Interpreting specs such as these found on many marketplace sites can be a bit of an art form, especially when no datasheets are provided.
When in doubt, the best idea is to ask the seller. If they cannot provide the information, then you have two choices: take pot luck, or find another seller!
That being said, here's how I read the information you are being provided in this case:
- Input Voltage: Dc 12V – 3A
.. is I believe the spec of the power supply that is provided, not the strip itself. In other words, capable of delivering a total of 36W at 12V.
- Lamp power: 12V 4.8 Watts
.. is most likely per meter i.e. each 60 LEDs. That is a pure guess based typical power requirements and how specs for similar LED strips are presented. You can divide/multiply this out by the actual number of LEDs you plan to use.
If you are planning to power a strip with batteries, the main challenges will be battery life and how many LEDs you have the power to drive.
Trying to use 2x9V does not seem like a very good idea. While you could probably drive at least a few LEDs with 2x9V, you will be wasting a lot of their already-limited (~500mAh) capacity in the step-down to 12V. And a step-down with resistors is going to be tough to get right. If you cut the strip and can squeeze say 0.5A out of the batteries, that's still 3W to burn off. A voltage regulator would be better if you really must.
AA or even C, D or 6V lantern batteries will be able to drive more LEDs for longer, and you can avoid losses in voltage regulation.
Firstly, to find the Vf for the LEDs your best bet is to find a similar component from a major supplier (RS, Farnell etc.). I've found a 5630 Red LED from RS components, whilst this isn't 100% accurate it will give a you a relatively close starting point.
A lot of these strips call for a 12V supply because it's simple and the majority of people using these strips don't put as much thought in as you have. If you wanted to keep the same brightness of the strips as when they first arrive then it's a relatively simple operation.
simulate this circuit – Schematic created using CircuitLab
Your strips may have a different setup to this (1 resistor per LED, less LEDs in series etc.) but the principle remains the same. To keep the same 'brightness' then the same current will need to be flowing through each branch of LEDs.
Now I've just used 100mA based on the datasheet of the 5630 red LED I linked. To calculate the current flowing through your LED branch you need to find the total resistance per branch. Using datasheets to get an estimated voltage drop of the LEDs you can calculate how much voltage is being dropped across the resistor(s) and use Ohms Law to solve for I.
The downside is, for maximum efficiency you will need a different power supply for your red LED strip and your blue/white LED strip.
This is the case if you want the LED strips to have the same brightness that they arrived with, if you want to match the brightness of the different LED strips this will be a little more difficult without the original datasheets.
Using the datasheet I linked we can see that the luminous intensity of the red LED is typically 4.2cd, if we want to match this to a green LED with a luminous intensity of 2.8cd we need to look in the datasheet of the red LED and find two graphs. Forward Current vs Forward Voltage and Luminous Intensity vs Forward Current.
Using our example value of green being 2.8cd, our 4.2cd red LEDs luminous intensity needs to be at 66% of its current value (2.8 is 66% of 4.2). In the graphs from the datasheet, we can now calculate the forward current required to have the red LEDs luminous intensity at 2.8cd and also the forward voltage.
Now that we know what the forward voltage of the LED required for 2.8cd we can adjust our first schematic accordingly
simulate this circuit
I may have gone a little bit off track at some points and rambled a bit too much but hopefully I've given you some sort of help.
Edit in response to questions in comments:
All of the voltage that is dropped across the resistor is essentially wastage. The closer to the voltage source to the combined Vf of the LEDs, the smaller the resistor required and therefore less power wastage.
I completely missed this earlier, but thinking now it's become obvious. If you power the LED strip with 12V. Using a multimeter, measure the voltage across the resistor in the LED chain. You can then subtract this measured value from your 12V supply and figure out the combined Vf of the 3 LEDs. Then just divide this value by 3 to get an approximate average Vf per LED.
In order to determine the 'highest light value' you will need to know what the maximum forward current is which unfortunately is in the datasheet. From other 5630's I've looked at they seem to all be around 150mA maximum forward current, unfortunately this might be a case of "Try 100mA, did it die? Try 150mA, did it die?". It's likely the supplier drives them well below their maximum value.
In terms of margin of error to calculate for. A batch of LEDs will have varying Vf values, this is why I suggested finding the combined Vf for 3 and taking an average. I doubt there will be significant differences between the LEDs branches. In order to play it safe however, I'd always pick the standard resistor value that's above what you calculate, for example 44.43R you would use a 47R resistor.
Best Answer
I had a flickering issue with LED strips and I solved it by adding a 1000uf capacitor between positive and negative as close to the LEDs as I could, in your case you want it close to the transistor. This may not be your issue, but it is a good practice anyway. Not all power supplies respond to changes in current draw quickly and the capacitor acts as a buffer.
I would put the negative side of the capacitor where the transistor is connected to ground and the positive side where the LED is connected to positive. Make sure the capacitor is rated for the voltage you are using.
In the diagram above placing it on the two power rails next to the LED should be good.
These capacitors are polarized, make sure the negative side goes to negative and not the other way around.