I am having this A80604 – LED Driver IC for driving a string of 9 LEDs. Each LED has a forward voltage drop of 3.1V and a forward current of 300mA through each LED. So, I need to generate a maximum of 30V for this LED string with forward current of 300mA. My input voltage range is 10V to 16V
I tried to search in google but unable to find. I want to understand – What is the operational difference between this LED driver and a general boost converter?
How does this LED driver help to output the 30V for the LED string when my input is 12V? Is its operation similar to a boost converter?
I have read that for LEDs, we drive the required current through it, and then the appropriate forward voltage appears across the LED. So, does this LED driver work by producing the 300mA and the output voltage of 30V would increase/decrease according to the LED forward voltage drop?
Please let me know on how an LED driver operates and how is it different and advantageous from a normal boost converter IC.
Best Answer
There isn't much of a difference but to get a blow-by-blow absolute definitive list of differences is not going to happen because Allegro will protect their IP and they won't disclose what the bottom end of the LED string truly connects to.
But, you can make certain inroads into a comparison providing you understand that IP is usually a guarded secret. Consider a fairly conventional boost converter like this one: -
The important pin to fixate on is FBX (red-lined). That pin "wants" to be at 1.6 volts and so it will adjust the duty cycle of the converter so that 1.6 volts appears on FBX. This means that the current through the 15.8 kΩ resistor MUST BE 101.27 μA.
And this means that the voltage across the 226 kΩ resistor is 22.89 volts.
And, it therefore follows that the voltage from the top of the 226 kΩ resistor to 0 volts is 22.89 volts + 1.6 volts = 24.49 volts. And, give or take a little bit of handwaving over resistor tolerances and the tolerance on the FBX "must-be" voltage, the output is pretty much 24 volts as stated on the schematic.
The important thing to note here is that whatever value the 226 kΩ resistor is, the output voltage will be manipulated (via PWM) to force a current of 101.27 μA through the lower resistor (15.8 kΩ).
So, now replace the 15.8 kΩ with a 15.8 Ω resistor and the current that the chip wants to take into that resistor needs to be 1.6 volts ÷ 15.8 Ω = 101.27 mA. Now, if you replace the 226 kΩ resistor with a string of LEDs you get this schematic: -
And that current that flows down those LEDs is determined by the FBX "must-be" voltage of 1.6 volts and the lower resistor (now at 15.8 Ω).
So, to get 300 mA, just make the lower resistor 5.333 Ω and what you basically have achieved is a constant current LED driver.
Of course, the Allegro chip has the 5.333 Ω embedded inside it and it probably isn't that value but something a lot smaller and there'll be an amplifier behind it to make the voltage up to whatever is the internal equivalent of the LT3957's FBX voltage but, essentially, it's the same bar a few bells and whistles.
Also note that this analysis can apply to a buck converter too.