Make a 5v and 12V power supply with a PC power supply
On an old functional PC power supply, 5 V and 12 V can be recovered easily.
The problem is when this power supply is disconnected from a computer, then it no longer works, the fan no longer turns.
To remedy this, you just need to know the trick…
Locate the ATX 20 connector (old model), which comes from the power supply unit, and which will be plugged into the support of the PC motherboard.
An electrical loop must be made between pin 3 and pin 4 of this connector.
If the figures are not written on this socket, proceed as in the photo, on the side of the socket blocking mark, counting from the left.
****
With a voltmeter, check the voltages on the molex connectors that go to the peripherals.
Between black wire and red wire: 5 V
Between black and yellow (or orange) wire: 12 V
Make a 5v and 12V power supply with a PC power supply
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Thanks for the info, it could be useful. Is there no 3.3 V (or something else?) Somewhere for the processor or is it managed on the motherboard?
Amperage level by cons? How is the power distributed between 5 and 12V? We have the maximum power of the power supply but not the detail.
Amperage level by cons? How is the power distributed between 5 and 12V? We have the maximum power of the power supply but not the detail.
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Do a image search or an text search - Netiquette of forum
If there is also 3.3V (orange wire normally) and maybe even -12V (but I may be confusing with the old generation)
By cons the big concern with this solution is that you must charge the + 5V to have stable voltages.
The + 12V is "the image" of the + 5V (it does not have its own regulation) and if the 5V is not loaded, the voltage in + 12V is sometimes far from the 12V
By cons the big concern with this solution is that you must charge the + 5V to have stable voltages.
The + 12V is "the image" of the + 5V (it does not have its own regulation) and if the 5V is not loaded, the voltage in + 12V is sometimes far from the 12V
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Ah that's good to know ... thank you Forhorse
But must the charges be proportional? Or a simple charge of 1W on the 5V is enough to have 200W on the 12V?
But must the charges be proportional? Or a simple charge of 1W on the 5V is enough to have 200W on the 12V?
Last edited by Christophe the 05 / 02 / 10, 18: 22, 1 edited once.
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Do a image search or an text search - Netiquette of forum
I learned the complement for the load resistors to add on the output of uses on =>
http://radioman64.e-monsite.com/pages/m ... fixes.html
Wirewound resistors of powers 15R, 56R, 150R approximately.
http://radioman64.e-monsite.com/pages/m ... fixes.html
Wirewound resistors of powers 15R, 56R, 150R approximately.
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If we take the example of 5V, I'm really not sure that 90mA of charge is enough to ensure voltage stability.
90mA on a power supply which can deliver 30A it only makes 0,3% of load, what a fart of fly.
And then why put a resistor of 5W when the dissipated power will not even be 1/2 W?
I don't know where you got this pattern from, but it seems a little esoteric to me.
90mA on a power supply which can deliver 30A it only makes 0,3% of load, what a fart of fly.
And then why put a resistor of 5W when the dissipated power will not even be 1/2 W?
I don't know where you got this pattern from, but it seems a little esoteric to me.
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The solution I cannot give it since I have no idea of the value of the current which it is necessary to draw so that the tensions are stable.
Either way, a switching power supply does not perform optimally when empty. And a load of 0.3% of the nominal current, for me it is "empty"
I have already seen sites that talk about putting a car headlight lamp, is it enough? is it "too much" (waste)? I do not know.
Either way, a switching power supply does not perform optimally when empty. And a load of 0.3% of the nominal current, for me it is "empty"
I have already seen sites that talk about putting a car headlight lamp, is it enough? is it "too much" (waste)? I do not know.
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