How to Make a IC 555 Inverter - Push Pull, Full Bridge, PWM

The post explains how an ordinary IC 555 astable multivibrator could be used to make an inverter without involving complex stages.

The idea was requested by Mr. ningrat_edan.

The Design

Referring to the shown diagram, a single IC 555 can be seen configured in its standard astable mode, wherein its pin#3 is used as the oscillator source for implementing the inverter function.

Using IC 555 as the Astable Oscillator

In an astable mode the pin#3 of IC 555 generates an alternating high/low pulses at a particular frequency rate depending on the values of the resistors and capacitor across its pin#7, Pin#6, 2 etc.

As per the diagram this pin#3 oscillating is fed to a couple of BC547/BC557 buffer driver stages, where one of the stages receives an oppositely oscillating frequency due to the inclusion of an extra BC547 inverter stage.

This ensures that when the left side BC547/BC557 responds to a positive pulse from pin#3 of the IC, the other BC547/BC557 stage is inhibited from the pulse, and their bases are grounded with the aid of the intermediate BC547 conduction.

The above flip-flop functioning of the two BJT stages in turn allow the attached power mosfets to conduct alternately and activate the associated transformer winding with a push-pull current from the battery.

The response allows the transformer to generate the required AC across its secondary winding and implement the intended IC 555 inverter circuit functioning.

The zener diodes at the gates of the mosfets introduce a slight delay time or dead time between the mosfet conduction and inhibits any possibility of both mosfets getting engaged together even for a fraction of a second.

Simplified Version for the above IC 555 Inverter Design

The above shown design can be actually made simpler by removing the BC547/BC557 buffer stages as shown below:

simple IC 555 inverter circuit

Technical Specifications:

Power Output: Unlimited, can be between 100 watt to 5000 watts

Transformer: As per preference, Wattage will be as per the Output Load wattage requirement

Battery: 12V, and Ah rating should be 10 times more than the current selected for the transformer.

Waveform: Square Wave

Frequency: 50 Hz, or 60 Hz as per country code.

Output Voltage: 220V or 120V as per country code

Video Clip:

Waveform Image:

IC 555 inverter waveform image

IC 555 Full Bridge Inverter Circuit

The idea presented belowcan be considered as the simplest IC 555 based full bridge inverter circuit which is not only simple and cheap to build but is also significantly powerful. The power of the inverter may be increased to any reasonable limits y suitably modifying the number of mosfets at the output stage.

How it Works

The circuit of a simplest full bridge power inverter explained requires a single IC 555, a couple of the mosfets and a power transformer as the top ingredients.

As shown in the figure, the IC 555 has been wired as usual in the an astable multivibrator form. The resistors R1 and R2 decides the duty cycle of the inverter.

R1 and R2 must be adjusted and calculated precisely for getting a 50% duty cycle, otherwise the inverter output may generate unequal waveform, which may lead to unbalanced AC output, dangerous for the appliances and also the mosfets will tend to dissipate unevenly giving rise to multiple issues in the circuit.

The value of the C1 must be chosen such that the output frequency comes to about 50 Hz for 220V specs and 60 Hz for 120V specs.

The mosfets can be any power mosfets, capable of handling huge currents, may be upto 10 amps or more.

Here since the operation is a full bridge type without any full bridge driver ICs, two batteries are incorporated instead of one for supplying the ground potential for the transformer and in order to make the transformer secondary winding responsive to both positive and negative cycles from the mosfet operations.

The idea has been designed by me, however it has not been yet tested practically so kindly take this issue into consideration while making it.

Assumably the inverter should be able to handle upto 200 watts of power easily with great efficiency.

The output will be a square wave type.

Parts List

R1 and R2 = See Text,
C1 = See text,
C2 = 0.01uF
R3 = 470 Ohms, 1 watt,
R4, R5 = 100 Ohms,
D1, D2 = 1N4148
Mosfets = see text.
Z1 = 5.1V 1 watt zener diode.
Transformer = Asper power requirement,
B1, B2 = two 12 volts batteries, AH will be as per preference.
IC = 555

Sine Pulse Width or SPWM IC 555 Inverter Circuit

SPWM waveform stands for sinewave pulse width modulation waveform and this is applied in the discussed SPWM inverter circuit using a few 555 ICs and a single opamp.

In one of my earlier posts we elaborately learned how to build a SPWM generator circuit using an opamp and two triangle wave inputs, in this post we use the same concept to generate the SPWMs and also learn the method of applying it within a IC 555 based inverter circuit.

Using IC 555 for the Inverter

The diagram above shows the entire design of the proposed SPWM inverter circuit using IC 555, where the center IC 555 and the associated BJT/mosfet stages forms a basic square wave inverter circuit.

Our aim is to chop these 50Hz square waves into the required SPWM waveform using an opamp based circuit.

Therefore we accordingly configure a simple opamp comparator stage using the IC 741, as shown in the lower section of the diagram.

As already discussed in our past SPWM article, this opamp needs a couple of triangle wave sources across its two inputs in the form of a fast triangle wave on its pin#3 (non-inverting input) and a much slower triangle wave at its pin#2 (inverting input).

IC 555 Pinouts

Using IC 741 for the SPWM

We achieve the above by using another IC 555 astable circuit which can be witnessed at the extreme left of the diagram, and use it for creating the required fast triangle waves, which is then applied to the pin#3 of the IC 741.

For the slow triangle waves we simple extract the same from the center IC 555 which is set at 50% duty cycle and its timing capacitor C is tweaked appropriately for getting a 50Hz frequency on its pin#3.

Deriving the slow triangle waves from the 50Hz/50% source ensures that the chopping of the SPWMs across the buffer BJTs is perfectly synchronized with the mosfet conduct ions, and this in turn ensures that the each of the square waves are perfectly "carved" as per the generated SPWM from the opamp output.

The above description clearly explains how to make a simple SPWM inverter circuit using IC 555 and IC 741, if you have any related queries please feel free to use the below given comment box for prompt replies.


A deeper investigation reveals that the slow triangle waves must have  a frequency of 100Hz and not 50 Hz for creating correctly dimensioned SPWMs, this may be done by using a frequency doubler stage bewtween pin#2 of the IC 741 and the 50Hz from pin#6/2 of the center 555 IC.

Designing a Compact Ferrite Core IC 555 Inverter

The above concept is based on an iron core transformer, in order to convert it into a compact ferrite cored IC 555 inverter circuit, the iron transformer could be replaced with a ferrite EE core transformer consisting of 9 + 9 turns for the primary, and 300 turns for the secondary, using 0.8mm wire for the primary and 0.3 mm wire for the secondary.

This will also need the 555 frequency to be increased to around 50kHz, instead of the 50Hz assigned for the orin core transformer.

IC 555 Pinout

IC 555 pinout details, ground, Vcc, reset, threshold, discharge, control voltage

IC IRS2453 Pinouts

IRS2453 pinout details

For a Ferrite Core Compact Version

Also, for implementing a ferrite core inverter, the output from the transformer will need to be rectified using a bridge rectifier and the resulting DC fed to a 50Hz full bridge or H-bridge processor, as shown below:

In this processor circuit the IC 555 functions as an adjustable PWM generator while IRS2453 constitutes the full bridge driver circuit, together the stages execute a pure sine sine wave waveform for the proposed IC 555 inverter circuit.

For more info you may feel free to express your queries through comments.

Need Help? Please send your queries through Comments for quick replies!


Please sir what is the Watt of this inverter ?
Swagatam said…
Will depend on the trafo wattage and the battery AH capacity.
Please can u help me and tell me how may i know the transformer WATT because i have UPS transformer but i don't know it WATT?
Davis Kakumba said…
hi swagatam for the first diagram what is the rating of the transformer
Swagatam said…
apply mains AC across its mains side and then using an AC ammeter measure the current directly across the other side wire ends...then multiply the reading with the voltage specification of those wires
i am greatfull so much and i am very appritiate for your site
eric maduku said…
pls sir, can you help me with the diagram of IC 4047 and fets 1kw diagram since thats the only ic i can get around. Thanks
Swagatam said…
Hi Davis, it can be as per your own desired specification....
Swagatam said…
Eric, you can try the following design

you can replace the BJTs with FETs, and use 100 ohm resistor with their gates
chucks ichoku said…
sir what are the appliances does the circuit above will power ?
chucks ichoku said…
sir can the circuit above power a fan without a noise???
chucks ichoku said…
I av 12v 0 12v to 230v AC, will it work on the cirvuit above? sir
Swagatam said…
you can connect a capacitor with the fan to reduce noise
Swagatam said…
It will power most of the equipment which use internal SMPS
sir can i connect the above circuit with 6-0-6 ? and how the watt be ?
Swagatam said…
it can be used if the battery is also 6V
I am using a simple transformer. When I press ctrl+E, but tell me that how can i calculate these values like primary inductance, secondary inductance etc. Can you suggest me some calculator for this purpose?. My requirements are below: primary voltage: 220V, 50Hz secondary voltage: 12V, 1A
john omaba said…
Is the citcuit square wave or sine wave? If square wave where to attach chopper to it and make it modified sine wave? Thanks
Swagatam said…
I have already discussed how to design an inverter transformer in this website, you can refer to that article. you can find it through the search box above or by Google search..
Swagatam said…
it's square wave, add chopper to the bases of the NPN/PNP
chucks ichoku said…
sir what is dis chopper all about. is it a circuit. if yes can u upload it, let me take a look at it. thanks
Swagatam said…
you can learn the details here
gogutu said…
Domnule Swagatam Majumdar,va rog sa-mi spuneti daca acest invertor poate fi folosit la o pompa electronica pentru centrala termica,atunci cand reteaua de curent a cazut.pompa consuma 40w.daca nu ce modificari trebuie aduse.Scuze ca nu vorbesc engleza.pls raspundeti.tank you
chucks ichoku said…
hello sir swagattam, I designed an inverter, it is working perfectly. but my problem is '' the heatsink ther was not fan connected to it and it is hot so much. I need an automatic fan circuit for cooling of the heatsink whenever it start heating the fan will automatically blowing, and whenever it stop heating the fan will automatically stop.... thanks
Swagatam said…
da, acesta poate fi folosit pentru orice aplicație preferată ca pe alegerea ta
Swagatam said…
hello chucks, you can try the following concept
Please Sir help me as you are doing to us. please guide me how i am going to calculate battery mAH . Assume that the battery is 12v 100mAH and connected with 500watt inverter if i connect 7 phone chager each is 6 watt and 9 watt CFL how many are hours may the battery's charging take before the it finish
Swagatam said…
I guess you meant to say 100AH battery??

you simply have to divide the battery AH with the load current, and multiply the result with 0.6, that will give you the approx battery back up time.
If i get u for example if my current is 9 watt and my battery is 100AH THE ANSWER IS 100/9=11.1*0.6=6.6 hours ??
Swagatam said…
It should be load current, not load wattage....your procedure was correct but in place of 9 watt you must apply current.

current can be found by dividing 9 watt with load voltage.
thank sir may GOD expand ur idea more again
is current the same with amp?
amps = 1W / 230V
= 0.004348A ??
Swagatam said…
yes that's correct. current is measured in amps.
My another Qestion Sir, for instant if i want to modify an low watt inverter less said 50watt please what and what do i going to change or to add. may i going to change height voltage transformer or i am going to add more transistor in pararal please which one should i going to do??
ningrat_edan said…
hello sir,
I'm very grateful you have deigned to answer my questions by posting an inverter circuit NE555 ic that works on ferrite transformer, I really hope you do not get bored if I will be asking a lot more about this circuit.
I pray that God bless your family
Swagatam said…
what modifications do you want to make??
Swagatam said…
You are welcome Ningrat!!you can feel free to post new comments
ningrat_edan said…
Thank you sir,
I am having trouble finding diodes past 10 amperes, .. is there a best solution for it
Swagatam said…
you can try 6A4 diodes, see if they work alright?
ningrat_edan said…
sir, this circuit produces high frequency ac voltage 50kHz whether it can converted into low frequency ac voltage between 50Hz - 60Hz
Swagatam said…
Ningrat, the Rt, Ct values in the second circuit determines the 50Hz frequency for the load.

you can get the formula in the datasheet of the IC
ningrat_edan said…
if we change the value of Rt and Ct to get 50Hz whether it could be applied in the ferrite transformer?
I mean a string of additional output transformer that will turn into a 50Hz 50kHz frequency
Swagatam said…
ferrite transformer is supposed to be connected with the first diagram which has 50 Khz frequency, the lower circuit needs to be set at 50 Hz....

the load or appliance can be directly connected the lower circuit mosfet.

the 50 khz from the top circuit is fed to the second circuit through the bridge rectifier
ningrat_edan said…
if the output of the circuit is converted into a dc voltage using a diode bridge and then a dc voltage is converted back into an ac voltage what it could be?
series such as what can turn into a dc voltage low frequency ac voltage in this issue?
Swagatam said…
ferrite core can be operated only with high frequency, that's why initially high frequency is used for stepping up the battery voltage to 220V and then it is converted back to low frequency for our home appliances.
ningrat_edan said…
series such as what is used for converting high frequency to low frequency
Swagatam said…
the four mosfets are toggled at the low frequency 50 Hz for executing the 50 Hz AC on the load
ningrat_edan said…
I'm very sorry sir, too many of my questions got me so confused
actually i need a simple circuit using a transformer ferritte inverter that can be used to supply domestic appliances such as water pump, fan, TV and other types of induction motors.
The circuit may be able to use ic TL494, NE555 because the component is very easy to be in my area, there are some circuit which uses components that are very difficult in the can in my area wide ic3524,3525, IR2153, and other types
ningrat_edan said…
a pressing issue, why I wanted to make this inverter because in the area I frequent power cuts in 4 hours
and if it's quite possible it will greatly help me
Swagatam said…
Ningrat, you can try the above explained IC 555 based ferrite circuit, it will certainly serve the purpose as desired by you....
Swagatam said…
or you can try the following design also
ningrat_edan said…
Hello, sir
Can you help me again, I have to stand by charger circuit using ferrite transformer would I plug the inverter circuit that I have made.
I had been using stand by charger which uses iron transformer I think if the circuit uses ferrite transformer would facilitate.
Thank you so much my regards
Swagatam said…
Ningrat, Please give the technical specifications of your charger
ningrat_edan said…
I need a circuit standby automatic charger for charging 12volt battery with a maximum current of 2 amperes, this circuit will be installed in the inverter.
Automatic question is:
1. The automatic cut off the charging current when the battery is in full condition
2. The inverter will turn on automatically when power outage
3. The inverter will automatically turn off when the power flame back.
Thank you so much my regards
Swagatam said…
Ningrat, you can try the second circuit from the following article

the charging input can be acquired from any suitable power supply source...
ningrat_edan said…
Thank you very much sir,
sir if I could replace the transistor BC547 and bc557 with other types may BD139 and BD140 or s8550 and s8050? if it can do I need to change the value of the other components?
my regards
Swagatam said…
Yes you can but it will be unnecessarily costlier....
Swagatam said…
no other components will need to be changed.
ningrat_edan said…
I tried to replace the transistor with s8550 -s8050 and a series of work thank you very much sir
my regards
ningrat_edan said…
maybe you have recommendations dc converter circuit 12volt to 300volt DC and converted into 220volt AC
Swagatam said…
that's great Ningrat, congrats to you
Swagatam said…
you can try the circuit explained in this article, you can replace the IC with the above 555 circuit stage
ningrat_edan said…
Hello, sir
I hope you are always healthy, ..
sir if i can send an email to you because I want to discuss a scheme and I find it difficult to attach the scheme in this forum
Swagatam said…
Hello Ningrat, for a long term consultation you can signup and start posting here


or for brief discussion you an use my email admin @
chucks ichoku said…
hello sir, pls I need ur assistance
i'm designing a digital display circuit, ther was a component I can't find in ma country such as 2N3053 transitor. is ther any other transistor use to replace and also gv me the same configuration it does???? thanks
Swagatam said…
I have already replied to the same comment in the other post
ningrat_edan said…
if the email you have received from my
Swagatam said…
Ningrat, I did not get any email,

please send it again or post it in the following forum


in this forum I am the admin.
chucks ichoku said…
hello sir, I do not see ur reply in post, please reply me here so that I will carry on my project thanks
Swagatam said…
you can use a 2N2222 instead
Swagatam said…
Chucks, you can use 2N2222 in place of 2N3053
ningrat_edan said…
hello sir,
Best wishes
sir, can you help me make an inverter circuit with an ac voltage pulsed out of 500watt, so if out burdened by 500watt light the lamp may blink and blink rhythm can be set using a potentiometer with a frequency of at least 50Hz maximum out 100hz.
thank my regards
Swagatam said…
hello ningrat, sorry I could not properly understand your requirement and the application need of the circuit
ningrat_edan said…
Sorry sir if it is so contrary to you due to certain factors, actually I wanted to create a series of electrofishing to output a pulsed ac voltage, ...
once again please forgive me if my request is contrary to your
my regards
Swagatam said…
Ningrat, you mean to say you want a 220V generator circuit having a variable frequency adjustment??
eric maduku said…
Hello sir, can you pls help me with 12v dc to 60v dc 100w circuit diagram please. Thanks
Swagatam said…
Hello ERic, you can try the second circuit concept from the following article
Asad Ahmed Miki said…
sir, i have made this on proteus but haven't got the desired output yet could u check it plz ?? i have to make an 100 watt inverter for my university project .need ur help sir
Swagatam said…
asad, sorry I cannot verify a software interpretation, but I can guarantee that this inverter will perform as specified when built practically without errors..
Asad Ahmed Miki said…
thanks for ur reply sir, hope u r doing well.i have some questions .which circuit should i use ? what is the transformer & battery ratings ? is that a sine wave or square wave inverter??
Swagatam said…
if you integrate the second circuit, then it will become a sinewave.
transformer and battery rating can be as per your inverter power requirement...
please sir help me as urgent,
I have build the first circuit but before testing i will like to insure the curret polarity of the zener diode, as you said the mosfit will conduct base on the negative and positive pulse of the 555 timer, so by looking at the circuit you can see that the zener diode is facing positive all of them to the mosfit, that is at a positive pulse one of the mosfit will not conduct. or the are as it is. Help
Swagatam said…
Kabir, the zener diodes in the first circuit are correctly placed.

the positives from the emitters of the BC547s will enter through the zener cathodes and reach the mosfet gates.
Swagatam said…
zener diodes do not work like rectifier diodes.
Sir, I have noticed my problem, that IRF547 In the circuit, I have replaced it with IRFP450, which prevent the circuit to work; later I just refer back to your 555 timer DC to DC boost converter, I just eliminated that 10k and 9v z from the circuit and the circuit is now working perfectly. But I tested it with 12v battery and achieved 320v DC out put. Sir my question is when I use 60v input from the solar panel, the output voltage will change or not? My ferrite core primary winding is 50 tune with 5 strand 20swg and the secondary 200 tune with single 20swg. You can see that I have altered your specification OF SWG in the circuit; this is because I determined the output load of my inverter and maximum current for the wire gauge, that 0.5mm seem like can not carry high load of current, and that 1.00mm can not carry high input current. But if I when wrong base on my assumption please correct me.
Swagatam said…
Hi Kabir, the 9V zener could be reduced to some lower value, say 4.7V etc because the zeners help to achieve a dead time which is important for the inveter's safe functioning...anyway for a 60V input, the tafo will need to be dimensioned accordingly and the winding could be altered to get any desired output from the design.
instead of using single thick wire it's better to use many thin wires in parallel for a much efficient response....instead of 20SWG, try a few strands of 30SWG in parallel.

....and make sure that the IC 555 gets 12V for operating and not 60V.
rest everything may be OK.
Swagatam said…
instead of 5 strands of 20SWG try 10 strands of 30SWG etc...
Thank you Sir,

Sir, if I change the capacitor value at the output of my Ferrite trsf,that is from 22uf/400 to 2200uf/450 What will be the effect of my output; its capacitance change or there should be a modification in the output load?
Swagatam said…
Kabir, The output response will improve if the filter capacitor value is increased, but the output voltage might increase because of this which will need to be corrected using the PWM adjustment. the output load will not require any change.
THANKS FOR YOUR REPLIES :And That is very correct Sir.

I have change the Zen diode from 7v to 3.5v, and it successfully drive my IRFP450, and for the Trf winding I have change the SWG to 30swg 10 (strand) 5o tune for primary and made 40 tune 5 (strand) for the secondary with thinking that the output voltage will reduce, but nevertheless the voltage remained as its (230v dc WHEN I APPLY 12V INPUT) And if I move to 24v the output increase to 430. and my intention is to Apply direct 60v from solar panel.that is at this stage when I apply 60v the output will be up to thousand voltage. but in your 555 timer booster, your primary tune is 50 and the secondary tune is 200. and that is exactly for my first winding and achieve 320v dc when I apply 12v, that is why I change the secondary winding to 40 tune with intention to reduce the voltage. HELP

Swagatam said…
the primary winding should be preferably equal to the supply voltage...for 60V you must use 60 turns. for reducing the output voltage you can reduce the the turns in the output winding.
Swagatam said…
buck converter is not required, reduce the output winding appropriately.....
May Allah bless you and protect you from unforeseen menace amen.

Thank you Very Much.

in the full bridge side, there is 2 pot, please which one is for setting the frequency, and how to set the other one for duty cycle? already I have frequency meatre .
Swagatam said…
you are welcome, and may God bless you too!
Swagatam said…
Kabir, both the pots are for PWM setting, the adjoining 1K resistors and the C2 cap can be tweaked for adjusting frequency.
Sir, so at which frequency for the 555 timer suppose to give to the full bridge state for acquiring 5HZ output?
Swagatam said…
It will be the same frequency, that is 50Hz which needs to be set for the IC 555, preferably at a 50% duty cycle.

this duty cycle can be changed for adjusting the RMS value of the output AC
Sir, how can I get 12v out of 60v from solar panel to drive my circuit, as you used 33k 10watt R in the full bridge side and get 15v? I have used 33k 10watt but the supply voltage is 60v, therefore I got 15v out, when I connect to the circuit the voltage dropped to 1v and the circuit did not function, I suspect the supply voltage differences that result in to this, Help.
Swagatam said…
Kabir, you can do it using a BC546 emitter follower transistor as shown in the following figure:
Sir,I have refer to the link; If I replace the zener with 15v, does the resistor 10k 1/2watt need to be change or not? or that 9v 1watt zner will give me 15v out put?
Swagatam said…
Kabir, the output voltage at emitter will be around 0.6V less than the zener value.

the 15K can be used for 15V zener also....this resistor determines the emitter current and a safe operating current for the zener.

I have change the 1k and c2 and successfully achieved 50HZ; but the problem is that the frequency is not staying at 50HZ fixed, its fluctuating to 55HZ When ever power is switch off and switch on again, when you measure again you will see it change from 50HZ to around 55Hz. therefore you have to reset it again. Help.
Swagatam said…
Kabir, where did you measure the frequency? please check it at the 555 IC pin#3....or you can replace the preset with a fixed resistor for permanently fixing the value.
Sir what is the effect if to say I leave it around 50Hz to 55HZ for the 555 timer at pin3 and the IC IRS2453D kept at 50HZ?
Swagatam said…
Kabir, which 555 IC you are referring to? the first 555 or the PWM 555 in the lower diagram?
Swagatam said…
It should be set to minimum 200Hz, 55Hz will not give correct result
Reitz said…
Dear Swagatam
I have a question about the centre tap of the primary winding. I understand how and why the transformer is working. Noting the following from your design explanation:
" The zener diodes at the gates of the mosfets introduce a slight delay time or dead time between the mosfet conduction and inhibits any possibility of both mosfets getting engaged together even for a fraction of a second."
As the mosfets cannot conduct together, would it then be possible to use this circuit on a coil without a centre tap, connecting the drains of both the mosfets to one side of the coil and the other side of the coil to +12V? I'm asking because I'm contemplating a 12VAC circuit to drive a de-magnetising/degaussing coil. I have a core from a old 12VDC auto solenoid handy which I want to use. Probably would have to up the frequency somewhat.
Thank you for a great site. I learned a lot from reading the comments and your replies
Swag said…
Thank you Reitz,

The method that you have imagined can be simply applied using a single mosfet and a single output oscillator such an IC 555, however that will not produce a push-pull effect, which becomes rather the main purpose of using full bridge networks.

by using a single mosfet, single winding topology you can eliminate the center tap mess, but the output from the secondary will be a one-sided waveform, and not an AC waveform.

I hope you got the point.
Reitz said…
Dear Swagatam,
Thank you so much for your reply. Yes, you are so right. The lift didn't go to to the top floor before, but now my H-bridge light is finally switched on.
Regards, Reitz
Swag said…
You are most welcome Reitz!
Satyavijay Durgule said…
Please tell how zener diode works to achieve dead time.
Swag said…
zener diode ensures that the transistor conducts only once the trigger voltage crosses the zener value, until then the transistor remains switched OFF
Satyavijay said…
Thanks for your very prompt reply. Your circuit are very useful and helpful. Your ideas use simple circuits but it does very fantastic job.
Swag said…
You are most welcome satyavijay, I'm glad you found it useful
Michael said…
Dear swagatam,

In the second circuit, I don't understand how the IC 555 can causes the circuit to generate a pure sine wave, that is, how it generates the correct PWM. Can you explain that?
Swag said…
Dear Michael, since no trafo is used a pure-sine would be difficult to obtain, however you can add an LC filter across the output to actually carve the waveform into a sinewaveform, you can refer to the following similar concept which will help to understand the waveform transformation at the trafo output side.
Michael said…
Dear Swagatam,

Thank you so many for your prompt answer, and for all you do in this site. Are you saying that the output of the second circuit is a square waveform? (you wrote in the description of the circuit that it generates a sine waveform, which I interpret as a PWM image of a sine waveform) ?
Swag said…
Dear Michael,
PWM is standard digital method for creating pure sinewave in all sinewave inverters. But ideally it is an SPWM which interprets into perfect sinewave at the output.

However that may be true only if a transformer is used or through the use of inductors and capacitors.

Having said this, even without inductors or filters When an SPWM is used the load will ultimately interpret the input as a sinewave, although a slightly crude one.

Therefore, the second circuit should ideally have an SPWM input instead of a uniform PWM, as shown in the diagram.
Good day sir,

Please the circuit shown above for 555 timer inverter if I convert it DC TO DC Boost, by replacing with ferrite transformer, and make it at 12v input to achieve 310v output, sir you can see at the time of running the H-bridge inverter the input voltage to dc to dc will ten to decrease to say 11v, likewise the rectified DC of 310v will ten to decrease also, so in this case how do I make a feedback voltage with 555 to increase the pwm automatically to achieve exactly 310v at the state of 11v decrease from 12v input to dc to dc boost?
Swag said…
Kabir, you may have to employ the concept which is used in the last diagram from this article

build the IC1, IC2 and the 741 stage only, and replace this over the existing IC 555 pwm in the above article.

the pwm from pin#3 then can be used for controlling the inverter output.

the 741 input can be acquired by creating a resistive divider across the inverter 220V output such that around 12V is sent to the 741 input and this must correspond to 220V across the inverter output, now if the 220V tends to rise, the PWM will be forced to get narrower and vice versa.
BUT how dose the ofset Adjust work in the given link and how to adjust it ?
Swag said…
you can correct the offset null issue by connecting a 10K preset across pin#1,5,4 of the IC, connect the outer leads of the preset with pin# 1 and pin#5, and connect the center lead with pin#4

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