voltage dividers

[tomwarrior]

i know that when building a voltage divider, its not the exact values, but their perportion to each other that gives the predetermined output voltage. ex; if r1=100ohms and r2=300ohms its the same output voltage as r1=1000ohms and r2=3000ohms. my question is about the current when applying ohms law(I=Vin/r1+r2). is the number i get for current the amount of current being consumed by the divider or the amount of current available to the project im feeding? how do i choose what size resistors to use? i know that using a 100ohm and 300ohm resistor gives me .03 amps and 1000ohm and 3000ohm gives me .003amps. what im trying to do i reduce the 12 volts comming from my voltage regulator to 9 volts, but the whole amps thing has me confused. thanks in advance.---tom

[niftydog]

I=Vin/r1+r2

That equation tells you the current through the voltage divider when nothing else is attached to it.

To determine the current "available" from the voltage divider is a fairly tricky process. This is because as soon as you start drawing current from the divider, the voltage changes. This is because the act of drawing current from the divider is equivalent to placing resistance in parallel with R2 - which changes it's value which in turn changes it's current draw which in turn changes the voltage! It's a bastard isn't it!?

To maximise the current draw from a divider without affecting the voltage much, you need to use low value resistors. However, there are two problems with this approach. The first is that the lower the value, the higher the current and this is no good for battery powered gear as it drains the battery quickly. The second is that this higher current draw means the resistors need to dissipate more power (wattage) and too much power can cause them to heat up and even fail.

So, it's all about finding a compromise between all of these factors.

Now, you say you want to reduce 12VDC from the regulator to 9VDC? Because that sounds like you're dealing with a power supply, I would warn against using a voltage divider to acheive this task. Again, your major issue with this approach is the power handling capability of the resistors. I would suggest that perhaps replacing the 12VDC regulator for a 9VDC regulator - or just adding a second regulator (9VDC) to do this task. You still need to be aware of the power dissipation, but the regulator IC is going to give a more consistent result and it will handle the current draw better than "normal" resistors.

P = IR
So, make sure those 1/4W resistors aren't trying to dissipate 1W!

[KORGULL]

niftydog wrote:

P = IR
So, make sure those 1/4W resistors aren't trying to dissipate 1W!

Isn't it P=IV? or P=I(squared) times R?

[R.G.]

what im trying to do i reduce the 12 volts comming from my voltage regulator to 9 volts, but the whole amps thing has me confused. thanks in advance.

Well, the first thing is - don't do that with a voltage divider if you're trying to power effects with it. Instead, use a three terminal regulator like the 7809.

Resistor voltage dividers only work well for supplying voltage to loads that are an insignificant amount of current compared to the voltage going through the resistor divider itself. Nifty is quite correct - the external load current looks like another resistor in parallel with the lower resistor, so you can't arbitrarily pick R1 and R2 to get the right voltage without knowing that load. In fact - if the load is well known, you only need the top resistor, the load is its own R2. And yes, the voltage changes if the current in the load changes. It's only good for constant loads, or loads that are well below the voltage divider's own current. This is only practical for low current loads. Powering effects - which is what I intuit that you're wanting to do - is a sufficiently high current that you need to do it another way.

Zener diodes work, but they're tricky for beginners to apply correctly. Three terminal regulators are simple and easy.

[niftydog]

Isn't it P=IV? or P=I(squared) times R?

Yes, sorry, yet another brain fade! I think I may have been thinking about my home alarm system at the time! (PIRs on the brain...)

[tomwarrior]

thanks everyone. ill use a variable voltage regulator since i dont have access to a 9 volt regulator.