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# How can I know which resistor to chose?

Is there any math calc to know which resistor will low the voltage to a desired voltage?

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Is there any math calc to know which resistor will low the voltage to a desired voltage?

If you are looking to drop the voltage then you need a voltage divider. Otherwise, choosing a resistor will depend on the resistance of the components you are using it with and the application.

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Thevinin's Theorem!

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Ohms law. Look on the color coding on the resistor and then you use ohms law to figure out if it fits your needs. ( if you dont know ohms law, look it up on wikipedia)

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Yes shure, but is there any calc to do to know wich resistor will low the voltage, for example, I have 9v and I want it reduced to 3,7v , how do I know which resistor to use?

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As far as I know, one resistor on its own won't do that. You would need to resistors to form a voltage divider and then you would use the formulae in the link I already provided. The point between the two resistors should come to 3.7V. Or better, you can do as alex suggested and get a variable-controlled voltage regulator. You should be able to pick one up at Radioshack.

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Mmm.. radioshack is not an option for me, I live in Argentina, and I don't really know if I could find a voltage regulator.. I was using some resistors to drop down from 12v to 3v, is that wrong? It seemed to work

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Using "voltage dropping resistors" was a common way to "regulate" voltage before voltage regulators were common. But it's a very inefficient way to drop voltage. They work because the device you're powering is itself a "load" on the circuit, and draws current. Since that inserted resistance is fixed, varying the current draw has to change the voltage, too (it's an "ohms law" thing.) Unless you know the approx load of the device itself, you cannot know the voltage drop (even if you know the value of the resistor.)

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So, there's no such thing as a math procedure to know approximately how to drop from a determined voltage to another right? With load you are refering to voltage or energy loaded in the battery? Sorry, I'm new in electronics and I'm not an english native. Thanks

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Yes, but only if you know the "load."

In it's most simple form, the load is equivalent to another resistor. It's more complicated, but we'll pretend your device (which we know nothing about) creates a "resistive load."

To find the approx load, you'll need an ammeter. Most cheap VOMs (Volt-ohm meter) are ammeters, too. You can find VOMs for less than $5 USD.

To drop 12V to 3V:

With the ammeter connected in series between the 3V and the device you are powering, take a reading. You are measuring the current (which is amps--

A, or 1/1000 of amps, which is milliAmps--mA.)For example, let's say the device draws 15mA (0.015 A) at 3V.

Using Ohm's Law: R = E / I , or

resistance = voltage / currentThe "resistive load" of the device is 200 ohms ( 200 = 3.0 / 0.015)

You can low plug that value into the voltage divider calculation randolfo referenced (see the pic below of a divider.)

To solve for R1 (the unknown "voltage dropping resistor")

R1 = ((R2 * Vin) / Vout) - R2

((200 * 12) / 3) - 200 =

600 ohmsfor R1.I.E., the voltage between R1 and R2 should be 3V.

This only works if the load (the device, R2) is resistive, and is at best an approximation if it's not.

Current draw (amps, milliamps) MUST be measured at the rated voltage. If the device runs on 3V normally, it must be read at 3V.

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get a voltage regulator, so you can have precise control

see this instructable

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