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
Question:
- 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.
Answer:
- 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.
the datasheet doesn't give a definitive forward voltage
It does: The 1.15 Volts typical, 1.5 Volts maximum specified at 20 mA is the LED's forward voltage specification: It never is a precise voltage, LED voltages can vary greatly between units even within a single batch.
For calculating resistance for minimum current, in the absence of input Vf versus current graphs in the datasheet, use the minimum forward voltage rating for calculations: Vf falls with decreasing current through the LED, but not linearly as the question indicates.
So, for 1 mA: (12 - 1.15) / 0.001 = 10.85 kOhms, use a standard value resistor of 10k.
Note that even if one uses the maximum Vf from the datahsheet, the results are 10.5k, not much difference.
Note:
With the CTR of the device specified as 200% minimum, 400% typical, a 1 mA input current would allow a nominal 2-4 mA of Collector Current to flow at the output if there were no other current limiting on the output circuit. This may not be sufficient to hold open the driven relay.
The input current will need to be increased to meet the required output current for the relay. Hence the above 1 mA "desired" current as specified in the question may well be invalid and irrelevant.
Best Answer
Ohm's law applies only to the resistor, so you need to use the voltage across the resistor when calculating current. That means you need to subtract the LED forward voltage from 12V before dividing by the resistance.