Wide forward voltage -- no resistors needed?

[rod_mullen]

I would like to make a flashlight/toy that has multiple LEDs. The forward voltage on the 5mm LEDs is 2.5V-6V ... a pretty wide range compared to others, with forward current of 25mA. If I want to power 5 of them with 3 AA's in parallel, I will have an output of 4.5V, correct? In that case, I will be safely within the forward voltage range, and should get pretty good battery life because even as the AA's get down to 1.1V (or close to dead) there will still be enough power to light the LEDs.

Because I can't ever exceed 6V with 3 AA's, is it safe to wire up this circuit without any resistors, thus lowering cost and boosting efficiency? Is such a wide vf range actually a correct specification? If it is relevant, the LEDs are color-changing and have an IC inside them.

Please excuse my lack of knowledge, as this is my first ever project and this terminology is all very new to me.

Thanks!

 

[rod_mullen]

I just realized this thread should have been in the "custom flashlights" forum...does a mod want to move it?

 

[parametrek]

We'd need to see a data sheet for the LEDs.

For consistency and much improved battery life, you could use a simple linear current regulator like the NSI450XX. It is a simple two pin device you use exactly like a resistor, but will take any voltage input and limit it to 15/20/25/30mA.

 

[rod_mullen]

these are the specs from the wholesaler:

model: YB-3120B1RGB-2-ZM
Forward Voltage (V) : 4.5V
Forward Current : 20mA

What I may have seen before was that the wholesaler sells LEDs that may have any voltage within the previously stated range. However, I ordered them directly from China so I've seen specs on other Chinese sites saying vf 3.5-4.5V, 6V, etc, and 25 or 40mA for the same exact model of LED. So I guess there's no way for me to say. Can I determine it by using a meter?

Let's just pretend they are exactly vf 4.5V as stated on the package. In that case, 3 AAs would provide exactly that and I wouldn't need any resistors, right? I mean what's the worst that can happen...I blow up a 20-cent LED? The battery won't blow up, right?