# Powering sound board



## Indyandy (Sep 7, 2005)

My sound board uses 3 button cells for the 4.5v. I want to replace them with a 9v battery. What I don't remember, is what formula to use to get the right size of resistor to knock down the voltage. Can any one help me out? Thanks


----------



## The_Caretaker (Mar 6, 2007)

You can use this voltage divider calculator http://www.raltron.com/cust/tools/voltage_divider.asp


----------



## lostskeleton (Aug 30, 2011)

That voltage divider will work when nothing is connected to it but when you actually connect the board you add a resistance to R2 and this throws this entire thing out of wack and will not equal 4.5V.






The problem is the loading effect.

If we know how much current the board takes then we can get the equation spot on and have the voltage divider work correctly.

Honestly I would just get a battery holder that will hold three AA batteries as they are 1.5V. Much easier and no calculations required. If you are hell bent on 9v and a voltage divider you should find out what the current draw is for the board and we can figure out the correct values really quick. Let me know


----------



## lostskeleton (Aug 30, 2011)

Hey just out of curiosity, what are the part numbers for the button batteries you are using right now? We could take the current rating on those and replicate that for the voltage divider and get a number for your resistor values. Its too bad that board does run on 5V then we could just say get yourself a 7805 voltage regulator and you would be done


----------



## pshort (May 6, 2008)

Aargh!!! The only approach that makes any sense here is to use three AA (or AAA) batteries that lostskeleton mentioned in post #3. If you want to use the 9V battery, you need some sort of active voltage regulator, not a voltage divider.

The voltage divider approaches are NOT going to work well because the current used by the sound card is not constant, but goes up and down on an instantaneous basis. This will throw the voltage divider calculations out of whack and making the voltage on the sound board go up and down (i.e. totally out of regulation).


----------



## lostskeleton (Aug 30, 2011)

http://search.digikey.com/us/en/products/SBH-331AS/SBH-331AS-ND/275299

Here is a link to a battery pack that holds three AA batteries and has an on/off switch to boot!


----------



## ouizul1 (May 22, 2010)

lostskeleton said:


> http://search.digikey.com/us/en/products/SBH-331AS/SBH-331AS-ND/275299
> 
> Here is a link to a battery pack that holds three AA batteries and has an on/off switch to boot!


Radio Shack has them with on/off switches...but they're four battery jobs. Easy to convert to three batteries by putting a shorting wire in place of one of the batteries.


----------



## Indyandy (Sep 7, 2005)

I was hoping I could wrap this project up tonight. I only have a double aa battery holder or a 9v battery clip. My local radio shack has nothing (unless you want a phone). I am using a 50 second usb voice module. Calls for 3) AG10 batteries. Operating current < 50 ma. Standby current < 5 uA. Thanks.


----------



## lostskeleton (Aug 30, 2011)

I did the math on it and if it and I highly recommend you go the three battery route.
If you design the divider to operate at 50mA and it always is at 50mA then you would be just fine. But if you do the math on the circuit when it goes into standby mode you will quickly realize that all of the sudden the sound module that once had a nice 4.5V going accross it all of the sudden has 8.18V dumped into it and we don't really know if it can handle that. Thats just one reason it isn't a good idea.


----------



## asterix0 (Nov 5, 2008)

I suggest a 7805 voltage regulator. It costs $1.79 at Radio Shack (part 276-1770). I just used on to drop 12v to 5v for an MP3 player. Here's the data sheet:

http://www.national.com/ds/LM/LM340.pdf

It has 3 pins. The left one goes to the positive pole of your battery. The right one goes to the positive input of your circuit. The middle pin goes to the negative pole of your battery. When you then connect the negative pole of the battery to the negative input of your circuit, the circuit will receive a steady 5v. (I know you wanted 4.5 but the circuit probably wants 5. Three batteries are simply being used to approximate 5v).


----------



## lostskeleton (Aug 30, 2011)

Yeah if you want to give the 7805 a try that I mentioned earlier you could go for it. I have no idea what sound board you are using so I can't say it wont hose the board. Do you have information like the model number of the board or a link to it? It is totally possible that the circuit can operate in a range of voltages from something like 3.3 to 5V and then asterix0 would be totally correct. Let me know!


----------



## Indyandy (Sep 7, 2005)

50 second recording module:
Go down 3 rows and it is in the middle.


----------

