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LED Drivers/LCD Bias
LED Drivers/LCD Bias Forum
TLC5927 running hot!
I have a question about TLC5927 driver. I use TLC5927 to drive an LED digit. Each of the 16 outputs is connected to 4 LEDs in series (except 2 of the 16 outputs that are not used). I am using 750 Ohm resistor for Rext to set Iout=25mA. LED Vf=2.1V. Therefore total voltage drop over LEDs = 4 x 2.1V = 8.4V I am using 12V power supply, so that leaves 12V-8.4V=3.6V over the TLC5927. My question is why is TLC5927 is running REALLY HOT in this configuration? Datasheet states that outputs can handle up to 17V (20V max)???
The maximum output voltage capability is necessary when the LEDs are off because then the voltage on the output will increase. In addition the output voltage capability does not mean that the IC is able to handle the power dissipation without additional cooling.
The power inside the IC in your configuration is: Pic = 16 * 3.6V * 25mA = 1.44W!
As long as your outputs are still on, this means the part can handle the power because the IC temperature is below the overtemperature shutdown threshold, but the IC is getting hot, which is normal for such a high voltage on the outputs. You need 0.6V for 25mA on OUTx pins and therefore when you reduce the 12V to 9V, the IC will not get as hot.
On page 6 you will find the power dissipation capability of the IC. The PWP package is capable of the highest power dissipation, but you need to connect the PowerPad to a big area on your board to help the IC dissipate the power.
Brigitte, thank you for your prompt response!
I see what you mean. Too bad I'm using the worst package as far as heat dissipation goes - DBQ! Can't really put any heat sinking on it...
Also, I did try it with 9V and of course it doesn't get nearly as hot.
I just wanted to use 12V power supply since it is much more widely available than 9V. :(
You could add a resistor to each of the strings to drop some of the voltage and power outside of the IC package. Basic equation is
Rser = (Vled - n*Vf -Vds)/Iled
Assume the Vds will be ~1V.
Putting a resistor in series with an output would limit the current to the LEDs on that output. This doesn't make sense, since the current is set by Rext.
That is the beauty of the Constant Current LED Outputs. The current will be the same no matter the load. The only key being the voltage drop across the components between the LED voltage and the device. With 25mA per output, and needing a 2V drop, add (2/0.025 = ) an 80 ohm resistor. The drop across the resistor will be 2V (make sure you take power into account - 2V * 0.025 = 50mW) when choosing the resistor power.
OK, so if I understand right, using a 12V power supply and 4 LEDS in series on an output, with LED Vf=2.1V, there is a voltage drop of 8.4V over LEDs leaving 3.6V to dissipate in the driver. Rext is set to limit output current to 25mA. So if I install a resistor (3.6V/.025A=144 Ohms) in series with the LEDs the 3.6V would now drop across the resistor not the driver. If I use a smaller resistor, say 120 Ohms to drop just 3V that would leave only 0.6V to dissipate in the driver which shouldn't be a problem (0.6V*.025A=0.015W*16=0.24W). And the string current will still stay .025mA because even though the smaller resistor would cause more current to flow using conventional driver, the TLC5927 still limits that current to 25mA. Did I get this right?
The only draw back is that one of the advantages using a constant current driver was to eliminate the use of resistors.
Your understanding is correct. There are two points to keep in mind during your development. First, the VF of your LEDs is 2.1V. You need to consider variation between the VF of different LEDs and lot to lot variation. If your variation is small, then your calculations are correct. If the variation is large, you need to mind the resistor value. This brings up point number two. The minimum Output voltage of the LED Driver must be met for your intended 25mA. If you have a lot with VF of 2.4V then you would have 9.6V plus 3V (120 * 0.025). That is greater than 12V. At 25mA, you would also need VOUT of about 0.6V.
The resistor size needs to take into account any VF variation and plan for the highest VF. If VF variation is 2V to 2.2V, then the R calculation would be based on 2.2V.
12V - 4*2.2V - 0.6V = 2.6V
2.6V / 25mA = 104 Ohms. Using a 100 ohm resistor would give a 2.5V drop. Now, consider the worst case dissipation at VF = 2.0V.
12V - 4*2.0 - 100*0.025 = 1.5V
1.5V * 0.025A * 14 channels = 0.525W. Based on the worst case DBQ board conditions (99.6C/W) your die temperature would be roughly 52C above ambient if running 100% ON.
And, yes, the DBQ has the highest thermal impact. Using the PWP in a solid thermal designed board, your increase would be about 1/3 (18C).
OK, great. Dick, thank you for your input. I am aware of LED bin variations.
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