# Help with finding the right resistor

1. Jul 27, 2015

### MattG2826811

so i need to reduce a dc 12v to dc 5v. do you know what type of resistor i should use? also could you please suggest an online calculator or an equation so that i can find the resistance i need in the future.

2. Jul 27, 2015

### billy_joule

https://en.wikipedia.org/wiki/Voltage_divider

Note that:

A voltage regulator like the LM7805 might be a better solution depending on your application.

3. Jul 27, 2015

### davenn

using a resistor is far from a good way to do it. This is because a resistor or resistive voltage divider as billy_joule suggested will also limit the current and that may be a bad thing for what you are trying to achieve

billy_joule's second suggestion is the best .... use a voltage regulator and ensure it can handle the current your project needs
And yes there are ways to boost current requirements via a regulator

for current requirements up to ~ 0.75A ( 750mA) a 7805 reg is ideal
for higher currents and much more efficient operation use a switching DC-DC buck converter
eg ....
http://www.ebay.com.au/itm/DC-DC-Bu...13?pt=LH_DefaultDomain_15&hash=item27f68c89dd

it can take a wide input voltage range and has an adjustable output and is capable of 3 Amps

cheers
Dave

4. Jul 28, 2015

### meBigGuy

I like both previous posts and third the suggestion that you drop the voltage with a voltage regulator.

You need to compute the power dropped in the regulator and make sure it can dissipate the heat (that would be true for a resistor also).

The power is 12-5 = 7V times load current. If 100ma. then that is 0.7 watts. Thermal resistance junction to air (to-220 package) is 65C per watt. so the junction will be 65 * 0.7 = 45C above ambient without a heat sink. Max junction temp is 125C. How hot will the environment be?

Another way to look at it. Say a 50C ambient, that leaves 125C - 50C = 75C. 75C / 65C/W = 1.2 watts 1.2 watts/7V = ~160ma. Above that you need a heat sink.

5. Jul 28, 2015

### MattG2826811

thanks for the tip i think i will use the lm7805