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### Topic: Drive LEDs with ULN2803 and TLC59711 driver (Read 272 times)previous topic - next topic

#### Grumpy_Mike

#15
##### Jun 22, 2016, 09:10 pmLast Edit: Jun 22, 2016, 09:21 pm by Grumpy_Mike
Quote
causing same or higher voltage drops and power dissipation than the darlington drivers above 100mA.
But you can not run a ULN2803 at 100mA per channel due to thermal considerations. See:-
http://www.thebox.myzen.co.uk/Tutorial/Power_Examples.html

Anyway can you show your working how 0.85V at 100mA, dissipates less power than 5R at 100mA?
I make it 0.0.85W for the ULN2803 and 0.05W for the shift register. Maybe I am wrong?

#### Chagrin

#16
##### Jun 22, 2016, 11:05 pm
...and back to OP's question...

Why is the TLC59711 insufficient for your project? What kind/how many LEDs are you trying to power on each channel that 60ma and 14V is not enough? Be specific.

Also note that TI has quite a variety of chips in addition to the TLC59711 that will support more voltage or current per channel. These shift-register-style chips are linear mode, mind you, and if you're trying to use more than 100ma per channel then you should probably be rethinking your project.

#### jwistrom

#17
##### Jun 23, 2016, 10:10 am
Hi
I am trying to power groups of parallel connected LEDs. I want to control one group from one output pin and one group may contain from 3 to 7 parallel connected LEDs. If each LED requires 20mA, 60mA will not be sufficient but I guess the arduino's 5V will be fine (regarding voltage). Why not use the TPIC6b595? It seems to suit my requirements

#### DrDiettrich

#18
##### Jun 23, 2016, 10:13 amLast Edit: Jun 23, 2016, 12:38 pm by DrDiettrich
You are right, the break even point is (slightly) above 100mA, and with the thermal restrictions and precautions event this continuous current cannot be reached, for both chips. Thanks for your link to the power calculation, I was not fully aware of the consequences for the current of 8 drivers in a package.

But I miss one detail from the calculations, the difference between bipolar and FET transistors. FET resistance increases significantly with temperature, and the power consumption is P = I² *R. This means that a not so optimal FET will produce more heat, what increases its resistance, what again produces more heat...

When I looked at the worst case, of about 10 Ohm at 125°, the power dissipation per driver is almost the same (0.1W) for both chips at 100mA. When we also take into account a heat resistance of 90°/W for the TPIC, this will limit its total power dissipation to 0.9W, as opposed to 1.1W for the ULN.

Now the same power calculation for the TPIC as for the ULN in the link, please correct me if I'm wrong:

With a case temperature of 50°C, junction 125°C, and 90°/W the allowed total power is (125-50)/90=0.83W.

With Ron=10 Ohm we get I² = 0.083, I=0.29A total, 36mA per driver.
Even with Ron=5 Ohm we get I² = 0.166, I=0.41A total, 51mA per driver.
This were significantly less than the 0.71/8=88mA for the ULN at 50°C case temperature.

If I'm right, a continuous current of 8*100mA would cause a temperature difference of at least 290°C, with a required TPIC case temperature of -160°C :-(

Correction: Wrong resistance :-(
The power per transistor can be 0.83/8 ~ 0.1W.
That's 0.1A at 10 Ohm, 0.14A at 5 Ohm.

Sorry for the confusion :-(

#### TomGeorge

#19
##### Jun 23, 2016, 10:18 amLast Edit: Jun 23, 2016, 10:21 am by TomGeorge
Hi,
Quote
parallel connected LEDs.
I think you need to clarify, that you do not just have a group of parallel connected LEDs.

You are using a LED strip, that is fitted with individual LEDs and their current limit resistors.

So you cut them off a roll.

Tom....
Everything runs on smoke, let the smoke out, it stops running....

#### jwistrom

#20
##### Jun 23, 2016, 10:25 am
I can not use a LED strip because I need to customize the space between the LEDs. I have tried to use a LED strip but I would prefer to use individual LEDs

#### DrDiettrich

#21
##### Jun 23, 2016, 10:28 am
I am trying to power groups of parallel connected LEDs.
Parallel LED connection requires proper distribution of the current, i.e. one resistor or current source per LED. For serial connection instead a single current source is sufficient, that's why drivers like TLC59711 allow for a high supply voltage, required to drive multiple LED in series.

#### jwistrom

#22
##### Jun 23, 2016, 10:35 am
I am thinking about doing something like this:
https://youtu.be/6fVbJbNPrEU

But instead of one LED to each output PIN, I want 3 to 7. Should I go for another approach?

But sinking the current since the TPIC6b595 doesn't support sourcing

#### jwistrom

#23
##### Jun 23, 2016, 11:48 am
I have tried to study the tech. sheet of the TPIC6b595 but I'd lika to know the difference between these specs:

- ID Continuous drain current, each output, all outputs on, TC = 25°C: max 100mA
&
- Continuous source-to-drain diode anode current: 250mA

If the max current is 100mA, this means I can have 5 LEDs in parallel, right?
Which of the above specs is relevant?

#### DrDiettrich

#24
##### Jun 23, 2016, 12:47 pm
The diode current is for reverse operation, when the output (drain) voltage happens to drop below zero. This can happen with inductive loads, but not normally with LEDs.

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