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I've been tearing down LED light bulb to see how they were driven and found out that most of them use this very simple circuit:

enter image description here enter image description here

This is my understanding of the circuit:

  • 220V AC Input (I'm un Europe)
  • 220V DC after the rectifier bridge
  • Each LED drops about 6V of voltage. There are 32 LEDs so 32*6 = 192V
  • The constant current driver regulates the current and drops the remaining 28V
  • (There are actually 2 drivers in parallel but it makes no difference to the concept)

Is that correct? If I want to make a custom circuit where I drive 1 single 36V LED (like a COB), how would I best drop the voltage from 220V to 36V? Using massive resistors?

Thanks everyone!

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  • \$\begingroup\$ That, or capacitive dropper on the AC side. How much power are we talking about? If more than a few W, you’ll be looking at an SMPS instead. \$\endgroup\$
    – winny
    Commented Mar 28, 2022 at 21:50
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    \$\begingroup\$ Note that 220VAC rectifies to AC*\$\sqrt{2}\$ = 311VDC. \$\endgroup\$
    – rdtsc
    Commented Mar 28, 2022 at 22:04
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    \$\begingroup\$ 9, 12 and 15v LEDs are more common. In this case, probably 9v. You can't use a linear regulator like that on 36v, so you need to use a switching regulator. \$\endgroup\$
    – user1850479
    Commented Mar 28, 2022 at 22:12
  • \$\begingroup\$ @rdtsc - Only for small currents. For larger currents, you get a more-or-less sawtooth with a peak value somewhere near 311. The amplitude of the sawtooth will depend on the current and the value of the capacitor on the output of the bridge. \$\endgroup\$
    – WhatRoughBeast
    Commented Mar 29, 2022 at 0:16
  • \$\begingroup\$ Unless you are ok with your LEDs flickering (and low quality products certainly do flicker), the capacitor needs to be large enough that the linear regulator will never enter drop out. That means a DC output with no more than at most ~20V of ripple, and preferably less since not everyone has exactly 220v line voltage and not all LEDs will have exactly the same forward voltage. \$\endgroup\$
    – user1850479
    Commented Mar 29, 2022 at 0:33

3 Answers 3

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At 36V, a 6W diode array would require that linear regulator to dissipate more than 45W. The maximum you can dissipate in a lightbulb like package is probably less than 10, at least if you want reasonable lifespan. Plus a 6W light bulb that need 51W is just a bad design. That circuit you found only works if you have a voltage drop across your LEDs that sums to close to rectified mains, which in this case is about 310V. If those are 9V LEDs, your 32 diodes sum to 288V, meaning that regulator has to drop just 22V, probably less at higher current. There is no way to adapt that design to a 36V forward voltage.

If you want to drive a 36V LED (which is probably a questionable idea in a light bulb), you will need to use a different type of power supply. One option is a transformer that steps the voltage down to slightly more than 36V and then adding a rectifier and linear regulator. Another option is a switching regulator in place of a linear regulator.

What are your actual goals here? Do you know how to safely design a power supply that runs off of mains power? That is not necessarily a trivial thing to do, at least if you want your device to be safe.

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Is that correct? If I want to make a custom circuit where I drive 1 single 36V LED (like a COB), how would I best drop the voltage from 220V to 36V? Using massive resistors?

There is a better way. Use less resistors (or different resistors) to get about 36V of drop, then if needed change the current limit with Rext

enter image description here Source: SM2082EAS datasheet with mods

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  • \$\begingroup\$ I don't follow, the Rext resistor only limits the driving current of the constant current driver, not the voltage, am I correct? @Voltage Spike \$\endgroup\$
    – coriv
    Commented Mar 28, 2022 at 22:18
  • \$\begingroup\$ It's not realistic to do that in a light bulb given the thermal constraints. That linear circuit works because the diodes drop at least 290v. At 36v that circuit will melt at the current required. \$\endgroup\$
    – user1850479
    Commented Mar 28, 2022 at 22:20
  • \$\begingroup\$ I thought you wanted to go down to 36V for Vcc \$\endgroup\$
    – Voltage Spike ♦
    Commented Mar 28, 2022 at 22:28
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The 220 VAC rectified into a capacitor would have peak voltage of 311V.

The average would be determined by the capacitor value and current draw.

You can't assume a 6V drop per LED, unless you measure it.

If you want to drive a single 36V LED, you would use a constant current power supply, not resistors.

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  • \$\begingroup\$ Do you have an example of a constant current supply IC that would work in this scenario @Justme ? \$\endgroup\$
    – coriv
    Commented Mar 28, 2022 at 22:17
  • \$\begingroup\$ No, partly because I don't know any, and partly because asking and suggesting products to buy is off topic. But you can just buy ready made modules, instead of building your own from scratch. \$\endgroup\$
    – Justme
    Commented Mar 28, 2022 at 22:21

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