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Why Don't All LEDs Light Up When Connected in Parallel?
Building a "chandelier" out of a bunch of different LEDs and powering them from a single powerful driver sounds like a great plan to save money and time. It's just a pity that this plan usually doesn't work, and you end up having to carefully choose an LED driver, sometimes even using two drivers simultaneously.
The worst-case scenario looks like this:
1. We assemble LED circuits on a heatsink;
2. Connect the assembled structure to the driver;
3. Plug the driver into the mains;
4. A flash, and half of the LEDs go into the trash. It sure was bright, though...
Flick the Switches in the Simulator
In the widget above, I put together a classic parallel connection circuit. We have a cool, reliable driver that honestly tries to output its 700 milliamps. And there are two channels. On the first one, there are three white LEDs; on the second — three red ones (660 nm).
If you close the switches one by one — everything works perfectly. But if you turn them on simultaneously (i.e., wire them in parallel), a current imbalance occurs: the white diodes fade to zero, while the red ones take the entire 700 mA current and shine at full power.
Why didn't the driver split the current evenly, 350 mA per branch? Physics is heartless. Current isn't divided "fairly"; it's divided by resistance. Whoever doesn't resist gets more. The total voltage drop across the red LEDs is lower, which means significantly lower resistance, so all the current passes only through them.
Current is Lazy! Or the Path of Least Resistance
An LED is not a resistor from a school textbook. It's a tricky semiconductor. It has a parameter called Vf (Forward Voltage) — voltage drop. Roughly speaking, this is the voltage threshold that must be applied to the diode for it to open and let the current flow through it.
The crystals in LEDs are not uniform. A red diode needs about 2.2 Volts to open. A white or blue one needs about 3.2 Volts. In our circuit, there are three diodes in a row. This means the red branch asks for 6.6 V, and the white one asks for 9.6 V. This is clearly shown in the widget.
Imagine that current is water in a pipe, and voltage is the pressure needed to break a plug at the end. The driver starts to increase the pressure. As soon as it reaches 6.6 V, the red plug blows out. The water (current) rushes through the newly formed hole.
The driver sees that the water is flowing and stops increasing the pressure. The system voltage freezes at 6.6 V. But the white diodes need 9.6 V! For them, this pressure is practically nothing; their P-N junction remains tightly closed. As a result, 100% of the current flows through the red branch. The whites sleep, the reds burn. With small imbalances, for example, 8.8 Volts and 9.2 Volts, the imbalance between the circuits will be less severe, and a minor portion of the current might pass through the chain with higher resistance. In that case, one group of LEDs will shine brightly, while the other will be very dim.
The Thermal Runaway Effect
What if we take two absolutely identical white LEDs and connect them in parallel? They both want 3.2 V! In theory, the current should divide exactly 50/50.
In a perfect vacuum world with spherical diodes, that would be the case. In reality, crystals are never 100% identical. Even if you pull them from the same reel, one might have a Vf of 3.19 V, and the other 3.21 V.
The diode with the slightly lower threshold (3.19 V) will open first and take a bit more current — say, 55% instead of 50%. What happens next?
- More current = more heat.
- LEDs have a negative temperature coefficient. As they heat up, their resistance (and Vf) drops!
- The opening threshold becomes even lower — for example, 3.15 V.
- Now this diode takes 70% of the current.
- It heats up even more. Resistance drops further. It takes 90% of the current.
This process is called Thermal Runaway. In a truly critical situation, the first diode flares up like a supernova and burns open. The entire driver current abruptly shifts to the second diode, which instantly goes to meet its maker due to the shock. A classic domino effect. Again, this is mostly from the "spherical vacuum" realm: in real life, with proper cooling, even Chinese LEDs with roughly the same voltage operate stably, as long as you don't run them at currents close to the LEDs' maximum limit.
The Golden Rule: When IS it Okay to Wire in Parallel?
You might rightfully point out: "But in branded quantum boards, tons of diodes are connected in parallel, and they work for years!" Yes, that is true.
Connecting LEDs in parallel to a CC driver is permissible only under strict adherence to three conditions:
- Perfect Binning. All LEDs in parallel circuits must be absolutely identical. No mixing of red, blue, and white. Furthermore, they must come from the same production bin (ultra-precise voltage sorting).
- A Single Heatsink. Thermal runaway is mitigated because all diodes are rigidly soldered onto one massive aluminum board. If one crystal starts heating up more than the others, the heat instantly dissipates through the aluminum, warming the neighboring diodes. Their temperature equalizes, their Vf equalizes, and the current balance is restored.
- Strict Underdriving. Engineers never pump maximum current into parallel matrix assemblies. The diodes are loaded to a maximum of 85-90% of their rating. Even if one diode takes a little more current, this imbalance won't kill it because there is still a safety margin left.
Conclusion: If you are soldering high-power diodes (1-5 Watts) on individual star PCBs, or assembling a multi-spectrum daisy chain for a grow box — calculate the voltage on the LED strings with an accuracy of 0.1 Volts. In certain cases, standard rectifying diodes can be used to balance the voltages.
