Multiplexing LEDs question

Regarding multiplexing LEDs, I've been learning a little bit about refresh rate and duty cycle. If you want to turn on an LED for a shorter amount of time each cycle and/or keep it off for a longer time, I've been told that you need to supply more instantaneous current . So do you alter how much current you supply by changing the resitor in front of the LED? What are some of the general rules to apply for this type of application?

Yes, you are right, that is what you have to do.
That is because the light intensity from an LED is about proportional to the average current going through it. So you need to make sure the average is about the same, to keep the intensity the same.

Since the average is Iave=Ipk*ton/T, if you reduce ton you need to increase Ipk accordingly.
However, you must keep Ipk below the maximum given in the LED datasheet.

Thank you VVV. One thing, what is "ton" in that equation?

Ton = T on = time on = part of period which the led is on

Thanks rkidaira.

What are some general rules for what size resistor to pick. Is it better one way or the other - longer on time, bigger resitor vs. shorter on time, smaller resitor? Should you always try to have them on as long as possible and design around how many are multiplexing? Either way, do you draw the same amount of current?

Hi !

The way you use multiplexing depends on several factors:
- The maximum current handled by the led
- The number of leds multiplexed
- The multiplexing frequency
- If there is a transistor switching the leds

For example, if I use 4 seven segments led displays at 50Hz, switched by transistors, and the maximum current for each led is 20mA (average), power supply 5V, we can do the following configuration:

50Hz = 20ms period, so you shoul divide this time between the 4 leds (one for each same segment of each display). Then each led will be on for 5 ms and off for 15 ms. The duty cycle is 25%. You can drive each led with 20mA/25% = 80mA to get the maximum brilliance (compensating the time the led is off). Since red leds have 1,7V drop voltage, and if the Vce saturation is 0,5V, the colector resistor should be: (5V - 1,7V - 0,5V)/80mA = 35 ohms. I suggest you do not run the leds in the limit. Choose 47 to 68 ohms instead, even 100 ohms.

Other tips: try to use high brilliance leds, so you will save power and the leds life will be longer.
Remember: the time each led is on depends on the number of leds. Don´t stress the leds using high current for relatively long on periods. You cannot multiplexing leds with low frequencies (below 40Hz), because a flickering effect appears and probably your leds won´t support the high current when they are on.

Hi!

You said that the time each led is on depends on the number of leds. It seems that it also depends on the amout of brightness you want. It may seem silly of me to ask, but is it usually a given that you design around getting the maximum brightness? Also, let's say you made a design for multiplexing 10 LEDs according to your formula. You still have enough time to get maximum brightness by putting more current through for less of a time slice, right? If so, it seems you could also use the same resistor used in your design for 10 if you had, let's say, 5 LEDs multiplexing, but you would keep the same duty cycle for 10. So, is either way better or preferred?

You said that the time each led is on depends on the number of leds. It seems that it also depends on the amout of brightness you want. It may seem silly of me to ask, but is it usually a given that you design around getting the maximum brightness?

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Of course you can design it using less bright, the light environment where your equipment is going to be used determines your necessity for brightness. The use of filter mask (colored transparent plastic) in front of the led displays will help the visualisation.

Also, let's say you made a design for multiplexing 10 LEDs according to your formula. You still have enough time to get maximum brightness by putting more current through for less of a time slice, right?

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Correct. But you are not willing to put 1000mA through the led even for very short periods. I suppose that the leds can´t handle so high currents. Keep the current within reasonable values. I think multiplexing 10 leds is too much, and you scanning frequency should be higher to avoid the flickering caused by a ratio on/off of 1 / 9.

If so, it seems you could also use the same resistor used in your design for 10 if you had, let's say, 5 LEDs multiplexing, but you would keep the same duty cycle for 10. So, is either way better or preferred?

Click to expand...

It can be done. But you are wasting 5 cycles with no leds on, although i see no further problem.

rkodaira said:

Correct. But you are not willing to put 1000mA through the led even for very short periods. I suppose that the leds can´t handle so high currents. Keep the current within reasonable values. I think multiplexing 10 leds is too much, and you scanning frequency should be higher to avoid the flickering caused by a ratio on/off of 1 / 9.

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I've been told 15-20 is the max you can reasonably multiplex. After that, you don't have enough time to hit a suitable refresh rate

rkodaira said:

It can be done. But you are wasting 5 cycles with no leds on, although i see no further problem.

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Yeah, I don't see any issues with it either. That's what I'm trying to figure out. I mean, if you design with a smaller resistor vs. a larger resistor, do you save money in the design? Is the power consumption the same?

There is no free lunch and there is no power consumption savings.
That is because the intensity ot the LED is proportional with the average current going through it. But the same average current means the same average power consumption.

Take an example: an LED is bright enough with 5mA. Suppose you want to multiples 4 LEDs. Then you need a 20mA peak through each of them. So each LED sees 20mA for 1/4 of the time, 0mA for 3/4 of the time. (The average current for each LED is 20mA/4=5mA; still 5mA, for the same brightness).

Now, the total input current is 20mA for the first 1/4 period, going throught the first LED, 20mA for the second 1/4, going to the second LED, 20mA in the third quarter, for the third LED and finally 20mA in the last quarter of the period, to power the last LED.
In other words, the input current is 20mA all the time.

But if you did not have the multiplexing, with all the LEDs on continuously at the same intensity, your current would be: 4LEDs*5mA=20mA.
Exactly the same current! No savings!

In fact, since you will have some dead-time between LEDs, you will actually lose more power with multiplexing.

VVV said:

There is no free lunch and there is no power consumption savings.
That is because the intensity ot the LED is proportional with the average current going through it. But the same average current means the same average power consumption.

Take an example: an LED is bright enough with 5mA. Suppose you want to multiples 4 LEDs. Then you need a 20mA peak through each of them. So each LED sees 20mA for 1/4 of the time, 0mA for 3/4 of the time. (The average current for each LED is 20mA/4=5mA; still 5mA, for the same brightness).

Now, the total input current is 20mA for the first 1/4 period, going throught the first LED, 20mA for the second 1/4, going to the second LED, 20mA in the third quarter, for the third LED and finally 20mA in the last quarter of the period, to power the last LED.
In other words, the input current is 20mA all the time.

But if you did not have the multiplexing, with all the LEDs on continuously at the same intensity, your current would be: 4LEDs*5mA=20mA.
Exactly the same current! No savings!

In fact, since you will have some dead-time between LEDs, you will actually lose more power with multiplexing.

Click to expand...

Hi,
Sorry to intrude, but I only partly agree with you. The main reason to multiplex LEDs is to save chips and real estate on the pcbs, not power. By multiplexing you reduce the number of drivers and registers (and money). You will also reduce power lost in the static drivers. Using PWM you can also make intensity adjustment depending on ambient light, and save power.

In your last paragraph: I can't see why you loose power due to dead-time between LEDs? If you don't use power you can't loose it.

TOK

Nobody disputes the advantages of multiplexing. But the question was about the power savings when multiplexing LEDs. And I have shown, the power is the same if you want to keep the intensity constant (not including the driver currents).

I agree, the last paragraph was not formulated too well. What I meant was that due to the dead times, you will be forced to increase the peak currents more, to compensate for the lost time. However, you will operate in a non-linear region, such that you may need to increase the peak currents (and thus the average) even more, to actually obtain exactly the same brightness.

You can actually save power by mulitplexing. This is becuase the human eye acts as a photometer that partially acts as a peak detector and partially as an integerator. So by pulsing an LED, you actually see a brightness level more than the brightness of the same LED current fed with the average DC value. How is that.

Techie, I am sorry, I do not agree with that statement.
As far as I know, the human eye perceives the LED brightness the same way, regardless of the individual pulses and only based on the average of the LED current. That is because the eye "integrates" what it sees.

What you are saying aplies only when the frequency is so low that the eye sees the individual pulses. In that case you will see each pulse very bright, but at such low frequency the multiplexing cannot be done.

Have a look at Dallas / Maxim applicaiton note 1883.