LED driver BCR420/1 implementation

Question

I've been trying to design a circuit to control 23 LEDs dimming using a LED driver, but I don't quite understand how they work. Put for example the model BCR420/1:

Datasheet: BCR420/1 Datasheet

I have 23 LEDs to control, with Vf = 3.3V and If = 30mA.

If I wanted to control all 23 LEDs in a single string, It would require Vs to be at least Vs = 1 (voltage drop of the IC) + 3.3*23 = 76.9V, which is too much.

Now, I could use two alternatives:

• Use several LED drivers to control fewer LEDs.
• Use parallel strings with several LEDs in each one.

In case of using parallel strings, it would require Iout = 30mA * Number_of_strings and then I'd have to make sure the string with uneven LEDs had the appropriate voltage. But how could this be done? What's the real meaning of Vout?

So my main questions are:

1. What is the meaning of Vout?
2. Could I use several strings of LEDs instead of an LED driver for a set number of LEDs?
3. If I used several drivers, could the same PWM signal be used for all of them?

Answer 1

1. Vout is the voltage at the OUT pin. It will vary depending on how many LEDs you string together and the supply voltage.

2. If you use multiple strings (in parallel with a single driver chip) the current will almost certainly not balance equally. The string with the lowest total LED Vf will take most of the current.

3. You should be able to use several driver chips in parallel and use a common PWM signal. Per the datasheet you would need to ensure that the PWM signal can provide N x 1.2mA. With N being the number of parallel driver chips. The 1.2mA (typ) is from the datasheet table 2.3.

Answer 2

1. If Vout drops below about 1.5 V, the IC will be unable to keep the current constant. You need to allow some variation in LED voltage drop from part to part and over temperature. +/- 5% would be a conservative number.

VdropLedMax = 3.3 * 1.05 = 3.46 V

VdropLedMin = 3.3 * 0.95 = 3.13 V

If Vs = 24 +/- 1 V, and 6 LEDs, minimum Vout ...

VoutMin = 23 - 6 * 3.46 = 2.24 V <-- Good

Now for power disipation, calculate the maximum Vout:

VoutMax = 25 - 6 * 3.13 = 6.22 V

Power dissipated in the IC = 6.22 V * 0.03 A = 0.18 W <-- Probably OK

Whether 0.18 W is acceptable will depend on factors beyond the scope of this question. Note that I have ignored the power in R3, this could be subtracted (~ 20 mW).

2. Use one series string of LEDs for each IC. You may need several driver LEDs to accomplish your goal.

3. The MCU may be able to drive more than one IC, you will need to know the maximum current available from the MCU GPIO. Worst case add a buffer chip.

^{simulate this circuit – Schematic created using CircuitLab}