# rheostat's and resistors! (More electronic helps)

Registered User regular
edited July 2010
Hey everyone, I've asked quite a few electronics questions on here and I've been getting a lot of help so I figured I'd ask one more (hopefully my last for this project).

My situation:
I need to use a rheostat (of some sort) to figure out just how much to drop my current in order to dim a set of LEDs to certain level of brightness. Once I get them dim enough I need to set up a resistor for a more permanent solution.

What I need help with:
I need to drop my voltage from 12v dc (from my car) down to between 6-8v(or somewhere about there, not sure).

How I need to do it:
This is where I need some help. I got the part where I use a rheostat in place of a resistor to drop the voltage, then I use a multimeter to measure the new voltage. Once I have the new output voltage how do I figure out what resistor i need to use? And another question is... how do I know what kind of rheostat I need? I mean, obviously they aren't going to be the same for every application right?

If anyone can help me out that would be AWESOME.

Cptn Pants on

## Posts

• Registered User regular
edited July 2010
Cptn Pants wrote: »
... use a multimeter to measure the new voltage. Once I have the new output voltage how do I figure out what resistor i need to use?

Your multimeter will have a function to measure resistance. It's most commonly labelled with "R", "Resistance", "Ohms" or capital omega "Ω".

Take the rheostat out of the circuit - you cannot measure resistance in circuit accurately. Place the two probes of your multimeter on the terminals of the rheostat that you used in circuit.

Cptn Pants wrote: »
And another question is... how do I know what kind of rheostat I need? I mean, obviously they aren't going to be the same for every application right?

You are correct.

In your case, you'll ideally want a "linear" rheostat - the other most common type is a "logarithmic" rheostat, which is great for audio volume applications, but probably not as good for what you want to do (actually, is the brightness of the LED vs current logarithmic? I don't know!)

Other names for "rheostat" are "variable resistor", "potentiometer" ("pot" for short), or "trim pot".

I don't exactly remember what current you were feeding to your lights, so I'm not able to give you a solid recommendation on exactly which pot to buy. Do your lights have a current rating?

I will give a rough guess though, based on a few quick searches, I'd say that every 3 LEDs take about 30mA. So for the 33 LED ring you linked last time, it would take 11 x 30mA = 330mA, while the 39 LED ring would take 390mA.

Since you want to drop 12V to around 7V, you're going to want (12 - 7)/0.330A ~= 15 ohms, or (12-7)/.390A ~= 12.8 ohms.

This calculation is actually totally bogus, because those current ratings are for full brightness, and you're deliberately dimming them, so the current would in actuality decrease.

But they show you'll want something in that range (e.g. this). You might actually end up needing something slightly higher, so maybe get two or three of those, and put them in series.

Most rheostats have three terminals - make sure that you use the middle terminal (the "wiper" terminal) and one of the outer ones (as opposed to only using the outer terminals), because that is the point that changes resistance as you tweak it.

Once you've figured out what resistance you need, make sure you buy 2-5W resistors (not the tiny 1/4W resistors, because they'll burn out at the currents you're at).

ecco the dolphin on
• Registered User regular
edited July 2010
Sounds good buddy, I'm gonna drop from 12v automotive to between 6v and 8v. The lights themselves are rated at 12v DC but there is no amp rating (i think). So I'm seeing a lot of different values of pots there so I'm not sure which to pick up.

One set The other

Thanks again!

Cptn Pants on
• Registered User regular
edited July 2010
Right now I'm using 3 of those 3 watt rheostats, does that mean I need a 3 watt resistor or a 9 watt? Also, if I bump up to 4 or 5 rheostats will it stay at 3 watts or does it go up with each?

Cptn Pants on
• Registered User regular
edited July 2010
Cptn Pants wrote: »
Right now I'm using 3 of those 3 watt rheostats, does that mean I need a 3 watt resistor or a 9 watt? Also, if I bump up to 4 or 5 rheostats will it stay at 3 watts or does it go up with each?

The best way to be sure is to grab the multimeter, and:

1.) Measure the voltage drop across all the rheostats with the LEDs powered on
2.) Measure the resistance of all the rheostats

Use this equation:

Power = Voltage * Voltage / Resistance

So if you measured 5V across a 18 Ohm resistance, you go:

Power = 5V * 5V / 19Ohm = 1.39W

So you'll need a 2W or higher resistor.

In your case, if you put more and more resistance on, the power dissipated in the resistors will go down.

ecco the dolphin on
• Registered User regular
edited July 2010
Cptn Pants wrote: »
Right now I'm using 3 of those 3 watt rheostats, does that mean I need a 3 watt resistor or a 9 watt? Also, if I bump up to 4 or 5 rheostats will it stay at 3 watts or does it go up with each?

The best way to be sure is to grab the multimeter, and:

1.) Measure the voltage drop across all the rheostats with the LEDs powered on
2.) Measure the resistance of all the rheostats

Use this equation:

Power = Voltage * Voltage / Resistance

So if you measured 5V across a 18 Ohm resistance, you go:

Power = 5V * 5V / 19Ohm = 1.39W

So you'll need a 2W or higher resistor.

In your case, if you put more and more resistance on, the power dissipated in the resistors will go down.

Probably gonna be my last 2 questions on this and I can call this project done.... probably, unless I run into any Eff ups.

1: You said measure the voltage drop all the rheostats with the LEDs on, I assume you mean on and dimmed correct?

2: If you look at your math there you had 18 ohms and then 19 in the equation, I assume that was a typo? Or is that one of those, better safe then sorry increase deals?

Cptn Pants on
• Registered User regular
edited July 2010
1.) Yup.

2.) Yes, definitely a typo. Well spotted!

ecco the dolphin on
• Registered User regular
edited July 2010
E= IR

Voltage = Amperage * resistance.

voltage/amperage = resistance needed in the circuit.

Slo on
• Registered User regular
edited July 2010
So here's my result so far. I got a drop of 10.2 volts. So from 12.4v down to 2.2v, and 51.8 ohms, so if I did the math right I should need a 2.08 watt at 51.8 ohm resistor right?

Cptn Pants on
• Registered User regular
edited July 2010
Your meter is reading 10.2v across your rheostats which you're trying to replace with a resistor, correct?

The math checks out for the values you've given.

Slo on
• Registered User regular
edited July 2010
D'oh! I think I messed up again. So the new output voltage (meaning the voltage coming out of the rheostats) is 2.2v. The voltage DROPPED by 10.2v, so I messed up... I did the right math but with the wrong values. So I re-did the math again with the correct values and I got 0.0934_ at 51.8 ohms. So i guess I need 1/8 watt resistor at 51.8 ohms eh?

Here are all the values I have:
Source: 12.4 volts
Needed: 2.2 volts
Drop: 10.2 volts
Ohms rating across all Rheostats: 51.8 ohms.
Resistor: 1/8th watt at 51.8 ohms?

Cptn Pants on
• Registered User regular
edited July 2010
Note: I believe you can get 12VDC input transformers that have variable DC output, these are a slightly more expensive but far more efficient method of dropping the voltage, and probably a better solution for a long term/permanent installation.

Ruckus on
• Registered User regular
edited July 2010
Cptn Pants wrote: »
D'oh! I think I messed up again. So the new output voltage (meaning the voltage coming out of the rheostats) is 2.2v. The voltage DROPPED by 10.2v, so I messed up... I did the right math but with the wrong values. So I re-did the math again with the correct values and I got 0.0934_ at 51.8 ohms. So i guess I need 1/8 watt resistor at 51.8 ohms eh?

Here are all the values I have:
Source: 12.4 volts
Needed: 2.2 volts
Drop: 10.2 volts
Ohms rating across all Rheostats: 51.8 ohms.
Resistor: 1/8th watt at 51.8 ohms?

Lemme draw a quick diagram to make sure we're talking about the same thing:
+12V (Source, [B]Point A[/B])
|
|
Resistor
|
|
[B]Point B[/B]
|
|
LEDs
|
|
Ground


Using the multimeter, with:

1.) The negative (black) probe on Ground, and
2.) The positive (red) probe on Point B

you measured:

A.) 2.2V, then your resistor is dropping 10.2V, and you would end up needing with a 2W resistor.

or

B.) 10.2V, then your resistor is dropping 2.2V, and you would end up needing a 1/8W resistor.

?

Either way, both cases seem a bit off to me, but then I expected the LED rings to take waaaaaay more power, so maybe I'm just being silly.

ecco the dolphin on