Potential Divider calculation

[Johnson777717]

Good day all. I'm wondering if someone wouldn't mind helping me with some math, or at least give me a push in the right direction. I realize that a lot of folks don't like to just give the answer because it defeats the purpose of learning. With that said, can you please help me to learn how to calculate this.

Here's what I want to do:

1. I have a + 12 V and Ground regulated supply.
2. I want to create a potential divider from the 12 volt supply in order to reduce the voltage to 9V.
3. I want to be able to supply only 560mA for the 9V supply (the 12 volt supply will be capable of approximately 1 amp total with a fan already drawing a max 200mA, which leaves me 800mA or so to play with.)

So far, for a potential divider, I've calculated:
V2 = (R2 / (R1 + R2)) * V1
where V2 is the output voltage, V1 is the input voltage.

I've also discovered that I need a R1:R2 ratio of 1:3 in order to obtain a 9V output for V2.

I'm stuck on how to provide current limiting to approximately 560mA...therefore this is my question...

How do I limit the current (or calculate for) to approximate 560mA using a ratio of 1:3?

Thank you very much! I appreciate your help.


Thank you

 

[Exo]

You cannot use a resistor divider to power something. The output voltage will vary with a verying load. You need a voltage regulator such as an 7809

 

[Johnson777717]

alrighty, thanks for your help Exo. I was trying to get away without having to regulate down again.

What do you think about using a zener?

 

[Exo]

You can use a zener to generate the voltage , but for a 560mA load you will need to buffer it, with a transistor for example...
a integrated voltage regulator would probabely be easyer and smaller...

 

[Nigel Goodwin]

As you've already been told, this isn't a practical solution for your problem, but one reason which hasn't been mentioned is the general 'rule of thumb' for potential dividers.

This 'rule of thumb' is that you provide five times the current through the potential divider that your load needs - in this case that would be 2.8A, rather too much for your 1A regulator!.

What are you feeding that needs 9V at 560mA?, and is the current constant (in which case you would only require a single resistor dropper), or does the current requirement vary - and if so, how much?.

 

[Johnson777717]

thanks for your replies folks.

What I'm trying to do is setup a regulated 12 volt power supply and branch off a 9v supply.

A fan will run off of the 12 volt supply. The fan is used for cooling a large heatsink. The 9 volt supply will power a multimeter circuit which was originally created to run from a 9V battery. So I'm basically trying to match a 9V battery with a regulated supply.

I wanted to try to stay away from an additional linear regulator for the 9V supply in an effort to minimize space and the addition of a heatsink for the 9V regulator. I mistakenly figured that I could use a divider to branch from the 12V supply. I figured since the 12V supply has already been regulated that I could use a divider or a zener. Nevertheless, I have learned a great deal just from what you folks have suggested.

As you have stated, running an additional 9V regulator off of the 12V supply would probably be easier.

Finally, the 9V supply will be a constant current when the multimeter is on and running, but I seriously doubt that 560mA will be consumed. I just wanted to match the original battery setup. I don't know the actual consumption at this time, probably somewhere in the neighborhood of 50-100mA.

 

[checkmate]

Just get a 9v regulator. Don't worry about the current as chances are that it won't be constant anyway. Just ensure the current rating is more than the peak instantaneous current draw. For a multimeter circuit, current draw shouldnt be too high. Any TO-220 package regulator would do.

 

[grrr_arrghh]

for the price of the regulator and associated components, you might as well...

 

[samcheetah]

This 'rule of thumb' is that you provide five times the current through the potential divider that your load needs

what a waste of energy!!!!!!!

anyway thanx for tellin'

 

[Nigel Goodwin]

what a waste of energy!!!!!!!

anyway thanx for tellin'

No body has suggested that potential dividers don't waste enegery 8)

That's why you generally don't use them for things like this.

The reason for this 'rule of thumb' is quite simple, five times the current through the divider will help to keep the output voltage 'reasonably' constant as the current drawn by the load varies. If the load is constant, you don't need a potential divider, just a simple feed resistor (with the load forming the bottom half of the potential divider).