# Series and parallel LED using

I am planning to build 4 LED-strips using two different 'white' settings (1+1 strip x 4). The LEDs will be driven by a 24V constant voltage dimmable RGB-DALI driver (up to 5A per channel). My idea is to control the two strips (2700K and 5000K) using the R and G channel and the PWM mode for dimming. I drew and calculated the necessary circuit:

The parts are both from this series. (Warm white LED 2700K, Cool white LED 5000K). The resistors are standard 1206 SMD. My calculations for a target illumination of ~500lm yield (for 2700K):

$$\ 20 \times 23...26.5lm = 460...530lm\$$

In order to satisfy the 24V I need to split the strip into 4 paths with 5 LED each.

$$\ 5 x V_{led} = 5 \times 2.8...3.3V = 14...16.5V\$$

$$\ R_{pre} = \frac{V_{in} - 5\times V_{led}}{I_{LED}} = \frac{24V - 5\times 2.8...3.3V}{60mA} = 125...166\Omega\$$

Choosing a 180$$\\Omega\$$ resistor gives an effective current of

$$\I_{path} = 41.6...55.5\leq I_{forward} \$$

For all four strips I would come to

$$\ P_{max1} \sim 4W \$$

Now my questions:

• Are there more parts (filter, overvoltage protection) necessary to build an LED application? It seems too simple for decent light source.
• Is my calculation correct? (Heat and prematurly degradation should not affect my circuit, since I am under the recommended forward current).
• Can aging both in the driver and the LEDs produce a state where the forward voltages cant be produced?
• Should an LED in one path fail the entire path is lost, but no current is induced in the other paths since the driver regulates the draw. Correct?

A constant voltage LED driver is attractive for its ability to run multiple strings in parallel, but this comes at a cost.

The first is the series resistor in each chain. This is required here to account for the variability in the diodes' $V_F$, but it accounts for up to 40% of your power consumption $\big((24V-14V)/24V\big)$. This is more heat, which must be a consideration.

Heat may cause premature failure, which is a another concern; you touch on open-circuit failure, which will shut down that path, but you should also consider a short-circuit failure. Under constant voltage, if one LED short-circuits then the others will see a higher voltage, drawing more current and generating more heat. This could cascade, via heat transfer, down one chain and across to others; worst case would end with the entire supply voltage being applied to the series resistors. Your supply won't help, since $24V/180\Omega*4=0.53A$ is still well within its capability; a fuse could work but would be tricky, especially across temperature, because the worst case current is only about double the operational current.

A constant current driver eliminates all of this. Any variation in $V_F$ is automatically accommodated, and the current directly controls the light output. An open-circuit failure will still take out that entire chain, but a short-circuit failure will have no effect on the brightness or power consumption (heat) of the other elements. But most important to my thinking, the series resistor is eliminated, making the entire product more efficient and decreasing heat dissipation.

A constant current driver requires one channel per chain, but the chains can be longer and the other benefits are probably worthwhile.

• Would I also need a overcurrent protection like edn.com/protect-power-led-strings-from-overcurrent ? If one path fails the other ones will get the current equally.
– v3xX
Commented Jul 23, 2020 at 4:18
• That overcurrent protection is for if you put more than one chain on a single CC channel. I would recommend 1-to-1, so that the CC driver handles this. Commented Jul 23, 2020 at 11:09
• Active overcurrent protection may help your original CV circuit, to handle the shorting failure mode. But really, that is just hacking it to behave more CC-like; better to start cleanly with CC. Commented Jul 23, 2020 at 11:16
• Can I use the same LED? Do I need a driver with exactly 60mA? Isn't it bad for the LED that is driven constantly at its maximum power?
– v3xX
Commented Jul 23, 2020 at 17:07
• You will want something close to 60mA, primarily because the element gets less efficient as you drive it harder (diminishing returns). It's not at all a problem to drive it at full power so long as you keep the temperature down. Managing that heat flow is a major concern, though. That's why most LED products use a thermal (metal core) PCB, basically soldering the LED's directly to a heat sink. Commented Jul 23, 2020 at 18:52