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There are many instructions "How to Convert a Computer ATX Power Supply to a Lab Power Supply". Often they suggest they suggest to place to staus LEDs to indicate it the power supply is turned on or in standby-mode or without electricity. The purple wire is on +5V if the power supply is connected to electricity (this includes standby and power-on) and the gray wire is on +5V if the power supply is turned on:

ATX connector (in German)

(Sorry, this is in German, there is no other public domain picture available.)

Instead of having a green and a red light at the same time turned on (e.g. doing it this way), I wanted to a red LED for standby and a green LED for power-on (this could be even nicer using a two-color LED). To do so, I used a NPN transistor (a BC547B, PDF datasheet) and some resistors to build a NOT gate:

circuit

It actually works: when you connect the power supply, the red LED lights up and when you connect the green wire to ground (not in the circuit diagram) the power supply turns on, the greed LED lights up and the red LED goes off. However, there is a strange voltage on the gray wire, which I cannot understand (and it might be the cause why the LEDs are not very bright). Between the gray wire and ground there are 2.5V. How does that happen?

Note that I verified that the gray wire is on +5V if I disconnect the right part:

part of the circuit

I am not very sure about the the resistor R2. In the beginning I used 1kOhm. But after playing a bit on the breadboard, I noticed that the green LED gets a bit brighter if I increase the resistor. I cannot explain that, maybe I made a mistake somewhere...

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  • \$\begingroup\$ Well, the only word that's actually in German in that illustration is "Masse", and I think everyone can figure out what that is :) Otherwise, there's always pinouts.ru's ATX overview, which - despite the Russian domain name - is in English. \$\endgroup\$ – Flambino Jan 6 '14 at 20:24
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It seems that the grey wire has a high output resistance so the more current you source the more voltage drop you get, although I can't explain why it is able to provide about 13mA to LED1 with no drop and then drops to 2.5v with a few more mA.

Anyway, the circuit I suggest used two transistors, a PNP and a NPN each driving one led.
The benefit of the circuit is that it only draws a couple of mA (depends on the gain of the transistors) from the gray line so you will get no drop there, the main current for the leds is supplied from the purple supply line.

There is no need for a base resistor because the base-emitter current is limited by the resistors connected to the emitter.

enter image description here


I found this ATX12V Power Supply Design Guide linked from the wikipedia page for Power Good Signal and it describes the specs of PWR_OK (grey wire)

enter image description here

Based on these specs the current you can source is negligible so I don't know how you manage to get 5v while sourcing more than 10mA (as shown in your circuit with only led1 connected). I have also seen this wiring scheme from the page you linked which also connects the anode of the led to the PWR_OK pin and sources several mA.
I assume that the specs about sourcing are the minimum and designers actually use output circuit with more current sourcing ability.

If I was to base a circuit on these specs then I would use a pull-up resistor (R4 below) to drive the base of the NPN or even better a mosfet.
The PNP circuitry doesn't need to be changed, sinking 1mA will be enough and is within specs.

enter image description here

enter image description here

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The 2.5V on your Gray wire, which I assume is PWR OK. Could be because of the sink caused by the resistors R1 and R2. These resistors are grounded though the diodes of your LED and the B-E junction of the transistor. If PWR OK has an internal pull up you would end up with a voltage divider.

I like Flambino's circuit, it would still form a voltage divider so your PWR OK signal would not go to 5V, if I'm right, but things should work fine...

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Complete electronics noob here, but I'd say you should connect the green LED to the base of the transistor rather than drive the transistor in parallel with the LED. I'm guessing the current is being split between the transistor and the LED.

schematic

simulate this circuit – Schematic created using CircuitLab

(errata: The +/- symbol for the purple wire should of course be reversed - can't edit the diagram, unfortunately)

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The PWR_OK signal (grey wire) is only meant to source a small amount of current. The 5V standby signal (purple wire) should be able to handle more. If you can spare a diode (marginally cheaper than another transistor), try out this circuit.

ATX Power LED Indicator Circuit

You may have to adjust the resistor values depending on the characteristics of your transistor and LEDs. I simply chose the values you had previously. The big difference here is that the standby signal sources the bulk of the power for each LED. The PWR_OK ends up being used only for a low current signal.

With the grey wire low, the transistor will be in cutoff mode. There will be no conducting path through the green LED or the diode. Therefore, the only path for current will be through R1 and the red LED. Then, when the grey wire is pulled high, the transistor will be conducting. So long as the transistor is in saturation mode, Vce will be low, <0.7V. The drop across the diode is assumed to be about 0.7V. Therefore, so long as the red LED has a forward voltage greater than 1.4V (probably less than that), it will be off. Then, current will be flowing through resistors R1 and R2, the green LED, and the diode.

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