((( Curious about this resistor )))

Thread Starter

///OSS

Joined Mar 27, 2010
6
I've been using these voltage regulator setups and one type has a standard ceramic film 10 ohm 5% resistor and I think its 1/8 watt on this LM317T another one of these setups (and the one in question) has a slightly different resistor on the adjuster pin, and looks like a 6.8 ohm (Blue grey gold gold) but why is it PINK? and am I reading that correctly? is it a metal film? or just flame proof coating?

and lastly, just to make sure, am I looking at a 1/8 watt or a 1/4 watt? and does it make a difference in this particular case? I basically want to make more of these configurations for 12V uses powering LEDs.

the 6.8 ohm one measures around 3mm so its kinda safe to assume its a 1/8 watt?

Here's the image for reference

http://img.photobucket.com/albums/v129/EVeRcyCLinG/004 OSS SHOOT/IMG00159-20100327-2025.jpg

Thanks for your help!
 

Thread Starter

///OSS

Joined Mar 27, 2010
6
The photo you have posted shows the 317 wired as a current regulator (not voltage).

When wired as current regulator than the resistor determine the current that will flow in the circuit with the following equation Vref/Resitor value. Since the V ref is 1.25 volts (it can drift sligtly from component to component).

So in your case you have 6.8 Ω resitor which results in : 1.25/6.8 ≈ 184 milliAmps.

Now the heat dissipated by the resistor (which is in series with your load) is given by the following formula : W = I^2 * R so in your case .184^2 * 6.8 ≈ 0.23 watts. The rule of thumb suggest to double this value so use 1/4 watts.

Alberto

Thanks! and why is that resistor Pink? can I just buy any standard ceramic resistor (cream colored) for this application? or is that resistor some metal film variant?

Sorry for the noob questions 0_0 lol
 

SgtWookie

Joined Jul 17, 2007
22,230
Yes, it's a 6.8 Ohm resistor, 5% tolerance.
That pink body color doesn't mean much.
Generally, tan bodied resistors are carbon film, and blue bodies are metal oxide film. The old carbon resistors with cylindrical bodies were generally dark brown. Power resistors can be a wide variety of colors.

An LM317's Vref is nominally 1.25v (difference between the OUT and ADJ terminals); the regulator attempts to keep Vref reasonably constant by sourcing current from the OUT terminal. If the resistor is connected from OUT to ADJ, then Vref/6.8 Ohms = 1.25/6.8 = 183.8mA constant current (nominally) will be available at the ADJ terminal.

In this configuration, there is nominally a 3v drop across the regulator.

Power dissipation will be 1.25*.184 = 230mW. The rule of thumb is to use a resistor rated for double the power dissipation for reliability, so a 1/2W resistor should be used.

[eta]
Well, Alberto beat me to it ;)
 

Thread Starter

///OSS

Joined Mar 27, 2010
6
So its safe to assume that this design will run hotter than it needs to be? a 1/2 watt is more appropriate?

I have 3 setups for 3 different led arrays, a 6.8ohm, 8.2 ohm and 10 ohm. all look like 1/8 or 1/4 so I should up them to 1/2 correct?
 

SgtWookie

Joined Jul 17, 2007
22,230
Gee, I thought I just gave you the formula for calculating current and power dissipation?
Iout = Vref/Ohms; where Vref = 1.25
Conversely;
Ohms = Vref/Iout
And power = voltage x current - in this case, Vref x Iout.
 

Thread Starter

///OSS

Joined Mar 27, 2010
6
Well now I understand why they used 1/4 instead of 1/2.. There's a space constraint, as that setup has to fit inside this small little plastic box. What are the adverse effects of me just using the 1/4 W instead?

Think it's a bad idea?
 

SgtWookie

Joined Jul 17, 2007
22,230
Ask yourself where is the heat going to go?

Sounds to me like you have a melt-down in the making.

You probably need a buck-type current limiter.
 

SgtWookie

Joined Jul 17, 2007
22,230
It hasn't been explained yet, but resistors are rated for power dissipation when they have air and space around them. Heat can be dissipated in any/all of three ways; conduction, convection, and radiation.

If you're going to bury this thing in a little plastic box, that heat is going to have to go somewhere, or it'll build up inside the box. Since you will be removing radiation from the picture, you'll have a little convection oven with very toasty little parts inside.
 
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