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Datasheet

The TLC5940 drives 16 channels of LEDs at constant current using PWM. I assume the 120mA limitation per channel as described in the datasheet has to do with how much power the HTSSOP package can dissipate.

I'm trying to power LED strips of varying lengths on each channel of the chip, and these LED strips have current-limiting resistors integrated in them. Since the current is being limited by the resistors in the LED strips, can I go above the 120mA-per-channel limitation in the datasheet, since the dissipation is happening in the strips, instead of in the chip?

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  • \$\begingroup\$ You can, but you'll tend to burn it out. \$\endgroup\$ – Spehro Pefhany May 24 at 19:24
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No, voltage is distributed that way, not current. Current is the same for things in series so 120mA is passing through the IC either way so heating is the same since nothing changes as far as the IC's I and R.

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  • \$\begingroup\$ Moreover, I would advise to take 100mA as real limit. And check about power dissipation. You may need a heatsink. The problem is that LED strips are paralleled circuits and the longer, the more current they take, If you could put LEDs in serie, then you could drive up to 5 x 3V in serie with the same number of amperes. You would have 5x more light output while staying within the limitation. But power dissipation would have to be taken into account seriously. \$\endgroup\$ – Fredled May 24 at 20:21
  • \$\begingroup\$ But the voltage and thus the power dissipation does change. The TLC5940 is meant to be used as a current-limiting LED driver with no resistors in series. So the power dissipation according to the datasheet, depends on the voltage being dropped across the chip. Ex. if I have 5.5V powering a 3V LED, then 2.5V must be dropped across the chip and Pmax = 2.5 * .120. But if I have 12V powering a 12V LED strip with current-limiting resistors integrated, then there is zero voltage drop across the chip and power dissipation is (nearly) zero. \$\endgroup\$ – Ben Winter May 26 at 1:18
  • \$\begingroup\$ @BenWinter I see what you mean and that reasoning is okay, but it is stil unlikely the chip itself can handle much more than 120mA due to its design \$\endgroup\$ – DKNguyen May 26 at 1:57
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These drivers switch the LED inputs hard to ground (or, as hard as they can), so the power dissipation on each internal switch is only determined by the \$R_{DS_{on}}\$ (the switch resistance in ON-state) and the input current:

\$P = R_{DS_{on}} \cdot I^2 \$

As you can see, the voltage over the switch in Off-state does not matter at all for the power dissipation.

So the answer is no! Exceeding the specified current limits will degrade/destroy your driver.

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  • \$\begingroup\$ I understand what you're saying, but the datasheet has a different equation listed for power dissipation under section 11.3. It consists of the power from the device's baseline current draw, plus the power dissipated across the channel outputs, which is calculated by Vout * Imax (times some other things <1.0 like greyscale and PWM dimming). If I'm using a 12V power supply to power a 12V light bar with resistors, then Vout = 0, since the entire 12V is being dropped across the LED strip, and not the chip, in the on-state. So isn't power dissipation actually zero? \$\endgroup\$ – Ben Winter May 26 at 1:06
  • \$\begingroup\$ Ah I see, I honestly did not realize the constant current capability. Well in that case you are right somehow, the power dissipation will actually be lower when you drop most of the voltage on an external resistor. But still, the datasheet's absolut maximum rating for output current is still only 130mA, and you really shouldn't exceed the absolut maximum ratings. \$\endgroup\$ – jusaca May 26 at 17:31

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