Convert 9V to 10 separate 4.5V

Hi I am a newbie of sorts, still not fully grasping the electrical principals, but getting there. Basically I have 10 sound recorder circuits that run off 4.5V (3 1.5V batteries) each, I'm using them in a project and would rather have them run off one battery rather than maintain 30 batteries. How can I divide a single 9V battery to 10 4.5V circuits. I realize the power won't last as long but that's not really the issue. Any help or links would be greatly appreciated.

Thanks,
Dan

Just a note, I guess it doesn't have to be a 9V battery it could be any set of batteries that is more maintainable, or a power supply. The main thing I don't understand is how to divide the power between multiple autonomous circuits and the math to calculate that.

The easiest approach would be a voltage divider but its not very efficient.
http://en.wikipedia.org/wiki/Voltage_divider

A little 9V battery can supply hardly any current.
If the circuits draw exactly the same tiny current continuously then 5 circuits in parallel can be connected in series with another 5 circuits in parallel and can be powered from a 9V battery.
It is extremely simple but if you don't understand I won't sketch it for you.

So the 5 circuits in series parallel to another 5 circuits will act as a voltage divider, taking the 9V down to 4.5V, but as you mentioned, the battery probably won't be able to supply enough current for all of the circuits. I think, however, that a 9V adapter might be able to. I am going to see at home how much current each circuit draws and try to figure it out. Thanks for the help on that divider though, that makes sense.

Hi Dan,
Do your recorders have aux power in jacks? If so, what's the voltage & power (or current) required?

What's the manufacturer and model of your recorders?

You say they operate using three 1.5v batteries - what size batteries?

They don't have an aux power in, they are just little key chain things that have a circuit for recording 15 seconds of audio and playing it back. I was just gonna take the two wires from the batteries and hook in there. The batteries are the little silver discs, watch-type batteries.

Unfortunately the packaging says nothing in the way of how much current is required to operate one of these so I am going to have to measure that somehow. Here's what it looks like: http://www.dealextreme.com/details.dx/sku.2878

Oh, OK. I'd be surprised if they drew more than 80mA current.

Why don't you try hooking just one up with, say, three 1.5v "C" or "D" cells in series with a DMM (Digital Multimeter) set to the 200mA range, and trying out the various functions to get a good idea of what the power requirements actually are?

If you don't have a DMM, Harbor Freight sells a couple of models really cheap when they're on sale - under $4 - and they're surprisingly accurate for the price. Or, you could pick up an inexpensive analog multimeter from Radio Shack (around $10 I think), which might actually be better for this exercise, as rapidly scrolling numbers on DMMs can be hard to read when the load is varying, where you can use an old-fashioned analog meter to get a pretty good guesstimate of what the average and peak readings are.

I'm doubtful that running two in series across 9v will work very well, now that I've seen what they are. It's likely that one of a series pair will get "fried" due to overvoltage. But running 10 of them wired in parallel on three alkaline "C" or "D" cells wired in series should last for quite a while. Note that you should wire a small capacitor (say, 10uF) across the + and - battery connections at the recorders along with the remote battery wires - that will minimize the effect that other recorders might have on the supply.

Battery cells for watches supply hardly any current.
Use three AAA alkaline cells instead to power 10 recorders for a long time.

Thanks I'll try AAA alkaline next. Referring to Sgt Wookie's post I'm trying to measure the current that the circuit draws but I don't think I'm doing it correctly or something as the circuit gets tweaked out. Basically I have the multimeter set to 200mA and I'm connecting the positive lead to the positive and the negative to the negative coming out of the battery pack. I get numbers around 180mA when it's recording and 50mA when it's playing, but I don't know if these numbers are correct since I'm not getting correct functionality on the circuit. Is there a better way to test the current? I've tried moving the common lead around to different points and it doesn't seem to affect the fact that the circuit gets messed up (audio record and play back are squelchy).

I'm genuinely surprised the current draw is that high.

Are you handy with a soldering iron?

If so, connect a ceramic or tantalum capacitor of about 0.1uF across the recorder's battery terminals. A 30 watt iron should work pretty well. Might be hard to solder to, as many terminals are nickel plated. You might need more capacitance than that.

The recorders' circuits are digital, and as such are very electrically "noisy". The combination of the length of your external battery leads plus the length of the meter's leads makes for an inductor large enough that the voltage level at the battery connectors is pretty unstable. The engineers who designed the recorder planned on the batteries being inside the recorder itself. You need to compensate for the external wiring by using a "bypass" capacitor. Typically, bypass capacitors are 0.1uF per IC (integrated circuit). Since your recorders are so small, I'm making the assumption that inside the unit is indeed one IC.

The audio circuitry is an additional load, which is why it may require more capacitance. In that case, you might need to add somewhere around 10uF to help smooth out the power transients - for the larger one, a 10V electrolytic will be OK.

If a test with the 0.1uF cap is less than successful, it's important to use both the larger (10uF) and the smaller capacitor (0.1uF, or 100nF) in parallel. Large capacitors are good for taking care of low frequency "noise", or transients. But for the higher frequency transients, you need to have a low value capacitor.

Try the above - you will need to do this anyway (add caps, that is) for each recorder in your array; best to find out now as to how much work you are in for.

I'll bet that your current readings won't change significantly, but your recorder will start to function normally.

If you understand, proceed and report back your amended current readings. If you're unsure, please ask more questions.

Fortunately I had a 10uf and a .1uf handy. Unfortunately when I connected them across the leads the current rating is a constant 160-180mA. I just hooked each one from the negative to the positive of the battery terminal (is that right?) The circuit ceases to function if I have the capacitors and the multimeter hooked up and i get that contant 180ma reading. Without the multimeter it works fine. I don't know if I'm doing it wrong. Are there other ways to measure the capacitance? I'm thinking, since running these off one battery pack isn't absolutely necessary, I might just attempt to run a few of them off 3 AAAs, and if that fails I might just leave them all with their own battery packs.

Well, don't give up just yet

Seems odd that having the meter in there would cause so much trouble - but let's deal with it.

Do you have a low-value precision resistor available? Say, 5 Ohms or less?

You could measure current by placing the resistor in series with the battery circuit, and using your DMM to measure the voltage drop across it.
I=E/R, so if you used (for example) a 5 Ohm resistor and when the recorder was running you measured 0.6 volts across it:
I=0.6 / 5
I=0.12 Amps or 120mA
5 Ohms might cause enough of a voltage drop to make the recorder operation unreliable; that might be what's happening when you have the meter in the circuit (particularly if the probe connections are dirty or damaged)
Test your probe's connections by setting your DMM to the lowest resistance setting, and short the probes together. If you read more than an Ohm or two, you've found the problem.

DanRilley said:

Thanks I'll try AAA alkaline next. Referring to Sgt Wookie's post I'm trying to measure the current that the circuit draws but I don't think I'm doing it correctly or something as the circuit gets tweaked out. Basically I have the multimeter set to 200mA and I'm connecting the positive lead to the positive and the negative to the negative coming out of the battery pack. I get numbers around 180mA when it's recording and 50mA when it's playing, but I don't know if these numbers are correct since I'm not getting correct functionality on the circuit. Is there a better way to test the current? I've tried moving the common lead around to different points and it doesn't seem to affect the fact that the circuit gets messed up (audio record and play back are squelchy).

Click to expand...

Just to clarify a bit.
You didn't put the positive from the multimeter on the positive on the battery AND the negative from the multimeter on the negative on the battery while measuring current? (in parallel?)

That would probably blow the fuse in the multimeter. The multimeter should be in Series with either the negative or positive leads from the battery to the recorder...

Actually that was it Cornelius! Duh. As I said, I'm just getting into this stuff, it did seem a bit wrong to be shorting out the battery! So I hooked the plus lead of the dmm probe to the + of the battery and the - lead of the dmm to where the + of the circuit connects and now I'm getting about 5mA while playing and about 25mA while recording. The numbers move pretty fast so I think it could be +- 10mA, but I think these are good numbers, sound about right?

So in that case, 3 AAA batteries should be fine. If they're supplying 1250mA of current, (which is what I found on google)? Even if they were all recording at once, which they won't be, that would only draw 250mA.

Oops!

Yep - measure voltage in parallel, current in series. Good thing the batteries couldn't supply more current than that.

Check out this page:
http://www.allaboutcircuits.com/vol_1/chpt_11/3.html
Note that a carbon-zinc "D" cell typically provides 4.5 amp-hours @ 100 mA.
Usually, there isn't a whole lot of difference in price between various sizes of batteries. However, you get far more AH capacity in the larger sizes than the small ones.

Something that isn't so obvious is when the current demand on a battery is increased, more voltage is dropped across the internal resistance of the battery, which generates waste heat, and results in the battery being less efficient.

You might think that a battery which can supply 100mA for a 1 hour period would last half of that time if the current were doubled to 200mA. Actually, it would last far less than that due to the power wasted on the internal resistance.

While you could get by with AAA or AA batteries, you'll be replacing or recharging them far more frequently than you would if you used "C" or "D" cells. Unless you have tight space constraints, go with the larger cells; your wallet will thank you.