Too bad, because what you understand is wrong. the Vf cannot be the same at 10mA and at 15mA, which is what @djsfantasi seems to be saying, though I know that he knows better.

Bob,

That’s not what I was saying, but I get that it appears so.

 

Okay, the error in his calculation is that he used the Vf measured at 10.3mA to calculate the 100Ω to get 15mA, and that is not valid because the Vf at 15mA is different and unknown.

I agree with you that it is probably close enough, but he asked whether it wrong , so I told him. The actual resistor to get 15mA will be less than 100, as I showed with my example.

 

Bob: earlier you had mentioned using some LEDs in series. With all LEDs the same, and using the same voltage source, if you have two LEDs in series and a resistor is chosen so that they are the same brightness as one LED, wouldn’t that mean less power is being wasted by the resistor with two LEDs as opposed two using one LED with a larger resistor? And wouldn’t this result in longer run time? Thanks

 

John: earlier you had mentioned using some LEDs in series. With all LEDs the same, and using the same voltage source, if you have two LEDs in series and a resistor is chosen so that they are the same brightness as one LED, wouldn’t that mean less power is being wasted by the resistor with two LEDs as opposed two using one LED with a larger resistor? And wouldn’t this result in longer run time? Thanks

Yes, it would result in more run time but not for the reason you stated. But you’d need more voltage from your power supply. Such as 7.5V to 9V. So the same power source may not work because the voltage it supplied is not enough.

When you put two 3V LEDs in series, you’d need 6V plus a little more for overhead to limit the current. In a series parallel circuit, current in each string remains the same as required by each string while voltages add. Hence 2x3V=6V So the savings in current comes from the fact that you’d require 7x the current with two LEDs in series versus 14x in current for one LED.

With 7.5V, let’s calculate a target* resistance for 10mA of current.

R = (7.5-6) / 0.01 = 150Ω​

As we have seen, a variety of resistors will work. As the resistor value goes up, current will go down and the LED will be dimmer. (LEDs are current controlled)(Current is I=V/R You can see how what I said I’d true).

But I said “target” resistance, so I’d start there and experiment with larger resistor. And I say “target” resistance for another reason. Which I believe is not critical for you. The forward voltage of 3V is not the same for different currents as Bob has stated. But the difference is so small it becomes irrelevant. Just test with different resistors to get the desired light level. (Or use a 5k potentiometer wired as a variable resistor and dial-in a resistance and measure the resistance for the desired light level).

 

John: earlier you had mentioned using some LEDs in series. With all LEDs the same, and using the same voltage source, if you have two LEDs in series and a resistor is chosen so that they are the same brightness as one LED, wouldn’t that mean less power is being wasted by the resistor with two LEDs as opposed two using one LED with a larger resistor? And wouldn’t this result in longer run time? Thanks

I discussed that in post #69.

 

I discussed that in post #69.

Thanks, Bob, I guess back then I didn’t quite understand your logic. You described using 12v with a resistor for each of 3 LEDs looking back that makes sense now.
I was playing with the idea now of using my 6v supply with just one smaller resistor and 2 LEDs in series; it looked like it would work and be more efficient too, with less resistor power loss and still allow maximum brightness or less as desired?

 

Yes, it would result in more run time but not for the reason you stated. But you’d need more voltage from your power supply. Such as 7.5V to 9V. So the same power source may not work because the voltage it supplied is not enough.

When you put two 3V LEDs in series, you’d need 6V plus a little more for overhead to limit the current. In a series parallel circuit, current in each string remains the same as required by each string while voltages add. Hence 2x3V=6V So the savings in current comes from the fact that you’d require 7x the current with two LEDs in series versus 14x in current for one LED.

With 7.5V, let’s calculate a target* resistance for 10mA of current.

R = (7.5-6) / 0.01 = 150Ω​

As we have seen, a variety of resistors will work. As the resistor value goes up, current will go down and the LED will be dimmer. (LEDs are current controlled)(Current is I=V/R You can see how what I said I’d true).

But I said “target” resistance, so I’d start there and experiment with larger resistor. And I say “target” resistance for another reason. Which I believe is not critical for you. The forward voltage of 3V is not the same for different currents as Bob has stated. But the difference is so small it becomes irrelevant. Just test with different resistors to get the desired light level. (Or use a 5k potentiometer wired as a variable resistor and dial-in a resistance and measure the resistance for the desired light level).

Okay, thanks for explaining it, I’m still getting back into all of this, you all have been really helpful.

 

Thanks, Bob, I guess back then I didn’t quite understand your logic. You described using 12v with a resistor for each of 3 LEDs looking back that makes sense now.
I was playing with the idea now of using my 6v supply with just one smaller resistor and 2 LEDs in series; it looked like it would work and be more efficient too, with less resistor power loss and still allow maximum brightness or less as desired?

6V is not enough for two LEDs in series, just like 3V is no enough for 1.

 

I'm re learning from all of this so I decided to do some testing this morning since I got some smaller resistors yesterday along with a more accurate digital multi meter.. still don't have the 4.5v C cell containers yet.
My two sets of C cells are now down from 6.2v to 6v. I am using the same 0402 SMD white LEDs with two of them in series and I got the following results:

100 Ohm .41V 4.1mA LEDs 2.78Vf each --too dim for anything

47 Ohm .3V 6.38mA LEDs 2.83Vf each --- dim, but maybe for street lights

2x 47 Ohm // = 23.5 Ohm .21V 8.9mA LEDs 2.86Vf each --- Pretty good, could use for darker interiors.

I don't have any smaller resistors, but maybe a 5 Ohm would be pretty bright with 2 of these LEDs in series on 6v?

I have several dioramas using a 3V C battery pack each with 4-5 same LEDs in // and they have been working fine, but I do want to get some small resistors on them for LED longevity as suggested by you guys.

Thanks,