Sum toroidal transformer outputs

I`m planning to buy a toroidal transformer for driving an LED. The LED needs 36 V and around 1 A.

I found a toroidal transformer which has a 230V Input and two 18V 0.83 A Outputs. Can I simply sum the outputs by connecting both positive outputs and both negative outputs to my LED?

No, you cannot "just" anything with a transformer, until you are sure about the make-up of the secondary windings.

For example a 18-0-18 make-up, which is common in transformers for lower power amplifiers, means it has 2 windings of 18V, but they are already put in series inside the transformer. Yet again, if you have {18V} {18V} as separate windings, it is not uncommon to have two wires of the same colour for each winding. If you want to put them in parallel you will need to make sure you connect the right two wire to each other. If you connect them in reverse your transformer will see a very strong short-circuit and 1. consume a lot of power with no output, 2. eventually burn out.

Also be aware that a transformer has no "plus" or "minus". What comes out of the transformer is an AC voltage, meaning it goes from + on one wire and - on the other, to the other way around at 50 to 60 times per second. LEDs don't like reverse voltage at all. So connecting an AC to a LED will always degrade its lifespan, even if the AC voltage is under its breakdown limit.

So you need to rectify the AC voltage. Some care may have to be taken by not taking the full 0.83A as well, but with design margins on the transformer, that's probably less important.

If you just rectify your LED will still blink. If you want to minimize that, the schematic below is the safest bet. If you know what's going on with the transformer you can save on one rectifier by combining the windings before rectifying, assuming they are exactly the same otherwise. If you are not sure, just use the two rectifiers to be absolutely sure you don't make mistakes. The chosen capacitor gives reduced effect on the blinking, but does increase the effective DC voltage. In this case by about a factor 1.33. If you double the capacitor the blinking will be gone entirely and the effective DC voltage will be 1.4 times the AC voltage.

simulate this circuit – Schematic created using CircuitLab

So as drawn: Vdc = 18V * 1.33 = 23.9V (with load!, without load it's the same as the next one!)

With double capacitance: Vdc = 18V * 1.4 = 25.2V

Please post a link to the transformer and LED you are looking at.

No, the arrangement of wires won't work, but it is likely there is a way to get 36V.

However, it is only providing 0.83A, which is too little, assuming the specification of the LED is accurate.

Further, it is providing AC, and the LED most likely needs DC. So the transformer voltage needs to be rectified (using a bridge rectifier) and smoothed using capacitors.

There LED is likely to include circuitry which ensures it has the right level of current, but without a link to the part, we can't be certain.

the circuit above, as is, will not work, Combine the Windings into a series configuration , giving you 36V, then use one rectifier, add a cheap linear chip , then It should be OK to drive an average 13 x 13mm or 19x 19mm LED , these are normally 33-36V at 250ma/8W to 700ma/25W but will fall short of 1A/40W

• Most "cheap linear chips" will not tolerate $36\sqrt 2 \ \text V$ and won't tolerate a 20 V drop at 1 A giving > 20 W power dissipation. Note the question is seven years old and has an accepted answer. Welcome to EE.SE. Commented Jan 31, 2021 at 23:38