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I'm trying to make a transformer-less LED driver with some high power LEDs and found I this circuit with a 130 mA output:

enter image description here

Can someone explain for this circuit:

  1. Do the voltage changes depend on the load? If so, what is the load of the LED?
  2. I looked up in the datasheet that the LED requires 3.3 V forward voltage and 300 mA to make it to 1 W of power. What will happen if I use 3.3 V with 130 mA? Will the LED turn on?
  3. What will happen if I add a 3.3 V Zener diode parallel to the LED? Will the current change?
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    \$\begingroup\$ Transformerless lighting circuits are somewhat dangerous and mostly used in very low cost, disposable devices that are heavily insulated from the user. Using one is not a good idea. Building your own is an even worse idea, especially if you do not understand how they work. You should buy a proper power supply and then design the low voltage part yourself. It will be safer and your lights will last much longer. \$\endgroup\$ Dec 4 '21 at 17:36
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Do the voltage changes depend on the load?

Generally, yes. But I am not clear what you mean by "the load" in this circuit.

If so, what is the load of the LED?

The LED does not have a "load". The LED is the load of another circuit.

What will happen if I use 3.3 V with 130 mA?

That is not consistent with the specifications of the LED. If its voltage drop is 3.3 V at 0.3 A, then its voltage will be less (such as 3.2 V) at 0.13 A. Regardless, that value will change with temperature. So, to answer your question, you can't "use 3.3 V with 130 mA". You can only limit the current to whatever value you wish (that doesn't exceed the specs for the LED) and then the LED will drop whatever voltage it needs to drop.

What will happen if I add a 3.3 V Zener diode parallel to the LED?

In parallel with one LED? Anode to cathode? If so: nothing will happen: the voltage drop of the LED is lower, so the Zener diode will not conduct any current.

Will the current change?

No.

Hey, let me offer a suggestion: instead of discussing your proposed solution, please focus instead on the problem you wish to solve.

  • If your problem is that the LEDs are too bright, then ask this instead: "How can I modify this line-powered LED lamp to make it dimmer?"
  • If your problem is that the LEDs you picked cannot handle the current that the circuit provides, then ask instead: "How can I modify this line-powered LED lamp to make it compatible with lower-current LEDs?"
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Do the voltage changes depend on the load? If so, what is the load of the LED?

The "capacitive dropper" circuit is approximately a constant current source, not a constant voltage. If you change the load, then the voltage across the output will change.

I looked up in the datasheet that the LED requires 3.3 V forward voltage and 300 mA to make it to 1 W of power. What will happen if I use 3.3 V with 130 mA? Will the LED turn on?

LEDs will work at well below their rated current. They will still light even at 1mA, perhaps even 0.1mA. But you can't "use 3.3V". The current through the LED determines the voltage it will drop. It might be as low as 2.8V at very low currents.

What will happen if I add a 3.3 V Zener diode parallel to the LED? Will the current change?

You have 10 LEDs there. Putting a zener across one of them might drop the current through the LED a bit. But I can't think of any reason why you would want to do it. If you put the zener across the whole string of 10 LEDs, they will all go out.

But whatever it is you're trying to do, a zener diode probably isn't the answer.


This looks suspiciously like it's an "X-Y question". An X-Y question goes something like this...

  1. I want to do X
  2. I don't know how to do X
  3. If I could do Y, then it might help me to do X
  4. I don't know how to do Y, either
  5. I will ask someone how to do Y.

See The XY Problem


Big hint: If you want to change the current through the LEDs, change the value of capacitor C1. It's that capacitor that determines the current that flows through the whole circuit.

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  • \$\begingroup\$ A capacitive dropper is NOT anything like a constant current source. The current through the capacitor is sinusoidal. As you add any voltage offset to the circuit (LEDs) you distort the current flow considerably, but it's still not a constant current source. At best you can consider the peak current (at 100 or 120 Hz) to be relatively constant for a relatively constant mains voltage. \$\endgroup\$ Dec 6 '21 at 19:42

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