I've go a small scale strobe light and from the output I'm using a power transistor to amplify it and Then I have two sets in series of 5 LEDs in parallel with one 200 ohm resistor after the transistor (9v Vcc). The problem is that is I use it with one LED its very bright, but when I use the Dual 5LED parallels the LEDs are very dim and I need them to be as bright as possible. Probably very simple, but I could use some help with how to go about it.
Electronic – Getting 10 LEDs in dual parallels to be brighter
currentled
Related Solutions
Put the capacitor in parallel with the LEDs and leave the resistor where it is.
When you power up, the voltage across the cap is 0 V. It will slowly charge up to match the voltage at the top of the LEDs.
You don't show any resistor values or mention what type of LEDs you are using (what colour/Vf?) but if you put the LED on the emitter side you have to include the ~0.6V drop across it and the resistor, which means it will see a maximum of roughly 3.3V - 0.6V - (I_LED * R_LED). Let's say you are using a 100Ω resistor, and the LED has a VF of ~2V, then you will have (3.3V - 0.6V - 2V) = ~0.7V across the resistor, which means you will only get around 0.7V / 100Ω = 7mA through the LED.
This may be better shown with a couple of examples, first we'll look at the emitter side:
Simulation:
This shows the base switching from 0 to 3.3V every 100ms.
As you can see, the highest voltage seen at the top of the LED + resistor is only ~2.5V, so allowing for ~1.8V drop across the LED we only have ~0.7V left for the resistor. So we get a maximum of 0.7V / 100Ω = ~7mA.
Now let's look at the collector side:
Simulation:
Here we are switching the base from 0V to +3.3V every second (no reason for the time difference, just set up that way)
Now we have almost the full 3.3V across the LED + resistor (minus a few 10's of mV for the transistor saturation voltage) so we get a higher current. If we assume 1.9V for the LED (the Vf will rise a bit for a higher current, then we have (3.3 - 1.9) / 10 = ~14mA, which is what we are seeing.
So, remember that the emitter voltage will always be around 0.6V - 0.7V above the base voltage (when base emitter is forward biased) So for example, if the base is at 0V then the emitter is at ~0.6V. If the base was at 1V them the emitter would be at ~1.6V.
EDIT - now we know the LEDs are 3.2Vf nominal, a 3.3V supply makes things a little awkward, ideally you would have a bit more headroom.
However if you study the datasheet (not given) then it should have a IV curve so you should be able to calculate things from this. The 3.2Vf value will probably be given for something like 20mA, for say 10mA it may be 3V, so you can work out the resistor value to give you roughly your desired current.
Best Answer
You will have a difficult time lighting LEDs in parallel reliably. A diode's relationship between current and voltage is subject to manufacturing variation and temperature dependence. When you put LEDs in parallel, the voltage across each must be equal, but the current may not be, because of these variations.
Importantly, a hot LED will draw more current at the same voltage than when its cold, so any LED which begins to heat unevenly will draw more current away from its parallel siblings, making it hotter, thus drawing more current. This sort of positive thermal feedback is called thermal runaway and is one reason why paralleled LEDs are avoided in all but the cheapest of cheap chineese flashlights.
That said, if all your LEDs are dim, it's likely because they are broken or not supplied with sufficient current. It sounds like you tried one LED with a resistor, and it was, then added more LEDs in parallel with the same resistor, and it was dimmer. If we just assume those LEDs share the current equally (ignoring the instabilities just mentioned), then they must share the current available. So, if you have 5 LEDs in parallel, each will have one fifth the current as there would be with just one LED.
Here's a simplified example, with just two LEDs in parallel, compared to just one LED:
Here we assume that the voltage across an LED is always 1.5V. We know the battery is 5V, so the voltage across the resistors must be 3.5V. By Ohm's law, the current through R1 or R2 must be \$ 3.5V / 175\Omega = 20mA \$. On the left, this current has only one place to go, through D1. But on the right, this current is split in half.
Again, don't parallel LEDs like this. The problem is that between manufacturing variation and temperature dependence, we can't know if the LEDs will split the current 50%/50% or 10%/90%. If you size your resistor to supply enough current for five LEDs, and one ends up hogging all of it, it will probably get too hot and fail.