Question about converting a 9v Battery Gadget to AA's.

Hi. I have a Boss DS-2 distortion pedal, but my problem is, it drains the 9 volt batteries like a semi-trailer. So I basically wanted to convert it to AA batteries. Technically speaking, I'll need 6 AA's to make 9volts, but 9 volt batteries have a different current don't they, or will that not matter in this application. I'd also like to insert an on/off switch because it drives me crazy that the only way to remove it, is by pulling out the 'input' cord (guitar to pedal). Please respond if you know the answer.

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gmoon10 years ago
FYI--Current draw is a characteristic of the device itself.

The effect pedal will draw the current it needs, regardless of how much the current the source can supply. You can only force more current through a device if you supply a higher voltage than the device's rating.

You could connect the pedal to a supply that can source 100 amps--doesn't matter, if the voltage is correct.

So sure, you can wire the AAs together, and yes--they will last a lot longer.

guyfrom7up gmoon10 years ago
yeah, for example if you took a 1 farad capacitor at 12 volts. And short itnitll create a huge spark. But if you touch both electrodes you wont get shocked because your body has too much resistance for 12 Volta to pass a noticable amount of current.
another example: i have a 12 volt bettery (7ah) put you fingers on the leads, and nothing happens. but when my pliers got too close.... btw, guyfrom7up, what are you planning to do with that super cap? make really really really low ripple dc?
well, I currently don't have a 1 farad cap (I plan on buying a 50 farad, 2.5 volt one soon, they're about 5 or 6 bucks). I do have 4x 33,000 uf, 100 volt capacitors. I'm colaborating with kruser to make a spot welder using some of them, and I'm using one of them for my powersupply for my cnc machine.
cool, but isnt there already a welder on the site? plus, where did you buy the 100 volt ones? i could make a good rail gun with 4 of those in series.
I was just googling around and I found my way back to this. Ebay, lol
gmoon guyfrom7up10 years ago
That's a decent analogy, since larger capacitors are potentially a very high-amperage current source (for a very short time.) But you can still release that current slowly through a resistive load.

And a charged high-voltage cap (say 350V) larger than 10 or 20uF will shock you--it's not storing as much current as the megacap, but it's higher voltage is driving that current.

Another way of thinking: A 5V microcontroller may draw 3mA; and a 5V relay may draw 70mA. But both can be connected to the same 5V supply--the amount of current they draw is "built in" (inherent) to each device.

But connect each to 10V and you let the "magic smoke" out--the high voltage has forced more current through than the devices can handle...(by it's nature, a relay may survive longer, of course.)

There is the rare fixed-current power supply; but don't worry about that if you're replacing batteries....
tech-king gmoon10 years ago
not only resistive; an inductive load could also act as ballast. and, you don't even need 1 farad. approx 2000 uf, at twelve volts, can actually melt the tip of a screwdriver placed across the leads.
gmoon tech-king10 years ago
Sure, any load will do (although you're describing how induction can store current, similar to a cap.) It's just easier to pretend for the moment that loads are resistive only.

If you assume that, then you can plug the values into Ohm's law--which illustrates why increasing the voltage forces more current through:

I = E/R (increase voltage (E) and current (I) increases also.)

It also explains why supply voltage falls if the load draws more current than the source can supply...

E = I * R ( if current (I) can't rise because the supply is inadequate, then voltage (E) has to drop in order to satisfy the law...)
Noodle93 (author)  gmoon10 years ago
Thank you. I don't really know much about the current.
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