Hey guys, first time post on the board.
Just wondering. It sounds like he needs a circuit to keep the voltage steady. Why can't he just use a zener diode?

 

Hey guys, first time post on the board.
Just wondering. It sounds like he needs a circuit to keep the voltage steady. Why can't he just use a zener diode?

I think he is trying to use the +5V output of a PC power supply as a lab supply, or in some non-PC project. The +5V output of a PC supply is typically not well-regulated (or worse) unless they have a large load on them. In his case, he has determined, or believes, that he needs a half amp load to stabilize the supply. A zener might work for low-current applications, but if he heeds it to work for loads from zero to 25 amps or so, then a zener is not going to help.

 

Hope this comes across appropriately & sorry if it's long. Short just won't get the questions answered.

First, I enjoy figuring things out then modifying them for my need.

Google "ATX Power Supply Breakout." That $25 board is buildable. There are no sandbars. There is SMT.

In particular, notice the line coming off pin 8 (has to be - pin 7 is ground - Pin 8 is the POST line). POST is nominally very low current & while opinions vary, ain't much in voltage either (mine is @ 2V). Anyway, coming off that line there is a resistor, transistor, diode & resistor physically arranged as I put in the first drawings I posted. This all feeds in to the 5VSB. I don't know why. I don't know if it's attenuating the circuit or amplifying it. My guess is attenuating since an increase would send the 5VSB above parameters. But I would think any addition to the 5VSB would be a bad thing for a line out to a binding post since it would no longer be 5V.

So I avoided that area all together & just decided to slap an LED on the end as a "load." I do know in common-emitter and common-collector amplifier configurations gain is mostly closely associated with hysterisis. In the common-base circuit, it's different: the α ratio between collector current and emitter current is what's considered. So I figured that's what I was needing was for the circuit to attenuate that 5V, then it would have about a .7 voltage drop (temperature & P material aside), then .7 for the diode, a resistor, a light & wha-la! Unfortuneately, not quite so. Since I wasn't sure of the math (not the formula, the formula values) to compute the α ratio, I just used commonly found resistor values.

And it was working, but it didn't make sense. Why was that 10k resistor in pulldown & not in series to limit the current into the transistor? Why were my measured voltages so off? So, off I go to AAC to ask the world.

Another note about the breakout board referenced above: I can tell from the shadow cast in the screw mount holes it is laying flat. So the only things on the underside can't be bigger than jumperwires or more SMT. No sandbars.

Also note that no visible line comes out of the switch at the "on'" position. That line (which has to go to ground) must be underneath the board, along with the other grounds, jumpers, etc. Also, on top they've added diodes & resistors between the rails & grounds at the terminals - my guess is to prevent unwanted feedback from motors or servos.

I'm not trying to infringe on that website's business. I just want to modify the idea for my ATX panel (& I'd rather spend the money I save on a pizza & an evening on my workbench, lol). I can't do SMT & the reason I know what's on the board is by the labeling. No clue as to specifications.

On my breadboard it works well except that my design is a fluke - I keep happily destroying transistors which just so happen to short just right to keep the load on the supply so it works. I can take them out, put them in, play with resistors, switch on/off, & it works. But that's wrong - it's working because it's broken.

Which I guess means the original question has been answered - I wanted to know why/how it works. Well, it works because it broke just right, lol.

Now what I'd like to know is how to make it work correctly. I'm thinking maybe a mirror circuit? That's the nearest I've come to finding a transistor circuit with a single supply line (which is the case here). Maybe I should go back to tying the POST line into the 5VSB, but then I'd need to add a color change LED to show when it was in standby vs. panel hot. Of course there is always the CPU radiator with the UV liquid in clear tubes solution. My head hurts. lol.

As always, thanks for your valuable time & feedback.

 

I don't agree. As temperature increases the base-emitter voltage decreases and if it continues uncontrolled the transistor will blow due to overheat. This is the reason BJTs suffer from thermal runaway.

Yes, you are right. Sorry.

 

I lost the forest looking at the trees...

beenthere wrote...

I don't understand either circuit. Just attaching the LED to the 5 volt source through the 470 ohm resistor works fine. Both circuits are non-functional with respect to the transistor. They are just a LED, a resistor, and two diode drops to source.

That was the first response. I was so focused that the needed "load" was coming from the transistor that I missed what was happening. Others made similar observations, especially Audioguru who kept pointing out that the transistors had to be shorting, which was simply creating a series of diodes (voltage drops) to the LED. Anyway, the solution:

One superbright Led, Vf=3.5, Ic=20mA & 82Ω. System works fine.

At least I learned a lot of very useful information regarding transistors: how to test them, destroy (short) them, not short them, & the formulas for computing the correct resistor values when placing them in a circuit.

Again, thanks for the participation (& patience) as I worked through this one. Sorry it took so long to "latch on."

I gotta get that short between the headsets fixed...

 

As far as your resistor calculations for the LED:
Rlimit = (Vsupply - VfLED)/DesiredCurrent = (5v - 3.5v)/20mA = 1.5v/0.02A = 75 Ohms.
Since you're using an 82 Ohm resistor, your LED current would then be 1.5v/82 = 18.3mA (approx) instead of 20mA.
RlimitWatts = 1.5v * 0.02A = 0.03 Watts; double for reliability = 0.06 Watts. You could use even a 1/10 W resistor.

 

Hiya Sarge! Good to hear from you...

Of course you're correct but I used my measured Vs of 5.1V, so that's how I got the 82ohms (80 rounded up). I'm using a 1/4 watt resistor because that's what I had on hand, but I had also done the math as you've shown.

I don't understand why the web world advocates the use of those "hot" sandbars. One site even sells a breakout board with them attached in the middle! The ATX Design Guide specifies that "PS_ON#" to be an active low, TTL compatible signal so the motherboard can control it. So why not use TTL? If you're gonna tell folks to solder sandbars together or in the circuit, surely they're capable of handling a couple TTL components. lol, but I'm ranting.

I am gonna build that darn transistor switch just to prove to myself I can do it. But for now I'm putting LEDs in a 92mm PSU cooling fan. lol

Take care all!

 

Of course you're correct but I used my measured Vs of 5.1V, so that's how I got the 82ohms (80 rounded up). I'm using a 1/4 watt resistor because that's what I had on hand, but I had also done the math as you've shown.

I see - that's fine.

I don't understand why the web world advocates the use of those "hot" sandbars. One site even sells a breakout board with them attached in the middle! The ATX Design Guide specifies that "PS_ON#" to be an active low, TTL compatible signal so the motherboard can control it. So why not use TTL? If you're gonna tell folks to solder sandbars together or in the circuit, surely they're capable of handling a couple TTL components. lol, but I'm ranting.

Well, the thing is that the ATX form factor PSU's have a minimum load requirement for the +5V supply (usually around 2A) in order to ensure proper regulation of all of the supply voltages. If you measure the other outputs with less than the minimum load, you'll probably find that they are off considerably; mine was, until I met that minimum load requirement (particularly +12v/-12v).

I am gonna build that darn transistor switch just to prove to myself I can do it. But for now I'm putting LEDs in a 92mm PSU cooling fan.

OK. Connect 100 of those 3.5v 20mA LEDs each in series with a 75 Ohm resistor, all in parallel, and connect them to your +5v output. You'll then meet your 2A minimum loading requirement, and will also have PLENTY of light to work with.

 

I see - that's fine.


Well, the thing is that the ATX form factor PSU's have a minimum load requirement for the +5V supply (usually around 2A) in order to ensure proper regulation of all of the supply voltages. If you measure the other outputs with less than the minimum load, you'll probably find that they are off considerably; mine was, until I met that minimum load requirement (particularly +12v/-12v).

I've helped others that latched on regulated with a 750mA to 1A load, but lower end supplies need a large amount of draw to stabilize.

Above, it was stated "I don't see big resistors on the motherboard". Answer - All the ICs, resistors, and especially the onboard CPU power supply (ATX buck switching to provide 1-2.5V to the CPU hogs a lot), at least 5A when idle to stabilize the output, which is when the "power good" line goes high.

OK. Connect 100 of those 3.5v 20mA LEDs each in series with a 75 Ohm resistor, all in parallel, and connect them to your +5v output. You'll then meet your 2A minimum loading requirement, and will also have PLENTY of light to work with.

That would be a great use of dissipating the heat! Put the PS above your workbench on a light panel! Draw the power lines down to the bench in a wire loom then screw into a banana jack panel with Volt/Current meters for each. Might need a few hundred feet of 1/16" shrink wrap, but, meh, it's cheap.

 

The 10k does nothing has to effects on the circuit it is just a load to the source what for? asume .7v Vbe .7v diode drop asume 2.5v for the LED then 5v - all of it leaves 1.1v to drive the transx or 2.3ma if the transistor can come on with 2.3ma then it is gone blow up. other wise it may just slowly burn up.

 

I think he is trying to use the +5V output of a PC power supply as a lab supply, or in some non-PC project. The +5V output of a PC supply is typically not well-regulated (or worse) unless they have a large load on them. In his case, he has determined, or believes, that he needs a half amp load to stabilize the supply. A zener might work for low-current applications, but if he heeds it to work for loads from zero to 25 amps or so, then a zener is not going to help.

Ah I see. I had never considered not having a load on it would make such a difference.