Using the Power (Supply) of your Computer

[jafo]

...to power effects. Can it be done? PC power supplies (and like all good geeks, I have an impressive heap of obsolete rigs) put out 12V and 5V of switched-mode goodness, which suggests a few possibilities. Similarly, USB puts out, what, 5 volts? It seems to me that you could power a 4049 quite handily with a cannibalized USB cable, especially if you're after something to record directly via audio interface.

[Perrow]

Problem is that switched supplies are often noisy in the audio spectrum of things. You might need to filter that power more than would make sense.

[DavenPaget]

Usually i would cut off a laptop PSU's plug because they all use proprietary plugs , and then use filtering if there is lotsa noise ( the Ratshack 4A choke comes to mind , or the single diode / dual diode filter could be both .  + some parallel capacitance ) followed by a 9v regulator if i need it  ( mine all comes in 18V , handy if you need 9V regulated in a cinch or you want a 18V pedal , just deadbug a 7809 ) .
I have like 3 laptop power supplies in my bin and those 3 prongs are noiseless types , or those 2 prongs , just hope they run at 3MHz

[R.G.]

... of switched-mode goodness

As Perrow notes, "switched-mode" and "goodness" may be contradictory.

It takes careful design and not a little repetitive trying to make a switched mode power supply quiet in the audio spectrum. The RF speeds of the switching can leak through a surprising array of filtering and wiring and get into audio circuits.

[iccaros]

to add, if you have an effect that is also using a SMPS, heterodyne can take place.. And that is bad..

[jafo]

Alas... once again, my imagination leaps ahead of my knowledge. Ah well; there seem to be enough cookbook circuits for 9v and 12v here until I understand them.

How about the other idea, drawing power from USB? IIUC, USB isn't regulated, so I'd imagine you could burn out a mobo.

[iccaros]

Alas... once again, my imagination leaps ahead of my knowledge. Ah well; there seem to be enough cookbook circuits for 9v and 12v here until I understand them.

How about the other idea, drawing power from USB? IIUC, USB isn't regulated, so I'd imagine you could burn out a mobo.

USB will self Limit current, and will shut the port down when you try to exceed it.. I have dead shorted USB, and the port was still good, well the plastic around it melted..

[jafo]

USB will self Limit current, and will shut the port down when you try to exceed it.. I have dead shorted USB, and the port was still good, well the plastic around it melted..

Heh, how's that for a case mod!

Anyhoo, this sounds promising. Since a 4049 only uses (well, handles up to) 10 ma, it oughta be safe to use USB to power it at 5v, right? Or would I have to limit the current going into (or out of?) the 4049? I know this is elementary electronics, but I've never been able to get my mind around power supplies. 

If I understand thing correctly, the PS puts out a certain current regardless of what is drawn, so I'll need to limit the current going in... IIRC, USB allows up to 100 ma, so 5v / 0.1a = 50 ohms? That can't be right, not when an LED uses 1-2k on 9v. Or should it be based on 10ma, 0.01a, which makes it 500 ohms?

[PRR]

> If I understand thing correctly, the PS puts out a certain current regardless of what is drawn

No.

USB puts out 5V. The load sucks however much it wants. CMOS will normally be a few mA. A dead-short starts to pull infinite current. In USB, either the mobo chip shuts-down or (in older PCs) a 500mA fuse pops (these often re-set when cool).

_I_ think PC power supplies are good for PCs and very little else. You can get batteries and small wall-warts everywhere.

[jafo]

> If I understand thing correctly, the PS puts out a certain current regardless of what is drawn

No.

USB puts out 5V. The load sucks however much it wants. CMOS will normally be a few mA. A dead-short starts to pull infinite current. In USB, either the mobo chip shuts-down or (in older PCs) a 500mA fuse pops (these often re-set when cool).

Then why have I fried so much gear when plugging in power supplies with the same voltages but more amps than called for? And isn't this why we put resistors between power sources and LEDs?* I don't doubt that you're right, but it's hard to reconcile my experiences with the theory.

*cue the jokes about Dark- and Smoke-Emitting Diodes and Light-Emitting Resistors/Capacitors.

_I_ think PC power supplies are good for PCs and very little else. You can get batteries and small wall-warts everywhere.

Alas, I'm better supplied in PC old hardware than in free cash right now.

[iccaros]

And isn't this why we put resistors between power sources and LEDs?

Nope.. a LED has a resistance, a red led that uses 20ma, and runs on 2v V=I*R  (2/.02 = 100) gives us 100ohms of resistance, since these are parameters that it is made to run under we can safely say this is its on resistance.  now if you push 9 volts across that same 100ohms (9/100 = .09) 90ma amps it would draw.. So we are allowing the LED to run away with current.

If we drop the voltage to 2 volts, so we have to lose so we divide 7 volts by the known current draw of the led, 7/.02 = 350 ohm resistor is what you would need.  We limited the amount of current of a led by controlling voltage as resistance is fixed. it had nothing to do with power supply

Edit,
I am trying to explain at a simple level, so I understand that an led does not have 100ohm resistance.. and that @ 20ma most LED will burn out.. Just trying to break things into simple chunks..

[DavenPaget]

Beware , 20ma can be a quite huge amount for a modern LED , my clear blue LED's make me feel like the back of my eye is burning ... 

[jafo]

And isn't this why we put resistors between power sources and LEDs?

Nope.. a LED has a resistance, a red led that uses 20ma, and runs on 2v V=I*R  (2/.02 = 100) gives us 100ohms of resistance, since these are parameters that it is made to run under we can safely say this is its on resistance.  now if you push 9 volts across that same 100ohms (9/100 = .09) 90ma amps it would draw.. So we are allowing the LED to run away with current.

Well, that's kinda the question -- is the power supply pushing or is the device drawing? My intuition (which, alas, I've amply demonstrated is comically poor when it comes to electronics) is that it's the first. I've destroyed devices by using a higher-amp power supply, devices are rated for how much current they can handle, and lightning strikes are an extreme example of this sort of pushing.

Gah... but then again, it can also be explained by drawing -- devices want to grab as much juice as possible. Still, that wouldn't explain why greedy devices don't burn out power supplies (at least, not as often as beefy supplies overwhelm whimpy devices)... which is it? And why can't hardware be as simple and sensible as software?

[DavenPaget]

Power supplies push and devices draw , if the Rds_on is high .
There's always a balance , if the Rds_on is low , power supplies push and devices stop drawing , they blow if the I_dss capacity is too low .

[ashcat_lt]

For our purposes I think it's safe to say that the devices draw the current they need.  Since none of us was there when you burned out your devices it's hard to say for sure what happened.  Could be as simple as wrong polarity.

OTOH, unregulated PSUs only run their stated voltage when the device draws the stated current.  It'll put out a higher voltage with a lighter load.  If there was a serious mismatch, you could have just exceeded the voltage capacity of some components.  I've perpetrated some pretty big mismatches without hurting anything, though.

I always thought that this was a max current rating (thus the mismatches I've made) but apparently it's not.  If you pull a bit more current than what it says the voltage will probably sag before anything burns, but eventually if you go far enough it will burn out something.

[defaced]

Well, that's kinda the question -- is the power supply pushing or is the device drawing? My intuition (which, alas, I've amply demonstrated is comically poor when it comes to electronics) is that it's the first. I've destroyed devices by using a higher-amp power supply, devices are rated for how much current they can handle, and lightning strikes are an extreme example of this sort of pushing.

Gah... but then again, it can also be explained by drawing -- devices want to grab as much juice as possible. Still, that wouldn't explain why greedy devices don't burn out power supplies (at least, not as often as beefy supplies overwhelm whimpy devices)... which is it? And why can't hardware be as simple and sensible as software?

Voltage is "push"; it is called the electromotive force. Current is "flow".  The devices either allow the current to flow or don't based on their resistance.  That's why it's defined as "resistance" and not "allowance" or "drawance" or some other made up word.

Now philosophically speaking, if you want to say that the devices "draw" current based on that explanation or not is irrelevant, because in reality it doesn't make a difference.  V = I*R doesn't care about "push" or "draw" , all it cares about is force, flow, and resistance.  Because of this, we tend to get a little sloppy with our terminology.

Hardware is actually pretty simple once you get over the initial hurdles of fundamental concepts and vocabulary. 

I've destroyed devices by using a higher-amp power supply

Something else bit you in the ass.  It all goes back to Ohm's law: V = I*R.  I being what the load is allowing to flow, not what the power supply can deliver. 

[jafo]

For our purposes I think it's safe to say that the devices draw the current they need.  Since none of us was there when you burned out your devices it's hard to say for sure what happened.  Could be as simple as wrong polarity.

OTOH, unregulated PSUs only run their stated voltage when the device draws the stated current.  It'll put out a higher voltage with a lighter load.  If there was a serious mismatch, you could have just exceeded the voltage capacity of some components.  I've perpetrated some pretty big mismatches without hurting anything, though.

I always thought that this was a max current rating (thus the mismatches I've made) but apparently it's not.  If you pull a bit more current than what it says the voltage will probably sag before anything burns, but eventually if you go far enough it will burn out something.

Well, I definitely had the right polarity, and a higher wattage, on the principle that more watts is better in speakers and PC power supplies. I'm almost certain that I used a supply that was larger in both volts and amps, reasoning that the device would draw what it needed.

Still trying to reason my way through this... if only hardware made sense!

According to Judging from response #19 in this thread on powering Boss pedals, a power supply that puts out too many volts will fry a device, which seems logical. Said thread also states that you can supply more amps than needed, which ultimately explains why electrical outlets exist. What if a device tries to draw more volts or current than the supply can deliver? Nothing, or a fried supply, or possibly both depending on the disparity?

When a device says it can handle, say 10 ma (perhaps a 4049), does that mean that the supply is pushing, or that the device will take all it can get? It seems to be both, which doesn't make sense... or does it? The more I try to think in terms of pushing and pulling, the more muddled it all gets. Should I instead be thinking in terms of volts, amps, ohms, and watts for everything; and whichever device gets too much of any of these gets quite literally zapped? I think I should, but my intuition about hardware is so out of touch that I can't tell if this is a help or a hindrance.

[waltk]

Didn't read the whole thread, but here's my go at a simple explanation...

A regulated power supply - say 9 volts - supplies as much current as required by the device you connect it to - up to the power supply's rated current.  So the device draw whatever amount of current it needs - at 9 Volts in this case.

If the device doesn't limit the amount of current it uses - it likely burns up - as in an LED hooked up to a power supply.  That's why you put a current-limiting resistor in series with an LED.

If device draws a limited amout of current - but it's over the rated current that the power supply can provide, the power supply will get hot, or burn up, or (most likely) will pop a fuse.

[waltk]

To more directly answer the original question,  I've recycled desktop power supplies to power effects and amps without any problem.  Even though these are switched-mode power supplies, they are VERY well regulated because computers are kind of picky about getting the right voltages.  Motherboards also do have their own regulators, but the power supplies themselves are fairly quiet.

Most of the little wires that come out of a desktop power supply are either ground, or 12 volts, or 5 volts (and some have the opposite polarity -12v, and -5V).  There's a wire with a "power good" signal, and another one that must be connected to a load for the power supply to start up.

Check out this link: http://web2.murraystate.edu/andy.batts/ps/PowerSupply.htm
for one of the many examples of how to convert a desktop PC power supply for benchtop use.

My favorite use of desktop PC power supplies is to power a car audio amplifier (up to multi-hundred watts).  The best power supplies for this are the ones with a single 12 volt rail (as opposed to the more modern multi 12v rail ones).

[defaced]

Should I instead be thinking in terms of volts, amps, ohms, and watts for everything; and whichever device gets too much of any of these gets quite literally zapped?

We have a winner.  Devices will usually have maximum ratings on most of those.  This is the MFGs way of saying "if you violate any of these, we can't guarantee operation".  It doesn't mean that if you have a 10v part that at 10.000001v it will die, it just means that the part is designed to take up to 10v and will work, but can not guaranteed operation above that.  Same goes with other limits too.  Depending on the device, it may be more or less sensitive to overages, but first you need to know the rules before you can decide which ones to bend/ignore.

What if a device tries to draw more volts or current than the supply can deliver?

A device does not draw voltage (within the context of a load device, source devices like batteries/alternators/generators create voltage from chemical reactions or physical motion).  Voltage is a supplied across a device, and based on the device's internal resistance and other resistances in series with it, current flows.  If voltage is too great, you can force current to flow when it shouldn't.  This is what kills parts when they are run over voltage, electrostatic discharge, lightning strikes, etc.  Similarly, if resistance is too low, and too much current flows, you can over heat a part and cause it to fail.  Balance is the name of the game, which goes back to the first part of this post: obey all the limits of all of the devices in the circuit.