Power Dissipation

From the datasheet:

Explanation

• V_cc is +5V for regular Arduino's.
• I_cc (Supply current) is listed in Electrical Characteristics on pg 4 of the datasheet:
  
  I'll assume you're using R_iref = ~2k, so I_cc < 25mA. (Data transfer is usually 8MHz, and assuming outputs are on).
• V_out is the voltage going into the tlc pin, which would be the voltage going into the LED minus the forward voltage drop across the LED. For example, +4V --> LED --> TLC, then Vout = 4 - (forward voltage drop across LED, usually 2.1V, see the LED datasheet) = 1.9V. I'll assume you're using +5V, so +5V - 2.1V = 2.9V.
• I_max is the calculated from the R_iref resistor, I_max = 39.06 / R_iref. I'll assume R_iref = 2k so I_max = 39.06 / 2000 = 20mA.
• DCn is the dot correction value for channel n. Unless you're setting the dot correction, it's 63.
• d_pwm is the current channel setting, (aka GS PWM value), divided by 4095. For example, Tlc.set(0, 4095) means d_pwm = 4095/4095 = 1. Lets assume d_pwm = 1.
• N is the number of channels set to d_pwm. For example, if we know that we're not going to turn on more than 5 channels at once, then N = 5. Lets assume N = 16, or we might turn all channels on at some point.

Example Calculation

Here's our assumptions from above:

• V_cc = 5V
• I_cc < 25mA (Looking up value with R_iref = 2k from table above)
• V_out = 2.9V (assuming +5V --> LED with Vf of 2.1V --> TLC)
• I_max = 39.06 / R_iref = 20mA. (Assuming R_iref = 2k)
• DC_n = 63 (unless you explicitly set this)
• d_pwm = 1 (whatever the maximum channel setting will be, eg Tlc.set(0, 4095) -> 4095/4095 = 1)
• N = 16 (maximum number of channels that will be on at once)

P_d = (V_cc * I_cc) + (V_out * I_max * (DC_n / 63) * d_pwm * N)
= (5V * 25mA) + (2.9V * 20mA * (63 / 63 = 1) * 1 * 16) = 1053mW

Let's assume that the chip is in an environment with a ambient temperature T_a = 30C. Then by power dissipation table (DIP is the package that fits on a breadboard), the maximum power dissipation is 2456mW - 19.65mW * (T_a - 25C = 30C - 25C = 5) = 2357.75mW.

So from before, if R_iref = 2k, the LED voltage is 5V, and all the outputs are on, the chip generates 1053mW and we should be fine.

Now assume that we lowered the LED voltage to 4V (so +4V --> LED --> TLC). Then
Pd = (5V * 25mA) + (1.9V * 20mA * (63 / 63 = 1) * 1 * 16) = 733mW

So if you're afraid of burning out TLCs, lower the LED voltage. But make sure it's enough voltage to turn on the LEDs! It should be a bit above the forward voltage drop from the LED datasheet.