# Help me understand current through LED(s)

#### robotDR

Joined Mar 17, 2020
80
If I have a string of 4 series LED and 1 series resistor.

That resistor will set the current. Let's say it sets the current for that path at 20mA.

So if I measure current at the cathode of the last LED, it will 20mA. If I measure current at any point, it should be 20mA

If I look at the luminosity vs current graph of an LED, it gets brighter as forward current increases.

Can I use 20mA to determine its brightness or do I need to divide by 4 (4 series led) to get a value of 5mA to determine brightness?

And why would I not need to divide by 4. Is it because I need more voltage to power an additional led which means 20mA at either 2, 4, 6, or 8v which is more power at each step?

So for an LED driver like : IS31FL3746A-QFLS4-TR
https://www.digikey.com/en/products/detail/lumissil-microsystems/IS31FL3746A-QFLS4-TR/9759696

If each current sink pin can only handle 34mA but I can put 11 led on each current sink pin, does that mean each LED can get the full 34mA?
I want to drive 126 white LED and this seems like a great driver.

On page 5 it says for I_LED: "Average current on each LED: 3.03mA" while I_OUT:"Maximum constant current of CS_y: 34.5mA"
Help me understand that.

#### crutschow

Joined Mar 14, 2008
34,078
Can I use 20mA to determine its brightness or do I need to divide by 4 (4 series led) to get a value of 5mA to determine brightness?
The current is the same in all series connected LEDs, independent of the number of diodes, so the brightness of each diode is the same, based upon the 20mA current.
The only difference is that the voltage drop is the sum of all the diodes' forward drop.

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#### robotDR

Joined Mar 17, 2020
80
No.
The current is the same in all series connected LEDs, independent of the number of diodes.
The only difference is that the voltage drop is the sum of all the diodes' forward drop.
Ok so for the led driver i linked above, why does it say 3.03mA average per led, when it's 34.5mA to the current sink pin?

#### Papabravo

Joined Feb 24, 2006
21,030
All elements in a series circuit will experience the same amount of current. The brightness of an LED will depend on the current, but there is likely to be a maximum limit to the current beyond which there will be too much heat and the device will fail. Once you let the magic smoke out, they will be dim forever.

I can't find this statement on the datasheet:
On page 5 it says for I_LED: "Average current on each LED: 3.03mA" while I_OUT:"Maximum constant current of CS_y: 34.5mA"
Help me understand that.

EDIT: I followed the link in Post #1. On the one I saw it was 8.33 mA average current per LED string, and that number comes from taking 34.5 mA divided by the duty cycle of 4.14. There are 4 SW lines and some allowance for deadtime between the enables on SW1 through SW4. No current flows through the strings where the SW line is LOW. Look at Figure 11. on p. 18 for a timing diagram on the multiplexing.

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#### crutschow

Joined Mar 14, 2008
34,078
Ok so for the led driver i linked above, why does it say 3.03mA average per led, when it's 34.5mA to the current sink pin?
34.5mA is the maximum LED current.
3.01mA average if for the specified PWM and current setting.

#### robotDR

Joined Mar 17, 2020
80

In this case, how much current can I run on each LED? In this array it looks like only ~3.0 mA because they aren't in series.

#### Papabravo

Joined Feb 24, 2006
21,030
The datasheet gives you the duty cycle which will always be slightly larger than the number of SW lines. It is 10.125 in the datasheet fragment from Post #6 above. For the amount of time an SW line is high, each enabled LED will see 34.96 mA and 0 mA for the time that the SW line is not enabled. The average of those two is Ta-da! 3.45 mA.

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#### robotDR

Joined Mar 17, 2020
80
The datasheet gives you the duty cycle which will always be slightly larger than the number of SW lines. It is 10.125 in the datasheet fragment from Post #6 above. For the amount of time an SW line is high, each enabled LED will see 34.96 mA and 0 mA for the time that the SW line is not enabled. The average of those two is Ta-da! 3.45 mA.
Hm ok that seems to make sense.

But if there are 11 parallel led cathodes on that CS pin, that limits each led to about 3mA of current because the pin has limit of 34mA. So what if during pwm there are 2 or more on for a fraction and each at 34mA that would be bad.

#### Papabravo

Joined Feb 24, 2006
21,030
Hm ok that seems to make sense.

But if there are 11 parallel led cathodes on that CS pin, that limits each led to about 3mA of current because the pin has limit of 34mA. So what if during pwm there are 2 or more on for a fraction and each at 34mA that would be bad.
LOOK AT THE TIMING DIAGRAM. One and Only SW line can be active at ANY point in time. Each CS line will have either zero current or the programmed current for whatever SW line is high. That's it. Don't overthink it. These are individual LEDs in a matrix -- not strings.

#### Papabravo

Joined Feb 24, 2006
21,030

#### robotDR

Joined Mar 17, 2020
80
No and no. You are not understanding these parts in any meaningful way.
Ok I understand the timing of the service lines now.

so really, I can’t have all on at the same time at 100% duty cycle.

but part number IS31FL3209-QFLS4-TR
Is 76 mA per channel with no sw lines. It’s not a multiplexing driver.
I could run say 18 channels each with 7 series leds with 20V+ On the high side right? This way all could get 60mA. Just sucks I have to run that high voltage.

#### robotDR

Joined Mar 17, 2020
80
LOOK AT THE TIMING DIAGRAM. One and Only SW line can be active at ANY point in time. Each CS line will have either zero current or the programmed current for whatever SW line is high. That's it. Don't overthink it. These are individual LEDs in a matrix -- not strings.
So using all SW lines, your duty cycle is already ~ 10% out of the box. Doesn't that kind of suck? I get the matrix benefit but if you can only use 10% and THEN dim from there if you like? That seems pretty weak.

#### Papabravo

Joined Feb 24, 2006
21,030
So using all SW lines, your duty cycle is already ~ 10% out of the box. Doesn't that kind of suck? I get the matrix benefit but if you can only use 10% and THEN dim from there if you like? That seems pretty weak.
I suppose it depends on your ultimate purpose and the type of LEDs you want to use. It looks like a feature rather than a bug. LEDs and the human eye have some persistence that this takes advantage of. What rate do you suppose it would take to make the detection of "flicker" in the individual LEDS almost impossible to perceive?

#### djsfantasi

Joined Apr 11, 2010
9,154
In a series circuit, current is constant across all components while the voltage drops add. In a parallel circuit, voltage is constant across all paths/components and the current draws (of each path) add. In a series circuit, voltage drop may differ through each component; in a parallel circuit, current draw of each path may differ. If you remember the first two rules, the last two can be derived by remembering the parameter which is additive can differ.