I bought this IC (http://www.ti.com/lit/ds/symlink/tlc59116.pdf) to power a few RGB leds. I was under the impression that it could drive up to 100ma per channel, constant current. However in practice I can only get around 180ma total before it self shuts down.

Maybe I misinterpreted the datasheet, and 100ma is the global maximum? Can somebody read it and tell me the correct interpretation, since I'm not really good in reading datasheets yet?

  • 1
    \$\begingroup\$ It was quite tortuous to follow the link to the data sheet. Maybe you can replace it with the TI link directly. Having not looked at the data sheet, what supply voltage are you feeding the chip and what voltage is it putting onto the LEDs? It may be like an analogue voltage regulator where output current won't necessarily make it warm unless the input power voltage is more than a couple of volts above what the diodes need? \$\endgroup\$
    – Andy aka
    Apr 5, 2013 at 23:10
  • \$\begingroup\$ Yea, if the voltage is higher than needed, than the internal transistor has to dissipate it. The closer you match VCC to the LED VF, or Series LEDs, the easier it is the transistor to sink the current. \$\endgroup\$
    – Passerby
    Apr 5, 2013 at 23:33
  • \$\begingroup\$ @Andyaka yep you're right, I switched to 7.5V for the LED (which has a voltage drop of 2.8V) and now it works well. Why is that? \$\endgroup\$
    – luca
    Apr 5, 2013 at 23:46
  • \$\begingroup\$ @Passerby it's actually the opposite, I had to increase voltage. Intuitively you're right but the lower (closer to 2.8V) the worse! \$\endgroup\$
    – luca
    Apr 5, 2013 at 23:47
  • 1
    \$\begingroup\$ @Luca, that doesn't make sense. The datasheet shows that the output voltage only needs about 1v to work right (Figure 9, page 13). Since you have 7.5 vcc, 2.8 vf, that means you are putting 4.7v * 100ma = 0.47w per channel on the internal transistor (Considering 1 led per channel). With 5 RGB leds, that's 7.05 Watts of heat in that chip being wasted. \$\endgroup\$
    – Passerby
    Apr 5, 2013 at 23:54

1 Answer 1


First, the TLC59116 should be able to drive 120mA Per channel, ideally with a small VLED - VF voltage of VOL, to minimize heat issues. The higher the "unused" voltage, the less current can be driven per channel before thermal shutdown happens.

A TI employee responded to a user with the TLC59116. More info at the link


  1. The TLC59116 (TSSOP package) gets hot and nearly goes into thermal shutdown when we run 16 strings of LEDs at 50 mA. Is that to be expected? We think we can get away with using the IC because we will be using the quad package which has higher dissipation.


  1. This depends on the voltage you are using on the LEDs. If the LED anode voltage is 10V and the string consists of 2 LEDs with a forward voltage of 2V each, the IC has to dissipate 16* 6V * 50mA = 4.8W. The power dissipation capability of RHB package is better than PW package because RHB has a thermal pad that has to be soldered to the board. Please make sure that you have a good thermal connection to the GND plane on your board.

So you need to match your LED VCC to the Forward Voltage of each led on that channel. Don't have too much left over voltage, as the internal transistor will need to dissipate that into heat.

From the Datasheet:

Figure 9 shows the output voltage versus the output current with several different resistor values on REXT. This shows the minimum voltage required at the device to have full VF across the LED. The VLED voltage must be higher than the VF plus the VOL of the driver. If the VLED is too high, more power will be dissipated in the driver. If this is the case, a resistor can be inserted in series with the LED to dissipate the excess power and reduce the thermal conditions on the driver.

As for more current, you can tie multiple channels together.

  • \$\begingroup\$ makes sense.. but then why the opposite is happening? If I put 5V, connecting 3 channels together, I get around 180mA. If I put 7.5V, it provides the 300mA I was looking for. \$\endgroup\$
    – luca
    Apr 5, 2013 at 23:51
  • \$\begingroup\$ @luca, 5v with one 2.8v led, that's 2.2v at the outx pin * 100ma, 0.22w disipated, you should be fine. Frankly, I'm not sure then. \$\endgroup\$
    – Passerby
    Apr 5, 2013 at 23:56
  • \$\begingroup\$ I will test more and see If I'm doing something else wrong, thanks for the clarifications \$\endgroup\$
    – luca
    Apr 6, 2013 at 0:02
  • \$\begingroup\$ @luca from page 12 of the datasheet as well (see edited post) \$\endgroup\$
    – Passerby
    Apr 6, 2013 at 0:13
  • \$\begingroup\$ @luca just as a side though, you don't have series resistors on each channel right? \$\endgroup\$
    – Passerby
    Apr 6, 2013 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.