Do I need a resistor to connect this LED lamp?

Question

I purchased this LED lamp on Aliexpress, the spec says it's 36V, 20W. I want to know how to connect it to my ebike 36V battery, in particular can I connect it directly or I need a resistor? I tried connecting it with a 2K resistor and it works, 10K and it also works but the light is much less bright. I think maybe it has a resistor inside so external one is not needed? When I asked Chinese support via messager on Aliexpress if this needs a resistor in the circuit, they just sent me a picture of the lamp with the polarity of contacts which does not help. I could just try and connect it directly to the battery but I am afraid it will go up in smoke and flames. On aliexpress this is also called LED matrix which is confusing because it is simply a multiple leds in a square form that only emit light all together. Thanks.

Answer 1

This appears to be a naked LED, for which you will need to control the current externally. You must not connect it directly to your battery.

A '36 V' LED on a '36 V' battery is not a great combination, as neither of those terms means exactly 36 V, they are both very, very approximately 36 V.

Connecting them directly together will only give you happiness by chance. In the best case you're likely to get too little light out when the battery is low. In the worst case, you will fry the LED when the battery is at a higher voltage.

Using a resistor will only work when the battery is fairly high voltage. When it's low you'll get too little light. You'll also waste a lot of energy in the resistor. Using a linear current source will also be useless when the battery voltage is low.

The ideal method of control will be a switch-mode constant current source.

As both battery and LED have a nominal 36 V, you can't simply use a boost or a buck type for which you need to ensure that the voltage ranges do not overlap, you need a buck-boost type.

At 20 W, have a good heatsink underneath it.

Answer 2

COB arrays like that are meant to be driven by a buck or boost LED driver. You can use a resistor too, but it will need to be rated for the (considerable) heat that would be dissipated, at least if you want to run at full brightness.

Answer 3

TL/DR: Use a 1kOhm resistor.

Rationale: This LED cannot be used as a bicycle headlight because it is the wrong type. In order to mount an optic suitable for a headlight, ie something with a 15° beam, you need a much smaller LED, much closer to a point source. This LED is way too large, it is meant to be used in floodlights. If you use it at full power on a bicycle, the following will happen:

The road will be very bright, right in front of the bike. You will see absolutely everything. Of course, your eyes will adjust to that brightness, which means...

The road will look very dark in the distance (and I don't mean "far", you won't see anything past about 10 meters). Without optics, there is no "throw" at all.

Everyone coming in the other direction will be blinded, because there is no optic to throw the light on the road instead of throwing it into other people's faces. If it is dark, either you will cause an accident, or someone coming in the other direction will crash into you, or they will express their dislike by kicking you off your bike.

In addition, you're not talking about the huge heat sink you'd need to mount this monster on your bike.

Therefore, use as an indicator light seems appropriate, at low power, with a resistor. It'll be visible from all angles.

If you want a light to actually see where you're going with your bike, get something with a tight beam and a bit of flood right in front. I've built one with a 6° spot lens and another LED with a 30° lens aiming a bit down. That gives good visibility in the distance, and a wider beam up close., without blinding other people. There are some nice lights for ebikes too.

Answer 4

You always need a proper means to limit the current through a LED arrangement. Connecting a LED to a voltage source directly always asks for trouble. Most LEDs do not have built in current limiting / constant current driving unless advertised as such.

You also have to know your power supply characteristics well enough to make good decisions on how to power your LED. In your case the 36 V e-bike battery is probably a 10 cell series connection of 3.6 V li-ion cells or 12 cell series of 3.3 V LiFePo4 cells.

I don't think a e-bike uses any circuitry to keep a stable 36 V output (could be wrong), in that case your battery voltage is changing from something like 42 V down to 30 V.

Using a simple resistor will not give you the best results for different reasons. First you have to select it in such a way, that on 42 V your LED doesn't get damaged. For that you'd need a 10.8 ohm resistor. This would waste around 3 W of power to heat - and you also need a resistor capable of handling this.

Next, the voltage of your battery will drop and you will get less and less light, which is probably not what you want.

As the battery lowest voltage is below the cutoff of the LED, at some point the light will just go out completely.

For this application a constant current buck-boost converters is necessary to achieve a constant light output over the whole range of the battery voltage.

Answer 5

If it is 36V and 20W it suggests you need to drive it with a constant current source around 0.5A. You can buy such drivers fairly cheapy, but most are optimized for mains input.

Answer 6

Typical specification for this kind of product is that the Vf will be between 30V and 36V at the rated 600mA.

Cheapest (not most efficient) way to go would probably be to buy a boost converter with CC/CV output (I see them for <$10 including shipping on Ali) and then add a resistor that would drop the excess voltage. Say 6V at 600mA, would be a 10 ohm 5W ceramic resistor. Then set the supply to limit at (say) 40V and to control the current for the desired brightness (600mA maximum). If you want less heat and less brightness, increase the resistor to drop a similar voltage. For example, for 300mA you would use a 20 ohm resistor. The resistor (and associated wasted power) is required because a boost converter cannot output less than the applied voltage, plus a bit.

As suggested in other comments you will need a heatsink (so the LED does not burn out) unless the LED is run at very, very low power and you'll want a lens and associated mounting hardware. You can buy an assembly with lens, heatsink and fan for around $30, maybe less I didn't look very hard. As you can see, the LED itself may be the least expensive part of this project.

Answer 7

A challenge is the required LED forward voltage vs. your battery. A series resistor doesn’t have enough overhead voltage to work with. And any inline current limiter (like a 2-transistor type) will steal a volt or so. Either way, you might find the light getting suddenly dim as your battery reaches its end of charge.

With this in mind, the LED will be happiest using a constant-current driver. Try to find one that is ‘buck-boost’ to deal with the fluctuating battery voltage. Or, choose a lower voltage LED (24V, say) and use buck-only.

On the other hand, you must not tie this LED directly to your battery: it will fry in short order. The ‘36V’ is a nominal forward voltage at rated power. What matters is that it’s 20W - meaning you absolutely need to limit the current to no more than 550mA.

Finally, at that power level your LED will need a heatsink.