How to drive an LED with an LED driver

Question

I found BCR421 driver and want to ask how to properly drive TSSF4500 LED. All the parameters seem to make sense if look separately, but where do I start to determine the necessary parameters?

It should be a continuous mode with voltage supply of say 3.3V. I would like to be able to adjust the current through the LED from 10mA to 100mA.

For that, I am planning to use the trimmer resistor in series with a 10 Ohm resistor in place of the Rext, where the 10 Ohm on its own will drive the Iout to about 90mA, and additional resistance from the trimmer will allow to drop the current to 10mA, according to the plot below:

1. Should I care about the Vout value? It seems to me like a chicken and an egg problem. Vout depends on the voltage drop across the LED and Rext, and determines the current Iout, according to the plot. It the same time, the current through the LED depends on the voltage drop across the LED, which also dictates the Vout that determines the Iout. Well, I can not seem to get out of this loop. So, what determines all the other?
2. How do I determine the current through the Rext, or the Vdrop? Is it correct to assume that for low Rext values the current through it is ~Iout?
3. What is the advantage of this circuit as opposed to simple resistor-diode in series (except the EN pin that allows fast switching)?

Answer 1

1. Yes, you should care. The BCR421 will try to maintain a (roughly) constant current at the collector terminal, however if Vout drops too low then it will not be able to maintain the constant current, as you can see from the graphs you posted. What will happen is that the output will saturate at a lower current than you want. To find out if this will happen, assume that the maximum desired current is flowing through the LED. Subtract the maximum possible LED voltage (which will occur at the maximum current) from the minimum possible supply voltage and compare with the dropout voltage of the BCR421 at that current.

   So the LED drops 1.35V typical at 100mA (1.6V maximum). Your supply voltage is 3.3V nominal, let's dock 5% for tolerance to give 3.13V. That gives the chip 1.54V to work with. The parameter VSmin indicates that typically 1.4V is enough at 25°C, and the graphs you posted indicate that above about 1.1V it's fairly constant current. So it's likely okay unless you have some extreme operating conditions.

2. The current through the external resistor when it is regulating correctly will be approximately Iout - 10mA. The internal resistor conducts the 10mA. It will be slightly higher because of base current (maybe 0.5%), but that's immaterial in this situation, given the +/-10% tolerance on the 10mA current. As @jippie points out in a comment, 100mA through a pot wiper is probably not a good idea, depending on the pot construction (okay for a wire-wound, but they are expensive). Minimum resistance may also be a problem. It's generally recommended not to put much current through pot wipers- they work better as voltage dividers than rheostats.

3. This circuit maintains a fairly constant current regardless of changes in the LED forward voltage (or differences from unit-to-unit when first assembled) and regardless of changes in supply voltage.

Answer 2

Should I care about the Vout value? It seems to me like a chicken and an egg problem. Vout depends on the voltage drop across the LED and Rext, and determines the current Iout, according to the plot.

What you want to do is make sure Vout is high enough (i.e. that your supply voltage is high enough) that you are on the flat part of the graph. Then the output current becomes dependent only on the resistor value.

How do I determine the current through the Rext, or the Vdrop? Is it correct to assume that for low Rext values the current through it is ~Iout?

You can assume the current through the two paralell resistors is approximately Iout. You are told that the internal resistor value is 100 ohm so working out the proportion that goes trhough each resistor shouldn't be too hard.

What is the advantage of this circuit as opposed to simple resistor-diode in series (except the EN pin that allows fast switching)?

The main advantage is that you can maintain a fixed current into the LEDs across variability in the characteristics of the LEDs and across variations in the supply voltage (which means you can use an unregulated supply).

Answer 3

(1) Yes within limits. If V(Led) + Vout(minimum) > 3.3V this will not work at 3.3V supply.

The graph tells you that Iout is roughly independent of Vout for Vout > 1V so take Vout (min) = 1V and add Vf (=1.3 to 1.6V). So Vout(min) + Vled = 2.6V < 3.3V and we're OK. (But running this LED at 1.5A, or running a white LED, would not work).

(2) Current in Rext = Iout - (Iout for Rext=infinity).
Practically, Current in Rext = Iout

(3) It keeps Iout more stable as your supply voltage and LED temperature varies. For example if running off a battery, Iout would be roughly constant until Vout fell below 1V (Vsupply < 2.6V).