# Converting Voltage

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**Converting Voltage**

How can I convert voltages down using resistors?

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You can lower the voltage at the load by putting a resistor in series with the load. That's basic circuit theory.

That's the theory. In practice, it's almost always a bad idea for any current above milliamps.

That's the theory. In practice, it's almost always a bad idea for any current above milliamps.

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That doesn't make much sense. What are you using for a power supply? Where did you get that 0.1 amp number from? When you change resistance, the amps change too.

We can't really help much with this without a lot more information.

We can't really help much with this without a lot more information.

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from what I understand a USB has a 5V current, and 100miliamp minimum output

Sorry for the confusion.

EDIT: ok i found the power source I am thinking about using and it has an output of 5V and 2000 miliamps

*Last edited by kmiller8; 05-16-09 at 08:43 PM. Reason: Found correct amperage.*

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A USB port only puts out 100 milliamps with a specific resistance load. If you change the resistance of the load, it won't be 100 milliamps any more.

So if the port has 5 volts DC, and you want 3 volts DC across your load, then you need a resistor that is equal to 2/3 of whatever the resistance of your load is. Then you need to make sure that the total resistance (the resistor plus your load) will keep the current within the acceptable range of the port. That may require another resistor in parallel with your load, which will change the resistance of the load which will in turn change the resistance requirements for the series resistor.

So you need to start by determining the resistance of the load and the acceptable current range for the port.

So if the port has 5 volts DC, and you want 3 volts DC across your load, then you need a resistor that is equal to 2/3 of whatever the resistance of your load is. Then you need to make sure that the total resistance (the resistor plus your load) will keep the current within the acceptable range of the port. That may require another resistor in parallel with your load, which will change the resistance of the load which will in turn change the resistance requirements for the series resistor.

So you need to start by determining the resistance of the load and the acceptable current range for the port.

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It is the amperage draw of the load not the total amps available that is used. Of course we are talking pure resistance on a DC circuit not even covering inductance on an AC circuit. That I have pretty much forgotten.

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So if the port has 5 volts DC, and you want 3 volts DC across your load, then you need a resistor that is equal to 2/3 of whatever the resistance of your load is. Then you need to make sure that the total resistance (the resistor plus your load) will keep the current within the acceptable range of the port. That may require another resistor in parallel with your load, which will change the resistance of the load which will in turn change the resistance requirements for the series resistor.

So you need to start by determining the resistance of the load and the acceptable current range for the port.

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ok, so how would i determine amperage draw of something that uses two AA's?

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It's impossible to answer that question without knowing the resistance of your load. Voltage, current and resistance are all interrelated. You cannot change one without chaning the others.

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**15**
Search "voltage regulator". It's a device that will change an input voltage to the required output voltage. 5 to 3 is very common.

Just don't try to jack that 2-amp (2,000 milliamp) supply into a USB connector with no current limiting resistor or you'll blow the next device in the chain.

Just don't try to jack that 2-amp (2,000 milliamp) supply into a USB connector with no current limiting resistor or you'll blow the next device in the chain.

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Thanks for all the responses, but the problem is that I have no kind of meter so that's why I was trying to solve it mathematically before putting a resistor on a power supply.

So i guess my real question is "How can I mathematically change voltage using resistors?"

And i'm guessing i'll need to know Voltage, Amperage(Load), and Resistance.

But that still leaves the problem of, when I drop Voltage, how much does Amperage and Resistance go up.

So i guess my real question is "How can I mathematically change voltage using resistors?"

And i'm guessing i'll need to know Voltage, Amperage(Load), and Resistance.

But that still leaves the problem of, when I drop Voltage, how much does Amperage and Resistance go up.

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**17**
So i guess my real question is "How can I mathematically change voltage using resistors

We are still guessing what you are trying to run. A light bulb might burn out at 5 volts but the tolerances on a cheap DC motor or such that it might run OK at 5V (just faster).

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I'd suggest a 3V DC wall wart (AKA power supply, power brick). You should be able to get one from Radio Shack or on line. On line such as at Ebay; may be cheaper even with shipping. power supply 3v, Computers Networking, Electronics items on eBay.com

*Last edited by ray2047; 05-17-09 at 02:14 PM.*

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"How can I mathematically change voltage using resistors?"

If the load is of near infinite resistance, and if your power supply can provide at least 100 milliamps and the resistors can handle that too, then you can put two resistors in series across the voltage source, one of 20 ohms and one of 30 ohms. Then you can put your load in parallel with the 30-ohm resistor.

I think, however, that you fail to appreciate all the variables involved. You cannot simply ignore the variables you cannot measure or do not wish to consider.

Are we answering a homework or test question? If so, I think you should complain to your instructor that the question is ambiguous.

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How did you get that though?, and no it's not homework, I just want to stop having to replace batteries in my 360 controller.

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Suppose the resistance of my load is "X". Now I add a resistor in series with the load of resistance (2/3)*X. The total resistance is now X + (2/3)X, or (5/3)*X.

The voltage supplied is 5 volts, applied over my load and the series resistor. So 5 volts over (5/3)*X ohms gives a current of 5 / ((5/3)*X), which is 3/X, based on Ohms law (I=V/R).

So if I is 3/X across my load of resistance X, then the voltage will be (3/X)*X = 3, again based on Ohms law (V=IR). Note that this is true regardless of the value of X, as long as it is not zero.

Now all these computation don't tell you if you're about to burn up that power supply. Nor does it tell you how hot that resistor will get or whether it will burn up. For these computations, you need to know the value of "X". You can buy a lot of batteries for the price of a new USB port. Warehouse stores sell batteries in packages of 48 for a pretty cheap per unit price.

The voltage supplied is 5 volts, applied over my load and the series resistor. So 5 volts over (5/3)*X ohms gives a current of 5 / ((5/3)*X), which is 3/X, based on Ohms law (I=V/R).

So if I is 3/X across my load of resistance X, then the voltage will be (3/X)*X = 3, again based on Ohms law (V=IR). Note that this is true regardless of the value of X, as long as it is not zero.

Now all these computation don't tell you if you're about to burn up that power supply. Nor does it tell you how hot that resistor will get or whether it will burn up. For these computations, you need to know the value of "X". You can buy a lot of batteries for the price of a new USB port. Warehouse stores sell batteries in packages of 48 for a pretty cheap per unit price.

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The voltage supplied is 5 volts, applied over my load and the series resistor. So 5 volts over (5/3)*X ohms gives a current of 5 / ((5/3)*X), which is 3/X, based on Ohms law (I=V/R).

So if I is 3/X across my load of resistance X, then the voltage will be (3/X)*X = 3, again based on Ohms law (V=IR). Note that this is true regardless of the value of X, as long as it is not zero.

Now all these computation don't tell you if you're about to burn up that power supply. Nor does it tell you how hot that resistor will get or whether it will burn up. For these computations, you need to know the value of "X". You can buy a lot of batteries for the price of a new USB port. Warehouse stores sell batteries in packages of 48 for a pretty cheap per unit price.

(I wish this forum had a "Thanks" button)