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Surge Protected Cheap Transformerless Hi-Watt LED Driver Circuit

Last Updated on January 2, 2024 by Swagatam 114 Comments

The increased number of complaints from the readers regarding burning LEDs associated with my earlier posted transformerless 1 watt LED driver circuit, compelled me to solve the issue once for all. The power supply section of the circuit discussed here remains exactly identical to the previous configuration, except the inclusion of the "switch ON delay feature" which has been exclusively designed by me and added in the circuit for rectifying the burning LED problem (hopefully).

Warning: All the circuits I have explained below are not isolated from mains AC and are therefore extremely dangerous to touch in an open and powered condition. A high level of safety precaution is recommended while handling these circuits.

Suppressing In-rush Surge in Capacitive Power Supplies

The complaints that I kept on receiving were undoubtedly because of the initial switch ON surge which kept destroying the 1 watt LEDs connected at the output of the circuit.

The above problem is pretty common with all capacitive type of power supply, and the problems has created  a lot of bad reputation to these types of power supplies.

Therefore normally many hobbyists and even engineers opt for lower values capacitors fearing the above consequence in case larger value capacitors are included.

However as far as I think, capacitive transformerless power supplies are superb cheap and compact AC to DC adapter circuits which requires little effort to build.

If the switch ON surge is tackled appropriately, these circuits would become spotless and could be used without the fear of any damage to the output load, especially an LED.

How Surge is Developed

During switch ONs, the capacitor quite acts like a short for a few microseconds until it gets charged and only  then it introduces the required reactance to the connected circuit so that the appropriate amount of current only reaches the circuit.

However the initial few micro second short condition across the capacitor inflicts huge surge to the connected vulnerable circuit and is sometimes enough for destroying the accompanied load.

The above situation can be effectively checked if the connected load is inhibited from responding to the initial switch-ON shock, or in other words we can eliminate the initial surge by keeping the load switched OFF until the safe period is reached.

Using a Delay Feature

This can be very easily achieved by adding a delay feature to the circuit. And that's exactly what I have  included in this proposed surge protected hi-watt LED driver circuit.

The figure shows as usual an input capacitor, followed by a bridge rectifier, until here everything's pretty common capacitive power supply.

The next stage which includes the two 10 K resistors, two capacitors, transistor and the zener diode form the parts of the important delay timer circuit.

When power is switched ON, the two resistors and the capacitors restricts the transistor from conducting until both the capacitors get fully charged and allows the biasing voltage to reach the transistor base, illuminating the connected LED after a delay of about 2 seconds.

The zener is also responsible for prolonging the delay for two seconds.

The 1N4007 diode across one of rhe 10K resistors and the100 K resistor across one of the 470uF capacitors  helps the capacitors discharge freely once the power is switched OFF so that the cycle can repeat enforcing the surge protection into action on each occasion.

More number of LEDs may be connected in series for increasing the power output, however the number may not exceed 25   nos.

Circuit Diagram

UPDATE:  A more advanced design is discussed in this zero crossing controlled surge free transformerless power supply circuit

The videos below show the LEDs illuminating after about a second on power  switch ON.

Complaints From the Readers (Resistors burn, transistor becomes hot)

The above concept looks great but is probably not working well with the proposed high voltage capacitor power supply.

The circuit has to be researched a lot before it becomes completely free from troubles.

The resistors in the above circuit are unable to withstand high current requirements, same is true for the transistor which also becomes quite hot in the process.

Finally we can say that that unless the above concept is thoroughly studied and made compatible with a capacitive transformerless power supply, the circuit cannot be put into practical use.

A Much Robust and Safe Idea

Even though the above concept failed to work it doesn't mean the high voltage capacitive power supplies are completely hopeless.

There's one novel way of tackling the surge issues and making the circuit failproof.

It's by using  many 1N4007 diodes in series at the output or in parallel to the connected LEds.

Let's have a look at the circuit:

The above circuit is yet to be tested for many months, so these are still early days, but I don't think the surge from the capacitor will be high enough to blow the 300V, 1 amp rated diodes.

If the diodes remain safe so will the LEDs.

More diodes may be put in series for accommodating more number of LEDs.

 Using a Power Mosfet

The first circuit attempt which seemed to be vulnerable itself to surge causalities can be effectively remedied by replacing the power BJT with an 1 amp mosfet as shown in the following diagram.
The mosfet being a voltage controlled device, here the gate current becomes immaterial and therefore a high value 1M resistor works perfectly, the high value makes sure that the resistor does not heat up or burn during the initial power switch ON. It also facilitates a relatively low value capacitor to be used for the required delay ON surge suppressing feature.

A little investigation revealed that the high voltage transistor in the first diagram is actually not needed, rather it can be replaced with a high current Darlington TIP122 transistor as shown in the following diagram.

The high voltage surge from the capacitor becomes ineffective against the high current specs of the transistor and the LEDs and no damage is caused to them, in fact it forces the high voltage to drop to the specified allowable safe limits of the LEDs and the transistor.

The TIP122 also allows the use of a high value base resistor thereby making it sure that it does not become hot or blow off in the course of time, it also allows the inclusion of a low value capacitor at the base of the transistor for implementing the required delayed switch ON effect.

Using a Power BJT

The above design further improves in terms of safety and surge suppression when used in a common collector mode, as given below:

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Filed Under: Power Supply Circuits Tagged With: Cheap, LED, Protected, Surge, Transformerless, Watt

About Swagatam

I am an electronics engineer and doing practical hands-on work from more than 15 years now. Building real circuits, testing them and also making PCB layouts by myself. I really love doing all these things like inventing something new, designing electronics and also helping other people like hobby guys who want to make their own cool circuits at home.

And that is the main reason why I started this website homemade-circuits.com, to share different types of circuit ideas..

If you are having any kind of doubt or question related to circuits then just write down your question in the comment box below, I am like always checking, so I guarantee I will reply you for sure!

Previous Post: « Remote Controlled Toy Car using 433 MHz Remote Modules
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Reader Interactions

Comments

Ashok Dhenge says:
July 25, 2014 at 6:24 am

what happen when supply of ac reverse because we cant chk neutral & negative every time at AC power plug? For safety what we can make provision? can we provide any diode?

Reply
Swagatam says:
August 8, 2014 at 5:20 am

or replace the 12V zener with a 30V zener, or put three 12V zeners in series.

Reply
Swagatam says:
August 8, 2014 at 5:19 am

remove the zener diode and check….

Reply
Ashok Dhenge says:
August 8, 2014 at 4:39 am

sir i made the last circuit but when i connect 4 LED it lights, but if i increase one more led it get off no lights. as you stated in above comment connect LED in series upto 10 no. right? ( we talk about last circuit in above article LED after TIP 122)

Reply
Swagatam says:
August 1, 2014 at 12:53 pm

you can connect upto 10 LEDs in series in the last circuit above and 100 LEDs in the following design:
https://www.homemade-circuits.com/2012/04/how-to-make-led-bulb-circuit.html

Reply
Ashok Dhenge says:
July 31, 2014 at 7:42 am

ok can we increase the no of LED ? if yes how? how can we replace 1 watt led by 5mm led

Reply
Swagatam says:
July 31, 2014 at 7:35 am

Ashok, a resistor as high as a 1M will NEVER burn, you might have connected some other value resistor. or may be the resistor that you have used is faulty or low quality….please check it

Reply
Ashok Dhenge says:
July 31, 2014 at 3:13 am

Sir i made this circuit succeful. Next i increase the no of LED upto 6 no but when i supply it 1 M resistor get burned i change resistor & decrease resistor upto 3 no circuit works ok so as increase in LED no there is change in resistor value? Now i made last one circuit, which circuit is cheaper & easy to assemble & small with 3 led no in each circuit for more than 10 no of circuit to assemble?

Reply
Swagatam says:
July 25, 2014 at 7:23 am

nothing wrong will happen, the input polarity is not critical

Reply
Ashok Dhenge says:
July 23, 2014 at 9:31 am

sir i made last one circuit but when i start the supply 100ohm/1/4watt resistor burn. So after that i remove that resistor and move the series of LED at position of 100ohm resistor as shown in 2nd last circuit but when apply supply LED get blew why happen this? please help me.

Reply
Swagatam says:
July 24, 2014 at 5:30 am

which LED did you use? only 1 watt should be used, remove 100 ohm use 22ohm instead, and replace the 1uF with a 100uF ….then check

Reply
Jayanath Ranasinghe says:
July 16, 2014 at 3:17 am

could you please introduce a LDR for any suitable circuit in the above to use as a automatic day night lamp.

Reply
Joseph says:
June 21, 2014 at 9:40 pm

Hello Sir,
I have built this circuit, but a few components are not available here in Paraguay at local shops. I used a 1uf 640V ceramic cap, and a 1uf 50V electrolitic Cap. I only hook one 1W LED up, but the first try, it blew… Do I need to have 3 hooked up? Also, do I need a resistor in series with the LEDs? For testing purposes, is there a resistor i can use as a dummie load for the LEDs / or a Diode? So that I can check outputs, and currents without shorting out the circuit or destroying lots of LEDs? (they are still about $1.25 each here)

Thank you

Reply
Swagatam says:
July 1, 2014 at 5:47 am

Hello Joseph,

that's very strange. I have built dozens of these units all have worked perfectly, even the most ordinary one without any surge suppression stage worked for months before bursting.

Are you sure about the capacitor that you are using? They must be rated at 400V and of very good quality

a squealing sound is again very strange because we do not have any kind of oscillator or frequency generating stages involved.

your input capacitor should be like these:

i01.i.aliimg.com/wsphoto/v0/1140853215/Free-shipping-Metal-film-font-b-capacitor-b-font-font-b-105j-b-font-font-b.jpg

please try replacing the TIP122 with MJE13055 and check the response using the last circuit.

Reply
Joseph says:
July 1, 2014 at 1:46 am

Hello again,
Is it ok to have a 50V cap instead of a 25V cap? I have the last circuit built, but there still seems to be no delay, although I am using 22uf at 50V.
Plugging it in, the LED did survive, but it was making an odd squealing sound, about the 5th time testing it i heard an arch sound (but did not see anything) and unplugged it immediately. Now the LED is very dim.

Any suggestions?

Thanks for your help

Reply
Swagatam says:
June 25, 2014 at 4:35 am

The delay effect is introduced so that the LEDs can be safeguarded from the initial switch ON surge, it could be a 1 second delay at the most for implementing the above.

The current is hardly 70mA for a 105/400V capacitor, and it's not responsible for the blowing up the LEDs, it's the initial switch ON surge that's doing so measuring with some other alternative won't help with a solution.

However if your LEDs are blowing-of each time, the issue could be something else because even with the surge coming the LEDs would blow up may be one in 10 times, never always.

please try the last circuit shown above, you may try it using 5mm LEds, and a 474/400V cap at the input, if it works alright, you may proceed with a 105/400V cap and 1 watt LEDs.

Reply
Joseph says:
June 23, 2014 at 8:36 pm

Thank you for your replies, I am sorry about the double post, from my phone (the first post) I couldn't tell that I completed the process all the way.

I do not believe I had experienced an On delay (how long of a delay are we talking here?) so I will try the larger cap. regarding my last question, besides just adding a 100 ohm resistor, is there a direct replacement for the LED so that i can take measurements without loosing more LEDs? I have done some searching but have really not found any good answers… If i short the output, what is the maximum current this circuit can supply? (do i have to worry about destroying components if i just measure the amperage across the outupt)

Thank you for your help, in the past I have been pretty good at building circuits correctly, but I do not have a breadboard to work with so its a little different for me.

Reply
Swagatam says:
June 22, 2014 at 9:22 am

Hello Joseph, I have answered your question in the previous comment (just above this comment)

You can use a 100 ohm, 1 watt resistor in series with the LED, however it's strange your LEDs are blowing of, this could have happened if the above mentioned protections were not taken, yet still the chances would be 1 in 10

Reply
Joseph says:
June 21, 2014 at 4:47 pm

Hello sir,
I have tried this circuit, but it blows the 1w led, do I need to use 3 leds in series like the schematic? Can I make this circuit run only a single 1w or 3 lw led? Also, for testing and not destroying leds (here they are quite expensive still) is the a recommended resistor I can use in place if the leds? Or should I use just a n4007 as the diode?

I forgot to mention I am using the circuit with the tip122, and I have 220v here
Thank you

Reply
Swagatam says:
June 22, 2014 at 9:16 am

Hello Joseph, are you getting the delay effect in your circuit, if not then there could be something wrong with the connections.

or try increasing the value of the 1uF capacitor to 22uF or any other higher value for getting a delay effect.

alternatively you can put the LEDs in the emitter arm of the TIP122 instead of the collector arm, this will make things extremely safe.

LEd count is not crucial…

You may also try the second circuit and use six diodes in series instead of the shown 12nos if you intend to use only one 1 watt LEd

Reply
Thilina Jayasingha says:
May 13, 2014 at 2:30 am

Sir, what circuit can i use for maximum brigtness1W led?

Reply
Swagatam says:
May 13, 2014 at 6:17 am

use your cell phone charger for driving the LED via a 6 ohm 1 watt resistor, you'll get awesome brightness.

Reply
Y0f4N says:
April 8, 2014 at 11:37 am

Hello Swagatam,
at your last circuit, you use 12 V for zener to drive 3 units LED with transistor so should be no issue if I use 3.7 V to power up single 3 watt LED. CMIIW
Rgds,
Yofan

Reply
Swagatam says:
April 8, 2014 at 2:59 pm

hello Yofan.

3.7V is OK but it should be minimum 3ah rated

Reply
Gopal Chauhan says:
April 6, 2014 at 4:37 pm

Hi Swagatam – I have made and tested the circuit that has 6 diodes per LED and it is working fine for last 3 days.

Regards
Gopal

Reply
Swagatam says:
April 7, 2014 at 5:38 am

that's great Gopal, appreciate it.

Reply
sachin says:
March 21, 2014 at 10:15 am

hi swagatam can we use inductor available in worn out cfl lamps to suppress surge current

Reply
Swagatam says:
March 21, 2014 at 12:34 pm

hi sachin, yes it can be done.

Reply
Swagatam says:
March 5, 2014 at 12:38 pm

I'll possibly check it in my free time ad let you know.

Reply
Abraham Antony says:
March 5, 2014 at 7:21 am

sir then can u check it practically and can u tell me why i am not getting 40 mA.

Reply
Swagatam says:
March 5, 2014 at 4:36 am

yes it should show 40mA or at least more than 20mA, …not sure until I check it practically.

Reply
Abraham Antony says:
March 4, 2014 at 2:34 pm

but in this link it is of 5mm led know i am having 8mm .5 waat led . so in the earlier circuit why it is not possible

Reply
Swagatam says:
March 5, 2014 at 4:39 am

In the referred link,the circuit employs two high voltage capacitor and resistor network for suppressing surge, this would help prevent your leds from getting damaged from switch ON surge currents.

Reply
Abraham Antony says:
March 1, 2014 at 8:29 am

sir when i reduced the input capasitor from 105 j to 474j i saw the brightness of the led become reduced.so now i have done 25 nos in series of .5watt so how much i can do parallely.

Reply
Swagatam says:
March 1, 2014 at 1:15 pm

Please try the configuration that's shown in the following link:

https://www.homemade-circuits.com/2012/04/how-to-make-led-bulb-circuit.html

You can use 25 bulbs here as well, and a 105/400V for C1 for greater illumination.

Reply
Abraham Antony says:
February 27, 2014 at 7:40 am

hai sir i have tried this circuit but my leds are going one by one i have put 1000uf and 470 uf capasitors and mje 13005a transistor why my leds are going can u tell me pls.

Reply
Swagatam says:
March 1, 2014 at 5:39 am

Abraham, reduce the input capacitor to 0.47uF/400V or 474/400V, this will solve the issue.

Reply
Abraham Antony says:
February 28, 2014 at 9:21 am

hai sir i had used 8mm .5 watt. 25nos in series .and 1 hour i had kept it on.after 1 hour 1by1 started blinking.in that circuit i had used 1000uf and other one 470uf and transistor mje 13005a .1st when i did that circuit 1st 10k resister was hotten then i changed 10k quater watt to 2 watt then there was no problem but the blinking of circuit is continuing and after 1-2 min the blinked led will go. can u pls tell me the reason.

Reply
Swagatam says:
February 28, 2014 at 4:11 am

Hi Abraham, how many LEds have you used and what's the specification of the LEDs?
Please mention these, accordingly I'll tell you the modifications for safeguarding the LEDs..

Reply
Jignesh says:
February 20, 2014 at 9:21 am

Hi Swagatam,
can u build the same circuit with use of battery, so that whenever supply goes, the LED should not get turned off for about 1 hour or less…

thank u in Advance

Reply
Jignesh says:
February 21, 2014 at 12:39 pm

Hi Swagatam,
thank u so much for the given design…..i'll soon implement it and tell the results…

Reply
Swagatam says:
February 21, 2014 at 4:30 am

Hi Jignesh,

You can make the following design:

1.bp.blogspot.com/-I2zlapFq6RM/UML_NxUJy2I/AAAAAAAAB3w/27ZLDjK9C9g/s1600/simple+modem+UPS+circuit.png

Replace the adapter with a capacitive power supply as above and connect LEDs across the shown output terminals

Reply
Avik Paul says:
February 11, 2014 at 6:20 am

MOSFET MLP1N06CL not available in my locality, have U any alternative…

Reply
Swagatam says:
February 12, 2014 at 5:31 am

any 30v, 1amp mosfet or nearby rated will work here.

Reply
Avik Paul says:
February 8, 2014 at 6:04 am

Sir,,
In your 3 rd circuit at this blog, the MOSFET MLP1N06CL which pin i connect where.
pin 1 is Gate
Pin 2 is Drain
Pin 3 is Source

Reply
Swagatam says:
February 9, 2014 at 6:13 am

The gate will go to the 1M resistor, source to the ground and drain to the LEDs.

Reply
Avik Paul says:
January 29, 2014 at 6:51 am

Sir,

So that my circuit output is 16v without load. is it any problem in my circuit.
sir i am used
1uf 400v ppc caps
4nos of IN4007 for bridge and 12nos in series
100uf 160v electrolytic caps
5 nos of 8mm White leds

PLZ help me out…

Reply
Avik Paul says:
February 16, 2014 at 4:38 am

because when i made this circuit with and without load i getting 16v dc 60ma where i used 100uf 100v capacitor and also like aruna Wrote on her blog,
but u said that

Swagatam MajumdarJanuary 30, 2014 at 11:37 AM

the voltage will be equal to the filter capacitor voltage rating, without any load connected.
as soon as you connect the load, the voltage will adjust as per the load specs.
5 leds would create a voltage drop of 3.3 x 5 = 16.5V
that's exactly what you are getting, remove them ad it'll be 160V.

if it's happening without load means there's something wrong with your component connections.
So I am not undrstand what U say….

Reply
Swagatam says:
February 16, 2014 at 4:14 am

Avik, The above statements are correct, i did not understand how you are relating them, please explain your question properly.

the diodes themselves are the load in the second circuit

Reply
Avik Paul says:
February 15, 2014 at 1:48 pm

hi sir,

as per your previous blog you said that

Swagatam MajumdarAugust 6, 2013 at 9:58 AM

Hi Aruna,

The total LED forward voltage drop value should match the total diode fwd voltage drop. 12 diodes gives 12 x 0.6 = 7.2V, this much voltage will suffice just two LEDs….for more LEDs you must go increasing the number of diodes as explained

so why now u say the output voltage without load is 160v and also i check it with your circuit diagram its same as it is..

Reply
Swagatam says:
January 30, 2014 at 6:07 am

the voltage will be equal to the filter capacitor voltage rating, without any load connected.
as soon as you connect the load, the voltage will adjust as per the load specs.
5 leds would create a voltage drop of 3.3 x 5 = 16.5V
that's exactly what you are getting, remove them ad it'll be 160V.

if it's happening without load means there's something wrong with your component connections.

Reply
Avik Paul says:
January 28, 2014 at 7:03 am

Ok Sir, Thanks for feedback now i understand.
so sir would you please tell me what is the output volt and current of that circuit you made and also what is total power consumption…
Sir please tell me…….

Reply
Swagatam says:
January 29, 2014 at 4:09 am

output voltage will be 310V DC and current will be around 60mA, power consumption will depend on the connected load

Reply
Avik Paul says:
January 27, 2014 at 7:29 am

Dear Sir,
I read your blog and also that link, in this calculation i got input capacitor's resistance but input capacitor give about 315v and 70mA but after filter capacitor and diode i got 16v 70ma so (315-16)V=299v is loss for filter capacitor and diode but i got 70ma current in output..

so that scenario is not clear for me…
So sir would you explain it…
PLZ…

Reply
Swagatam says:
January 28, 2014 at 4:10 am

Dear Avik, you must use a 400V filter capacitor, for example you can use a 10uF/400V capacitor as the filter capacitor then you would get the suggested 310V DC output

Reply
Avik Paul says:
January 23, 2014 at 12:30 pm

Hi,
I made your 2nd circuit with 1uf 400 PPC cap.
Output voltage 16.22volt
carrent .06 mA
so total .97 watt
Now question is how to calculate voltage and current loss in that circuit…

PLZ Suggest me……..

Thank for the circuit….

Reply
Swagatam says:
January 25, 2014 at 4:12 am

first you will have to calculate the reactance of the capacitor which will give the resistance of the capacitor in oHMS. This resistance value can be used for getting the max current limit from the cap by using ohms law. after this the total voltage (330v) should be divided by this current to get the value of the voltage drop that would occur with each mA consumption of the load.

you may refer to this post for more info:

https://www.homemade-circuits.com/2011/12/how-to-calculate-and-deduce-current-and.html

Reply
Avik Paul says:
January 22, 2014 at 8:07 am

Can i use 2.2uf 400v PPC capacitor in your 2nd circuit in this page.What the change i Need.

PLZ Suggest me……..

Reply
Swagatam says:
January 23, 2014 at 4:39 am

yes 2.2uF can be used in the second circuit

Reply
lahiru gunathilaka says:
September 15, 2013 at 2:09 am

how to connect 7 LED's.what are the changers.

Reply
asaad un says:
August 20, 2013 at 10:08 am

hi my name is abdullah i have 50 bright led 3.2volt and 1 watt power per led how to make a circuit on 220 volt AC .
plz reply

Reply
Swagatam says:
August 21, 2013 at 6:02 am

Hi, you will require an SMPS unit for powering the LEds, you may refer to the following post for additional info:

https://www.homemade-circuits.com/2013/02/make-this-1000-watt-led-flood-light.html

Reply
Swagatam says:
August 6, 2013 at 1:46 pm

NTC as given in the referred link.

Reply
Swagatam says:
August 6, 2013 at 4:28 am

Hi Aruna,

The total LED forward voltage drop value should match the total diode fwd voltage drop. 12 diodes gives 12 x 0.6 = 7.2V, this much voltage will suffice just two LEDs….for more LEDs you must go increasing the number of diodes as explained, or alternatively you can simply opt for a thermister which will get rid of this diode mess, as shown here:

https://www.homemade-circuits.com/2013/02/using-ntc-resistor-as-surge-suppressor.html

Reply
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