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### Topic: TLC5940 LED Driver power dissipation and connecting pins in parallel (Read 2064 times)previous topic - next topic

#### JBMetal

##### Jul 06, 2011, 10:09 pm
Hi

I have two questions in regards to the TLC5940:

1. Power dissipation:
The calculation looks like this:
P =   (Vcc x  Icc) + (Vout * Imax * (DCn)/63 * Dpwm * N)

What is the Vout value referred to here? Is it what is indicated in figure 6 on the datasheet as 55 ma (and lower) = 0.5V and 61ma = 1V (at 25C) or is it the actual voltage supply for the LEDs (max 17V as per page 3)?

2. Would it be possible to connect a 1W 350ma LED 'in parrallel' over 7 of the outputs to 'split' the current?

Datasheet:
http://www.ti.com/lit/gpn/tlc5940

#1
##### Jul 07, 2011, 12:02 am
Vout is the voltage drop for each channel, measured at the chip between the relevant pin and GND. The chip burns what isn't "taken" by the LEDs.

Yes, you can parallelize the outputs to drive one big LED.

#### JBMetal

#2
##### Jul 07, 2011, 07:41 am
Thank you.

So it would make sense to power the LEDs at a voltage as close as possible to the max Vf rating of the LEDs e.g. if I drive each set of 16 white LEDs with a Vf of 3.4V max at 4V, Vout would be at 0.6V?

Mmm ... any reason then why I can't attempt to approach Vout = zero and drive them at 3.45V for example?

#### jluciani

#3
##### Jul 07, 2011, 12:24 pm
The range of Vf for the LEDs is from 3.4V to 4V. Typically the Vf will be 3.4V but a 4V LED
is also within specification. Your Vout would be 4V.

On my 5940 LED driver board http://wiblocks.luciani.org/NB1/NB1A-LED-index.html
I add a separate adjustable voltage regulator for the LEDs. This enables the Vout to be adjusted
for different Vf LEDs. The power dissipation is divided between the regulator and the 5940.
The regulator can dissipate 1W at 25degC without a heatsink.

The schematic is in the datasheet.

(* jcl *)

#### JBMetal

#4
##### Jul 10, 2011, 07:57 pm
Sorry to have to bring this up again but the last two replies don't seem to tie up, at leat in my interpreatation.

Madworm says its the voltage drop measured at pin, which, as I interpret it is: 4V supply - 3.4V Vf on LED = 0.6 VOut

jluciani says its 4V, which I think he means is either max Vf for the LEDs or supply voltage .. ?
Problem with this value is that the calculation at 4V supply, with 30 milliamps (dot correction and pwm at 100%) looks like this:
P =   (Vcc x  Icc) + (Vout * Imax * (DCn)/63 * Dpwm * N)
P =   (5 x  0.130) + (4 * 0.025 * 1 * 1 * 16)
p =   0.65 + 1.92  =   2.57W

The package rating is 2.4W (at 25C)

Does this mean that the recommended constant output current for all outputs at 60 milliamps in the datasheet is unachievable, insofar the DIP package is concerned?  :~

#5
##### Jul 10, 2011, 08:51 pmLast Edit: Jul 10, 2011, 08:54 pm by madworm Reason: 1
I think TI would have a very hard time selling led drivers that have an internal voltage drop of 4V at just a few mA.

Just for comparison: the MBI5168 is designed to tolerate about 0.7V internal voltage drop at about 80mA.

"Vout = Vsupply - Vf" and nothing else.

If you want to feed "x mA" through your led(s), the graphs I/V graphs tell you that you have to expect a voltage drop of "z V" inside the chip. I suspect the graphs are for short-circuit conditions of the outputs (as shown in Figure 2). Therefore your supply voltage must compensate for that and be high enough, otherwise the current won't be reached. This also means that Vout will be higher, if your supply voltage is too high. Therefore is makes sense to adjust the supply voltage to be pretty close to the expected Vout and the expected Vf. Maybe 1V above the sum, so the regulator has some leeway to work. If you give it too much it will burn.

In my projects I've adjusted things so that touching the chip with my index finger doesn't cause pain. If it does, it is too hot.

Edit:

In the case of 60mA, the graph says Vout=1V. Now have a look at your LED's datasheet and find the Vf that matches 60mA as well. Your supply voltage should be at least the sum of those two voltages, maybe a bit higher, but not much. Any excess voltage will be turned into heat.

#### jluciani

#6
##### Jul 10, 2011, 09:53 pm
@JBMetal -- This is my mistake. Sorry about that. You're original equation
looks correct.

I was thinking of Vout as the voltage source connected to the
anode of the LEDs.

As you suggest if you set your LED voltage source to 3.45V you will
dissipate very little power in the LEDs. The issue is that the 3.4V is
a typical value and the mfg could ship you LEDs that have a Vf up
to 4V.

The other issue (which may not matter in your application) is cold-temperature
startup. The Vf increases as the temperature drops. It varies depending on
the LED. For white LEDs it is typically in the 2-3mV/degC range. If your LED voltage
source is set too low you will not be able to drive them to the desired
current on startup.

(* jcl *)

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