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HighFidelityGuy
06-08-2009, 09:50
Hi,

I want to have a go at making a cable burn-in device to speed up the burn-in process. I have some ideas of how to achieve this but there's one area I'm currently stuck on as I'm not an electronics genius.

I can produce a signal to send down the cable but I need something to go at the other end of the cable to act as a dummy load so that the current flows down the cable. This would need to be able to cope with several watts of power for several hours at a time, while outputting as little heat as possible and being as cheap to buy as possible.

I've heard of resistive loads and such but I'm not sure what products would be suitable for this.

Does anyone know what I could use for this?

Thanks.

StanleyB
06-08-2009, 10:36
This would need to be able to cope with several watts of power for several hours at a time...
It could be a case of 'hasta la vista, baby' for your cable... A few Watts might be good enough to soften up the soldered connection points.

HighFidelityGuy
06-08-2009, 12:16
Thanks Stan, that's a good point.
So I don't need to use as much power as I first imagined. That simplifies things.

Do you know how many amps a standard line level output produces at the standard line level voltage (about 2V RMS I think)?

Thanks.

StanleyB
06-08-2009, 12:19
Thanks Stan, that's a good point.
So I don't need to use as much power as I first imagined. That simplifies things.

Do you know how many amps a standard line level output produces at the standard line level voltage (about 2V RMS I think)?

Thanks.
I'll give you a clue:
2VRMS at 10 to 47K resistance at the input.

HighFidelityGuy
06-08-2009, 12:33
This takes me back to school. :lol:
Right, so I=V/R. So 2V into 10K Ohms = 0.0002A.
Watts=I x V. So 0.0002A = 0.0004 Watts @ 2V RMS.

Do I get a gold star or detention?

HighFidelityGuy
06-08-2009, 15:32
I've just remembered a much simpler equation: Watts = V^2/R.
That also gives me 0.0004 Watts but much quicker. :)

I'm just going to concentrate on interconnects at the moment to keep things simple.
I know that some Hi-Fi components have output voltages much higher than 2V RMS, so I was thinking of using a higher voltage to try and speed up the burn-in process. My reasoning for this is that I've heard that cables that are used in very low voltage situations like tone arms etc take much longer to burn in.

So I'm thinking of using somewhere between 5 and 12V RMS.
The important thing when it comes to cable damage is the current. I'm not sure whether the amount of current used will have an effect on the burn-in process. I'm guessing that as long as the current is at least as high as that in a standard line level signal then the cable will burn-in properly. Using a slightly higher current may speed up the process but I'll need to make sure I'm not using enough current to damage the cable. I guess I need to try and establish what the maximum safe value is for the current, including a healthy safety margin. :scratch:

Does anyone know roughly how much current it would take to fry an interconnect?

therockst4r
06-08-2009, 15:44
What gauge wire do your interconnects use?

HighFidelityGuy
06-08-2009, 15:53
Hi Anthony,

I'm not 100% sure but I'm guessing it's no more than 18 gauge US. So 1mm diameter or less.

therockst4r
06-08-2009, 15:57
At 18 gauge the max current is ~2.3amps if it is 20 gauge then that number goes down to 1.5amps max.
Hope this helps.

HighFidelityGuy
06-08-2009, 16:13
Thanks Anthony that's very useful. :)
I thought there must be a website somewhere that listed this type of info but I wasn't sure what to search for. Based on what you said about wire gauge I started Googling and found THIS (http://www.powerstream.com/Wire_Size.htm) website that confirms your figures. I'll add that too my favourites for future use. :)

I'll try to double check the diameter of the various cables I have to get an idea of what the average size is. That should allow me to work out the average maximum current. I'll then need to reduce this further as like Stan pointed out, the solder joints etc are likely to be the weak link and will no doubt not be able to handle as much current as the cable.

It certainly looks like the maximum safe current will be less than 1 amp.

I think I'll knock up a table that shows the relationship between Volts, Amps and Watts over varying voltages so I can try and find the sweet spot. Hopefully that will then make finding some kind of dummy load easier.

therockst4r
06-08-2009, 17:29
Personally I think I would run 0.5 amps through it.

tubehunter
06-08-2009, 19:00
hi,

how about a 100w light bulb at one end and then plug the other end into the mains 240v, show be about 0.4 amps.

duncan

therockst4r
06-08-2009, 19:07
A light-bulb doesn't sound like too bad of an idea...

HighFidelityGuy
06-08-2009, 20:16
hi,

how about a 100w light bulb at one end and then plug the other end into the mains 240v, show be about 0.4 amps.

duncan

Thanks but I think I'll leave that one for you to try. :mental:


A light-bulb doesn't sound like too bad of an idea...

It would certainly do the job but I want something a bit more hard wearing and professional really. I might try it for testing purposes though.

Barry
06-08-2009, 22:18
Hello Dave,

I ought to start by declaring my scepticism over the need to ‘burn in’ audio signal cables, but since it is a harmless pursuit and you might very well hear improvements, who am I to object?

To answer your question as to how much current to pass during the burn in period, I assume we are talking about low level signal interconnects here rather than speaker cables. In that case the greatest signal current is probably that which flows in the leads that connect between the preamp and the power amplifier. Let us assume that the least sensitive power amplifiers have a sensitivity of 5V with an input impedance of 50KΩ. This then implies an rms current of 0.1mA, so go for 10x this amount, which is 1mA. The dissipation in the load resistor will be 50mW.

Now what kind of current waveform should one use? Well, electrons are lazy little blighters: in a copper conductor of 1mm diameter, a current flow of 1mA means the electrons have a drift velocity of approximately 0.1μm per second. Using DC is not going to shake the electrons out of their lethargy.

The actual type of conductor doesn’t really matter either. The majority of the world’s copper is mined in Zambia, the USA and Chile. Whilst the electrons of American mined copper may well follow the Protestant work ethic of that country, the electrons in Zambian or Chilean sourced copper probably have a far more laid back attitude to life; so need an extra workout. Similarly for silver; over half of it comes from Mexico and Ecuador, with about another 80% coming from China and Australia. Best give the silver electrons a good work out as well, in case they fall back on their noble status and forget that a certain amount of noblesse oblige is required of them.

So it’s AC for the lot of them! And if its AC, it may as well be audio frequencies and not an easy going dose of 50Hz. The best regime would therefore seem to be a punishing course of handling a signal of 5V amplitude that repetitively sweeps from 10Hz to 10kHz over a say a 5 or 10 second period, and subject the cables to this for say 24hrs. And just to make those electrons ‘feel the burn’, make the signal a square wave.

I believe waveform generator chips are available from RS Components and I expect these are VCOs, so the triangular control voltage waveform can in turn be derived from another waveform generator chip or a 555-timer chip fitted with an integrating circuit.

Trust this is of some help – don’t blame me if the electrons end up making a claim for repetitive strain injury.

HighFidelityGuy
07-08-2009, 09:24
Hi Barry,

That was a very informative and amusing read, thanks. :lolsign:
I appreciate your willingness to help despite your quite understandable scepticism.

To answer your question about cable type, yes I'm just going to try this on interconnects first. I'll then look at speaker cables later if I have any success.

So we've established that about 1mA would be a safe and sensible amount of current to use by looking at the question from a couple of directions. Excellent. :)

When it comes to generating the signal, I want to use lots of different types of signal, including sweeps of different wave forms and different types of noise etc. Because of this and my lack of electronics design skills, I thought the simplest method would be to use something like an mp3 player loaded with wav files that I would make on my PC. I would then connect this to a basic low power stereo amp circuit to boost the signal by the appropriate amount. This would then be connected to the cable which would have a pair of load resistors at the other end, one for each channel.

Would that simple setup make the signal flow down the cables in a similar fashion to how it does when they're used in a hi-fi?

Thanks.

Mike
07-08-2009, 10:25
sweeps from 10Hz to 10Hz

:lolsign:

StanleyB
07-08-2009, 12:07
I would like to point out that the properties of the copper mined South of the Equator are not the same as those mined in the Northern hemisphere. Due to the earth gravitational pull and magnetic rotation, the molecular structures are in anti phase between the two.

NRG
07-08-2009, 12:40
Ah yes, good point Stan...that will mean HFG will have to burn the cable in both directions...or at least mark the cables directionality after the burn process so theres no confusion later ;) :doh:

Barry
07-08-2009, 14:27
:lolsign:

Oops ! I did in fact mean 10Hz to 10kHz. So much for typing these things in a hurry - spell checkers don't pick these things up.

Mea culpa ....

Regards

StanleyB
07-08-2009, 15:21
Ah yes, good point Stan...that will mean HFG will have to burn the cable in both directions...or at least mark the cables directionality after the burn process so theres no confusion later ;) :doh:
Alternatively, use the Right cable from conducting material mined in the South and the Left cable from the North.

Clive
07-08-2009, 15:31
Alternatively, use the Right cable from conducting material mined in the South and the Left cable from the North.
How about using wire from the pre-Hiroshima era?

StanleyB
07-08-2009, 15:33
How about using wire from the pre-Hiroshima era?
Was the earth's magnetic field different then:confused:?

Chippy_boy
02-09-2009, 15:28
I would like to point out that the properties of the copper mined South of the Equator are not the same as those mined in the Northern hemisphere. Due to the earth gravitational pull and magnetic rotation, the molecular structures are in anti phase between the two.

Very amusing. Shame copper isn't magnetic in the first place, LOL.

StanleyB
02-09-2009, 16:02
Neither is water, but if you pour it down the sink on the northern hemisphere it flows down the pipe in one direction, but in the southern hemisphere it flows down the other way;).

HighFidelityGuy
02-09-2009, 16:07
Neither is water, but if you pour it down the sink on the northern hemisphere it flows down the pipe in one direction, but in the southern hemisphere it flows down the other way;).

Ah but I believe this was disproved on QI. The direction of flow is determined by the diameter of the plug hole apparently. :confused:

StanleyB
02-09-2009, 16:19
It's factually correct, and has been shown on TV many times.

The Grand Wazoo
02-09-2009, 17:37
It's true. Don't believe a word Steven Fry's scriptwriter says.

Take a look at this..............

Pb69HENUZs8

Barry
02-09-2009, 19:46
Sorry Guys - it is a fallacy:

http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html

Under very controlled conditions and at least 30 degree above and below the equator there is a small difference, but a few metres each side of the equator - no.

I've tried the same myself in Sao Tome, Gabon and Kenya with no noticable effect.

Barry

StanleyB
02-09-2009, 19:58
There was a program not so long ago where two famous actors rode their bikes around the world. They carried out the experiment on TV when they crossed the equator. Distance was about 100m either side of the equator.
Might be available to download still with iPlayer.

Barry
02-09-2009, 20:07
It only needs a very small amount of bias to set the water flowing in whichever way you want.

Look very carefully at Chris's YouTube clip and you will see what I mean. Also the demonstrator is using three different pieces of apparatus. How do we know that the hole is identical in each case? Again it would not take much deviation from a smooth bore to induce a preferred spin.

Tony Moore
03-09-2009, 10:11
They did say at the end of that program episode (Long Way Down) that they'd been conned and it was in fact a setup.