough. I didn't derive the reconstruction transform since it was already in a book. Neither did I derive any of the basic ideal waveforms which were already in the ITT Blue Bible. The delay multiplier used to set rise, fall, and 50% width was another book value, simplified to faster calculation simplicity because the original was a math problem thing with more terms than needed. As to WHY of the energy distribution, that's up to any person who has the textbook formulas and math smarts to fool around with. I can't sum that up in one message. I doubt anyone can. I do know this: Using the formulas and the program, then setting up a test with careful adjustments of a pulse generator and using a well-calibrated spectrum analyzer, the numbers agree within the tolerances of the analyzer calibration. To me, and lots of others, that is all the proof needed. Beyond that, its too much time and nobody paying me to do this... Its like pushing the baby on the swing in the park, you only need to give it the occasional push or pull in the right direction. A 5f resonator gets has to go for 2.5 cycles in between refuelling from a square-wave (1:1) of frequency f. Use any analogue you want. I don't agree with the above, but feel free and I not going further on that... What do you mean by intuition here? My intuition suggests to me that as the rise and fall times get shorter, the energy available for the harmonics approaches that for the fundamental. In other words, as a square wave approaches perfection it For any ideal rectangular shape, the harmonic energies have a (SinX / X) locus. That's explained in textbooks also. Harmonics of a repetitive waveform Fourier transform will NEVER have more energy than the fundamental. That's also basic book stuff. If the rise and/or fall times are finite, the harmonics will drop their energy levels compared to the zero rise and fall time ideals. As the rise and fall times get longer and longer the harmonic energy gets less and less. By the time one gets to a sinusoid waveshape, there are NO harmonics in any Fourier transform, its all fundamental frequency (1 / repetition-period). Recess. Len Anderson retired (from regular hours) electronic engineer person |
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