# VERY simple dc Q

#### daml

Joined Jan 20, 2007
4
ok my source is 12 - 14vdc and i need it to be 5vdc. what resister value in ohms do i need to accomplish this and what formula is used to figure out the correct solution. second (mainly just a refresher Q) if i put a 1.5 amp fuse in line of a circut which is 25 amps, will only 1.5 amps pass through the fuse and pop only if the unit being powered up attempts to draw more than 1.5 amps? sorry if the questions seem very novice, but i am a novice to this and eager to learn.
Dave

#### thingmaker3

Joined May 16, 2005
5,084
Re: first question:
12v-5v is 7v, so your resistor must drop 7vdc.
If you know how much current your circuit will draw at 5vdc, then you can use Ohm's law: 7v divided by current in amperes = your required resistor value. If you know your circuit resistance, then you can use algebra: Rd=Rc(7/5).

Or you could use a fifty cent voltage regulator: http://www.fairchildsemi.com/ds/LM/LM7805.pdf

Re: second question:
The fuse will only blow if the circuit draws more current than the fuse is rated for.

#### fanie

Joined Jan 20, 2007
63
Hi Daml

You will need a regulator to bring the 12 to 24V down to 5V !

The fuse rating is just what it say - it will blow and go open circuit if you exceed it's rated current, it's used for overcurrent protection and not regulation.

I would highly recommend you check out a few of the online electronic courses so you get a basic idea - else it may cost you and create a health hazard !

#### daml

Joined Jan 20, 2007
4
ok so if i did my math right, i would need a 4.66666666666666666667 ohm dc resistor at 12vdc and a 6 ohm dc resistor at 14vdc to reach my desired current of 5vdc @ 1.5a. please correct me if im wrong

#### n9352527

Joined Oct 14, 2005
1,198
Resistor would only work if the circuit drew constant current. Further, have you calculated the resistor power rating? Or the heatsinking required for a linear 7805?

#### fanie

Joined Jan 20, 2007
63
Dave,

If you say what the 5V is for we may be able to give you better advice.

#### beenthere

Joined Apr 20, 2004
15,819
Hi,

The problem with resistive dividers is that the solution only works for one current draw - there's no load regulation. Using a 45 cent 7805 regulator is a better way to go.

#### thingmaker3

Joined May 16, 2005
5,084
1.5 Amp?!

Forget what I said earlier. Use an LM350 instead. And a heatsink.

#### mrmeval

Joined Jun 30, 2006
833
You'd need a hideously expensive resistor. One made from wire wound on a ceramic core about a foot long.

Get an LM7805 and be done with it.

#### daml

Joined Jan 20, 2007
4
First off i'd like to thank you all for your helpful input. New problems though... The LM7805 will not handle the required 1.5a rating. but i found an adjustable regulator, the LM317T. Below is a diagram of how i need to wire it up but there are obvious holes that are critical to get the output at 5vdc with 14vdc input. if someone could fill in the Blanks (R1 & R2) or provide a formula that would help me fill in the blanks i'd be estatic!

Thanks again for your time and effort!!

#### daml

Joined Jan 20, 2007
4
ok nevermind. with some extensive research i found this:

which has all the numbers i need. If i have anymore Q's ill be sure to post here!!

#### thingmaker3

Joined May 16, 2005
5,084
Vout=1.25v(1+R2/R1) just like it says in the image you posted...

So: 5v=1.25(1+R2/R1)=1.25+1.25(R2/R1)
And: 5v-1.25v = 3.75v = 1.25(R2/R1))
Therefore: 3.75/1.25 = 3 = R2/R1

In other words, any set of resistors with R2 being three times R1's value will be dandy. This is why the second image you posted has that ratio of resistors. (720/240 = 3)