OK, here we go. How the hell am I gonna wire and plug all of that stuff? Well, first, let's go back to the very nature of the LED. Once that is understood the sky is the limit (depending on how big your home is or how high your roof is).
The LED is solid state device and the word is actually true. It is simply impossible to kill, but for two things. Too much electrical current will fry it and too much heat will also destroy it. Let's first look at the current issue. The wattage? It has to do with both, temperature and current. I'll adress this topic in both discussions.
Well, why not jump into it right away. Take a look at the data sheet for a given 3W LED. You will notice that the test voltage is 3,6V and the test current is 500mA. Multiply the both together and you get 1,8W. Strange, isn't it? Not at all. All that means is that this given LED will sustain heat up to 3W and beyond that point, it'll fry. It is actually meant and designed to work at 3,6V and 500mA (1,8W) for optimal operation and last the stated 50,000h. Yes, you could push it near to it's limits, but you'll be hurting her bad.
So at this point, one must understand that the main factor to drive a LED is polarizing current. If you exceed that current, the maximun power and heat dissipation will be reached and the poor thing will rejoin her ancestors in the silicon kingdom. But how can one limit, or make sure that that faithful current is not reached? By using a constant current power supply, also knowned as a LED driver. But there are other ways.
A LED (and any given diode, as a matter of fact) has an almost vertical I/V curve (current over voltage). In other words, you polarize it with some insufficient voltage and nothing happens. You push the voltage up until you hit the "knee" of the curve and a small glow will occur. You push just a little further and she will light in full glory. But just a few mV volt more and the current will sky-rocket and the LED will burn. Thus the reason for limiting the current. Power supplys come in two flavors. Constant voltage, where the voltage will remain constant and the current will vary depending on the load and constant current that work just the opposite. Those are the ones used for LED. Now, let's design.
Let's assume that a blue, white or full spectrum (the both being intrinsicly blue LEDs) have a Vf (forward voltage drop) of 3,6V and a red LED has a Vf of 1,8V. Those figures, though not exact to all, are recognised as proper approximations. If I were to string one blue, one white and four red in series, I'll end-up with 3,6V+3,6V+(4X1,8V)=14,4V. So, my constant current LED driver will have to be able to push at least 15V. If we are dealing with 3W LEDs, it'll be set to 500mA. If we are dealing with 5mm small LEDs, it'll be set to 20mA. In both cases, the current must be respected according to the device in use and in both cases, the PSU will have to be able to push more than 14,4V.
What do we get? In the first instance 14,4V X 500mA = 7,2W and in the other 288mW. That simple. This is where the constant current PSU shines.
But if I were to use a constant voltage PSU, what would happen? Say a recycled laptop PSU @ 19V/5A. Everything would instantly burn. But there is a twist to it. And this is where we come on the cheap. Yes! Using salvaged laptop PSUs.
Let's think of the 3W LED string. It tallies-up to 14,4V. With my 19V PSU, I am 4,6V too many. So, since I must limit the current to 500mA, all I have to do is put a resistor in series. 4,6V / 500mA = 9R2. Yes, just a 9,2 Ohms resistor and bingo! Now, what power of a resistor? 9R2 X 500mA X 500mA = 2,3W or if you prefer, 4,6V X 4,6V / 9R2 = 2,3W. So, a 5W resistor would nicely survive.
But efficiency is quite low. 7,2W in the LEDs and 2,3W in the resistor. Let's work it better. Say that I had used a cool white LED in my string. What about adding a warm white LED? My string is now 18V @ 500mA (9W) and my calculated resistor becomes 2 Ohms @ 500mA (0,5W). That is much better, Ain't it?
Next time, we'll talk about construction and heat dissipation.
So long...
The LED is solid state device and the word is actually true. It is simply impossible to kill, but for two things. Too much electrical current will fry it and too much heat will also destroy it. Let's first look at the current issue. The wattage? It has to do with both, temperature and current. I'll adress this topic in both discussions.
Well, why not jump into it right away. Take a look at the data sheet for a given 3W LED. You will notice that the test voltage is 3,6V and the test current is 500mA. Multiply the both together and you get 1,8W. Strange, isn't it? Not at all. All that means is that this given LED will sustain heat up to 3W and beyond that point, it'll fry. It is actually meant and designed to work at 3,6V and 500mA (1,8W) for optimal operation and last the stated 50,000h. Yes, you could push it near to it's limits, but you'll be hurting her bad.
So at this point, one must understand that the main factor to drive a LED is polarizing current. If you exceed that current, the maximun power and heat dissipation will be reached and the poor thing will rejoin her ancestors in the silicon kingdom. But how can one limit, or make sure that that faithful current is not reached? By using a constant current power supply, also knowned as a LED driver. But there are other ways.
A LED (and any given diode, as a matter of fact) has an almost vertical I/V curve (current over voltage). In other words, you polarize it with some insufficient voltage and nothing happens. You push the voltage up until you hit the "knee" of the curve and a small glow will occur. You push just a little further and she will light in full glory. But just a few mV volt more and the current will sky-rocket and the LED will burn. Thus the reason for limiting the current. Power supplys come in two flavors. Constant voltage, where the voltage will remain constant and the current will vary depending on the load and constant current that work just the opposite. Those are the ones used for LED. Now, let's design.
Let's assume that a blue, white or full spectrum (the both being intrinsicly blue LEDs) have a Vf (forward voltage drop) of 3,6V and a red LED has a Vf of 1,8V. Those figures, though not exact to all, are recognised as proper approximations. If I were to string one blue, one white and four red in series, I'll end-up with 3,6V+3,6V+(4X1,8V)=14,4V. So, my constant current LED driver will have to be able to push at least 15V. If we are dealing with 3W LEDs, it'll be set to 500mA. If we are dealing with 5mm small LEDs, it'll be set to 20mA. In both cases, the current must be respected according to the device in use and in both cases, the PSU will have to be able to push more than 14,4V.
What do we get? In the first instance 14,4V X 500mA = 7,2W and in the other 288mW. That simple. This is where the constant current PSU shines.
But if I were to use a constant voltage PSU, what would happen? Say a recycled laptop PSU @ 19V/5A. Everything would instantly burn. But there is a twist to it. And this is where we come on the cheap. Yes! Using salvaged laptop PSUs.
Let's think of the 3W LED string. It tallies-up to 14,4V. With my 19V PSU, I am 4,6V too many. So, since I must limit the current to 500mA, all I have to do is put a resistor in series. 4,6V / 500mA = 9R2. Yes, just a 9,2 Ohms resistor and bingo! Now, what power of a resistor? 9R2 X 500mA X 500mA = 2,3W or if you prefer, 4,6V X 4,6V / 9R2 = 2,3W. So, a 5W resistor would nicely survive.
But efficiency is quite low. 7,2W in the LEDs and 2,3W in the resistor. Let's work it better. Say that I had used a cool white LED in my string. What about adding a warm white LED? My string is now 18V @ 500mA (9W) and my calculated resistor becomes 2 Ohms @ 500mA (0,5W). That is much better, Ain't it?
Next time, we'll talk about construction and heat dissipation.
So long...