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Topic: Multiplexed LEDs without resistor? (Read 2546 times) previous topic - next topic


I'm planning a 24x16 LED matrix using these:


The sheet says forward voltage is 2v and peak forward current is 175mA @ 10% duty cycle. The LED resistor calculator says I'll need 18 ohm resistors for this.

If my duty cycle will be 6.25% (1/16th instead of 1/10th), is it a bad idea if I don't use any resistor to limit current on the LEDs?



I would not take a chance on that one.
A program hickup can ruin your project, and that for a 1 cent component.
Use the components as calculations indicate and do not gamble with that.


Peak Forward Current: 175 mA, 1/10 Duty Cycle,  >>> 0.1ms Pulse Width<<<
How wide of a pulse were you planning?
Were you planning to run from an ATMega's 40mA ABSOLUTE MAX ( and not intended for long term use either) output pin, or something else?
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.


Thanks, I hadn't considered that. I want to use pwm but if I'm doing my math right I'll need > 625 hertz so I don't exceed 0.1ms / pulse. I'll have to check that out.

If I can get a refresh rate > 625 hz will it be OK to skip the resistors?

As far as driving the LEDs, I'm using a modified verison of Evil Mad Scientist's Peggy2: the rows will be driven by 4 ATMega pins demultiplexed to 16 lines via a pair of 74hc138s 3:8 decoders, which in turn connect to PNP transistors to supply enough current. The columns will be driven by 3 74hc595 shift registers, which will drive 3 ULN2803a Darlington arrays as the 595's can't supply enough power to continuously light 8 LEDs. (EMS's Peggy2 uses a STP16DP05 in place of the 595 + ULN2803, but I can't find that chip cheap enough to justify using it over the two chip solution).

Thanks again for your help.


If I can get a refresh rate > 625 hz will it be OK to skip the resistors?

No, it is NEVER ok to not use a resistor, see:-

The LED resistor calculator says I'll need 18 ohm resistors for this.

No it doesn't you are putting the wrong numbers into it.


Sep 12, 2011, 11:39 pm Last Edit: Sep 12, 2011, 11:41 pm by Chris Magagna Reason: 1
Hi Mike, I use this calculator:


Supply voltage = 5v, voltage drop = 2v, desired current = 175mA, it says resistor = 18 ohm.

Is this wrong?

As far as needing the resistors -- I was pretty sure I do based on that web page (and some postings from you) but I've seen several multiplexed LED designs that don't use them so I wanted to check.



While the figure might be right using such a low resistance value is meaningless. This is because the forward voltage drop is not that stable and repeatable from device to device. And it changes over time. The resistor bit is the linear part of the circuit and once you get below 50R or so it becomes an increasingly tiny part with the variations in forward voltage drop making drastic changes to the current. Try calculating the current with a resistor 5% lower than 18R, I assume you will use 5% resistors. Now calculate the current at the maximum and minimum forward volts drop given in the data sheet. Once you start trying to go over 100mA with an LED it is time to think about constant current drive.


Thank you for the information and advice Mike.


Mike et al,

I've been researching your advice on constant current drivers and think I can use 3 TLC5916's


in place of the shift registers, Darilington arrays, and resistors.

Since the chip can handle 960mA each LED should get 120mA if I use a 160 ohm resistor (if I did the math right).

Please let me know if I'm missing anything; thanks again for your help.


The resistor value looks right. However running a chip at the maximum current stated in the data sheet rarely works because what you also have to consider is the power dissipation in the chip. Often the maximum current figure is not obtainable because the chip overheats first. You can calculate the maximum current based on the maximum chip temperature you want to use.
See this for detailed:-


Hi Mike,

Thanks again for all your help on my projects. I've been reading your links (and some of the other great documents you've written) and now (hopefully) have a better understanding of how everything fits together.


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