Voltage and ohms law

[interoth]

I'm new to electronics so I am not sure as to how to achieve this.
What I don't understand is how to divide voltages, if I put a resistor in series with a voltage source, does it reduce the voltage or does it reduce the amperage?
Say if I wanted a 6 volt line from a 9 volt battery, how would I do it and how would I work out what resistor to use?

Sorry if I seem 'out of the loop'
Thanks in advance

[IanB]

There are so many questions in your post, some asked, some unasked. So you are not going to get everything you need to know in a short summary.

However, let's go with the basics.

A voltage when applied across a resistor produces a current through that resistor. The current is proportional to the voltage and inversely proportional to the resistance. If you double the voltage you double the current. If you double the resistance you halve the current.

Now, to divide voltages, you need more than one resistor.

Put two equal resistors in series and apply a voltage to the outer ends. A current will flow through the resistors, the same current through both resistors. Since both resistors are equal and the current is the same, the voltage across each resistor will be the same as the other. And these two voltages must add up to the total voltage. So by measuring at the half-way point between the two resistors, you have divided the applied voltage by two.

Now, it's not quite that simple, however. Because if you try to draw any current from the half-way point you will upset the balance of equal currents and now the voltage will move. But as long as you only try to draw a minute current (such as with a voltmeter) the divider will be fine.

Lastly, you need to know that a "9 V" battery is not actually 9 V exactly. It is approximately 9 V, give or take a volt or three up or down (one up, three down). So you can't get exactly 6 V using the voltage divider method unless you first make sure 9 V is exactly 9 V.

[interoth]

Thanks for the explanation, cleared a few things up.

[Hideki]

If you have a FIXED load that always draws the same amount of current, like a resistor, then you can certainly put another resistor in series to drop 3 of the 9 volts across it.
The total resistance increases, so the current will drop, but you will get your 6V across the load.

However, if the load is unknown or variable, then a resistor will not be able to always drop 3V, since the voltage across it always depends on the current drawn by the load.

To ensure 6V in all situations (within limits), the solution is using some kind of voltage regulator instead of a resistor.

[6502nop]

^ Pretty much everything IanB said, only I brought PNGs!

See attached files for a visual explaination, with math outlined.

Note that using just a resistor divider, as Hideki mentioned, isn't a good idea, as the loading at the point of output (the tap) will affect the actual voltage. If you want a solid 6V from a 7.5-9V battery, then a good 3-terminal voltage regulator with Low DropOut (LDO) is better.

If you just want to know how to do it with resistors, then look at my gEDA snapshots.

As an aside, interoth, any/every good electronics starter book - like the Forest M. Mims III series you can pick up at Radio Shack - will cover this. Also, just back out one level, and click on the sticky post about learning material/courses and follow some of those links (Talking Electronics, NEETS, etc.). Good stuff.

nop

[FenderBender]

You'd want a voltage regulator as others have said. Some search terms for Google might be "Linear power supply design" "7805 Schematic" "7805 Power Supply"

7805 is just a popular 5V regulator. If you needed 6V, you would use a 7806. They don't make fixed regulators in every voltage but there are quite a few options.

Just be warned, there are tons of projects like these all over the web...Almost too many, so try not to get lost in all of them..but it's really not all that bad...Not sure if this is just a question in theory or if you actually want to do it. But good luck.