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PC Power Supply Discussion Troubleshooting and discussion of computer power supplies

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  #21  
Old 04-06-2018
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Quote:
Originally Posted by ashiekh View Post
0db is not the lowest (its a log scale ;-)
OK. I'll bite. Just what is the lowest amount on a log scale? 1?
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  #22  
Old 04-06-2018
Stefan Payne Stefan Payne is offline
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Quote:
Originally Posted by twothreefive View Post
From the slickdeals post where this power supply deal was posted - a user opened one up:

https://slickdeals.net/f/11445547-ex...-amazon?page=3
Whoa, that looks way better than expected.
Didn't expect it to be DC-DC.

That looks fairly similar to the Plattform of the Xilence Performance A+.

Quote:
Originally Posted by Samueru Sama View Post
Wao, you don't see that everyday.
------------------------------------------
Why is the earth connection thru a ferrite core?

I assume this is in order to limit the initial peak current in case of a short.
No, its something like a Coil, so it acts somewhat like a coil.
So that has nothing to do with shorts.

It has something to do with high frequency interferences that are filtered by that thing...
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  #23  
Old 04-06-2018
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Quote:
Originally Posted by Stefan Payne View Post

No, its something like a Coil, so it acts somewhat like a coil.
So that has nothing to do with shorts.

It has something to do with high frequency interferences that are filtered by that thing...
Coil are made of a ferrite material. (well most).

I don't see why it would be there, by placing the coil between the earth you're decreasing the capability of the PC to filter noise, essentially lifting the earth connection for high frequency noise.

Hopefully this way it is better understood:



Because there's some inductance between the load and the noise, the amplitude of the coupled noise will be greater on the left example, and nothing on the right (well almost nothing, it would depend on the series resistance and series inductance of the wire, but you get the point).

And the load can for example be the input impedance of the audio chip of the motherboard, now the resulting noise will be amplified by it since it can't be completely filtered to earth due to the inductor.

I do hope it is there due to some short current regulation (like in the US that people place some small resistance between the earth and neutral connection at the breaker box to limit the peak short circuit current).

Last edited by Samueru Sama; 04-06-2018 at 08:30 PM.
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  #24  
Old 04-06-2018
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Quote:
Originally Posted by ehume View Post
OK. I'll bite. Just what is the lowest amount on a log scale? 1?
There is no lowest, one just keeps getting less...

.
.
10^2 = 100
10^1 = 10
10^0 = 1
10^-1 = 1/10
10^-2 = 1/100
.
.

https://en.wikipedia.org/wiki/Decibel

shows -ve decibels

It is just that the audible scale is such that the negative decibels are too low for human detection, so we rarely use them.

Last edited by ashiekh; 04-07-2018 at 11:58 AM.
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  #25  
Old 04-07-2018
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Also worth noting that the weighting makes it very important. There's a big difference between dB(A) and dB(C); the two most common dB SPL scales for human hearing references.
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  #26  
Old 04-07-2018
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Quote:
Originally Posted by Samueru Sama View Post
I don't see why it would be there, by placing the coil between the earth you're decreasing the capability of the PC to filter noise, essentially lifting the earth connection for high frequency noise.
I think at high enough frequencies the ferrite dissipates energy unlike an inductor.

https://en.wikipedia.org/wiki/Ferrite_bead

"
Ferrite beads are used as a passive low-pass filter, by converting RF energy to heat, by design. (Contrast this with inductors, which by design do not convert RF energy to heat, but rather offer a high impedance to RF.)
"

Without this the Earth wiring becomes one big antenna.
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  #27  
Old 04-07-2018
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Quote:
Originally Posted by ashiekh View Post
I think at high enough frequencies the ferrite dissipates energy unlike an inductor.

https://en.wikipedia.org/wiki/Ferrite_bead

"
Ferrite beads are used as a passive low-pass filter, by converting RF energy to heat, by design. (Contrast this with inductors, which by design do not convert RF energy to heat, but rather offer a high impedance to RF.)
"

Without this the Earth wiring becomes one big antenna.
It doesn't change what I'm talking about there, the ferrite bead adds inductance to the earth connection, doesn't matter how it dissipates it, it says it right there that it is used as low pass filter (AKA low impedance for low frequency and high impedance for high frequency). Again it is essentially lifting the ground connection for high frequencies.

This is a low pass filter:



Why? Because the capacitors reactance decreases with high higher frequencies (basically shorts all high frequencies to ground) while the inductors reactance increases with higher frequencies, essentially you have a voltage divider that gets bigger and bigger with higher frequencies.

For example here, lets say it is a 10 MHz signal, the capacitors reactance is 0.1 Ohm and the inductors reactance is 1000 Ohm for that frequency, that means that here:



The potential difference of the 10 MHz signal is 0.1/(1000+0.1) = 0.00009V (Nothing basically).

However if we add an inductance to the earth connection, the capacitors reactance is in series with the new inductors reactance, and now it is more than 0.1 Ohm in our formula, essentially making the resulting noise higher!!!

Last edited by Samueru Sama; 04-07-2018 at 01:04 PM.
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  #28  
Old 04-07-2018
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An inductor is energy conserving; ferrite is not acting as an inductive core for high enough frequencies.

I could be completely wrong, and it may be a learning opportunity for me.

Last edited by ashiekh; 04-07-2018 at 07:29 PM.
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  #29  
Old 04-07-2018
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Quote:
Originally Posted by Samueru Sama View Post


The potential difference of the 10 MHz signal is 0.1/(1000+0.1) = 0.00009V (Nothing basically).
Wouldn't the impedance of the capacitor and inductor have not real but purely imaginary value? If we do the operation "0.1/(1000+0.1)" we are treating the capacitor and inductor with real impedance which is not true. While the error on this calculation in particular is small, doing this the same way could result in more error in less extreme cases.

Instead, in this case, it should be along the lines of "-0.1/(1000-0.1)" which gives 0.0001V/V of gain. Obviously not a big difference here but would have an effect elsewhere.



Also seen here: https://www.allaboutcircuits.com/tex...ies-r-l-and-c/

Or if I happen to be misled in some way here, please inform me.

Edit: This also looks like something that would have resonance.

Edit 2: Well my diagram I realize isn't the same anyway. Without the load resistor it changes things a little. Mine assumes it is infinite resistance. Well, yours does too.

Last edited by turkey3_scratch; 04-07-2018 at 09:50 PM.
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  #30  
Old 04-08-2018
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Yes, it could resonate and Vout could then be bigger than Vin
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