# Using a voltage divider on LED IC driver output

Would using a voltage divider on the outputs affect the drivers operation?

The Driver is a 24 channel current sink IC that will be used for automotive interior lighting.

LED will have a Vf of 3.3v @ 20 mA. Using 12v supply to calculate power dissapation from the data sheet formula give a total of 5170 odd mWs which is larger then the IC can sink.

Tried to post the equaation but it gets messed up.

The whole application will have aprox 120 RGB LED but would like to avoid using a seperate 5v supply for the LEDs

TLC5947 data sheet

TLC5947 Output circuit diagram:

• How many LEDs does each output drive? If you can arrange for most of the 12V to drop over the LEDs (minus some margin for lower battery voltage and driver drop) you can limit the power dissipation in the chip a lot. – Wouter van Ooijen Jan 7 '13 at 12:04
• Each output will only drive 1 leg of an RGB led. – Daniel Jan 7 '13 at 12:08

I understand you want to be able to simultaneously feed 20 mA through each subpart of 120 RGB LEDs. That is a total of 7.2A at (maximum) 14.4V, for a grand total of 104W, of which 24W is dissipated in the LEDs. If you exclude switching voltage conversion the remaining 80W of heat must be dissipated somewhere.

A first step is to establish the maximum ambient temperature your chips must tolerate without going into thermal protection mode. Then you can use fig 11 in the datasheet to find the power a single chip can handle. Let's for the moment assume 2W. You need 120 * 3 / 24 = 15 chips, so they can handle 30W. That leaves 50 W to dissipate elsewhere.

You could insert a resistor in series with each LED. Assuming a worst case drop over the LED of 4V, 1V for the chip, and a worst case low battery of 10V leaves 5V for the resistor, which must hence be 5 / 0.02 = 250 Ohm. Calculating from the opposite case, 2.4V for the LED, 14.4V accu, this gives 14.4 - 2.4 - 5 = 7V for the chip, so it dissipates 7 * 0.02 * 24 = 3.4W. That's still uncomfortably high.

A better solution would be to use a regulated (switched!) 5V supply. Now the chip has to dissipate (assuming a worst case LED drop of 2.4V) 2.6V, for a total dissipation of 2.6 * 0.02 * 24 = 1.25 W. That's more comfortable. And you don't need the resistors (but in exchange for a single SPSU).

My calculations show how you can evaluate these two designs. It is up to you to supply the correct figures (ambient temperature, range of LED drop, accu voltage range, etc), redo the calculations, and evaluate the results.

• Of course thanks for accepting my answer, but note that it is considered good form on this forum to allow for some time (let's say a few hours) to give others time to post maybe even better answers. – Wouter van Ooijen Jan 7 '13 at 12:59
• Ok thankyou for the letting me know. I had used 85deg as an ambient temp because most of the ICs will be located in behind the dash allowing for engine temp and sun warmth. It more then likely will be easier from a intgtergration POV to include an onbaord 5v supply. – Daniel Jan 7 '13 at 13:01

A resistor (~350 Ohms, use 330 Ohm standard value) on each current sink pin, in series with the LED and then to Vcc (12 Volts) will work fine.

The resistors will drop part of the surplus voltage, thus decreasing voltage across (and dissipation within) the LED controller IC. This will not affect the TLC5947 operation, as each sink line within the IC uses a constant current regulator referenced against Iref: Refer to Functional Block Diagram on page 5 of datasheet. Thus the current sink function is essentially independent of voltage Vo, up to 30 Volts.

Using a single high wattage resistor tied to Vcc to drop voltage, however, may result in instability, as the voltage Vo seen by the sink lines will vary depending on Vf of individual LEDs as they get turned on, and also by number of LEDs on at any time.

Notes:

• Sufficient headroom is required over the LED forward voltage, for the TLC part to regulate current effectively. For calculation convenience, take this minimum headroom to be 1.7 Volts (+ Vf 3.3 Volts = 5 Volts).
• The individual resistors must be rated to dissipate sufficient power, i.e. for 20 mA and (12 - 5=) 7 volts drop, 1/8 watt resistors will be insufficient, 1/4 watt will be required.
• 330 Ohm at 20 mA will drop 6.6V. IMO that leaves too little headroom for a worst case situation (low battery, higher LED drop, higher driver drop - note that the figures in the datasheet are probably typical figures, not worst case!) – Wouter van Ooijen Jan 7 '13 at 12:42
• @WoutervanOoijen Worst-case headroom on TLC5947 is 0.8 Volts, I believe, but I agree with you... 330 Ohms is marginal. – Anindo Ghosh Jan 7 '13 at 13:00