Can I power a 6v circuit from a 9v battery?

[mattpocket]

Can I power a 6v circuit from a 9v battery if I put a resistor in the way before the power goes into the circuit? Is there a simple way of doing this? Also, if this can be done, would it also be safe to use a regulated 9v dc adaptor.

Matt

[d95err]

Putting a large resistor in series with the power supply will limit the current available to the circuit. This could cause a problems, depending on the circuit.

One solution is to use a 6V voltage regulator (7806). You can also create a simple voltage limiter using a 6V Zener diode.

[Gus]

you could use a few 1n400x diodes in series to drop the voltage in the +9 line.

[mattpocket]

The voltage regulator or zener diode sounds the best so far... how do I implement these? Do I just solder them in series with the power line before it hits the circuit? Do I need any other components to implement the voltage regulator?

Matt

[Gus]

Why does the voltage reg or zener sound the best?  You posted simple.

  The simple .7v or so drop of a 1n400x is in series with the load.  The zener(needs math to do right) or reg has a shunt (added current drain for both).  I assumed it was an effect.
  You did not give enought information for a good answer, does it need to be regulated, what is the current draw....

[d95err]

I agree with Gus that the diodes in series is the easiest solution (why didn't I think of that?). 3-5 diodes in series should get you near the voltage you want, depending on the type of diode used.

Since you plan to use a battery, a voltage regulator is probably overkill since the battery does not have any AC ripple. If you sant to use a badly regulated wall-wart, you may want to use a regulator though.

[mattpocket]

Its for a speak and spell. I prefer using 9v batteries and I would also like to use a regulated pedal power supply to power it.

I really dont know much about power supplies, so maybe the 1n400x is the route to go down.

I am afraid I dont know the current draw because I dont have the speak and spell (its in the mail) and I wasnt planning on buying the size batteries (I think they are C or D) that the speak is originally powered with. I was just hoping for a simple solution so I could go right ahead and use a 9v battery.

Matt

[MikeH]

Why couldn't you just put in a trimpot and make a simple voltage divider?  Send 6 volts into the circuit and dump the rest to ground.

[mattpocket]

Sounds good Mike

[ulysses]

check the circuit for devices that use the power and check the datasheets to see if they will fry under 9v.

they may offer different sounds with 9 volts.

ive done circuit bends with voltage limitation before.. my daughters nersery rhyme machine starts to sound like something out of a freddy kruger movie when the batteries start to get low

cheers
ulysses

[mattpocket]

I will give it a whirl and see what I get... lets just hope I dont fry it! haha

Matt

[remmelt]

Why couldn't you just put in a trimpot and make a simple voltage divider?  Send 6 volts into the circuit and dump the rest to ground.

Wouldn't that drain the battery really quickly?

[Gus]

Unless the pot is a wirewound of a good power rating the wiper and other parts of the trim will burn, any current not going to the circuit is a waste.

[mattpocket]

So what should I do? haha

[sfr]

http://en.wikipedia.org/wiki/Voltage_divider

All the maths you need for calculating the resistances needed for a voltage divider is there.  So you don't have to use a pot., if you were going to use a voltage divider.  As far as burning up pots, since we know the voltage and resistance, couldn't we calculate the current and then wattage, and choose a trimpot of appropriate power rating for the application? 

As far as eating up the battery faster, if we just chose higher rated resistances for the voltage divider, wouldn't that lessen the current used and extend the battery life somewhat?

[mdh]

Why exactly do you want to do this? If the thing really is designed to be run from C or D cells and you run it from a 9V battery, then at the very least, you're not going to get very good battery life. The 9V battery will last about 1/10th the time of the 4 C cells necessary to produce 6V (http://en.wikipedia.org/wiki/Battery_%28electricity%29; see the table of battery sizes and capacities about 1/3 of the way down the page: 500mAH @ 9V versus 6000 mAh @ 4*(1.5) = 6V). I haven't checked battery prices lately, but I don't think that 4 C cells cost as much as 10 9V batteries.

Now, if you want to DIY a wall-powered supply to run the thing, then you should probably just build a little regulated 6V supply. This could be as simple as a 9V wall wart, a 7806 regulator (as others have mentioned), and a few caps. Take a look at the data sheet for the 78xx series voltage regulators, and/or search GEO for "power supply basics".  If you can't easily get a 7806 regulator, then an LM317 would also work. You just have to look at the data sheet or (guess what: GEO!) to understand how to set the output voltage with resistors. The size of the caps necessary depends on the quality of the DC coming in from the wall wart. If there isn't much ripple (e.g., it's coming from a nice, clean supply), then you won't need any big caps. If it's just your basic unregulated wall wart, though, you might need a 1000uF or larger cap to smooth out the ripple.

[petemoore]

  I guess theory that 6v PS rating is good because the circuit works on that, and they can use 10v caps safely.
 But maybe there's an opamp in there thats rated for LV, I've never seen opamps that can't stand 9v, but they may exist...or an opamp or circuit that is tuned to work a certain way @6v.
 Looking up all the parts voltage ratings and then applying higher voltage...you'll probablynot cause damage.
 An lm317 does a nice job, pretty easy to get going, and may have other uses..

[mattpocket]

Thanks guys...

Looks like running the circuit from a 9v cell is out fo the question, but I would still like to be able to power it from a PS.

Looks like the regulator is the route to go down in the case of a powersupply.

Matt

[mattpocket]

How would I implement the regulator? Could I use the GGG power supply layout in the gallery and just swap the regulator and the input?

[mdh]

Could you link to the layout you refer to? It's easier that way, b/c then we don't have to try to hunt it down. I'm not sure what you mean by "swap the regulator and the input." But if the circuit you refer to uses a 78xx voltage regulator (7809, most likely), then you can just swap a 7806 in place of the 7809. If by "swap... the input" you mean change the input voltage, you don't really need to do that anyway, though you could. Basically, if the DC that the regulator sees at its input is high enough (I think it needs about 2V above the desired output voltage) and isn't ridiculously high, you're fine. So if you have a clean supply that runs at 9V under the load you would expect from the device, that would be a sufficient supply to the regulator. That said, if you only happen to have a 12V supply, you're not likely to fry anything with that. Seriously, though, look at the data sheet, it should answer all of your questions.