Resistor driven LED question

[terrik_zion]

I think my math may have confused me. Take a Cree XR-E Q5, 3.7v and giving it the maximum 1000mA. If I hook up 3xAA primaries in series (battery holder) I get a voltage supply of 4.5v, with fully charged batteries. Put in a resistor, dropping 0.8v (4.5 dropped to 3.7) divided by 1amp equals .8ohms (and then use the nest higher standard resistor value.)

Here's where I see problems, and tell me if I am correct or not. Once each AA has lost 0.1v leaving 4.2v supplied, it is still being resisted by 0.8v, leaving only 3.4v to drive the led, which is less than what it requires.

Is this just inherent to resistor driven systems, or am I missing something simple like changing amperage to compensate for lower voltage? Even so, more amps shouldn't be enough if adequate voltage isn't present?

I want to build a simple led light with this setup, but I don't feel like putting fresh primaries in every few minutes, I want this specifically to be a non_driver setup as well, just for the sake of experimentation. Any info helps!

 

[Lynx_Arc]

If you are not using rechargables (you said primaries, then said fully charged?) then you forgot the internal resistance of the battery in the equation, alkalines have a pretty high resistance when driven at 1 amp and lithium has a lot less resistance but still it is there. I recommend not using alkalines at 1amp drain levels instead use nimh or lithium primaries.

 

[terrik_zion]

Ah,, poor choice of words on my part, what I meant by fully charged primaries was fresh out of the box at 1.5v each, as oppsed to after they've been used and discharged somewhat to (for instance) 1.4, 1.3 1.2v, etc...

This is true, too, niMh would be a better choice for this application (and most likely will be once I buy some more).


Still curious though as to the voltage issue, since resistors are a set value and won't doctor voltage and amperage like a driver will. With, say 4 nimh AAs, 1.2v each, 4.8 in series, I would add a resistor to drop about 1.1v from the batteries to the led, but once each aa only has 1.1v, 4.4 total, that same resistor would drop the voltage by 1.1v, supplying the led with only 3.3v, insufficient for operation.

 

[Lynx_Arc]

nimh AAs are not 1.2v each, hot off the charger they can be as high as 1.5v but usually about 1.4v and after they cool off they can be 1.33v. the extra 0.13 to 0.2v adds up to 0.52 to 0.8v and can make a lot of difference in LED output. LEDs are dynamic in that the voltage/current varies they don't quite act like a resistor and the Vf ratings vary within a batch some too. I tend to take the easy route when I wire up LEDs by using a variable resistor I dial in the current I want and measure the resistance taking in account things such as runtime at what average level I desire because using resistors when batteries are fresh the output will be higher than when they are discharged and if you want it brighter you have to start at higher levels. Nimh is easier to deal with because the operating voltage range is more flat than alkalines under normal circumstance.
As far as a resistor dropping a certain voltage it will drop that at the start but as the batteries are depleted the voltage dropped will drop less but the voltage will be less too.... It will start bright and fade to nothing over time.

 

[terrik_zion]

Re: Resistor driven LED questionAh, okay thanks. I learn something new every wisdom i

nimh AAs are not 1.2v each, hot off the charger they can be as high as 1.5v but usually about 1.4v and after they cool off they can be 1.33v. the extra 0.13 to 0.2v adds up to 0.52 to 0.8v and can make a lot of difference in LED output. LEDs are dynamic in that the voltage/current varies they don't quite act like a resistor and the Vf ratings vary within a batch some too. I tend to take the easy route when I wire up LEDs by using a variable resistor I dial in the current I want and measure the resistance taking in account things such as runtime at what average level I desire because using resistors when batteries are fresh the output will be higher than when they are discharged and if you want it brighter you have to start at higher levels. Nimh is easier to deal with because the operating voltage range is more flat than alkalines under normal circumstance.
As far as a resistor dropping a certain voltage it will drop that at the start but as the batteries are depleted the voltage dropped will drop less but the voltage will be less too.... It will start bright and fade to nothing over time.

 

[Mr Happy]

I think my math may have confused me. Take a Cree XR-E Q5, 3.7v and giving it the maximum 1000mA. If I hook up 3xAA primaries in series (battery holder) I get a voltage supply of 4.5v, with fully charged batteries. Put in a resistor, dropping 0.8v (4.5 dropped to 3.7) divided by 1amp equals .8ohms (and then use the nest higher standard resistor value.)

Here's where I see problems, and tell me if I am correct or not. Once each AA has lost 0.1v leaving 4.2v supplied, it is still being resisted by 0.8v, leaving only 3.4v to drive the led, which is less than what it requires.

Here is where you have made a logic error. It is being resisted by 0.8 ohms, not 0.8 volts. (It is a resistor, not a revolter )

So the LED will be dropping 3.7 V, and the supply voltage will be 4.2 V, and the voltage across the resistor will now be 4.2 - 3.7 = 0.5 V. So the current will be 0.5 V / 0.8 ohms = 0.625 A. The LED will still light up, but less brightly.

However, as Lynx_Arc said you must factor the internal resistance of the battery into these calculations. You will not get these results with alkaline batteries.

Is this just inherent to resistor driven systems, or am I missing something simple like changing amperage to compensate for lower voltage? Even so, more amps shouldn't be enough if adequate voltage isn't present?

If the battery voltage is higher than the LED forward voltage then current will flow. Resistors cannot block voltage, only reduce current.

 

[terrik_zion]

nimh AAs are not 1.2v each, hot off the charger they can be as high as 1.5v but usually about 1.4v and after they cool off they can be 1.33v. the extra 0.13 to 0.2v adds up to 0.52 to 0.8v and can make a lot of difference in LED output. LEDs are dynamic in that the voltage/current varies they don't quite act like a resistor and the Vf ratings vary within a batch some too. I tend to take the easy route when I wire up LEDs by using a variable resistor I dial in the current I want and measure the resistance taking in account things such as runtime at what average level I desire because using resistors when batteries are fresh the output will be higher than when they are discharged and if you want it brighter you have to start at higher levels. Nimh is easier to deal with because the operating voltage range is more flat than alkalines under normal circumstance.
As far as a resistor dropping a certain voltage it will drop that at the start but as the batteries are depleted the voltage dropped will drop less but the voltage will be less too.... It will start bright and fade to nothing over time.

Thank God for knowledgeable people and cpf. Your advice is well received and understood. (Also, I learned something new about nimh battery voltage and resistors)

I could not find specific info on resistor behavior as supply voltage lessens before, so indeed the resistor will drop less voltage as the batteries slowly lose their voltage!

Thanks, this will make designing flashlights much more understandable!

 

[terrik_zion]

Here is where you have made a logic error. It is being resisted by 0.8 ohms, not 0.8 volts. (It is a resistor, not a revolter )

So the LED will be dropping 3.7 V, and the supply voltage will be 4.2 V, and the voltage across the resistor will now be 4.2 - 3.7 = 0.5 V. So the current will be 0.5 V / 0.8 ohms = 0.625 A. The LED will still light up, but less brightly.

However, as Lynx_Arc said you must factor the internal resistance of the battery into these calculations. You will not get these results with alkaline batteries.

If the battery voltage is higher than the LED forward voltage then current will flow. Resistors cannot block voltage, only reduce current.

Correct you are! I missed the .8ohms. Thank you as well!