9 volt power supply? mA?

[eadgber]

I just got a cheap Behringer TO800 pedal and it takes a 9V 100mA adapter.I looked at some adapters I saved and I do have a 9V 150mA. Do you think it would be ok to use it? 50mA over,but I'm just not sure if it really matters or not.

I have not tried it because I don't want to damage it. I wonder if it would affect it's tone too.

Thanks guys. I know it's a probably a dumb question but that's the type I usually ask.

btw. the jetking 21 mod I have from the guys on this forum is still the best gtr I have ever played!

dirtypool71 a t yahoo d o t com

[newey]

Don't take my word for this, as I'm just learning this stuff myself. Let someone else confirm this before you go breaking out the soldering iron.

But a resistor of the appropriate value, wired in series with the hot line of the wallwart, should step the current down to the appropriate value.

This assumes you can sacrifice the wallwart (and also assumes you can wire to the flimsy little wires). If the wallwart is needed for other things, an adapter cable (female jack to male plug, with the resistor in between, and paying attention to polarities) could do the trick.

Ohm's Law gives the calculation here.

By my figuring, a 30Ω resistor should do the trick (9V / .15A = 60Ω- that's what you've got now. 9V / .10A = 90Ω- what you need to get 100mA, difference is 30Ω)

HOWEVER:

1) I'm not at all sure I figured that right, so let others weigh in with correction if need be.

2) 9V is a nominal rating, so there may be a "fudge factor" to be applied.

3) You also need to be sure the resistor can handle the power (i.e. wattage).

Also, before you do anything, check to see that the polarity of the wallwart is the same as the jack on the pedal. These do vary, some are negative to the center, others are positive to the center. Both should be marked with a little semicircular symbol to show the correct polarity.

Now, you also asked whether you could just use the wallwart as is. That may be possible, but we'd need to see the circuit in question to be sure. Tone isn't the problem, possible frying of components is.

[sumgai]

Well, I guess you can teach an old dog new tricks! Nice effort newey, very nice.

However, there's one little teensy detail you left out, probably due to your learning curve not yet having encountered this particular tidbit. So here it is....


WARNING:   Professorial hat has been donned! Skipping this may be beneficial to your mental health - see Summary below (in red).

In general, and that means in all cases where electrons are being abused, a device asks for only as much current as it needs, and a supply delivers only as much current as has been asked of it. Knowing this, it's easy to see where you got out of whack - my answer on the LED thread where I stated that an LED must have an external resistor to limit the current flow. Well, that's the same thing, isn't it? - you ask. Yes and no. (An answer almost as lovable as a lawyer!)

In an LED, there is effectively no resistance to a supply, so the supply will assume that because there is a complete connection, it must supply power (both voltage and current). This is the exact description of a short circuit, I'm sure you'll agree. So it follows that if there's no resistance, then there's no Ohm's Law to get in the way, and thus we have a "full speed ahead" condition.... and that's when you get a "fuse condition", when the LED pops from having too much current run through it.

After the environmental damage has been cleaned up, you use an external resistor on your next LED, and as stated, it's sole purpose is to limit the current flowing through the rest of the circuit. (Kirchhoff's Law, part 1: The amount of current flowing through any one part of a series circuit will be the same as the total current flowing through that entire series circuit. (Mass quantities of Brownie points for anyone who comes up with Kirchhoff's other law.))

OK, but where does that comport with e-to-er's request? Simple - the device in question that's going to request current has some amount of resistance already built into it - no external resistor needed!

So what's next.... oh, we need to answer the original question. Now comes the fun part. Here, hold my beer and y'all watch this!

The last factoid we need in our arsenal is one that relies on Mr. Kirchhoff (as stated above). Since the device in question is requesting 100mA of current, that's all that's going to flow within the circuit. (Said circuit consisting of supply -> device and back again.) At this point, we could say that the circuit is self-limited or self-regulated to consuming only 100mA of current. But the supply is putting out 150mA! you exclaim. Oh? Really? What part of the first paragraph after my Warning escaped your attention?

For those who came in late (and spilled their popcorn all over the guy sitting next to them, 'cause it's dark in here), a device requests only as much power as it needs. Write that down on a sticky and paste it to your monitor bezel. Don't remove it until you're sure you can pass the test that's coming up at the end of the quarter!

To put the horse in front of the cart, we can say that a supply can put out as much power as a device wants, up to whatever it's rated to deliver at maximum demand (or politely speaking, at maximum request). Thus, a supply that can deliver some X Amps of current doesn't ram that full amount down the throat of a requesting device, it only delivers what's been asked of it. To be sure, a power supply is quite happy to do only a partial job - after all, the check's the same at the end of the week! (And we see it pay off as less heat being generated, which often translates to longer life.)

Summary answer:
A power supply needs to be capable of delivering everything requested of it - that's the only minimum requirement. All other capacity to deliver more current than requested is not going to be "forced out the door", so to speak.

Higher capacity than necessary to fulfill a request will never harm the requesting device. (That's assuming that the requesting device is able to self-limit its demands. If not, then "help" in the way of an additional component may be needed. An example would be an LED with an external resistor. At that point, we're essentially saying that the requesting device is a combination of the LED and the resistor.)

</lecture>


HTH





sumgai

[JohnH]

yes to what sumgai says. According to this, it only draws 30mA:
www.behringer.com/EN/downloads/pdf/TO800_P0488_M_EN.pdf

You can plug it into any 9V supply with the right connection, but I would only use one intended for guitar effects, ie properly smoothed and regulated

[sumgai]

Jul 4, 2010 15:53:49 GMT -5 JohnH said:

......
You can plug it into any 9V supply with the right connection, but I would only use one intended for guitar effects, ie properly smoothed and regulated

Also, not only are some wall-warts "dirty" compared to others, but some of them are not DC, they're AC! I have several such from Digitech and Boss, I'm sure there are others. Yes they're marked, but it might as well be in Braille for all the good that does, particularly when you're in a panic - that's when it's so easy to pick up the wrong "spare wart". The results would range from "it still doesn't work" to "uh oh, what's that smoke?"......

I color-coded all my warts with a small sticky dot, to avoid this very possibility.

HTH



sumgai

[cynical1]

For the Brownie Points...

Kirchhoff's law of thermal radiation

HTC1

[sumgai]

c1,

Ya know, I was expecting something a little more electronics oriented.....

And in fact you pointed to a page of just one of Herr Kirchhoff's Laws - the disambiguity page lists several more of 'em, including the one I wanted.

But what the hey, just the researching effort is worth a +1.

In point of fact, the Wiki page closer to what I was suggesting would be: en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws, but even thereon, I see what I'd consider a travesty of what the Laws are supposed to state. Let me elucidate:

The first rule implied thereon, the so-called KCL that says "the sum of currents flowing into that node is equal to the sum of currents flowing out of that node" is really a concocted contraction of what the laws should say (and what they did say, when I went to school).

1) In any series circuit, the current flowing through a single component is the same as the current flowing through all the components of that circuit.

2) In any parallel circuitry, the total current flow from the source is divided proportionally across each branch.

Let me just say that without an understanding of these two principles, it'd be damned difficult to grasp the usage of Norton's and Thevenin's Theorems, a pair of most useful methods for analyzing voltage drops (Thevenin) and current flow (Norton) through any circuit, no matter how complicated. To simplify everything down to "it's all the same, in or out" does an injustice to a very clever man who lived well before his time.
</rant>


BTW, the KVL that says "the sum of the electrical potential differences (voltage) around any closed circuit must be zero" is just a restatement of the KCL, using voltage instead of current. (Sort of like using both integral and differential calculus to get the same results....) This renders it worthless, in light of what I just said, above. I kinda hate to do that people who contribute to the WIKI project, but sometimes an Engineer's gotta do what an Engineer's gotta do.




sumgai

[eadgber]

Ok Thanks. I guess I'll wait and see how long the battery last in it first. I don't like having to unplug the input to turn it off though.I was thinking it would be nicer if I had an adapter I could plug into my power strip that has an off switch. I do that with some other pedals that have no on/off.

Jul 4, 2010 15:53:49 GMT -5 JohnH said:

You can plug it into any 9V supply with the right connection, but I would only use one intended for guitar effects, ie properly smoothed and regulated

I'll probably just end up buying the one made for it.

Anyway I searched the model# of the one I was thinking of using and this looks like it, except mine says Procter & Gamble Co. not Royal.
azsurplus.com/index.php?main_page=product_info&products_id=602&zenid=5afpgoel4v4b8hrecllkiunq16

[ashcatlt]

Don't forget Polarity!

This is where you'll end up destroying pedals. Most of the pedals out there (behringer included) use the Boss style power supplies with positive voltage on the barrel and negative inside. This should be indicated on the bottom of the pedal, or near the power jack as well as somewhere on the PSU. It may be a somewhat cryptic set of symbols. Make sure they match! If you're not sure, get some good (focused) pics and we'll try to sort it out.

Edit - the one you linked to is wrong! If yours has the same polarity, don't use it!

[sumgai]

e-to-e'r,

That's twice you've been notified about polarity, so John and I haven't tripled or quadrupled it.

If you already have the supply, and really want to use it, then you can build a polarity reversing doohickey (technical term, can't be easily translated). Just buy a plug of the same diameter, a jack (ditto), and a few inches of wire, if you don't already have them on hand. Solder the wires to each piece, but remember to connect the two sides opposite of each other - center goes to shell, and vice-versa.

The important thing here is that you can't let the two pieces short together, while things are plugged in. I recommend shrink-tubing (the kind that you blow a hair dryer over, to shrink it up tight) to keep things safely insulated.

If you don't already have it, then do as the nice ash recommends, and get one that's already wired correctly for your new toy. And if you have other toys and warts and such, make sure that either they all have the exact same polarity and voltage, or else start color-coding them so you don't blow things up. (Making sure that you're also meeting the minimum current requirements.)

Either that or buy one of those uber-expensive all-in-one power supplies that have a number of output connectors. The better ones can handle both polarities.


Oh, and one other thing, just for drill......

ash notes that the device in question requires a supply of 30mA, whereas you first stated that it needed 100mA. I see the problem here - what you read is that it requires a Behringer power supply, model number such-and-such (according to one's country). Behringer doesn't have supplies that go that low, their minimum capacity supply is the aforementioned 100mA unit. Which is overkill, but again, not harmful or anything. If you should happen to find a 50mA supply of the correct voltage and polarity, then it would work just as well. (Errr, providing that it's not a lump of crud......)

HTH




sumgai

[ashcatlt]

I missed newey's comment re: polarity, but I guess it bears repeating.

You don't actually need a jack and plug to change the polarity, if you're willing to make a permanent change to this thing. Just snip and flip, solder and tape (or shrink tube) and you're on your way.

I worry, though, that it doesn't really look like this thing is meant for audio applications. It isn't likely filtered particularly well, and the cable is likely not shield. All that could cause some serious noise issues.

All of this can be fixed, too, but starts to get more complex. Where are those pictures of my tuna can?

[sumgai]

ash,

Yes, re-working the output cable itself works, but what if the supply fails? Have to do the job again, eh?

The doohickey survives such failures, not to mention that it'll let you try other supplies that are also of incorrect polarity, should the need (or opportunity) arise. Plus, it also works for any other supply/device combination that needs to have the polarity flipped. Build a couple of them for spares, and you've covered not only yourself but your bandmates too.



sumgai

[chuck]

that is a great idea Mr Gai ... handy lil gizmos to have around.

that same idea would work for combining speaker cabs wouldnt it ?

as in two 2x12s that you want to wire like one 4x12 .... series / parallel

i know that is a subject for another thread , and if the theory is sound i will start one for further discussion

[sumgai]

chuck,

Have at it!

I'm sure you can find the correct Forum to start that discussion.


sumgai

[ashcatlt]

Yeah, sorry, I still kind of have that "need it now" syndrome where I can't sometimes be bothered to wait (or pay) for parts when I need a solution like this.

If one was going far enough to procure jacks and plugs and whatnot, it wouldn't take too much more to add a BAC across the thing to bolster the filtering action, then use a shielded cable after the cap to help reduce power supply noise even further.

[eadgber]

Good thing I looked close! the tip/sleeve of that 9v was backward of what I needed. Dudes it was marked on bottom of the t0800 and on the adapter I was asking about! I'm glad I asked here first! I'm a big ol' dumby haha!

See this is why I ask here before I do dumb crap. lol

[jcgss77]

Would a 7812 or 7805 power regulator help here, with filter cap, or will those not clean up the noise?

[ashcatlt]

I may have a higher tolerance for noise than some people, but I've never needed much more than a couple of caps. Most pedals that come with wallwart jacks will have at least som e PS filtering, but more is always better, right?

These pedals also usually have "polarity protection" in the form of a diode across the power jack oriented in such a way that it only conducts when presented the wrong polarity. This shorts the power before it can get through to more sensitive components. Of course, it works exactly long enough for the diode to explode! Then you have to hope it destroys at least one of the PS traces, or else other things will start popping real quick.