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16th August 2012, 11:18 AM  #26401 
diyAudio Member
Join Date: Jun 2007
Location: Blackburn, Lancs

They would either be classified and unless you work on the specific program you wouldn't know about them or they are not classified and you know about them. As John knows about them they cant be classified, or someone is breaking their security clearance, that is somthing the US dosn't like. A lot of this stuff is covered under ITAR, which is a PITA.

16th August 2012, 11:35 AM  #26402 
diyAudio Moderator

Or he's just repeating a story someone on the "team" told him (most likely, my guess).
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And while they may not be as strong as apes, don't lock eyes with 'em, don't do it. Puts 'em on edge. They might go into berzerker mode; come at you like a whirling dervish, all fists and elbows. 
16th August 2012, 12:58 PM  #26403 
diyAudio Member
Join Date: Jul 2003
Location: berkeley ca

Actually, I was directing this at Jneutron. Perhaps I might still get a response. I wouldn't expect most people to have any new info.

16th August 2012, 02:34 PM  #26404  
diyAudio Member
Join Date: Jan 2004
Location: away

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Part of my interest in low inductance resistors stems from trying to measure phase response of magnets. To do so requires absolute accuracy in measuring the current as it varies with time. This requires a resistor with two qualities: 1. As little inductance as is physically possible, and 2. no loop to trap external field. 1. All resistors will have an inductance associated with it. At any frequency, the endpoint reactance of the resistor is the sum of the IR drop and L dI/dt. The IR drop is what is desired to measure, the L dI/dt is the error. It is also 90 degrees out of phase when the resistor is driven with a sine. Once you have eliminated the simple inductive things, like a wirewound element and magnetic lead materials, you have only the current centroid (the center of the distributed current in space) to work with. Now, all you can do is try to cancel the magnetic field of the resistor's current. You can bring the return path of the current as close as possible, my earlier suggestion on using braid over the resistor to bring the return current over the body is the best one could do for an existing resistor. Or, you could create a resistor which has the return current passing through the primary current. The earlier picture had 20 ten ohm resistors in parallel between two flat plates, one half ohm total, and then another 20 ten ohm resistors interleaved between the first 20. Another half ohm, wired in series to give a 1 ohm resistor. This construct has the current centroids passing through each other. Externally, the magnetic field is pretty much gone. Internally, most of it is also cancelled.. For this resistor topology, the inductance can be calculated using massive parallelism of inductors because they essentially cancel one another. (while not absolutely accurate, it is quite close.) The simple calculationdon't worry, no calculus involved... ) A wire pair has inductance of approximately 200 nH per foot. My resistors are 1/4 inch long. Each pair has ~200/48 nanohenries, or 4.2 nH. I have 20 pairs in this resistor, so the equivalent inductance is 4.2nH/20, or 210 picohenries. Using 3 pieces of one sided perf board and 40 1/4 watt mondo cheap metal film resistors from digikey (P/N P10.0CACTND), I now have a 1 ohm, 210 picohenry resistor. (1 microhenry is 10e6, 1 nanohenry is 10e9, 1 picohenry is 10e12), my resistor has an inductance of 2 times 10e10 henries. Since the voltage is IR + L dI/dt, what is the error? At a current slew rate of 1 ampere per microsecond (dI/dt of 10e6), my resistor has an error of 2 times 10e10 times 10e6, or 2 times 10e4 volts, 200 microvolts. At 1 ampere per nanosecond, it will have an error of 200 millivolts. If you need lower errors, you simply increase the number of resistors. A normal resistor, if you returned the current right alongside it, would have about 5 nanohenries inductance, or 5 times 10e9. The errors scale accordingly.If you choose NOT to return the current next to the resistor, you will see much higher inductances. 2. If you use the resistor as a current viewing resistor (the essence of a resistor ladder divider), you have the additional problem of preventing pickup of external fields, like the current path to and from the resistor. My resistor eliminates that, note the twisted pair pickup leads vs the power ones. For example, my SWTPC's have the gain set via a 2.2k,100 ohm feedback divider. If I make the first part of my resistor using 20 pieces of 2k resistors, and the second part using 44k resistors, I have eliminated any dI/dt errors of the divider network in the feedback. jn edit: the tiger 250 is 3db down at 400Khz.. and output v of 70 volts, ladder current of 30 milliamps. If the feedback ladder has a microhenry of inductance associated with the 100 ohm resistor, a full power slew of 1 uSec will give a ladder slew dI/dt of 3 * 10e4, inductive error of 30 millivolts added to 3 volts of IR drop, 10% inductive error. At 10 usec 70 volt slew, 1% inductive error (both at full scale out). Remember, the inductive term is rate dependent, the percentage error calcs I'm doing here is based only on full scale, a 10 usec slew to 7 volts gives 10% error. Remember folks, math isnt one of my strong points, let me know if anybody finds errors...found two, will run outta edit time.. pps..I used this puppy (the tiger 250) for a 10 uSec slew to 40 volts into a 4 ohm non inductive load fed by 6 pieces of cat5e 15 feet long paralleled as a tline of 4 ohms, so I know the tiger can do at least that.. Last edited by jneutron; 16th August 2012 at 02:59 PM. 

16th August 2012, 02:51 PM  #26405 
diyAudio Member

Same here, and they do hear nuances of performance and instrument quality that I could not begin to hear. I expect the same old excuses will be made.
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16th August 2012, 02:54 PM  #26406 
diyAudio Member

Is that like being nearly pregnant.?
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"The question of who is right and who is wrong has seemed to me always too small to be worth a moment's thought, while the question of what is right and what is wrong has seemed allimportant." 
16th August 2012, 03:04 PM  #26407 
diyAudio Member

I suppose this construct could be used to minimize a speaker cable inductance, but in that case the capacitance would be maximized?
__________________
"The question of who is right and who is wrong has seemed to me always too small to be worth a moment's thought, while the question of what is right and what is wrong has seemed allimportant." 
16th August 2012, 03:11 PM  #26408  
diyAudio Member
Join Date: Jan 2004
Location: away

Quote:
As I said, I used 6 pieces of cat5e to make a 4 ohm t line, it drove a one meter long stripline made of two copper ribbons half inch wide separated by 1 mil kapton to the load, a bifilar heater embedded into a supercon magnet. Seeing a 50usec wide heating pulse through all that into a 4 ohm NI load in liquid helium with 10 usec rise and fall and no overshoot or ripple was a beautiful thing.. The boundary for speaker cables will be the equation: LC = 1034 DC. L in nH per foot, C in pf per foot. Typical dielectrics will be around 3. The only caveat to low z speaker lines it the case where the speaker unloads below the open loop unity gain point of the amp. As the speaker unloads, the amp will see the capacitance. If the speaker could remain constant impedance beyond the amp open loop unity gainpoint, the amp will see only the characteristic impedance of the cable. jn Last edited by jneutron; 16th August 2012 at 03:14 PM. 

16th August 2012, 03:24 PM  #26409  
diyAudio Member
Join Date: Nov 2008
Location: Oakmont PA

Quote:
At least one of the high end manufacturers uses networks to keep the input impedance flat. That way the loudspeakers tame some of the high end "artsy" amplifiers. 

16th August 2012, 03:42 PM  #26410 
diyAudio Member
Join Date: Jul 2003
Location: berkeley ca

Thanks Jneutron, your explanation for your resistor configuration was interesting.

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