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TransformerLess Power Supply

November 27, 2009 D.Mohankumar 36 comments

Transformerless power supply for low current applications. C1 is the X rated AC capacitor that reduces high volt AC. D1-D4 rectifies AC to DC and C2 removes ripples. R1 is the bleeder to remove stored current in AC when power is off. R2 limits inrush current. A Zener can be used in the output to get regulated DC.

Transformer Less Power Supply Circuit Diagram- Revised

Caution ! Circuit is lethal since it carries high volt AC. Do not touch or test when connected to Mains.Enclose in a shock proof case.

D.Mohankumar

Power supply, Transformer Power supply, transformerless

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Written by Tami on May 16, 2010. Reply

Thanks pls i want to know if the heat generated can make the capacitors go bad in a short time and can a transformerless power supply circuit of 12v and 100mA power a battery charging circuit of 12V 1mA. Thanks
- Written by D.Mohankumar on May 17, 2010. Reply
  
  Capacitor will not generate much heat.It will not be destroyed due to heat problems.Dielectric of the capacitor will be destroyed only if the line voltage goes above the rated voltage of the capacitor.It is better to avoid capacitor power supply for battery charging, if the battery is kept open.Generally these forms of power supply are not recommended to use in circuits that require human contact .If you want to charge a battery that is kept in an enclosure, use a series resistor at the output to get desired current. Use ohms law for the current.
  - Written by Sahil on May 31, 2010. Reply
    
    Sir
    i need 6v dc supply
    can u tell me how could i get 6v dc from above circuit i.e. Transformerless Power supply
    - Written by D.Mohankumar on May 31, 2010. Reply
      
      Connect a 6.1 volt Zener across the output
Written by Tami on May 18, 2010. Reply

Thanks for your support. The battery charging circuit in question can be used to charge a 6watt or 12watt battery. The input voltage ranges from 9V To 40V and the current is 1A. Thanks for your contribution
Written by Nick on June 1, 2010. Reply

I have a pumpmotor 240VAC / 50W that is controlled by a mecanical thermostat switch. I would like to indicate that it is running with a simple LED in serial with the pump somehow. The problem is that i only have very limited physical access only to the switch that controlls the 240VAC for the pumpmotor. How can i sifon off 5V/20mA for the led with som simple componets that will not significantly affect the pumpmotors capacity? I tried connectin a cheap battery eliminator 240VAC/5VAC but that made the pump slow. Im looking for a super simple solution and i dont want to use a battery that recuires changing.
I am wondering if perhapps a simple ferite ring with a few loops (<5) and a small resistor on the secondary side could be used as a transformer. How would one go about calcuating the number of loops to get 5 v/ 20mA. Would this affect the pumpmotor. Will I perhapps need an extra sparc protection on there realay since I have a serial tranformer. I have electronic schooling but it was to decades ago.
Written by D.Mohankumar on June 1, 2010. Reply

The most simple solution to light an LED is to connect a 47k 1 watt resistor in series with the LED . This should be connected across the phase and neutral after the pump switch.The brightness of the LED will be low since it is lighting only in the positive half cycle of AC.
Written by Nick on June 2, 2010. Reply

Thank you for your quick and wize reply. Im sure it would work if i had access to the neutral cable after the pump. 240V/47K equals 5,1mA thru the LED. But would it not be to much voltage for a “simple led” to handle?
I was schooled 20 year ago and have not practised electronics since. Back then the diodes we used “handled” quite low voltages i think it was 0,7V – 12V. Does modern day LED’s handle 240 V or how would the voltage be over/on the LED. Im a bit afraid it might blow up violently taking out an eye and a few brain cells
(murphys law :-) ).

Sadly I do not have access to the neutral wire after the pump. I want to put the indication diode at the switch (in serial somehow). Im sorry for the unclear problem description. What i would like, if it possible, is some simple cirquit thingie in serial with the switch. That would not affect the pump noticeably.

If i parallel connect three 10 ohm resistors i would get a 0,03 ohm resistor. 10+10+10/10*10*10=0,03 ohm.
In serial with the switch that would make 240VAC*0,03 Ohm=7,2V over the resistor(s). If i then also put a (diode and a 360ohm(7,2V/20mA) in serial) in paralell over with the 0,03 ohms. What would i get? What kind of Watt tolerance resistors would i need for this? I cant realy figure out the watt here and i dont want redhot resistors just to lite the LED. Would this be safe?

Here is a link to the boyler thermostat i want to put the indicator led in serial with. It’s basicaly a normal cirquit switch. http://www.ra.danfoss.com/TechnicalInfo/Literature/Manuals/04/ICPDP10F102.pdf

With grattitude
Nick
- Written by John on June 19, 2010. Reply
  
  Nick,
  
  You can loop your hot wire going to the pump through a CT (current transformer), the turns ration determines the current in CT winding, for example if your pump current is 1Amp then 100turns will induce 10mA in the CT coil, you can use this to light up a LED.
  You can try a small ferrite core and wind the magnet wire onto this.
  
  John.
Written by D.Mohankumar on June 2, 2010. Reply

Do not think too much theory. Just connect the LED with a 47 K or 56 K resistor( 1/2 watt) .LED will glow with half brilliance. But you have to connect it between the phase and neutral lines.
Written by tami on June 4, 2010. Reply

how can i calculate thecharge rate of a battery and to know the charge rate of my circuit. also the calculation to know the capacitor to use in rectification
Written by nouman nayyer on July 25, 2010. Reply

dear sir
sir i made it but with correct component but after few minutes the 100uf-25v capacitor overheated and destroyed and the voltage was also 33 volts dc…..
plz help me regarding this
thank u
Written by D.Mohankumar on July 25, 2010. Reply

The problem is with the capacitor C2. Use 100 UF 100 Volts capacitor and check output voltage. If required use a 12 volt zener across the output
Written by vikas on July 27, 2010. Reply

sir i have a small problem that i need a circuit without transformer which provides 12v 1 ampere current can u please give me suggestion
Written by D.Mohankumar on July 27, 2010. Reply

Transformerless power supply can give only 500mA or less current.If you want 1A,use a transformer
Written by japanjot singh suri on September 1, 2010. Reply

hello sir, i am doing electronics engg,your post seems to be very usefull to me and my frnds, sir as we work on simple projects, we require batteries,and the main problem is bulky transformers in the project, sir we require 6v 350 to 500 ma dc supply from 230v w/o transformer to charge 4v 1amp battery, and we also require 8v 500 ma supply to charge 6v 4.5 ah battery.

acording to your circuit i judged dat we could do dat using zeners of our required voltage, but how to calculate and adjust the output current which we would get

would be glad if u would help us
thank you :-)

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