Hello, guys! If you have any good ideas can suggest me, thank you so much. I have twenty LEDs that I want to light up. I planned out a few possible circuits, which I describe below along with a few questions. Would any of them not work, and why? Anything else I should be taking into consideration but haven't? The LED has the following specs:
Vmin: 2.8V
Vmax: 3.5V
If: 20 mA
From what I understand, the Vmin is essentially the same as Vd in this diagram, so the current will be 0 until it reaches that voltage. Is this correct?
If the LEDs are connected in series, that means that I'll need 20*2.8V=56V in total to get current passing through every LED. I calculated the approximate amount of power necessary (60V*0.02A=1.2W) and compared it to the amount that can be provided by a typical AA battery, which I've been told usually provide around 800mA of current (1.5V*0.8A=1.2W). Seeing as they're the same, I'm pretty sure it'll be possible to step up from a single AA to what's needed to light the LEDs. So far, I know of the SEPIC, which I'm reading up on right now. Is there a better way of accomplishing this? Possibly without needing inductors?
Another solution I've considered is connecting them in parallel, which I recall reading somewhere that it's a bad idea unless there's a resistor in series with each of the LEDs. Two AA batteries in series would be enough in this case, providing a 3.0V potential across each branch. Since the LED takes 2.8V, the resistor would take up the remaining 0.2V, and to get a 20mA current, I'll need a (0.2V/0.02A=10ohms) 10 Ohm resistor.
I could also plug it into an adapter which converts the AC electricity from the outlet to 15V (the current rating on the adapter means the maximum it can give and not the amount it'll always give, right?). In this case, I could separate the LEDs into four groups of five (which is convenient given their layout), each LED in a group of five is connected in parallel to each other, and the four groups are connected in series. That would give a voltage of (15V/4) 3.75V across each group of five, and since the LEDs in the group are in parallel, they'll each get 3.75V too. The current needed would then be (0.02A*5) 100mA. The voltage across the resistor in series with the LEDs is (3.75V-2.8V) 0.95V. The resistance needed would then be (0.95V/0.02A) about 50 ohms.
Thanks all!
Vmin: 2.8V
Vmax: 3.5V
If: 20 mA
From what I understand, the Vmin is essentially the same as Vd in this diagram, so the current will be 0 until it reaches that voltage. Is this correct?
If the LEDs are connected in series, that means that I'll need 20*2.8V=56V in total to get current passing through every LED. I calculated the approximate amount of power necessary (60V*0.02A=1.2W) and compared it to the amount that can be provided by a typical AA battery, which I've been told usually provide around 800mA of current (1.5V*0.8A=1.2W). Seeing as they're the same, I'm pretty sure it'll be possible to step up from a single AA to what's needed to light the LEDs. So far, I know of the SEPIC, which I'm reading up on right now. Is there a better way of accomplishing this? Possibly without needing inductors?
Another solution I've considered is connecting them in parallel, which I recall reading somewhere that it's a bad idea unless there's a resistor in series with each of the LEDs. Two AA batteries in series would be enough in this case, providing a 3.0V potential across each branch. Since the LED takes 2.8V, the resistor would take up the remaining 0.2V, and to get a 20mA current, I'll need a (0.2V/0.02A=10ohms) 10 Ohm resistor.
I could also plug it into an adapter which converts the AC electricity from the outlet to 15V (the current rating on the adapter means the maximum it can give and not the amount it'll always give, right?). In this case, I could separate the LEDs into four groups of five (which is convenient given their layout), each LED in a group of five is connected in parallel to each other, and the four groups are connected in series. That would give a voltage of (15V/4) 3.75V across each group of five, and since the LEDs in the group are in parallel, they'll each get 3.75V too. The current needed would then be (0.02A*5) 100mA. The voltage across the resistor in series with the LEDs is (3.75V-2.8V) 0.95V. The resistance needed would then be (0.95V/0.02A) about 50 ohms.
Thanks all!
