RE: TI and Würth Elektronik LED RoadTest+ - The Schema - 5/10

jc2048 — Mon, 04 Jan 2016 14:25:06 GMT

Don't do the pcb yet, this isn't going to work.

The LEDs won't light up (and the transistors couldn't switch them properly if they did).

You need to read up on Ohm's law and look carefully at how diodes behave.

Let's start with the transistors and pretend, for the moment, that your LED's were taking their design current of 350mA. The three groups of LEDs would take 1.75A (5 x 350mA), 2.1A (6 x 350mA), and 2.8A (8 x 350mA).

Absolute maximum for 2N2222A is 800mA, so all three exceed the capability of the transistor by at least twice. So switching them with this transistor is a non-starter.

Even if the current were much less, it wouldn't work as you expect. Let's say that there were just two strings of LEDs taking 2 x 350mA = 700mA. If you look at the datasheet for the transistor it says that the current gain falls to 40 at 500mA collector current. Let's pretend that it's the same for 700mA (not true, but good enough for this discussion), then you need 700mA/40 base current which is 17.5mA. But your base resistor only gives you about (5V-0.6V)/1000 ohms = 4.4mA (I'm assuming the uP IO is 5V, adjust accordingly if it's 3.3V), which isn't enough. You could decrease the size of the resistor, but then you'd be sourcing 18mA from your IO pin which it would manage but is a bit on the high side.

Anyway, to get around this you either need more gain and a heftier part (look at a darlington pair) or use an n-channel MOSFET.

Now to the LEDs.

Imagine the LEDs weren't there and were replaced with wire links. What is the resistor current? By Ohm's law, it's 12V/1000 ohms = 12mA. Let's put back one LED with a voltage drop of 3.2V. What's the current now? (12V - 3.2V)/1000 = 8.8mA. [That value isn't accurate - the forward voltage is at the rated current of 350mA and would be a good bit less for only 8mA - but the point is that the 12mA is the most you can ever get and adding diodes reduces from that.]

So, let's say we put three diodes in a string, run them from the 12V, and calculate the resistor value we need.

The resistor value to get 350mA is (12V - (3 x 3.2v))/0.35A = 6.85 ohms. The resistor heat dissipation is 2.4V x 0.35 A = 0.84W, so that would need to be a 1W resistor. [Note, while we're here, that each LED is about 1W, so the resistor is using about a quarter of the total power for the string, which is quite wasteful.]

However, the LED will warm up (how much depends on how good your heatsinking is) and the forward voltage will drop. That means more voltage across the resistor and the current will go up. A drop of 0.2v [for each LED] isn't unreasonable for the kind of heatsinking you'll manage (LED chip at 80C rather than the 25C at which the datasheet Vf value is measured). That would then give a current of (12 - (3 x 3.0))/6.85 = 438mA. So, you can see the resistor isn't regulating the current very well at all. It's also worth bearing in mind that the forward voltage is just a typical figure, and the actual LEDs you have may be different.

You also have the further problem that your supply isn't really 12V in a car (automobile). If the alternator is running, the regulator will hold the supply higher than that to charge the battery, and if it isn't running the supply depends on how run down the battery is. [Be aware,too, that you need good protection on automotive circuits - there can be hugh spikes on the supply.]

If you use an IC LED driver, it will measure the current through the LEDs and keep adjusting the voltage across the LEDs to keep that current at the required value. Any changes, because the LEDs differ from the typical value, or because they heat up, or because the supply voltage varies, don't matter [as long as they're within the range the driver can adjust for, obviously].

RE: TI and Würth Elektronik LED RoadTest+ - The Schema - 5/10

DAB — Wed, 30 Dec 2015 22:15:46 GMT

Nice start.

DAB