Before we learn the formula for calculating and optimizing the mains capacitor current in a transformerless power supply, it would be important to first summarize a standard transformerless power supply design.

The following diagram shows a classic transformerless power supply design:

Referring to the diagram, the various components involved are assigned with the following specific functions:

C1 is the nonopolar high voltage capacitor which is introduced for dropping the lethal mains current to the desired limits as per the load specification. This component thus becomes extremely crucial due to the assigned mains current limiting function.

D1 to D4 are configured as a bridge rectifier network for rectifying the stepped down AC from C1, in order to make the output suitable to any intended DC load.

Z1 is positioned for stabilizing the output to the required safe voltage limits.

C2 is installed to filter out any ripple in the DC and to create a perfectly clean DC for the connected load.

R2 may be optional but is recommended for tackling a switch ON surge from mains, although preferably this component must be replaced with a NTC thermistor.

In the entire transformerless design discussed above, C1 is the one crucial component which must be dimensioned correctly so that the current output from it is optimized optimally as per the load specification.

Selecting a high value capacitor for a relatively smaller load may increase the risk of excessive surge current entering the load and damaging it sooner.

A properly calculated capacitor on the contrary ensures a controlled surge inrush and nominal dissipation maintaining adequate safety for the connected load.

The magnitude of current that may be optimally permissible through a transformerless power supply for a particular load may be calculated by using Ohm's law:

I = V/R

where I = current, V = Voltage, R = Resistance

However as we can see, in the above formula R is an odd parameter since we are dealing with a capacitor as the current limiting member.

In order to crack this we need to derive a method which will translate the capacitor's current limiting value in terms of Ohms or resistance unit, so that the Ohm's law formula could be solved.

To do this we first find out the reactance of the capacitor which may be considered as the resistance equivalent of a resistor.

The formula for reactance is:

Xc = 1/2(pi) fC

where Xc = reactance,

pi = 22/7

f = frequency

C = capacitor value in Farads

The result obtained from the above formula is in Ohms which can be directly substituted in our previously mentioned Ohm's law.

Let's solve an example for understanding the implementation of the above formulas:

Let's see how much current a 1uF capacitor can deliver to a particular load:

We have the following data in our hand:

pi = 22/7 = 3.14

f = 50 Hz (mains AC frequency)

and C= 1uF or 0.000001F

Solving the reactance equation using the above data gives:

Xc = 1 / (2 x 3.14 x 50 x 0.000001)

= 3184 ohms approximately

Substituting this equivalent resistance value in our Ohm's law formula, we get:

R = V/I

or I = V/R

Assuming V = 220V (since the capacitor is intended to work with the mains voltage.)

We get:

I = 220/3184

= 0.069 amps or 69 mA approximately

Similarly other capacitors can be calculated for knowing their maximum current delivering capacity or rating.

The above discussion comprehensively explains how a capacitor current may be calculated in any relevant circuit, particularly in transformerless capacitive power supplies.

NABAJYOTI TALUKDAR says

Dear sir, in my case what is the o/p current.

Swagatam says

25mA

NABAJYOTI TALUKDAR says

Dear Sir,

I am making one ckt by using 474K 400 V capacitor and bridge rectifier .At Output I got 15 V DC.

So my question is

How much watt we can connect and how it is calculated on various Non POlarised capacitor.

Please define by using example.

Thank you .

Swagatam says

Dear Nabajyoti,

I have already explained the formula using the example of a 1uF capacitor…in your case it will be 0.47uF, because 474 = 0.47uF

watts will be = output V x I

arjit says

Can I make it without any zenner diode to power 12volt 25ma two way relay , or it can handle unfix voltage with near fix current ? ðŸ¤”ðŸ¤”

Swagatam says

yes you can do it…

kirams says

I wanted to know the efficiency of the power conversion here.

For eg, assuming it supplies 50mA @12V = 0.6 Watts, How much power it is drawing from mains.

If it is not stepping down the voltage, is it consuming 50mA @230V also making it massive 11.5 Watts, dropping efficiency to 0.6/11.5 =~ 4%, considering 0.8 Power factor, it may boost to 4.32%,

Is my calculation correct? or I am missing something?

Swagatam says

It will be near to 100%, since in this design there's nothing that may be dissipating electricity as heat, except the zener diode which could be quite negligible.

if the output is 0.6 watt, divide this with 220V for the input current, that's 0.6/220 = 0.0027 amps

however the capacitor must be rated correctly according to the load otherwise the efficiency might reduce due to heavy shunting by the zener

Swagatam says

It is not stepping down the voltage but it's restricting the current according to its reactance and thus resulting in lower consumption

Swagatam says

that's the difference between using a resistor and a capacitor…a resistor would dissipate all that it may be restricting through heat, whereas a capacitor would simply use only the amount it's rated to use.

Unknown says

if we need 24v what we can do ??

Swagatam says

use 24V zener diode

xcutionerz says

Greetings,

Dear Sir, it was informing following the capacitive reactance way in solving

for C.

For years I've use the energy based simplified derivative form:

CV=IT

where… using values from your example for a 1uF Cap:

C=Capacitance (F) __

V=Voltage in Vpk to pk = 2 x /2 = 2.828

I=Current in (A) = .069

T=time (period of ripple frequency) sec (full wave = double) = 1/100Hz

solving for C:

…dividing V on both sides to solve for C;

CV/V=IT/V

C = IT/V

C = 0.069 x 0.01 / (220 x 2.828) = 622.16

C = 1.109039475376109e-6 or 1.11uF

CooL thanks,

STeve

Swagatam says

yes that's another interesting way of solving the results, thank you for updating this valuable information

cheekin says

Can you please design a circuit for my battery operated (1.5v} wall clock to AC supply but will run to battery when AC supply is in off state. Thanks.

Swagatam says

I'll try to post it soon

Alberto Villa says

Hello Mr, I'm trying to use a microcontroller for control a relay.. The total consumption is around 220mA, because I have another components. So, I'm the typical voltage in my country is 120VAC. I need to stablish the current, Â¿how can i do it?

Swagatam says

hello, you can use the following circuit for your purpose….120V is not important.

http://www.homemade-circuits.com/2016/07/scr-shunt-for-protecting-capacitive-led.html

Alberto Villa says

Thanks Mr., Â¿what happens if i put two zener diodes in paralell?, and if I put an inductor after the bridge rectifier to stablish more the current.

Swagatam says

you can putting two zeners in parallel but it's ideally not recommended…..if you do so make sure you solder them very close to each other such that their terminals share the same solder pad and solder joint.

Alex Antony Edakkattuvayal says

Thank you sir ..

Alex Antony Edakkattuvayal says

Sir so in this case can I connect 225/400v capacitor and a .5 w 5.1 v zener diode regulator..and what should be the resistance value of zener series resistor and it's wattage

Swagatam says

Alex, the resistor can be avoided if a 1 watt zener is used, or arbitrarily you can try a 10 ohm resistor instead of the 50 ohm for the purpose with a 1/2 watt zener

Alex Antony says

Halo sir I constructed this circuit for the purpose of a 5 v 60 ma with 474 400 v cap and the filtering capacitor is 330 MFD 25 v and a zener diode and 50 Ohm 2 waat resistor as zener series resistor. The circuit is working fine..but it is not connected to a constant current load.when the circuit is off the output will have just 10 ma current and when the circuit is on it takes 60 ma. Is it a dangerous design ? please help me

Swagatam says

It is OK, but it can give a lethal shock if any of the wires associated with the circuit is touched since the circuit is not isolated from mains.

by the way a 474/400V will never produce more than 25mA

Bruno Araujo says

Hi, first thanks for your site and dedication. Is possible to get a transformerless supply with 3.3V 0.5A, and also get a pulse to detect the zero crossing from the input sine? It is for a dimmer porpuse. Thanks.

Swagatam says

Thanks, 0.5a is quite high and the capacitor value for achieving could be very high, you could try the circuit shown in the above article with C1 = 10uF/400V non-polar.

for 3.3V you could replace the zener diode with 6nos of 1N4007 connected back to back with cathode towards the ground.

the zero crossing detection could be achieved by using the opamp circuit from the following first circuit.

http://www.homemade-circuits.com/2016/07/how-to-make-zero-crossing-detector.html

swapan byapari says

Sir, please.. Tell me that is this circuit carry excessive power losses due to use a c1(capacitor) without using transformer to drop voltage to 12 volt dc from 230 volt ac. If I connect 12 volt 100 ma fan to circuit,then what will be the power consumption by circuit——

1. 1.2 watt or

2. 23 watt

Swagatam says

Swapan, there won't be power loss if the capacitor is correctly calculated…for your application you can use two C1 in parallel, and make sure to include an NTC.

Swagatam says

consumption will be 12 x 0.1 = 1.2 watts

Mohamed Rihawei says

Hello dear,

I need to know if I put 5* 1 micro farad capacitors , shall I use 6A diod for bridge or only 1A diode can be OK?

Swagatam says

Hello, using 5uF will generate 5 x 50 = 250mA current, therefore 1N4007 can still be used without much issues.

kumaran says

Sir,

I use 6nos.of smd led(spec. Forward voltage3.2v, current 120mA). I want to connect AC main(230vac) Please guid what range of capacitor (c1) I use and how to calculate.

Swagatam says

kumaran, you can use a 2uF/400V capacitor with an NTC or an MOV at the input mains side

kumaran says

sir,

i use a 2uF/400V capacitor, and i calculate current value is apprx.150ma,.please tell me sir, how to calculate output voltage..

Thanking you

Swagatam says

Kumaran, the output voltage will be exactly equal to the input voltage.

Syed Ameer Hamza Askari says

Sir what is pi?

Swagatam says

Syed, please refer to this article:

https://en.wikipedia.org/wiki/Pi

rolli laya says

hi, in your circuit above how can we accurately compute the value of the capacitor (C1)we will use with a known load such as an led with 20ma rating.

Swagatam says

hi, you can use the following method:

since we know from the above article that a 1uF capacitor allows 69mA, therefore:

1/C = 69/20

69C = 20

C = 20/69 = 0.28uF will be the required capacitor value for your 20mA application, but since .28uF may not be a standard value, you could probably go for a 0.33uF instead.

Tiago NET says

may be used motor capacitor AC in C1?

Tiago NET says

can i use motor capacitor?

Swagatam says

yes you can

hendra sana says

Hi sir, I want to asking something, I have made it I using 225 K/ 250 V nonpolar capasitor, for the first time I try it, it work 1 minute, but when I plug out and then try to plug in again the adobtor didnt work, Im using project board for this case, please give me information

Swagatam says

Hi Hendra,

Use a PCB, and assemble by soldering the components, if you have used a breadboard then there could be a possibility of a loose connection, so recheck the connections, and confirm the same.

also you should use a 400V capacitor for maximum safety….250V is not safe.

hendra sana says

yes I got the 400 v capasitor, but can I using resistor 1 M 2 watt and 56 ohm 2 watt with zener 1 watt for cicuit above? I cant find resistor 1m 1 watt, please help,,,

Swagatam says

1M/2 watt is OK for R1, actually any resistor above 100k will do for R1

any resistor between 10 ohm and 50 ohm will work for R2

hendra sana says

HI sir I got some problem, my adabtor has burnt, I dont know whats wrong, flash happen and burn my adabtor, my house mains also extinguished. what does commonly make this happen? you can see the picture

https://twitter.com/cahyadi_suara/status/642889259943260161/photo/1

Swagatam says

hendra, you might have done something incorrect in the connections.

don't use a breadboard…solder the parts exactly the same way as shown in the diagram over a general purpose PCB or a veroboard,

shubham bansal says

sir, can we simulate this circuit on multisim software?

because i think, in this software the available capacitor is electrolytic capacitor.

and we are using 105/440v PPC capacitor.

if we cant simulate on multisim software so please guide me on which software we can simulate this circuit.

and how can we calculate the value of voltage and current after the bridge rectifier.

Swagatam says

Hi Shubam, I never use simulators so I won't be able to suggest much about these softwares.

the voltage after the bridge is solely determined by the zener rating, if you remove the zener it will jump to 330V DC, although the meter would show the reading that could be equal to the filter capacitor's voltage rating…. but that would mean the capacitor being subjected to a significant electrical stress….so the zener is a crucial part for controlling the output voltage to the required levels

manpreet singh says

Thanks for ur respons sir 1 more think i want to know how to calculate the loading resistor value in the ckt.

Swagatam says

According to me you can use Ohm's law for this:

R = V/I, V = 330, and I = LED current

pradeep says

sir is load resister and R2 are same ? if not how to calculate it ? how v=330 after bridge rectifier ?

Swagatam says

pradeep the load resistance is not crucial, it should a low value resistor just to restrict the switch ON surge.

the peak value of 220V RMS is around 330V…after rectification the RMS turns into peak value in the form of 330V DC

pradeep says

thanks sir for reply,

my basic knowledge about analog electronics is not so strong, i want to know that if i will get 220V RMS value after C1, then what is purpose of C1. how and where ac voltage is dropping.

Swagatam says

the capacitor restricts the 220V current to a very small level, when a load which has voltage requirement much lower than 220V but current much higher than the capacitor output is connected, then the load acts like a shunts or shorts the capacitor output until the 220V volts drops to the load voltage level.

Saqib Mehmood says

Sir as I know that c1 control current then if current is 15 ma and my out put current requirement is also 15ma then are u sure that at the out put we will get 15ma

Swagatam says

yes that's a sure thing…the current control will be constant but during switch ON it can increase in the form of a instantaneous spike.

manpreet singh says

Thank you sir but Iam using 5730 smd .5 watt led Forward Current is 150 mA and according to the above calculation 1uf/400V at 250V/50Hz able to provide 69 mA current.

and the other issue I have LED heating to much for this I remove loading resistor and put 22 ohm resistance in the series it work for 4 hrs.

Swagatam says

you may use 2.2uF capacitor but make sure its 400V rated and is non-polar, and also don't forget to put an NTC thermistor in series with the input of the capacitor as described here:

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

R1 should be 1 Meg Ohm or any other higher value

manpreet singh says

I am using .5 watt smd led and polyester film cap is 225j

Swagatam says

try a 1uF/400V PPC caapcitor

manpreet singh says

hello sir I am using the same ckt but driving 24 led in series C1 is 2.2uf or R1 is 470E but cap C1 get blast after few time can u help me

Swagatam says

hello manpreet, what's the wattage of the LEDs?

…..by the way the capacitor must be nonpolar and rated at 400V

Naresh Jain says

Lot of thanks

Naresh Jain says

sir please explain:

Xc = 1/2 x 3.14 x 50 x 0.000001

=Â 3184 ohms approximately

on solving above formula it is 0.0000785 not 3184

Swagatam says

Naresh,

the numerator 1 is for the entire denominator, calculate in the following manner:

1 / (2 x 3.14 x 50 x 0.000001)

(***)(impetuos)(***) says

Hello Mr, I was reading your interesting post and I was wondering about the C1 value, I'm trying to get work a 60led string (3,2V 150mA) but the simulation doesn't work, my calculation is for 120V 60Hz and the C1 value is 3,3uF, also the C2 I use 15uf 250V but is not working at all, I have a severe wave of VDC I mean is not continue , what could be wrong? Also I tried to get a voltage doubler but the signal on VDC is the same as your circuit simulation

Swagatam says

Hello Mr. here are the calculations:

Since the supply is 120V you will need to make two strings of 30 LEDs each and connect them in parallel, each having a 22 ohm series resistor

Therefore the total current requirement would become 300mA.

As per the calculations a 1uF caapcitor at 60Hz/120V would produce 45mA current, therefore for achieving 300mA we would need 300/45 = 6.66uF capacitor

For Z1 you may include a 120V zener diode or simply eliminate it.

C2 could be a 15uF/250V no issues.

vasudevan v says

Swagatham Majumdar,

If I need output 3 V DC or 6 V DC what value of C!, C2 , R1 and R2 to have in the circuit. Pl explain

Regards

Vasudevan V

Swagatam says

Vasudevan, voltage can be controlled only by using zener diodes or volatge regulator ICs, C1 decides the current not the volatge, 1uF/400V will allow you to get around 70mA current.

R1 can be 1M it's not relevant to the output, R2 is also not crucial can be removed if the current specs of the output matches with the C1 max capacity.

An NTC or MOV is a must and should be considered.

Hz021 says

GREAT!.. thank you Mr Swagatam Majumdar. ðŸ™‚

Swagatam says

you are welcome Hz!

kostas says

Firstly i would like to thank you for your schematic it was really usefull and your explanation is great !

But i would like to make a question about the current. I did make the circuit and i used a 7805 also in the output and i did supply an other pic microcontroller circuit with voltage. BUT i measured the current betwen the AC power and the C1 and it was always AC 220mA (it's maximum because i use 3,3uFcap at 250V) and when i measure the DC current after the 7805 it was draining only DC 47mA.

Can you explain me what happens ? Because if leave the device open for long time the cap the diode bridge and the 7805 get hot.

I would provide you with a schematic but i didn't know how to post the Picture here.

Regards, Kostas.

Swagatam says

Thank you kostas! If you ask me about your design I would suggest that it's not recommended and is dangerous to use a capacitive power supply for sensitive circuits, I would advise you go for any smps adapter instead.

as for the current difference, the issue may be due to the zener diode or the 7805 which are itself draining all the current and allowing only 47mA to pass to the circuit, try removing the zener and check, or you may want to try a transistor common collector stage instead of the zener and the IC for achieving the required regulation without dropping much current