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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.

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2 Answers 2

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What is the operational difference between this LED driver and a general boost converter?

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: -

enter image description here

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: -

enter image description here

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.

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    \$\begingroup\$ Thank you very much for this clear and wonderful explanation! \$\endgroup\$
    – user220456
    Commented Mar 31, 2021 at 13:57
  • \$\begingroup\$ Just a question, what happens when no current flows through the LEDs. (I am not sure, whether this scenario can happen, just trying to understand). In that case, how would the output voltage be and what would the voltage at FBX pin? \$\endgroup\$
    – user220456
    Commented Mar 31, 2021 at 14:02
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    \$\begingroup\$ The output voltage will rise until current does flow through the LEDs or the maximum output that the chip can produce without burning is reached. It's a feedback loop and the output voltage will rise to whatever level is needed to force that current into the LED string and down through the lower resistor. $$$$ Imagine what happens to the output voltage of a regular boost converter when the upper resistor goes open circuit; smoke and fumes. \$\endgroup\$
    – Andy aka
    Commented Mar 31, 2021 at 14:04
  • \$\begingroup\$ In that case, the output voltage will rise which will also make the voltage at the FBX pin to rise (following the output voltage). Am I correct? \$\endgroup\$
    – user220456
    Commented Mar 31, 2021 at 14:10
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    \$\begingroup\$ If the upper resistor is open circuit (or the LEDs are disconnected), there can be no current into the lower resistor and the FBX voltage will remain at 0 volts and the loop will not stabilize and there will be smoke and fumes. However, the Allegro chip (as a specialist LED driver) will probably have something to "catch" that scenario and prevent device failure. \$\endgroup\$
    – Andy aka
    Commented Mar 31, 2021 at 14:31
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A boost converter usually is set to output a specific constant voltage, to drive loads that need regulated voltage, and the current the load takes can vary up to a specified limit.

A LED driver usually is set to output a specific constant current, to drive LEDs that need regulated current, and the voltage the LEDs have can vary up to a specified limit.

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  • \$\begingroup\$ Thank you for the answer. So, an LED Driver sets a constant output current. In the process of setting the output current, how does the LED Driver decide on how much the output voltage has to be? How does it arrive at the output voltage? Because, the LED String would always have 300mA of output current, but I can have any number of LEDs (say 4 or 8 LEDs) in the LED String? In this case, due to difference in LEDs, how does it know to adjust? \$\endgroup\$
    – user220456
    Commented Mar 31, 2021 at 12:44
  • \$\begingroup\$ Could you tell me - what pin helps to decide the maximum output voltage and how to control the maximum output voltage for the above LED Driver? \$\endgroup\$
    – user220456
    Commented Mar 31, 2021 at 12:47
  • \$\begingroup\$ The driver does not set the voltage. It sets the current and the voltage ends up being whatever the LEDs happen to need to pass the fixed amount of current, unless there are too many LEDs and the maximum output voltage the driver can provide is reached. \$\endgroup\$
    – Justme
    Commented Mar 31, 2021 at 12:47
  • \$\begingroup\$ that's wrong , the driver adjusts the output voltage so that there is a constant current \$\endgroup\$
    – roaibrain
    Commented Apr 12 at 4:25
  • \$\begingroup\$ @roaibrain What you wrote essentially means it does not set the output voltage but the current. The "output" that it measures the feedback is not the voltage over LEDs but the current through the LEDs, maybe converted to voltage with a low-value shunt resistor, but it's still the current you as the user of the module see to be constant. Hence, as a whole, the driver really sets the current. \$\endgroup\$
    – Justme
    Commented Apr 12 at 5:34

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