Thread: New to relays and ATX PSU

Hmmm.. I'll have to search and see if you're hookup is right.

But assuming it is... is your relay requiring more current than can be supplied to it?

EDIT:

Wait, I just re read that... what is the tolerance for your relay? If you're only getting 4.75 volts... that might not be enough for a 6 volt relay... I'm not sure.

I thought so too, but I think my main question is.. is it normal for pin B to be dragged down to 0V because the -ve end of the coil is connected to ground? Shouldnt pin B maintain at 4.75V at least even if its not enough for the 6V relay?

I think P-ON is a signal line, it's not designed to have any sort of load on it.. if you put a load on it, you'll exceed the maximum power that the circuit is designed to output and voltage will fall to match the increased current drain.

If you want to detect if your motherboard is on or not, why don't you connect a relay to the standard 5V power rail? That's high when you motherboard is on, and off when your motherboard's off, and it'll supply more than enough current for your relay.

(Or do you also need to detect when your PSU has power, but your computer is switched off? If you still need to monitor P-ON then you'll need to build a transistor circuit to drive your relay)

starfox: yeah, the last line in your post. Basically, the mobo has to send the signal.. not derived from the PSU itself.

If your PSU and Motherboard are hooked up correctly, your PSU will never have power on the 5v, -5V, 12V, -12V, 3.3V rails EXCEPT when the motherboard is also on. To do what you want, the easiest thing is to drive your relay off of one of these rails, as starfox correctly stated. You also need to make sure that your relay switch on voltage is BELOW the line you will drive (this may or may not be true for a 6v relay your trying to drive with 5v). Fortunately 5v and 12v relays are also quite common.


The ATX 2.01 specification requires power supplies to pull up the PS-ON signal through an internal resistor. As such, the motherboard need only drive this signal low to turn-on the the power supply or float the signal to turn it off. It is not intended to supply any current. In your case where you are using this line to drive your relay, the line is unable to supply the current required and thus the voltage adjusts downward accordingly. This is the expected behavior.

If you can find out how much current PS-ON on your motherboard can sink, then it might still be possible to do it your way. You just need to hook your relay coil up between a constant 5v source (+5v stby) and the PS-ON signal. When the motherboard is off, PS-ON is at 5v and thus no current will flow through the relay coil. When the motherboard is on, PS-ON is pulled low, and current will flow through the relay. AGAIN, MAKE SURE THAT YOUR MOTHERBOARD CAN SINK THE REQUIRED CURRENT BEFORE YOU DO THIS!!!!!

The problem is that my mobo is not hooked up correctly =P The +12 and +5 and the +3.3 are giving me headaches.. they are on even when the mobo is off. In fact, this is what I'm trying to do.. I want to turn +12 and +5 off depending on the signal from the mobo.

I want the relay connected in such a way that if the PSON signal is 0V, then my circuit gives +12, +5 etc. But if the PSON signal is 5V then the +12 and +5 rails would be 0.

Rando, from your last suggestion, are you suggesting I use the PSON as the ground for the relay? Its interesting but because people are saying that the PSON wasnt meant to have load, I'm assuming the connection of PSON to ground isnt very strong either. Is that what you meant by "MAKE SURE THAT YOUR MOTHERBOARD CAN SINK THE REQUIRED CURRENT"? Because I dont know what sink the required current means hehe.

Okay how do I check if the mobo can sink the required current? Its an Epia VIA.

Look like PSON is a logic output most likely its within TTL spec.

This signal normally pulls low when active, meaning the load must be connected to +5V and PSON. If you connect an LED (through a resistor) one leg is on +5V and the leg on PSON then you should see it light up.

As been said it depend if PSON can sink enough current to drive a relay. If you really wanna do this then you can get a "sensitive" relay or a reed relay...those relays can operate with low current. Just remember not to overload the relays switch...they are normally small relays.


You said you want VOUT = +5V when PC is on and VOUT = 0V when PC is off?

What exactly are you going to do with the VOUT? if you only gonna feed it to logic gates then there other way to achieve this.



In simple terms :

SINK = when the output is pulled to 0V. The other end of the load must be connected to +V for it to work.

SOURCE = when the output is chucking out +V. The other end of the load must be connected to 0V.

Of course the output has a limit of how much current it can pull down to the ground (sink) or how much current it can supply (source).

Thanks ricky, some terms are quite new to me. So you say a sink is when there's a +V on one end and you try to make it drop to 0V? And source is the other way around, like making a 0V point go to +V?

How do you determine the limit of sink/source current? How do you play with this value? (I think I'll ask more in detail later on this question, because I have another problem like this)

Okay, the Vout, I dont want it 5V, I want it open-circuited (infinite ohms).
The Vout will not be driving anything, it will be connected to a chip. The chip operates ONLY if that pin connected to the Vout is not 0V.

I actualyl achieved this with a very simple circuit

I used an npn transistor. Base si connected to a 10K resistor which then connects to +5V PSON. The emitter is connected to ground.