Current limiting resistor

[zachtheterrible]

Just wanna make sure that there's nothing more to this before I go blowing up my 1W LED, if I have 4.5 volts, and I have a 1W LED (330mA, 3.3v rated), I just use ohm's law to calculate a resistor of 13 ohms, correct?

 

[Russlk]

13 ohms is safe, but I calculate 3.6 ohms.

 

[audioguru]

They recommend a constant current circuit or current regulator, don't they?

The 3.3V of the LED is typical. What if it was a little low at 3.0V? Then its current from a resistor would be higher.
If the 4.5V is from a battery, then it is actually 4.8V when new. The current from a resistor would be higher.

I don't remember what kind of cooling your LED needs to dissipate 1W safely.

 

[captainkirksdog]

Zach,
This has always worked for me:

Vsupply-Vled/Iled

So, 4.5-3.3/0.330=3.63 ohms

For safety, (the LED's, not yours) round up, not down. A 3.8 ohm carbon composition resistor would be my choice. And remember, dissipated power is a function of the series current, i.e. make sure your current-limiting resistor is rated for 1-2 watts.

 

[Someone Electro]

Use a single transistor and diode constant curent source.It will make the led work at maximun brightnes no mather the input voltage(If voltage is high enugh)

For a schematic www.google.com

 

[Hero999]

Seriously, use a constant current source, it depends on how much you're willing to pay but you can make one from a LM317 regulator but it's wasteful as you'll need a higher voltage or you could use a switching regulator which is harder to build and more expensive.

 

[zachtheterrible]

Thanks, I forgot about that little equation. I was actually using a boost dc-dc converter to run the circuit off 3 volts. It worked great, but it had two problems: It was using an AMP of current! Highly inneficient. I was using a high current inductor too. The other problem was since the circuit was in a mag-lite, whenever the maglite got banged against something, the circuit would turn off but still keep drawing current. So I decided to go with a 3-cell maglite and use a current limiting resistor. Worke great!

Are there voltage regulator circuits that are efficient. i know that lm317 converts the dropped voltage into heat, its linear if I recall correctly?

 

[stevez]

ohm's law

Zach-for future reference (so you only have to remember Ohm's Law) look at the LED/resistor this way.

You have 4.5 volts supply and an LED that is rated at 3.3 V/330 ma. You'd like 330 ma of current to flow when 4.5 volts is applied to the LED/resistor.

In a series circuit the flow of current is the same thru all components so in this case it will be 330 ma.

In a series circuit the total voltage drop across both elements is the total applied voltage - in this case it's 4.5 volts. Each component in the series circuit will have a voltage drop across it and the sum of the voltage drops across each component will total the applied voltage. You've said that 330 ma will flow when the voltage drop across the LED is 3.3 volts. That means that the voltage drop across the resistor is 1.2 volts. Since we said that the current flow thru all elements is the same (in a series circuit) then we know the resistor has 330 ma flowing - 1.2 volts and 330 ma is a 3.64 ohm resistor.

This may seem trivial compared to just cranking thru the formula but I though it might help to understand.

 

[Nigel Goodwin]

Zach-for future reference (so you only have to remember Ohm's Law) look at the LED/resistor this way.

You have 4.5 volts supply and an LED that is rated at 3.3 V/330 ma. You'd like 330 ma of current to flow when 4.5 volts is applied to the LED/resistor.

In a series circuit the flow of current is the same thru all components so in this case it will be 330 ma.

In a series circuit the total voltage drop across both elements is the total applied voltage - in this case it's 4.5 volts. Each component in the series circuit will have a voltage drop across it and the sum of the voltage drops across each component will total the applied voltage. You've said that 330 ma will flow when the voltage drop across the LED is 3.3 volts. That means that the voltage drop across the resistor is 1.2 volts. Since we said that the current flow thru all elements is the same (in a series circuit) then we know the resistor has 330 ma flowing - 1.2 volts and 330 ma is a 3.64 ohm resistor.

This may seem trivial compared to just cranking thru the formula but I though it might help to understand.

I've been using ohms law for 40 years or so, most of that time probably averaging once every day or two - but you've now completely confused me!

 

[stevez]

confused

Nigel -ok, I am not sure what's unclear - just breaking things down for Zach.

 

[zachtheterrible]

Thanks for explaining it that way steve, I understood

 

[zevon8]

Stevez, I liked the explanation, fell into the category of teaching someone to fish instead of just feeding them. ( thumbsup )

-sidebar, not hijacking a post-

Zach...did you ever get around to fixing that 'scope? Nice instrument, hope it was salvageable.

 

[Oznog]

Frankly, you're seriously doing this the wrong way. You need a special circuit and probably a specific IC to do this because you want a current mode boost converter. Zetex & Supertex make some. It will be much better regulated and much more efficient.