Hi
I am interested in driving multiple sets of parallel connected LEDs using the ULN2803 (to get enough current for the parallel connected LEDs) together with a chain of TLC59711 (to enable many sets of parallel connected LEDs). I am new to this but I wonder if you could provide me with some guidance or a schematic like the one I have attached. The attached schematic contains only the arduino and the ULN2803. How would I go about to incorporate the TLC59711?
Regards
Johan
Hi
I am interested in driving multiple sets of parallel connected LEDs using the ULN2803 (to get enough current for the parallel connected LEDs) together with a chain of TLC59711 (to enable many sets of parallel connected LEDs). I am new to this but I wonder if you could provide me with some guidance or a schematic like the one in the link below. The linked schematic contains only the arduino and the ULN2803. How would I go about to incorporate the TLC59711?
Link to schematic: http://www.haberocean.com/2014/11/simple-circuit-for-controlling-led-strip-using-arduino-mega-and-uln2803/
Regards
Johan
You throw away a lot of the circuitry of your TLC59711 if you add a ULN2803, basically the constant current bit. How many LEDs do you want to drive? Note with the TLC59711 you can put up to 4 LEDs in series.
Basically you need to replace the LEDs in the TLC59711 circuit with pull up resistors and then connect the TLC59711's output to your ULN2803. You need a current limiting resistor with every LED and you connect the LED / resistor combination in parallel.
Note also the ULN2803 while you can switch 500mA per channel the total current of on LEDs at any one time is only about 640mA. This is due to thermal considerations ( the chip will get too hot ).
Hi,
(to enable many sets of parallel connected LEDs)
Any reason you have the LEDs in parallel groups?
Each LED in parallel will have to have its own current limit resistor.
What is the supply voltage for the LEDs.
Thanks.. Tom.. 
Hi
I want to make a word clock. My idea was to group each word into parallel connected LEDs (longest word is five letters, swedish clock). Since it will be a lot of words I thought I'd use a chain of TLC59711 and since each group would require more current than the I/O of the arduino can handle I thought I'd use the ULN2803s
Is it better to use the 74Hc595 (74HC595 Shift Register - 3 pack : ID 450 : $2.75 : Adafruit Industries, Unique & fun DIY electronics and kits)?
Well I would use the TPIC6B595 power shift register so there is no need for any more chips.
Is the TPIC6B595 kind of like the 74Hc595 and ULN2803 built into one?
jwistrom:
Is the TPIC6B595 kind of like the 74Hc595 and ULN2803 built into one?
Yes but it uses FETs not transistors and so generates much less heat.
I wonder about voltage requirements and current limiting in the suggested circuits. When the voltage drop on a typical (red) LED will be about 2V, up to 3V for other colors, and a row driver has a voltage drop of up to 1.5V, such a circuit cannot be powered by 3.3V, and even at 5V there remain only 0.5V for a current limiting resistor or other current source, in the worst case.
So how can the LED current be limited in a practical circuit?
I also wonder about the high ON resistance of the TPIC transistors, of 5-10 Ohm. This would mean a drop of up to 1V at 100mA, close to the voltage drop on the ULN darlington transistors. So the power dissipation of both technologies is about the same, or worse for the TPIC at higher current.
Since power FET can have an ON resistance as low as 0.01 Ohm, wouldn't the use of discrete transistors be preferable?
Grumpy_Mike:
Yes but it uses FETs not transistors and so generates much less heat.
I dare to disagree. But if so, can you explain why? I can't, based on the data sheets.
jwistrom:
Is the TPIC6B595 kind of like the 74Hc595 and ULN2803 built into one?
Yes, only vastly better.
The ULN2803 is essentially obsolete.
DrDiettrich:
I dare to disagree. But if so, can you explain why? I can't, based on the data sheets.
I am talking about the TPIC6B595, the data sheet clearly shows open drain FETs for the output.
Grumpy_Mike:
I am talking about the TPIC6B595, the data sheet clearly shows open drain FETs for the output.
The data sheet clearly shows that their ON resistance is > 5 Ohm, causing same or higher voltage drops and power dissipation than the darlington drivers above 100mA.
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?
...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.
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
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
Hi,
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.
What are the specs of the roll, are they 12V, more info on the LEDs please.
Tom....