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SCR Shunt Circuit for Protecting LED Drivers

SCR Shunt Circuit for Protecting LED Drivers

The post presents an effective method for protecting capacitive LED driver circuits through an SCR shunt regulator circuit and explains how it can prevent filter capacitors from blowing of and LEDs from getting destroyed.

The remedy was requested by Mr. Max Payne.

Protecting High Watt LEDs

Please suggest regarding 3W, 5W LED bulb circuit design fault. the LED bulb running very cool. The electrolyte capacitor (47uf to 100uf , 50v rated) keeps explode. without any reason. and wasting all LED's but smd resistor (474 & 560), MB10S, 105j400v has no problem, This is the problem pcb of the cheapest 3W, 5W LED set.

How can i solve the problem permanently or Pls suggest me few cost effective designs ... I noted that these explosion happening when i operate induction cook-top.


The Design

A filter capacitor in a capacitive power supply will mostly blow-of due to an insurge of voltage higher than its rated value. For example if the filter capacitor breakdown voltage is 50V, and if the surge voltage exceeds this limit could instantly cause the capacitor to explode or gas.

A rather simple solution to protect the vulnerable parts in a capacitive transformerless power supply such as the capacitors and the LEDs is to introduce a shunt regulator circuit as indicated in the following diagram:

Circuit Diagram


SCR Shunt Circuit for Protecting Capacitive LED Driver


Here the SCR forms the main element for curbing the initial high voltage, high current surge and to ensure a regulated constant supply regardless of the input fluctuations.

If we recall, this concept has been already discussed in one of my earlier posts titled how to make a high current transformerless power supply circuit

Circuit Operation

The idea here is to switch ON the SCR or a triac as soon as the input supply tries to surpass the stipulated safe voltage limit of the circuit, which in turn depends on the indicated zener diode voltage.

Referring to the above SCR shunt circuit for protecting capacitive LED drivers, whenever the input from the mains capacitor tends to go above 12V, the zener diode conducts fully, triggering the SCR, and causing its anode cathode leads to short circuit the supply instantaneously. This response becomes enough to stop the voltage to rise any further and ensure that it's limited within the assigned 12V range.

The stage comprising the SCR, the zener and the 1K resistors act like a high current zener diode  and is employed here since the input current is much higher and beyond the capacity of a normal 1 watt zener diode, however for low current transformerless power supply circuits the SCR and the 2nos 1K resistors may be eliminated and only the 12V zener could be used for the intended regulation.

Using High Voltage Transistor

Another effective shunt regulator could be built using a high voltage transistor and a variable zener diode, as shown below:

LED driver shunt regulator


About the Author

I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. I am also the founder of the website: https://www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials. If you have any circuit related query, you may interact through comments, I'll be most happy to help!

24 thoughts on “SCR Shunt Circuit for Protecting LED Drivers”

  1. Hi swatam
    I tried the first circuit. Every thing is cool fine, but 12v 10w led is blinking due to ripples. I tried with 1000uf electrolite capacitor, but the proplem is still same. How can i solve this issue

    • Hi Raees, It could be due to insufficient current. A 12V 10 watt led will require 0.8 amp current, so the input capacitor will need to be around 16uF/400V, which value did you use in your set up?

      • Thanks for reply
        I understand. I used 2.2 uf pp capacitor. Actually i require 300mA current, so i should add more pp capacitor to obtain exact currentagain thanks a lot. May u live long sir. Your valuable suggestions are always regarded.

  2. Hello swagatam

    We looking for a topology that will provide a constant 12 vdc output at 1 amp, over a 5 to 1 input voltage range, we can do it with a switch mode, however the transformer and the size of the circuitry becomes very large, a transformer less topology looks interesting, our input voltage is 120 to 600 v @ 60hz, any ideas?


    • Hi Andrew, according to me the exact same configuration could be tried, except the shown SCR which must be replaced with a 1kv SCR such as the TYN1012RG.

      The input 5uF capacitor will also need to be replaced with a 20uF/1000V capacitor for achieving 1 amp output

  3. Hello swagatam,

    here in this circuit X rated capacitor used to step down the voltage ….but for the 5uf 400V capacitor how much voltage will get drop down ….this has any calculations ?? how much voltage will get before bridge and after bridge….

  4. I am trying to use MB10s with an 8mm 2Chip LED so that I can use the LED without worrying for the polarity but the brightness is very less. What can I do to increase the brightness. Input voltage is around 5V AC

    • rectifier bridge will drop 0.7 + 0.7 = 1.4V, so 5 – 1.4 = 3.6V, that looks fine if the LEd is the normal 3.5V rated

      by the way the MB10 is rated only at 0.8A, and will be extremely hot at this full current level.
      what is your LED amp rating??

      you may have modify the bridge current accordinly

  5. Thx swagatam. nice work. limited space in smd board to implement that, instead i added C1 on another side parallel to existing one & R1 series to LED line, but I'll test that circuit, and Happy Ramzan!

    Iwanna post some interesting Electronic articles here, pls gv me ur gmail id.

    • you are welcome Max, you will need to include only the 4 components, the SCR, the two 1 K and the zener, rest are already present in your board…and these can be made into SMD and easily accommodated in the existing board.
      the other simple remedy is to use an inductor in series with the supply…because inductors can be quite good at suppressing switch ON surges.

      you can definitely post innovative articles in this site…it will be greatly appreciated.

    • Hello SWAGATAM,
      I continue here, now i have another question, can i remove 400V 5uF and 1M Resistor ?, take too much space on my pcb 🙁
      Should i use another way to get what i wish ? (5v 200mA)
      Is there another way like high voltage regulator (LR645) ?
      Best regards.
      Oops, i mess something, i need also 3,3v , so i use two ams1117 regulator behind 12vdc , but i only use 5v and 3.3v

    • Hello SWAGATAM,
      LR645 and DN2450 is also too expensive.
      I have to find a way wich is low coast and not energy consumer (…)
      Now, do you think a can find a way with LR8N8 ? and find a way to get up to 30mA
      I don't really know how to do this .
      Best regards.

    • Thanks , very interesting !
      May i disturb you for any other question about that ?
      where should i post , in the new link ? or this one ?

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