# 9v volts battery

#### anchim

Joined Feb 25, 2004
5
HI Guys! First time on board. I have a question hope someone here can help me.
is it possible if I wanted to use 9v battery and make it into 3v and 5v. what are the parts do I need and how to would be very appreciated. Thanks you all in advance.
Anchim

#### Dave

Joined Nov 17, 2003
6,969
Hi

Are you wanting the 3v and the 5v at the same time?

You can make a simple voltage divider and utilise the ouput across the resistors. Here is the explaination about voltage dividers on All About Circuits. If you want any help about selection of resistor values or how to come to a suitable configuartion post back

#### anchim

Joined Feb 25, 2004
5
Thanks for the quick reply Dave. Yes, I need 3v and 5v at the same so I can select it by a switch. Is there diagram or pic for dummies to follow, cause I don't have any background in eletronic. I read thought the link, but don't know how to use it. Once again, Thanks for the help.
Anchim

#### Dave

Joined Nov 17, 2003
6,969
Ok, no problem.

I will post back more thoroughly tomorrow as I am off to work in the next 20 minutes, also it gives me time to think about it. There is a more sophistocated way of doing such a task but this is a good way to become aquainted with basic electronics. B)

Forgive me if I have misinterpreted your level of electronic ability, but this link will explain some of the basic concepts of voltage, current, resistance and power. It may help you understand some of the information on the previous link.

Will post tomorrow

#### anchim

Joined Feb 25, 2004
5
Thank you Thanks you!

#### Dave

Joined Nov 17, 2003
6,969
Sorry I didn't get back to you last night :unsure:

Anyway I am having trouble uploading a design of a voltage divider that will allow you to switch between 3v and 5v from a 9v battery. I am not sure of the nature of the circuit to which this will be connected so I have only suggested typical resistor values that will give you desired results. We can taylor these to suit your circuit later.

I will post back with where you can see the design when I sort out the uploading of it :unsure:

#### Dave

Joined Nov 17, 2003
6,969
Just to let you know I haven't forgotten about you. I just can't seem to get a design onto the web, my personal web page just don't work! <_<

If I can't get it uploaded by tomorrow, I'll just have to give you an in depth explaination. Don't worry its nothing complicated.

Sorry for the delay

#### anchim

Joined Feb 25, 2004
5
Thank you so much for your time and effort trying to help me. I am very appreciated that. Thanks.
Anchim

#### anchim

Joined Feb 25, 2004
5
Dave Dave Dave. Where are you?

#### Dave

Joined Nov 17, 2003
6,969
Hi anchim,

Sorry about the delay, I can't upload the drawing of the circuit so I will have to verbally explain it to you.

I have assumed that the current drawn from the battery is 20mA, which is not an unreasonable value, but you will have to assess this against what size your load is (your load being the appliance that you are switching the voltage down for).

You will require three resistors wired in series, i.e. end you end. The positive end of the battery to the top of R1 and the negative side to the bottom of R3:

|
|
|
|
/
\
/Resistor R1
\
/
\
|
|__________Line 1_____________________
|
|
/
\
/Resistor R2
\
/
\
|
|__________Line 2_____________________
|
|
/
\
/Resistor R3
\
/
\
|
|
|
|___________Line 3_____________________

Resistor R1 = 200 Ω
Resistor R2 = 100 Ω
Resistor R3 = 150 Ω

You need then to connect one end of you load to the bottom of resistor R3 at Line 3, and the other end should switch between the top and bottom of resistor R2, Lines 1 and 2. When the switch is set to the top of resistor R2 (Line 1), the voltage to the load will be 5 Volts and when it is switched to the bottom of resistor R2 (Line 3)the voltage will be 3 Volts. Just as you wanted.

This is a cheap method that will give fairly accurate results. The better the tolerance on your resistors the better your accuarcy of the circuit. Not knowing how large you load is, I have just guessed at the required current, but it is very simple to step you resistor values up or down to give the same voltage to your load using Ohm's Law:

Resistance = Voltage/Current.

You will have to do this for all resistors. The best thing to do is post back with the current required for your load and I'll work it through for you.

Also I'm not sure what value resistors are available on the market, so you will have to check at wherever you purchase them. Just remember if you double the value of resistor R1, then you must double the values of the other two resistors to obtain the same results, this will cause the current to half.

I know this is very difficult to visualise, but think of it in terms of the right hand diagram on this link. The only differnece to your network is that you will have three resistors and you are switching either side of the middle resistor to obtain two voltages out at the Load. Don't worry about all the calculations on that page I've done all that for you.

I'm sorry I've not been able to post up a picture, I will have a word to a mate of mine to see if he could host it for me. Please post back if you have any questions or can't understand something and I'll try and explain it better.

Dave

#### Dave

Joined Nov 17, 2003
6,969
If you wish to read up on Voltage Divider networks, here is a superb page from TPub.