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Question about a 9v dv adaptor

matrixcs

matrixcs

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May 21, 2009
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hi guys,

I need your help. I have a 9V DC adaptor and here is the specs

Model: SPA-2000
INPUT: 100-240VAC
60/50Hz 15W

OUTPUT: 9VDC
Current: 2000mA Max

My question is, how can i convert it's current to a 100mA what component will I add because it's hart to find a 100mA 9vdc here in the philippines..

Any help??
 





first off what are you going to be powering with this wall wart ? most will put out a higher voltage untill you load them some. it could be possible to just use a single resistor to only alow the current to reach what you need.
 
im trying to power my guitar gadget it needs a 9vdc 100mA.. but i got a 2000mA one
 
ROTFL

No, wait, you're serious, right ?

Ok ..... it's current, not voltage ..... the voltage of your power supply is fixed, and if it say 9V, it's 9V, but the current that you take from it, depend from the resistance of your load (I=V/R) ..... this mean that your power supply is rated 2000mA cause it can give you until 2000mA, not that it push 2000mA in your device .....

So, as far as the voltage is the same (9V), you can safely use it for your device ..... it just adsorb from the power supply the current that it need, and no other ..... is the opposite situation that cannot work ..... if, as example, you had a device that need 1000mA, and a power supply that give only 200mA, you cannot use it, cause or the power supply don't give you enough current and its voltage decrease, or, if it don't have protections, it can become overheated until it burn .....



Edit: sorry for the laugh, i tought it was a joke question ;)
 
i think most electronic devices, at least good ones, have built in resistors and electronic components. as long as the volts are what are needed it should be fine. i don't picture an electronic guitar having no components inside.

lasers are different and need a driver because you're starting from scratch. (edit)but in your case, it would be more like if you bought a laser pointer, then got two batteries, and then asked how would u drop the max current of the battery to what i need. it's already been done because you bought an electronic item.

now if the guitar is home made and you're trying to put one together then that's different.. some people might be able to help you. i hope i have helped clarify things.
 
denton808 said:
lasers are different and need a driver because you're starting from scratch. .....
Click to expand...

denton, no, sorry .....

It's just matter of physics laws (Ohm law, in this case) .....

Let me try to explain, and sorry for the bad english ..... the voltage that a psu (power supply) push out, is a fixed value (except when you overload it and the voltage decrease, ofcourse), if it's 9V, it push out 9V ..... the current for which a psu is rated, is the maximum current that you can adsorb from the psu at this voltage, but it depend from the output voltage and the resistance of the load that you hook to the psu .

So, if your 9VDC adapter is rated, as example, 1A, you can safely use it for any load rated to work at 9VDC, and that can adsorb until 1A ..... if the load need 100mA, this mean that his internal resistance on the power line is 90 ohm, and with 90 ohm of load, you can adsorb only 100mA from the psu, no matter if it's rated 100mA, 500mA or 1A ..... same for any other current request form devices ..... if you build a guitar effect circuit, and your circuit absorption is , as example, 200mA, it adsorb just 200mA (except if it burn and go in short circuit, ofcourse, but this is a different case :p)

For the lasers, the things are a bit differents ..... we use for laser diodes, drivers that work as current limiters, cause for its structure, a laser diode work in current mode, not in voltage mode (same as the common leds, after all) ..... a laser diode have an intrinsic working voltage (called forward voltage, or FV), that is due to the dropout of the junction, and a power emission that is proportional to the current that pass through it when the supply voltage is more high than the FV, but the step in the curve is very "hard", and the internal resistance of the LD is very low (few ohms, and also it change when the temperature of the junction change), when it's in the working point, that is much more easy and convenient to use a current regulator, instead a voltage regulator .....

Old drivers was using voltage regulation, but they had a power feedback for the control (using the photodiode inside the laser diodes) ..... other than this, the circuit is a bit more complex, and the actual BR diodes don't have any photodiode inside, so the current control is still the better solution, for the LDs .....
 
Last edited:
It's simple... Your adaptor is voltage suply. That means that it gives 9V constantly. Current through adaptor depends on input resistence of your guitar gadget. This input resistence has to be more than 4.5 ohm cause with this resistence you get 2000mA current:

R=V/C=9V/2A=4.5ohm

I also play guitar and I can say I doubt that this guitar gadget has less than 4.5 ohm input resistence. So, you would be able to use your adaptor without any problem.
 
ok i dunno how you took my post, but i was trying to explain it to him using the idea of every day devices instead of all kinds of technical jargon.

simply put... i was trying to tell him powering a laser diode (i assume that's what he was comparing it to in his mind because he was worried about the current and how to regulate it.) is different from powering his guitar. when it comes to the laser diode he's powering the actual object that emits the light (from scratch); his guitar is more like a CD player where his plug powers a circuit that powers the laser in it and every other component. so the whole electronic component is built in and will regulate the current in his guitar.

i don't see how what i said was wrong that a laser from scratch needs a driver. the idea of a driver having a set working voltage to power the laser diode whether 2.25 over the forward of the diode or a 3-5v buck boost like the microflex, is the same idea of his guitars circuit. (the required voltage to run it in either situation would be stated in the device requirement, just like if u sold a laser with a linear driver you'd say 7.5 volts for BR and if it was a microflex it's 3-5v.(the labels on the device would say 7.5v and 3-5v respectively, just like how his guitar says 9v, so ohm's law isn't needed for him to feel safe that it won't fry his guitar.) i was trying to compare and contrast the two similar ideas so that he could see that a 9v plug rated at 2amps is safe.
 
His guitar gadget doesn't need current regulations. Laser diodes draw as much current as they can get, because they are a diode, not a resistor. Most electronics will only draw the current it needs, it wont draw the full 2000mA (2A) unless you have shorted it out somehow.
 
@ denton: then i have interpreted your post in the wrong way (happens, with different languages :D) ..... my mistake, sorry.
 
Post #4 is right on the money...

to clarify.....

1) your Guitar "gadget" requires/needs 9V DC @ 100ma (minimum
required) to function correctly.

2) your Adapter can supply 9V DC .... so we are good here..

3) you adapter can not supply MORE than 2000mA (the maximum
available) but can easily supply LESS.

Your adapter will work fine with your Guitar "gadget" if the data
you supplied is correct..:cool:


Jerry
 
well guys thanks for the answer but apparently i tried using the adaptor on my guitar gadget. It was a Behringer BDI-21 V-TONE (BASS) sadly i tried adjusting the switch for the polarity tried everything the adaptor is not working with my gadget. then when i use a normal 9v alkaline battery the gadget works perfectly.

when i contacted the behringer support they said it only works with a 100mA 9v dc adaptor. hmmm :(
 
No, this is not normal (or they are just trying to sell you their own psu at high price as "spare part", LOL)

But you need to check also the diameter of the power plug, especially the internal one, and the polarity ..... most of these guitar boxes, as example, have the positive at the external of the plug, not at the internal contact, as usual ..... and there are at least 3 different "pin" diameters, for these sockets, corresponding ofcourse at different diameters of the central hole in the plug, so, if the pin is too big, the plug just don't fit and you see it immediately, but instead if the hole in the plug is too big for the pin diameter of the socket, it still fit, but the inside don't touch the metal, and it can't work .....
 
agree w/HIMNL9. you know it'd really suck if they are the kind of people that make weird plug sizes that aren't common proportions and so you HAVE to buy theirs -_-

but yeah as he said.. u should check your pin and polarity before you buy theirs.
 
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