Balancing LED current in light fitting.

[Diver300]

I've had one of these:- https://www.screwfix.com/p/lap-twin-5ft-led-batten-55w-6300lm-220-240v/115pp for a couple of years. It could be that the specification has changed a bit as the one I have is labeled 60 W and actually consumes 62 W.

It's always had a bit of 100 Hz modulation in the light.

Today I noticed that parts of the light weren't working. I opened it up and found that the LEDs were driven by a switch-mode power supply. It has no input capacitor, and the power factor is 0.94. I think that the power supply is current limited. There was a 100 μF, 160 V electrolytic capacitor on the output of the power supply. The LEDs are in eight strings, each of 41 LEDs in series. The LEDs are on fibreglass PCBs, slotted into an aluminium extrusion.

One LED in each of three of the strings had failed open circuit. I replaced them, and the light works. I've added a 560 μF capacitor which reduced the ripple on the LED voltage from about 5 V ac to about 0.7 V ac, and the 100 Hz flicker has gone.

I think that the lamp suffers from the currents differing in the 8 strings, as there is nothing other than the resistance of the LEDs to keep the currents the same. Two of the sections that failed were in the middle of the lamp, so presumably the warmest part, and I think that they were taking the most current. The power supply seems to be constant current, so when the one string fails, the current in the other strings goes up. I think that the three sections failed in quick succession.

Would it be a good idea to add circuitry to keep the currents the same in each string?

The overall LED voltage is 120 V, so the current is around 500 mA and there is about 63 mA in each string. When I added 50 Ω in series with one string, there was about 2.3 V across the resistor, so the current had reduced to 46 mA. I shorted one LED in a string, and the current increased to around 80 mA. I estimate that the entire string has a resistance of about 150 Ω, so to improve the balance significantly using simple resistors I would have to add maybe 330 Ω in series with each string. Those resistors would dissipate around 1.3 W each.

Is there a better way to stabilise the current, or is there no point?

 

[rjenkinsgb]

I'd try a 50 Ohm in series with each string, to start with, to keep the power loss down a bit.

If it's a constant current driver, once all strings drop the extra few volts the current should be near equal again & rather better than the original setup.

 

[Diver300]

I'd try a 50 Ohm in series with each string, to start with, to keep the power loss down a bit.

If it's a constant current driver, once all strings drop the extra few volts the current should be near equal again & rather better than the original setup.

I was thinking that 50 Ω wouldn't make much difference when the whole string is already around 150 Ω, but I guess that any resistance would improve the balancing a bit.

 

[rjenkinsgb]

I was thinking you only need to reduce the effect of the difference between strings.
Also, then having a series resistance allows you to check the voltage across each for actual current, and you could then adjust the values individually to equalise it, if any string was a long way out.

 

[Diver300]

I added 50 Ω resistors in series with each section. The voltage across them is between 2.92 and 3.30 V, so if nothing else, I now know that the currents are within about 15 % of each.

There is still the potential problem that if one string fails, the same current will be spread between fewer strings.

The larger capacitor has resulted in a long turn-on pause of around 3 seconds. It's not a problem as I have two other lights of a different design in the garage, but I am slightly curious as to why the delay is that long. The power supply produces about 500 mA. 500 mA would charge the 660 μF capacitance up to 120 V in about 160 ms, and the start-up time is much longer than that.