Pure Sine Wave Inverter Circuit Using IC 4047

A very effective pure sine wave inverter circuit can be made using the IC 4047 and a couple IC 555 together with a few other passive components. Let's learn the details below.

The Circuit Concept


In the previous post we discussed the main specifications and datasheet of the IC 4047 where we learned how the IC could be configured into a simple inverter circuit without involving any external oscillator circuit.

In this article we carry on the design a little ahead and learn how it can be enhanced into a pure sine wave inverter circuit using a couple of additional ICs 555 along with the existing IC 4047.

The IC 4047 section remains basically the same and is configured in its normal free running multivibrator mode with its output extended with the mosfet/transformer stage for the required 12V to the AC mains conversion.

How the IC 4047 Functions


The IC 4047 generates the usual square waves to the connected mosfets creating a mains output at the secondary of the transformer which is also in the form of square wave AC.

The integration of the two 555 IC to the above stage completely transforms the output into a pure sine wave AC. The following explanation reveals the secret behind the IC555 functioning for the above.

Referring to the below shown IC 4047 pure sine wave inverer circuit (designed by me), we can see two identical IC 555 stages, wherein the left section functions as a current controlled sawtooth generator while the right hand side section as a current controlled PWM generator.

The triggering of both the 555 ICs are derived from the oscillator output readily available across pin#13 of IC 4047. This frequency would be 100Hz if the inverter is intended for 50Hz operations, and 120Hz for 60Hz applications.

Using IC 555 for the PWM Generation


The left 555 section generates a constant sawtooth wave across its capacitor which is fed to the modulating input of the IC2 555 where this sawtooth signal is compared with the high frequency signal from pin3 of IC1 555 creating the required pure sine wave equivalent PWM at pin#3 of  555 IC2.

The above PWM is directly applied to the gates of the mosfets. so that the square pulses here generated through pin10/11 of IC4047 gets chopped and "carved" as per the applied PWMs.

The resulting output to the transformer also causes a pure sine wave to be stepped up at the mains AC secondary output of the transformer.

The formula for calculating R1, C1 is given in this article which also tells us about the pinout details of the IC 4047

For the NE555 stage C may be selected near 1uF and R as 1K.




Assumed output waveform




An RMS adjustment could be added to the above design by introducing a pot voltage divider network across pin5 and the triangle source input, as shown below, the design also includes buffer transistors for improving mosfet behavior

4047 sine wave inverter with RMS adjust

The above pure sine wave inverter design was successfully tested by Mr. Arun Dev, who is one of the avid readers of this blog and an intense electronic hobbyist. The following images sent by him prove his efforts for the same.


More Feedback


Inspiring response received from Mr. Arun regarding the above IC 4047 inverter results:

After completing this circuit, the result was amazing. I got full wattage by the 100 W bulb. Couldn't believe my eyes.

The only difference i had made in this design was replacing the 180 K in the second 555 with a 220 K pot to adjust the frequencies accurately.

This time the result was fruitful in all respects... On adjusting the pot, i could get a non disturbing non flickering full wattage glow in the bulb, also the 230/15 V transformer connected as the load gave a frequency in between 50 and 60 ( say 52 Hz ).

The pot was adjusted gently to get a high frequency ( say 2 Khz ) output from pin#3 of second ic 555. The CD4047 section better calibrated to get 52 Hz at the two output terminals....

Also I am facing a simple problem. I have used IRF3205 mosfets at the output stage. I forgot to connect the safety diodes across the drain terminals of each mosfets...

So when I had tried connecting an another load ( say table fan ) in parallel to the given load ( 100 W bulb ), the glow of the bulb also the speed of the fan was reduced a little and one of the MOSFET was blown due to the absence of the diode.

The above 4047 sine wave inverter circuit was also tried successfully by Mr. Daniel Adusie (biannz), who is a regular visitor of this blog, and a hardworking electronic enthusiast. Here are the images sent by him verifying the results:

Sawtooth Waveform Oscilloscope Output



Illuminating a 100 Watt Test Bulb




The following images show the modified waveforms at the output of the transformer as captured by Mr. Daniel Adusie after connecting a 0.22uF/400V capacitor and a suitable load.

The waveforms are somewhat trapezoidal and are far better than a square wave which clearly shows the impressive effects of the PWM processing created by the IC555 stages.

The waveforms could be probably even further smoothened by adding an inductor along with the capacitor.

Showing an near Sinewave Oscilloscope Trace  after PWM Filtration











For connecting more mosfets in parallel (to increase wattage), the PWM integration could be made at pin9 of the IC 4047 for achieving the same. The connection details are shown in the following diagram:



Intersting feedback received from Mr. Johnson Isaac who is one of the dedicated readers of this blog:

Good day
In your post, Pure Sine Wave Inverter using 4047, in the second I.c stage (ic.1) you used 100 ohms resistor in between pin 7 and 6.,
Is that correct? I use to think an astable multivibrator using 555 pin configuration should have the 100 ohms between pin 7 and 6. Also, the 180k variable between pin 8(+) and pin 7. Pls check the pin connection and correct me pls. Because it oscillate sometimes and it doesn't sometimes also. Thanks,
Isaac Johnson

Solving the Circuit Issue:


In my opinion, for a better response you can try connecting an additional 1k resistor across the 100 ohm outer end and pin6/2 of IC1

Johnson:

Thank you very much for your response. I actually constructed the inverter you gave in your blog and it worked.

Though I don't have an oscilloscope to observe the output waveform BUT I bet readers its a good one cos it operated a fluorescent tube lamp in which any modified or pwm inverter can't power on.

See the picture sir. But my challenge now is when I add load, the output flickers sometimes. But am happy its a sine wave.


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Comments

Rashid Ansari said…
Dear Sir,
Thank you for a nice and simple schematic.
By the way what are the values of capacitors which are
connected to pin 1 of both 555 IC's ? You mentioned them as
C and C only.

Thank You.
Rashid
Swagatam said…
Dear Rashid,

Thank you! Both Cs can be selected as 1uF/25V
Muhammad Akhlaq said…
Sir, can more number of FETs be added for higher power ?
Swagatam said…
Yes, more FETs can be added in parallel for more power
musa kizito said…
Please. Can you add the output waveform of the circuit. I really want to built this inverter this weekend. Thank you sir.
Swagatam said…
I have updated the article with the waveforms, please check it out
yashu said…
Please help me
saheed banjoko said…
hello mr swagatam i really have to confess to you, since i've come accross this site i've been addicted to it, thanks for being a mentor and also a teacher to me, i realy appriciate the good work and effort you've put toward educating people like me thanks once again.
Swagatam said…
It's my pleasure Saheed!
alex said…
hello Mr Swagatam I have been following a lot of ur other circuits and built a few so far. How could u incorperate a output voltage correction circuit in this design because i know that with varing loads we will get a voltage drop here. Thank u for ur support
Swagatam said…
hello Alex, it can be done by adding a opto couper circuit across pin#5 of the right hand side 555 IC, quite similarly to the method discussed in one of my previous posts.
alex said…
ok thank u Mr Swagatam I am going to build this circuit until Mr Robin comes up with the report on the other design with the 4017 I really want to get one of these circuit s going Sir. Thank u for ur support to us
alex said…
hello Mr swagatam i am trying to understand ur principles every day more and more. Please tell me if I have it wrong . Lookin g at this circuit is it possible to use sg3524 in place of the 4047 and send pin 3 which is the oscilator out from the sg3524 and feed the two 555 ic just the same way as how u send the pin 13 of the 4047 to these two ic.. Would we get the sine wave effect with this ic.Thank u Sir just want to know if I am learning or I am wrong on the principle
Swagatam said…
Hello Alex

yes IC 3524 can be also applied similarly, however IC 4047 fits better here due to its simpler configuration.
alex said…
thank u Mr Swagatam I want to make sure I am unerstanding the principle because sine wave inverter circuits are so hard to find online and every body discourage us from building them but u make it look so simple in ur designs well I think the filtering part might be the hard part. Still waiting on Mr Robin to come back to us witht the finish product of ur 4017 sine inverter. Thank u Sir
Swagatam said…
Thank you Mr. Alex, let's wait for Mr.Robin's response.... or may be you can take it ahead from here and finish the project with your efforts.
Swagatam said…
That's correct Robin,

I'll correct the design as soon as possible.
alex said…
hello Mr Swagatam I seee u have changed the above circuit by adding another 555 should we build it this way now instead of the first circuit . could u explain this new addition . Thank u Sir
Swagatam said…
Yes, we need to build it in this way. The chopping pulses needed to be of much higher frequency than the triangle pulses, that's exactly what's been modified in the new design.
alex said…
ok thank u Mr Swagatam I built the top section lastnight with the 4047 and that section works only that I cant really set the output voltage to the exact amount i need because ther is no control pot for this >. Monday I will build the other section and tell u the result Thank u Sir
alex said…
Mr swagatam what is the value for c and for resistor r thank u
Swagatam said…
That's great Mr. Alex, wish you the best!

R = 1K, C = 1uF
alex said…
Thank u Mr Swagatam for ur help I am getting the parts together to build. I hear they talk about efficiency of sine wave inverters being difficult to achieve thats why we should buy instead of building what is the efficiency of this inverter and would we need a filter after the transformer . If we need a filter could u add for us please. Thank u Sir
alex said…
Mr Swagatam I built this circuit but I am having problems with it . The section with the 4047 works I get my ac voltage out ok . When ever i connect the bottom section ( the two diodes to the gates the circuit shuts down and no ac commes out..I even loose the drive signal from the 4047. I dont have a scope so it is difficult to show u any wave form. All I can say is I am getting about 91Hz from pin 6&7 junction with the collector junctuon from the left ic that signal that goes to right ic pin 5. I get about 552hz from pin 3 of the middle ic and about the same from the right ic pin 3 just dont know wher to go from here. So as long as i lift the gate drive from pin 3 of right ic the circuit comes on again. Please help me here Sir. Thank u
Swagatam said…
Mr Alex, Where did you hear this? Please show me any link where this is discussed.
Modified sine wave inverters are actually controlled square wave inverters, so they are as good as a square wave inverter.
Swagatam said…
Please check the following things in your inverter design:

Check frequencies at pin5 and pin2 of IC2 (extreme right 555)

Pin5 should show = 100Hz approx
Pin2 = high frequency...2kHz will be OK, adjust 180k resistor for this.

The voltage at pin5 and pin3 of the above IC should be approximately equal.
alex said…
Thank u Mr Swagatam I will check tomorrow. What component would adjust the 100hz frequency at pin 5 of ic2. Thank u Sir
Swagatam said…
Hello Alex, P1 of 4047 IC will adjust the 100Hz frequency at pin5 of IC2.
alex said…
Thank u Sir I will check again and let u know
alex said…
hello Mr Swagatam I had a try at the circuit again but had some problems. I got the 100 hz at pin 5 2khz at pin 2 . The 4047 section as a inverter works but as soon as i connect the diodes to the gates the output voltage just comes up to about 20v and goes right down immediately dont know if something is wrong with the circuit diagram. please help us here Sir.
Swagatam said…
Alex, check the output voltage at pin3 of IC2 555, it should be around 30 to 50% of the the battery voltage meaning around 3 to 5V.

Connecting the diode to bases of the mosfets will drop the output voltage but not to 20V that's too low.

please check and tell me about the above readings.
alex said…
ok Mr Swagatam I think I will give this circuit a break for now and start to build the other circuit that Mr Robin is persuing in that his own seem to working to a certain stage sir I will let u know the results
Swagatam said…
may be because the triangle wave peak is not equal to supply voltage, it's only 2/3rd of supply.
Swagatam said…
the 555 at the left will produce triangle waves having peak voltage = 2/3 of the supply only as per its internal specifications
Rithwik said…
Sir,
Prior doing this project i had just tested the working of a simple inverter using IC Cd4047 as given in this link

1.bp.blogspot.com/-z9RRPSTi8zs/UbR4uKyuLRI/AAAAAAAAESo/rGu-sAhOqio/s1600/IC%204047%20inverter%20circuit.jpg

But it got failed. The problem is that when i am connecting the center tap of the power transformer to the battery positive a spark appears and battery charge goes down to zero voltage and no ac output.



I am using a 300 W 15V power transformer with 5 primary taps and 4 secondary taps. The 15-0-15 voltage has been taken by
1st and 4th tap wires being +15V and -15V respectively and the 2nd and 3rd wires rapped together to avail 0v. The battery being used is a 12V 20 Ah battery

i have also tried with taps 1-3-4 and 1-2-4. But the same problem persists. So will you please help me in solving this issue.

Is there anything wrong with my transformer .... I have used the high power transformer and the battery without considering their specification, my only aim was to test whether the inverter functions properly, and then only proceeding to further high power design
Rithwik said…
Forgot to say sir. I have used IRF540N instead of IRF150 in the circuit. Also there is a slight change in the value of C1 and C2 used. I have used a 0.1 uf ceramic capacitor at the oscillating circuitry of the IC CD4047. Resistance R1 used is 22.7 K (adjusted by a potentiometer ). These values were obtained by solving the formula
f at 10th and 11th pin = 1/(8.8RC) for 50 Hz at pin 10 and 11.
Instead of 0.1 uf/600V b/w the secondary terminals, i have used a 1uF/450V Electrolytic capacitor.


So i want to know whether does these changes cause that problem....
Swagatam said…
The issue is no doubt with your circuit connections, either you have connected the mosfes wrongly or the transformer taps.

Try an ordinary small 12-0-12/220V transformer with your circuit and confirm whether your circuit connections and working are OK or not.

Once your circuit working is confirmed, next you can connect different taps of your actual trafo for the intended correct results.
Connect a 12V car headlight lamp in series with your battery positive...this will restrict any possible short circuit hazard and blowing of mosfets.
Ram kumar said…
Sir how a buffer stage can be made which solves the problem said by Mr. ALEX
Ram kumar said…
Sir please help me in identifying the connection of an optocoupler to this stage for automatic load correction. I found one in your previous article, but there optocoupler's +ve terminal is connected to the RMS voltage. I can't distinguish that from this situation.
Also can you add a selectable overload protection circuit in this schematic. Even an overload indicator is enough for me
Ram kumar said…
Sir what type of optocouplet i have to use for a 250 W inverter??''
Swagatam said…
Ram, no buffer stage is required accordion tome.....i would advise you to first successfully build the above shown inverter, once its working is confirmed we can go ahead with the opto stage.
Ram kumar said…
I am not getting 3 to 5 volts at pin#5 of IC2. Only 2.72 V
Also pin#3 voltage is 2.5V seen that is not equal to that at pin#5. What to do
Is this circuit a tested one???? ALSO TELL ME HOW TO MEASURE FREQUENCY AT RESPECTIVE PINS USING A DIGITAL MULTIMETER. I can see a square wave symbol in the meter, but don't know how to connecte the leads to measure hertz
Ram kumar said…
Sir what does it means 'D1 2V7'
Is it a zener diode????


Can i use a 4V7 zener. In my earlier doubt, i had forgot to say you that i have used a 4V7 zener in this place
Arun varma said…
Sir can i use a battery of the order of below 100 Ah such as 50 or 60 Ah with a 300 W transformer intended to produce the required wattage only ( less than the rating of the transformer ). ????.....
When i am connecting a single 20 Ah battery to the simple inverter using only IC CD4047 i am getting ac out, but when 3 or 4 batteries are connected in parallel no ac output....... when ac is present there is a humming noise produced by the transformer....... I have used the parallel connection schematics as explained by one of your earlier posts in which diodes are used
Swagatam said…
Did you study and understand the complete functioning of this inverter before building it?

The above circuit is a complex design and is only meant for experts hobbyists.

I would suggest you to first learn all the basics of inverter and then go for the practicals.

You should first try building a simple inverter.
Swagatam said…
4.7V zener will also work, don't bother about the lower 555 circuit, remove the stage and check only the 4047 inverter first.
If it works only then attach the 555 circuit with it.

you are getting 2.7V at pin5 and 2.5V at pin3 which are almost equal, so may be your circuit is working correctly.

remember your trafo voltage should be 50% of the battery voltage,means if battery 12V use a trafo rated at 6-0-6V
Swagatam said…
yes you can use lower rated batteries for operating the rafo with lower watage
check the voltage of your batteries when connected in parallel, the diodes mus be rated for carrying over 10amps, use two 6A4 diodes in parallel.....or may be you can try them without the diodes too, if all are above 12.5V
Ram kumar said…
Sir a 6-0-6 power transformer is not with me. Can i use the 10-0-10 tap of the one already have. What happened if 12 V tap is used.?????
Sir by measuring frequency in ac socket ( putting negative lead to N and positive to P aftet setting ti frequency range ) i got a value near 150 . Is it correct???? Actually should it be 50??'?
Sorry 4 the bad english
Ram kumar said…
Sir as you suggested at the last time, i have tested the single inverter circiit using only IC4047. At first time i got some ac output, but after sometimes nothing came out through transformer secondary. I saw a reduce in battery voltage when it had connected to the 0 tap. I am using a power transformer rated 20 to 25 A at secondary and batteriea are rated 20 AH 12V each* 3 no.s.
This time i didn't use any dioded for the parallel connection. But the result was the same with diodes. Please help me
Swagatam said…
12-0-12 trafo will be OK if you are not incorporating the lower PWM stage, if you want to include the pwm stage then a lower voltage trafo becomes essential.
You might have blown off the mosfets, or the IC, or your battery could be dead.
You can try using transistors instead of mosfets as they are easier to handle. you may take the help of this article to know how to connect the transistors:

https://homemade-circuits.com/2012/02/how-to-make-mini-homemade.html

connect 12v head lamps to your battery and check their intensity to confirm the battery condition, and for the inverter circuit use only one battery initially instead of using all the batteries together.
Swagatam said…
(7)555 is CMOS digital based IC with current output not more than 20mA, whereas ordinary 555 are analogue based however with output current as high as 200mA
zinnaboy2 said…
sir its this circuit a pure sine wave Pure Sine Wave Inverter Circuit Using IC 4047 and also what is the different between ne555 and (7)555 can i use ne555 for all the three ics? can irf540n be ok without problem?thank you sir i am waiting for your reply.
Swagatam said…
it's a pure sine equivalent circuit.

ne555 will provide more current (200ma)at the output and will also consume relatively more current during non-operative conditions and will need minimum 4.5V to operate.

7555 will produce not above 10ma at the output but will work with very low current and even at 3V supply.
Arun Dev said…
Sir i found the modification done to this circuit by Mr. Robin at earlier times as in the figure given in the following link.

www.google.co.in/gwt/x/i?gl=IN&wsc=hg&u=3.bp.blogspot.com/-7PTqKTB9Gvk/UmYcnX3OFiI/AAAAAAAAFiU/F0mIzZBOjNo/s320/inverter%2Boutput%2Bvoltage%2Bcorrection%2Bcircuit.png&hl=en-IN&ei=dXvRUuWYKcfliAfQv4HwCQ


I tried your basic circuit as well as this modified one. But the results were not fruitful and remained the same providing an ac output of around 20 V only.
Please read what changes i had done to this.....
Since i do not have an oscilloscope with me i was unable to make out an exact 2Khz by the first 555 in the modified version. But i have a simple digital frequency meter with me. So i could confirm the frequency of the last 555 ( which gives clock pulses to 4017 ) out to be exactly equals 200 Hz. I thought this 555 stage is a 50% duty cycle pulse generator ( since it employs diodes ) and the resistances 1 K and 3.4 K were replaced by 470 ohm and 10 K respectively and tweaked the 10 K pot to make the output wave form exactly equals to 200 Hz. After confirming this step I HAD USED THE SAME 555 STAGE to generate 2 Khz signal, but this time 0.05 uf capacitor was replaced by 0.5 uf one ( because of the formula for frequency of 50% duty cycle wave as f=1/(0.69RC) )
Now it is expected to produce a 2 Khz signal by the first two 555 ics. I have also made a change that 4047 was used instead of 4017 and the pulses from 3rd 555 stage were fed to pin#13 ( CLK INPUT ) of it. No more changes to say

Now i have some doubts...

1). The voltage at gate of each mosfet w.r.to ground had reduced to 1.3 V compared to the simple configuration in which no pwm section were used ( 5.6 V ). How to solve this issue sir ? thereby i am getting very low ac output
2). Will the frequency of pulses fed to the gates be 50 Hz even after feding the pwm stage ? ( i am getting higher values compared to the simple configuration not using pwm in which 50 Hz was shown )
3). How to use the 10 K pot in the 2nd 555 stage in the modified version ?
4). I have found a more better circuit provided by Mr. Theofanakis at the last page of that article given in this link

www.google.co.in/gwt/x/i?gl=IN&wsc=hg&u=2.bp.blogspot.com/-j_f23T5_4l4/UnxgdQS3eGI/AAAAAAAAFpY/0s7ogZHW1HA/s1600/modified%2Bsine%2Bwave%2Ba.JPG&hl=en-IN&ei=RYXRUvfoFubriAfH_4CQBw

Will it be more effective and can be used for even sophisticated applications ?



I am expecting your reply as soon as possible sir
Swagatam said…
Hi Arun,

The referred design has been tested and verified by Mr. Robin and by Michael so there's no doubt regarding its working.

The reason behind your circuit not working would be difficult to diagnose without seeing it practically,
I think you should proceed step wise, first check the circuit without the PWM diodes and then again by connecting the diodes and adusting the PWM mark space ratio.
If the 4017 version is also not working then surely there could be something hugely wrong with your circuit configuration.

You may refer to the following design to et a clearer idea:

4.bp.blogspot.com/-Y2aIrZugaFY/UselIcNJh1I/AAAAAAAAGG0/sxZw8y5fDCs/s1600/pure+sine+wave+inverter+with+auto+correction+circuit.png
Arun Dev said…
Thank you very much sir for replying.
I will inform you any changes ,will occur doing further verification of the entire circuit.
What is the part no. of the optocoupler being used with an LED for output voltage regulation in the picture given in the link you provided ? Can i make the combination manually. How it can be done ?
Swagatam said…
You can refer to the second image in this article, it shows how to build an LED/LDR opto-coupler at home:

https://homemade-circuits.com/2011/12/how-to-build-simple-electronic.html
biannz said…
sir in this circuit i want to ask you that in the 4047ic can i use (10k for R1) (10k for p1) (0.01uf for c1)?
sir please answer this question for me in fact in my country we don't have this kind of ic (7)555 i never seeing this ic before only we know or we have here is ne555 so can it be use in place of the two ic (7)555 ca i can see the left side of the hand is ne555
Swagatam said…
biannz, try googling "4047 inverter circuit"...you will come across many designs online where you will be able to get the exact values of P1 and R1

NE555 will also work
biannz said…
sir please it D1 a zener diode 2.7v?,sir please i have check 4047 on Google and a lot of them but sir there something i don't understand please explain it for me the 4047 it produce square wave so how come this inverter be produce pure sine?
the 555 stage need to be power? thank you sir.
Swagatam said…
D1 is a 2.7v zener diode.
the 555 stage is responsible for generating sine wave outputs from 4047.
555 is powered from the battery itself.
biannz said…
hi sir it been a while,sir please i build this circuit but unfortunately it don't seem to work sir any help? thank you
Swagatam said…
hi biannz, pls show me the picture of the built prototype including the connections that you have done, I'll try to help.
biannz said…
ok sir i have sent you the pic sir please try to help me build this circuit even if i survive building this circuit i will not go for any other inverter circuit i hope this circuit is good enough for me if you help me, sir please im appealing to you if it could be possible for you to send me the prototype if you have a time to design it thank you hope to hear from you.homemadecircuits@gmail.com
Swagatam said…
as expected you have only simulated the design and have not tested it by building it practically.

pls don't discuss simulated results with me.

procure the components, build it practically and only then confirm the results with me.
biannz said…
ok sir please here are the connection circuit
Swagatam said…
remove the 1N1448 connection temporarily from the mosfet gates and check whether the 4047 inverter is working normally or not.
It should, if it doesn't means your IC or connections are bad.

If it shows normal response, check the following things in the 555 section:

frequencies at pin2 and pin5 of IC2 and also the respective voltages at these points, tell me the results, i'll instruct you with further procedures.

biannz said…
ok sir thank you for your kindness im quite appreciate,sir i did as you said when i got 5v at pin5 of ic 2,when i remove the tow 1n4148 diode the circuit work so i got 152ac out put but there is some noise from the circuit i have sent you the video so please check it for sir why the noise? sir what is the per pose of the two 1n4148 diode and can this circuit be use without two 1n4148 diode?.
Swagatam said…
biannz, use 6-0-6 trafo if the battery is 12V, this will give 220V approximately even with the diodes connected.

you get pwm sinewaves only when the diodes are connected otherwise it's just square wave.

the diodes shape the mosfet gate triggers into digital sinewaves.

I am not so sure about the noise, it will be there somehow, try clamping the trafo hard.
biannz said…
sir please if i connect two 1n4148 diode the circuit don't work im confused i need your effort
Swagatam said…
it will not work if your 555 circuit is faulty, check the waveform at pin3 of IC2, also check the voltage at this pin.

use 6-0-6 trafo for 12V battery
biannz said…
sir please i checked all over the circuit everything is connected as it showed i didn't make any mistake sir the problem is the two diode if removed it work but if i connect it doesn't,sir pls is there any way to change the direction of the two 1n4148 diode? i used 6-0-6v transformer but not seem to work sir.
Swagatam said…
check the voltage at pin3 of IC2, it should be around 50% less than the battery voltage,

the lower 555 stage is creating the pwm waves which is fed via the diodes to chop the mosfet gate voltage causing them to conduct according to the pwms, thus forcing them to generate the sine waves over the trfo.

your 555 stage may be faulty or with tecnical issues.
biannz said…
sir i got 120.5v dc at pin3,p1 i use 100k,R1 39k and C1 10nf
Swagatam said…
120V from a 12V battery?? how's that possible?

Please note that the 1N4148 cathodes are only connected to pin3 of IC2 and nowhere else.

biannz said…
sir please in this circuit
300 Watts PWM Controlled, Pure Sine Wave Inverter Circuit with Output Voltage Correction you design it two times which one is correct and can be use? you detected ic1 ne555 and bd135 so which one it should be contracted?


sir please answer this two question thank you.

sir i have sent you a video and the circuit,i build it i got 4.8 frequency going it don't stable and i got 223AC at out put so sir can you please help me cos i don't think this can handle a load i put 220v AC fan but it couldn't
Swagatam said…
biannz, I did not uderstand what you are asking, where is BD135 in my design, please provide the link

the load handling capacity will depend on battery size (AH) and trafo size (Amps) if you dimension these two correctly,you will be able to make it handle the desired wattage loads.
Swagatam said…
...ok I i understood now....the following diagram is correct and more appropriate:

4.bp.blogspot.com/-Y2aIrZugaFY/UselIcNJh1I/AAAAAAAAGG0/sxZw8y5fDCs/s1600/pure+sine+wave+inverter+with+auto+correction+circuit.png
biannz said…
sir thank you.sir please tak a very look at this circuit for me the second circuit you said it is implemented by the inclusion of the LED/LDR opto-coupler stage. i build it it work so sir please should i set the frequency at the out put of tranf or at the pins of the ICS as you shown 180k preset? 300 Watts PWM Controlled, Pure Sine Wave Inverter Circuit with Output Voltage Correction


sir pls can i you IRF1404 in place of IRF540N?
sir it this circuit a pure sine wave?
can this circuit be build up to 5kv 5000W?
can 4n35 be replace the led and LDR?
it s possible to use 24v DC? thank you sir i will be happy if you answer all this questions one by one for me to understand it well
Swagatam said…
biannz,

LDR/LeD module is NOT for adjusting frequency, it's for adjusting the output voltage.
for frequency setting, do it by adjusting the IC1 preset.

check the datasheet of the mosfet, if it's wattage is beyond the desired load wattage, the you may use it.

first make it to handle 200 watts correctly then we can proceed for higher loads

only LDR will work, nothing else can be used.

24v can be used, but use 7812 for supplying the ICs, else the ICs will fry at 24V
biannz said…
ok sir thank you but sir there is ic two which of them should be adjusting frequency? can i do without the led and ldr?
Swagatam said…
The one which is associated with pin14 of 4017 is for 50Hz frequency.

the LED/LDR is only for automatic load correction.
biannz said…
ok sir but i want know that it this circuit a pure sine wave? thank you
Swagatam said…
it's not pure sinewave but has all the properties of a pure sinewave.
Arun Dev said…
Sir can a LED/LDR optocoupler as in the figure given below

4.bp.blogspot.com/-Y2aIrZugaFY/UselIcNJh1I/AAAAAAAAGG0/sxZw8y5fDCs/s1600/pure+sine+wave+inverter+with+auto+correction+circuit.png

be used as a output voltage regulation circuit for the current 4047 design ?
If it is possible, where should it be connected ????
Swagatam said…
Arun,
this design will be difficult for you,because first of all you'll have to make the LED/LDR optocoupler perfectly, which could be a lot difficult for you.

instead you can try the following circuit for the same application:

https://homemade-circuits.com/2014/01/automatic-output-voltage-regulator.html
Arun Dev said…
Sir,
Eventhough i am working in musical industry ( Music production ), i am a big enthusiastic of making such small small electronic projects.... I am used to go through almost every projects you posted in this blog.... So i have noticed the making of that optocoupler also.. That automatic output voltage regulator, i have also noticed... I am very much confident in building even complicated designs myself using the basic knowledges in electronics, i had achieved in my whole life.... But to complete and make the undertaken project working, valuable helps and suggestions from experts like you are necessary...

So could you plz tell me sir, to which point in this circuit, i should have to connect the optocoupler for solving the output voltage regulated issue ??
Swagatam said…
Arun, the connections will be exactly identical to the one shown in the original schematic, pin5 of IC2 is where it needs to be integrated along with the associated resistors
Swagatam said…
.....pin5 of the triangle wave generator IC could be also tried for the same.
Arun Dev said…
Sir, if i am doing so will there be any confusion arised to IC2 stage in comparing the signal arrived at its pin#5 with the high frequency signal ( say 2 Khz ) arriving at its pin #2.
Because, there is already a sawtooth / triangular wave reaching pin#5 from its generator.. So if i am incorporating the additional optocoupler stage, there is also an another signal to pin#5 from this... So how would that comparison be look like, means how the IC2 stage is gonna efficiently handle this situation without changing the sine wave nature of the inverter also incorporating output voltage correction feature.

I am also having another doubt,
Can any high frequency be used to compare in the IC2 stage ? Any problem with 2 Khz wave ?
Swagatam said…
yes, let's not disturb the sawtooth at pin5 of IC2, it'll produce wrong results.

I think the following idea is the best one:

https://homemade-circuits.com/2014/01/automatic-output-voltage-regulator.html
Swagatam said…
frequency at pin2 of IC2 is not critical, it can be of any value but not too high, 2khz is OK.
Arun Dev said…
Sir you mean, connecting the diodes( 1N4148 ) of the output voltage regulator to the gates of the MOSFETS ?
Two diodes are already connected to those terminals for sine wave operation, reducing the voltage at those terminals from 5.9 volts to around 3.0 V.... So if we are gonna connect the diodes in the regulator section again in the same place, will that voltage be reduced again to a small value so that the MOSFETs will be unable to conduct ???

As i have noticed from the design that the gate voltage when the PWM stage connected was only 3.0 V, Is that possible to think for a single transistor amplifier or anything like that to boost this voltage for maximum power output ?
Swagatam said…
Arun, yes that's right, connect the diodes from the regulator circuit too at the gates of the mosfets.
these diodes will only conduct when the output is above normal, otherwise they would be inactive.
PWM is introduced to carve the voltage as per the duty cycles of the pwms, so the voltage will reduce....but 3V is the average voltage that you are seeing, the peak voltage of the gate pulses will be always equal to IC supply voltage...measure the peak by using a suitable digital meter set at "peak" or "MAX" voltage at the gates.

it's your wish, if you want to use BJT stages you can do so.
Arun Dev said…
Sir,,
Could a high value capacitor ( say 1 uf / 400 V ) be able to suppress harmonics in ac instead of 0.1 uf one. In our area 0.1 uf / 400 V capacitors are rare.... I have more than 10 no.s of 1.5 uf/ 400 V capacitors with me.

So will the below given schematic be more useful ??
4.bp.blogspot.com/-o0KcHSbQKlE/UsQmcGY3NzI/AAAAAAAAGE0/LhlxgAGj7vM/s320/inverter+sinewave+filer+circuit.png
Swagatam said…
Arun 1uF would drop the output voltage a lot and drain the battery fast, so such high value is not recommended.

however a design by Motorola shows that it's possible to use such high value capacitors in conjunction with an inductor, you may see the following image and try it accordingly if you wish:

4.bp.blogspot.com/-khKat0UA8Jw/UITpCnNd7yI/AAAAAAAAA_A/8x-KAD2aHLE/s1600/simplest60+Hz+Inverter+circuit+diagram.jpg
Arun Dev said…
Sir Now i got exactly what those digital meter readings say exactly at gate terminals... I could found the actual peak to peak voltage by the method you suggested last time ( taking voltage across a 10 uf capacitor......... )...... THANKS FOR GUIDING ME THROUGH THE RIGHT PATH....

As i said before the voltage at the gate terminal reduces from average 6.0 V ( 12 V pp ) to average 2.8 V ( approx. 5.8 Vpp), reducing the power considerably..... But when i am connecting transistor amplifier stage before each gate terminal, voltage retains the peak level ( 5.8 V average ) and the power increases as in the case without PWM..... On checking the frequency at the gate terminals ( collector terminal of transistor ) i got only 52 Hz, but it was found to be 520 HZ something at the transistor base terminal, so actually functions as the stage without PWM...... So i have some doubts..

1). What z the reason for this mismatch ?

2). How to make use of these transistor amplifier stage efficiently to avail full voltage at each
gate terminal with PWM connected, producing maximum power output ?

2). What is the reason, in the original situation ( without the use of any transistor amplifier
stages ) behind the considerable reduction in voltage in gate terminals ?

I could understand, by reading your last comment, that the output voltage will reduce considerably if a 1 uf capacitor is used at that terminal and the battery will drain quickly. And also found the circuit you have suggested that could compensate for higher value capacitors being used in co-operation with inductors... So

3). Can you suggest me a method by which any valued inductor can be made by hand ?..... I
have copper wires with me
Swagatam said…
Arun, I think still you haven't understood the utility of the PWM stage

PWM is introduced to chop and reduce the final RMS so that it matches with the real sine AC RMS approximately.

A lower voltage transformer is intentionally used to counter this drop so that it can be raised to the desired levels by the trafo.

You cannot get 5.8V, the peak voltages from the IC will be always equal to the battery voltage under any case.

As far as the inductor is concerned I have no idea about the calculations, it could be quite complex.

By the way in your transistor amp where did you insert the PWM diodes?
Swagatam said…
you will need an oscilloscope for measuring the actual frequency, your meter will give wrong results due to the high PWM frequency insertion over the basic 50 Hz.
Arun Dev said…
I have introduced the transistor amplifier before the gate terminal, and the PWM diodes are connected to the base terminal of the transistor ( using a 10 K resistor ) along with the resistance from output terminal of IC4047..... ( i have send the image to your mail.... Plz check it out )..... Then output of the transistor ( collector ) is then directly connected to gate terminal....
Sir, my frequency meter could show high frequency waves at gate terminals ( say 520 Hz under PWM ) keeping the transistor amplifier disconnected... It could also show 52 Hz when PWM stage disconnected.....

The noticed thing is : when the frequency is checked at the collector terminal of the transistor without pwm, it was the same 52 Hz.... But when it is checked again under PWM ( 520 Hz seen at the base ), it shows the same 52 Hz reading..........
Swagatam said…
if you have connected the diodes directly with the BJT bases then it would work normally as before.

The frequency will be 50 Hz at the trafo output but the meter will not be able to pick it up due to harmonic content.

I have update the RMS control circuit above, you can check it out.
Arun Dev said…
Let me try to do the updation Sir.. I would let you know the results soon....

But what z about the reduction in voltage at gate terminals when PWM stage came to act ( without using any transistor amplifier stages ) ? Any solution for it, to maintain full voltage at the gate terminal, at any conditions ??
biannz said…
hi sir it been a while sir you have renew this circuit sir please what is the purpose of the 10kpots? sir please can you tell me the value of c and r with bc557 thank you.
Arun Dev said…
I forgot to ask you an another doubt :

In your last comment you said this,
" if you have connected the diodes directly with the BJT bases then it would work normally as before. "
So you mean, if i am gonna use Transistor amplifier section, it will only function like that stage without PWM feature ( Only Square wave AC output ) ???
Swagatam said…
the peak voltage from the IC outputs and at the mosfet gates will be always equal to the supply voltage that's applied at the Vdd of the IC.

THe transistor stage that you have used serves no purpose.

PWMs are made up of high pulses and zero pulses, during high pulses the diodes are completely blocking them, meaning during these periods the gates must be getting the full IC output voltage, how can the peak voltage drop then??

Only during the zero pulses from the PWMs, the gates are forced to become zero (grounded) through the diodes.

The oscilloscope images in the previously referred article by Robin peters clearly show this effect
Swagatam said…
if the pwms are removed the circuit becomes an ordinary square wave inverter.
Arun Dev said…
Then what may be the reason for the reduction in the voltage, which i had already shown to you Sir ??
Swagatam said…
Hi bianzz, the 10k preset or pot may be used for trimming the output voltage to the correct level. For example if suppose you are getting 300V initially, you can adjust the 10k pot until the outputs settles down to 220V, conversely if initially the output shows a lower level such as 180V, the same may be used for pulling it up to 220V
biannz said…
ok sir thank you sir plse in the 4047 i build this circuit but the problem is i got 4.7v at pin 10 and 4.9v at pin 11,sir pin11 going to the fets gate got very hot it almost burn
Swagatam said…
procure a multimeter which has a peak voltage detector or a MAX voltage detector option, the feature will enable you to see the max voltage at any point..... and the meter will also lock the result for proper reference.
Swagatam said…
or it would be even better if you could go back to basics and learn how CMOS outputs behave and what the 555 PWMs are made up of and how they work.
Swagatam said…
the fet could be faulty or wrongly connected, there cannot be any other reason...
biannz said…
ok sir i got it right now the circuit is ok but sir what nest for me to do to make it handle heavy load?i put 220 fan it work but the speed is very low and put 60w bulb it lit up well i have sent you the pic of it to this email address homemadecircuits@gmail.com including the wave form of it check it for me if the wave form is ok it is a trangular wave form or pure sine equivalent?
Swagatam said…
thanks Biannz, I saw the images, where did you check those waveforms if it's at pin5 of IC2 then it's fine, if it's at the output of the transformer then it's NOT fine...the output will not generate triangular waves rather an approximate sine wave.....connect a 0.22uF/400V capacitor across the output of the trafo and check again.
biannz said…
ok sir the wave form is at of ic2 pin 5, i forgot to test transf out put i well test it and send it to you, it very difficult to get 0.22uf/400v cap here only we have is 0.22uf/27v but still the fan could not speed up and there is much noise in the fan when it working thank you.
Swagatam said…
OK biannz, no problem.

0.22 is a not compulsory value, you can try with any nearby capacitor, but should be rated at 400V..
Swagatam said…
.....the triangle (sawtooth) waves look perfect and match with the specifications.
Arun Dev said…
Sir are these reading OK ?

1). Frequency generated by IC1 pin#3 and fed to IC2 = about 2.2 Khz

2). Frequency generated by CD4047 ( at pin#13 ) = 106 Hz corresponding to two 52 Hz opposite phased square waves at pin#10 and #11.

3). Frequency of the sawtooth wave form from leftmost 555 IC = 106 Hz

4). Frequency output ( across pin#5 capacitor ) of IC2 = 106 Hz, which is the PWM signal


Comparing the High frequency ( say 2.2 Khz ) with the sawtooth frequency ( 106 Hz ) i am getting the low frequency ( 106 Hz ) value as the PWM frequency...
Arun Dev said…
Sir the addition of the PWM controller stage ( 10K Pot ) you have done at present, doesn't change the output voltage of the inverter..... Tweaking the pot doesn't change its value.

Can this circuit be used to correct the Output voltage to 230 V as well as for voltage regulation
4.bp.blogspot.com/-9OIstOS8aAU/UsUgRWbQoQI/AAAAAAAAGFE/ZtvuDjK8-XE/s1600/inverter+load+correction+circuit.png
Arun Dev said…
Sir i have sent you a mail...hitman2008@....
Plz check it out
Swagatam said…
Arun, yes the linked circuit will surely work, but before that please try the new position of the 10k pot which can be seen in the updated diagram above.

All your readings are perfect, I can't see anything in them.
Swagatam said…
.............All your readings are perfect, I can't see anything wrong in them.
Swagatam said…
I'll check it soon, and let you know.
Arun Dev said…
1).Sir If i am not using a PWM stage , What is the real wave form fed to each secondary windings of the transformer other than the null tap ?
Is it a quasi square wave or just a square wave ?

2). Sir if i am using the PWM stage, will the shape of the wave get affected ?


3). Does the wave fed to each winding has -ve and +ve cycles or just a half cycle ?

Could u plz check your hitman inbox ?
Swagatam said…
Arun, without PWM it will produce square wave without RMS control

2) with PWM on, the shape of the wave cab be controlled to match sine wave RMS and properties.

3) yes the winding switch alternately in opposite directions resulting in +/- cycles at the output.
Arun Dev said…
At last by many trial and errors, without using any DSO, i got the maximum power output at the output terminal of the inverter..
But the circuit in which i made success is not this one. That is the PWM controlled 300 W pure sine wave inverter which you had posted in your blog earlier...

3.bp.blogspot.com/-7PTqKTB9Gvk/UmYcnX3OFiI/AAAAAAAAFiU/F0mIzZBOjNo/s1600/inverter+output+voltage+correction+circuit.png

On inspecting this circuit at initial stages, i got gate voltages equal to about 3.0 V ( Average ) only... I have tried many methods to amplify this signal to at least 6.0V to make the MOSFETs conduct to enable maximum power output, as i have told you many times. But all of them failed, have also told you about this.
At last i have studied the pin details of IC4017 in detail and carried out different practical tests on it... Then i could understand that,
the output voltage available from each output pin of CD4017 = 12V * Duty Cycle of the wave
So if i am connecting pin#10 ( corresponding to 25 % duty cycle ) to the reset pin ( Pin #15 ), i will get only a peak to peak voltage of 6 V.. From this, it is clear that seeing 3.0 V average at the output pins of the IC ( pin#2 and #7 )

So i have changed this connection and connected pin#4 ( by trial and error ) to pin #15. Now i got 6.0 V average at the consecutive pins pin#2 and #3.. But the earlier 52 Hz frequency has been increased to a higher value say ( 150 Hz something )....so frequency adjustment was done at the 555 stage which feeds the clock signals to the IC.. the feeding frequency was adjusted to 100 Hz...then i got exact 50 Hz signals at the above stated pins.... The voltage was 6.0 V average on both pins.... Next step was to ensure whether the signals at these pins are appearing in tandem....I have confirmed this by placing two LEDs at these terminals at 1 Hz frequency....... So everything was set up....
Praying the god, connected the transformer and fired the center tap using the 24 V source...... The result was awesome... Full power output at the inverter output.... No such considerable harmonic disturbances were detected.. Only a slight squeezing sound was heard... Then i have connected the PWM stage.... As in the case of the CD4047 inverter in this page, the gate terminal voltage also got reduced initially... But when i have tweaked the pot associated with the second IC555 ( PWM control ), i could make the voltage reach more than 4.0 V average, hence the power too.

So,,, i want to hear the valuable suggestions and corrections from your side to make this strategy into practical..
On measuring the frequency at the gate terminals with PWM connected, i got nearly about 865 Hz. The Power output in the Bulb was stable. It didn't flicker or make any noises... So can i proceed with this pin strategy ?
I will send you shortly, the current design pictures with output.
I am waiting here to hear your suggestions and comments....

With Regards,
Arun
Arun Dev said…
Sir in the last comment of mine,answered by you, i had actually asked you about the waves reaching the two secondary terminals of the transformer ( the terminals connected to each MOSFET drain points ), but you have answered for the output winding.... Think you will notice it.
Swagatam said…
Thanks Arun, that's great news!

I won't be able to comment regarding the 3.0V issue unless it's verified through a scope.

I think the culprit could be the triangle waves which are not equal to the supply voltage, it means even during the high pulses at pin3, the output would be less by 25%, which would sink the gate voltage by that much, so on an average this would be dragging the gate voltage to 3.0V average.

It would be interesting to see what happens if we replace the square wave and the triangle wave positions in the above design.....square from IC1 to pin5 of IC2 and triangle wave to pin2 of IC2
Swagatam said…
the alternate switching of the winding at the primary via the two mosfets results in a reverse/forward effect at the output winding.
Arun Dev said…
Sir, new problems are arriving...
In the current prototype i am working with ( using CD4017 ), i can hear a squeezing sound by the inverter transformer when PWM is connected, this didn't effect the 100W bulb, it didn't flicker or show any disturbing effects.
But the case changed when the bulb was replaced by a table fan ( 55 W only ). The fan rotated at maximum power when PWM was not used, but didn't rotate when PWM has been connected.... Then i checked the frequency at the output terminal of the inverter by connecting a 230/12-0-12 500 mA transformer as the load and checking its output frequency... I got around 1500 Hz.... About the same reading was detected at the gate terminal of the MOSFET..... I remember that, once you had told me that, the output frequency of the inverter will be exact 50 Hz even the gate terminals are fired using high frequency PWM waves... So why it is happening like this...
- Does the result means, the gate terminals are not being fed with the real PWM waves ?
- Then why did the bulb glow at that much high frequency ?

Arun Dev said…
Another problem is;
As i have already said you that i am getting 330 V+AC output voltage..
I have tried the output voltage regulation circuit..
I got 390 V something DC on processing with the BR. Then tweaked the pot and inspected the voltage at collector terminal...
BUT that voltage w.r.to ground was -ve voltages... Tweaking the pot gave variation from -7.0 V to -16.0 V...
The special thing noted is that ; WHEN I AM FEEDBACKING THE OUTPUT VOLTAGE TO THE PWM CONTROL STAGE, THE ANALOG VOLTMETER CONNECTED AT THE OUTPUT TERMINAL IS NOT SHOWING VOLTAGE MORE THAN 20 V..... thereby unable to adjust the output voltage
Arun Dev said…
Sorry Sir... Last comment was a mistake.... Actually i didn't connect the -ve ( ANODE-ANODE ) junction of the BR to the ground, that is why the voltage appeared -ve.... it is due to my carelessness.
Although the Output voltage doesn't respond on tweaking the pot... It remains at its high level...

4.bp.blogspot.com/-5GDTaart2mw/UmYX6r5TyzI/AAAAAAAAFiI/il_1gPepQVQ/s1600/inverter+output+voltage+correction+circuit.png

SO can i use the below given circuit for my application..
4.bp.blogspot.com/-9OIstOS8aAU/UsUgRWbQoQI/AAAAAAAAGFE/ZtvuDjK8-XE/s320/inverter+load+correction+circuit.png

In this circuit how the output voltage is getting stepped down to power the 741 IC ?
The only one diode along with the capacitor acts as the half wave bridge rectifier.
But, Will that resulting voltage not more than the maximum supply voltage of the opamp ?
I can't see any voltage adjustments for the resulting voltage to suit for powering the opamp here..
And also, Will the output voltage be suitable for this operation since it contains ripples compared to the full BR ?
Arun Dev said…
I forgot to ask you one thing.
Will that 220 k, 15 K resistance combination act like a voltage divider instead of a pot ?
I can't think so, since those resistances are placed prior to the rectification stage.. They are exposed to the direct AC. Think it would be better to place after rectification.
Swagatam said…
yes the resistor values are selected for getting around 14V, you can check it on any online "voltage divider calculator" software.

It doesn't matter whether it's at AC or DC, the voltage divider will produce the calculated voltage across the selected output points.

But yes, while calculating I should have used 310V as the input (peak) since it's this voltage that would be resulting after the half wave rectification, I used 220V as the input which is not correct....for getting 14V from 310V it should 10k instead of 15K
Swagatam said…
...the ripples will not have any effect of the intended results, according to me.
biannz said…
thank you sir but my problem is the out put i don't get correct wave form at the transf out put so it could not handle ac fan well i sent you some pic and video please check it for me, i need your help im almost done so please help to succeed thank you homemadecircuits@gmail.com
Swagatam said…
biannz, connect a 0.22uF/400V capacitor at the output of the trafo and check the waveform...also check by connecting a load to it.
Arun Dev said…
Sir.. Now the problem has been cleared. I got the real output of this circuit ... Now the voltage at gate terminals are not getting reduced as before, thereby getting the same power output under PWM ( very small deviation can be found ).... Frequency at the gate terminals is showing 260 Hz something... The output frequency of the inverter ( found by using a secondary transformer ) was also found to be in the range of 50 to 60 Hz.... The bulb as well as the fan both worked perfectly.
But two small problems remains unsolved;
1). While a load is working, disturbing humming noise is heard ( especially when fan rotates )..
So could it be possible to eliminate these harmonics completely by a second order low pass filter like this...

4.bp.blogspot.com/-o0KcHSbQKlE/UsQmcGY3NzI/AAAAAAAAGE0/LhlxgAGj7vM/s1600/inverter+sinewave+filer+circuit.png

Will there be reduction in inverter power if any of the filter is added at the output ?

2).. The output voltage is still above 300V.. I have tried the simple opamp output voltage regulator..

4.bp.blogspot.com/-9OIstOS8aAU/UsUgRWbQoQI/AAAAAAAAGFE/ZtvuDjK8-XE/s320/inverter+load+correction+circuit.png

But the problem is that the transistor collector terminal is showing no voltage when the opamp output goes high.. I have checked the circuit separately ( without connecting to the gate terminals of the MOSFETs )......checked removing the zener diode to the base and connecting the 10 K resistor directly....but no response.. Pin#2 shows the correct reference voltage of about 5.0 V... I could also change the voltage at pin#3 using the 10 K pot, also got the output pin#6 go high at a point.
For knowing whether the transistor responds to the opamp output voltage or not , i have checked the voltage between collector and the positive supply of the circuit... It always shows a constant voltage ( full supply voltage... No cahnge )

Noted things :
- Ground terminal of this output voltage regulator has been connected to the circuit ground
-The Ac voltage across the 15 K resistance in the Divider networrk is about 14-16 V.
-The Dc voltag created by rectification is also having the nearer magnitudes.

The fact is that when i am connecting the 10 K POT as in the circuit, the AC voltage found across the 15 K resistor is getting reduced to a low value ( say 6.0 V ), so the DC voltage too... Maximum of 6.0 V DC i am getting


Expecting your reply soon.
biannz said…
ok sir thank you but sir i forgot to ask you something that confused me, in the 555 stage on your left hand side there's resister connect to the base of bc557 is it 2k2 or 27k cos i can see it well and sir this is very important question i need you to answer it for me in the 555 stage on your rite hand there's 1uf capacitor non-polar can i use electrolytic capacitor like 1uf/25v or 1uf/35v tantalum? cos 1uf non p is difficult to get it here sir,thank you.
Swagatam said…
Arun, If the mosfet gates are not showing a reduced average voltage with the PWM stage connected it means the PWM stage is not working or has been rendered inactive.

For the overload protection circuit the cut off switching at the collector will be so rapid that you won't be able to read the switch OFF time there.

To visualize whether the cut off and self adjusting feature is working or not, just connect an LED in series with the base of the transistor, connect the zener diode also.

Now suppose you have set this circuit to cut off at around 250V, the red LED will switch ON when this level is crossed and the LED will keep glowing as long as the output keeps producing the high voltage.

Actually the LEd and the opamp switching would be oscillating ON/OFF at a very high rate but won't be visible to the eye.
However this rapid ON/OFf switching of the opamp/transistor will make sure that the output voltage is controlled at 250V and not allowed to exceed.

Now if the output drops below the 250V mark, the opamp output will switch OFf and the LED will also shut off indicating a normal or lower voltage from the inverter, below 250V

You must set and test the above things manually first and only after confirming the opamp behavior connect it with the final set up.


Arun Dev said…
What is the problem behind reduction in the Supply voltage in the automatic regulator circuit when that 10 K pot is being used ?

I have tried using preset as well as volume control, different valued resistances.. but nothing changed.... still the 15 V supply voltage is getting reduced to about 6 V initially connecting the pot. Further reduced on tweaking the pot... So supply to the IC becomes variable. Some times reaches 3.5 V..
The supply is getting reduced to 3.0 V even on connecting the 1 K resistance to pin#2...

I will send you the video to your email... plz check it soon
Swagatam said…
biannz, the resistor is 2.2K or 2k2, the capacitor can be any polarized type such as electrolytic or tantalum.
Swagatam said…
Arun, please tell me what is the logic behind measuring the supply voltage of the opamp.

You should measure and check whether the output mains is getting controlled or not.

By the way the opamp supply voltage will get proportionately reduced if the output mains gets reduced because the voltage divider input is derived from the mains, so it means the circuit is working.
Arun Dev said…
But.. the initially i have tested the circuit with the AC from wall outlet ( which is constant 230 V )...
So what variation to have occurred there ?
I have said that, the supply voltage is getting reduced to a low value ( say 2.5 V ) by connecting the pot or the resistor at pin #2 at this constant ac voltage...
How it is possible for the OPAMP to work properly at this low voltage ??
It has been also noted that the supply voltage reduction is less ( say 4.6 V from 6.0 V ) on connecting a 10 K resistor to pin#2 instead of 1K....
but i am fed up with that 10 K pot.
Swagatam said…
Arun, I think you have connected the preset wrongly otherwise there's no possibility of this happening.

the center lead must go to the relevant pin of the opamp.

you seem to get fed up without investigating the fault
Arun Dev said…
The connection was correct.... Center of the pot to pin#3 of OPAMP other terminals respectively to Sensing voltage and ground...
It z not the problem due to any wrong connections... Then why it is happening the same by connecting the 1 K resistance between the Supply voltage and pin#2 ?
Swagatam said…
where did you connect the output of the opamp, or the transistor collector through the diodes.
I really don't know how and where you have used the circuit, so please give all the possible details.
Arun Dev said…
Sir
The circuit has been set up initially as a separate testing mode, not incorporating inverter output. The input 230 AC voltage ( under which the inverter output has to become stable ) is given directly as in your circuit... THat 230 V has been taken from the normal house wall socket..
All the circuit components were placed as is on the given circuit... I have not connected the output of the circuit ( transistor collector through the diodes ) to any point.. It is kept free.
But i am getting wrong results even before connecting the transistor stage.... When i am connecting the pot as said before or connecting the 1 K resistance the changes in voltage are very discouraging ones.... I will send the complete pictures to your mail.. Dn't 4get to check it soon...
Arun Dev said…
I have tested the circuit with replacing the 10 uf capacitor with other capacitors also.... but the result was same.... The interesting thing noted that, when i am connecting larger capacitors ( say 470 uf/25 V as is found in many ordinary rectification stages ) the produced DC voltage is getting increased slightly from a low value ( say 2.8 V ) to its maximum...
biannz said…
ok sir thank you sir please take a look at this if it good to get this voltage at the pins,sir in the ic2 i got 9.31v at pin3 and ic1 i got 6.0 v at pin13 so sir it good?
Swagatam said…
Arun, I have modified the supply terminals of the diagram a bit, you can check it out here, please do it accordingly:

https://homemade-circuits.com/2014/01/automatic-output-voltage-regulator.html

Swagatam said…
the above modification in the diagram will solve the issue.
Arun Dev said…
Sir can you suggest any method to identify a sine wave and square wave generated by a circuit separately without using any graphical means such as OSCILLOSCOPE TRACES ??
Swagatam said…
it looks OK, but now adjust the 10K RMS pot, it should proportionately vary the pin5 average voltage and also the inverter output voltage.
Swagatam said…
Arun download "goldwave" free version in your PC, and then you can use this software for monitoring the waveforms on your PC screen,

But remember the input signals will need to be first converted to 1V PP before feeding your PC MIC socket..
Arun Dev said…
Could you please suggest a circuit for enabling the connection from high power output voltage from the inverter to the laptop MIC socket ?
Since i am not so much confident about myself for making such a design to be used with my laptop i am seeking for your help
biannz said…
ok sir,please sir if i connect the diode where the black mack is to the pin 6 and 7 of ic ne555 the sine wave work but it cannot adjust the RMS to get out put 220v A/C but if connect it as it is the sine wave don't work sir all i mean is that the 10k pot cannot be adjust the RMS if the 1n4007 diode is connected,
Swagatam said…
Try an online "voltage divider software" and select R1/R2/input prarameters appropriately for getting the 1V output.

After this you can make this configuration for acquiring the 1V output
Swagatam said…
when the capacitor C charges, the slow rise in the voltage will pass through the diode and go to pin5 of the 555 via the pot adjustment, so I cannot see any reason how the diode can stop this from happening.

check your diode polarity or may be it's faulty.

Anyway if you don't want the diode you can remove it and use the 10k preset directly in the shown position.
biannz said…
ok sir thank you sir please check the wave form i sent it to your email address homemadecircuits@gmail.com sir i think it shouldn't be like this any help? so that i can get the wave form correct with this wave form it handle small 220v a/c fan the speed of the fan is ok but the moment connect another load the fan stop waking i use 200ah 12v batt with 9-0-9 transf.
Swagatam said…
Biannz, it looks very impressive, but please let me know how you got these waveforms, for additions did you do for getting such waveforms.

did you connect a 0.22/400V capacitor at the output, and is the RMS = 220V???

You cannot get better waveform than this, because it's a modified sine wave so please don't expect to get a real sinusoidal type waveform.

Also you may try increasing the output capacitor to 0.33uF/400V and check the response.

I would be publishing the waveforms in the above article so that others can also visualize it.
biannz said…
ok sir thank you,so you mean this wave form is alright?but you said that it equivalent sine wave not modified sine wave cos modified sine wave don't look like this or? ok sir help me with this if i connect 220v fan it run very good but if add another load like 60w bulb the voltage reduce so the fan stop.
Swagatam said…
yes this wave looks alright since it's the outcome or the resultant of the PWMs applied at the primary.

The transformer is not able to support higher loads because its winding may not be rated to handle that much load.....use thicker wire for the secondary...and/or increase the AH rating of the battery.

The power handling capacity is not anyway related to waveform or PWM of the circuit.
biannz said…
ok sir please you said that inductor can be use so im asking you that where it should be connect and how many turns, this wave form is (2ms and 5 volt/DIV) sir i think i have succeed in this circuit so what i need is low battery cut off so sir can you please design one for me? thank you.
Arun Dev said…
Sir,
Have u deleted the Pure sine wave inverter using Traic from your blog ?
Are you in the middle of making a correct working model ???
Swagatam said…
Arun, yes I have deleted the article temporarily, but there's no easy way of switching the output with the PWM unless a full bridge driver circuit is used at the output as shown below:

https://homemade-circuits.com/2014/01/simplest-full-bridge-inverter-circuit.html
biannz said…
sir please it good to get 70HZ out put? cos when i set it to 50hz the more i load the more it going up it not stable,please sir if i want to connect inductor where it should be connect ,sir can this circuit handle 24v dc batt and how to connect it thank you sir.
Swagatam said…
one filter circuit is shown in the following circuit you can try it out:

https://homemade-circuits.com/2013/10/modified-sine-wave-inverter-circuit.html

for the low battery cut off you can try the folowing design:

4.bp.blogspot.com/-r70Wh0qXC78/U18yct62aFI/AAAAAAAAGws/gx2FCQRc5e8/s1600/automatic+battery+charger+circuit.png

use the lower free relay contact(N/O) for connecting with the inverter positive.
Swagatam said…
biannz, 70Hz is not good.
use 7809 IC for supplying the 555/4047 IC positives from the battery.

once you do the above, you can use 24V also without worry. use the 24v battery in the same position where presently the 12V is.
daniel adusei said…
sir

i want to add more fet how do connect 1n4148 diode for example if i want to add 10 most fets every each of them should connect with 1n4148 diode?
Swagatam said…
daniel,

instead of using the mosfet gates for pwm control, it's better to use pin9 of the IC 4047 for applying the pwms, ......do the following:

first disconnect pin9 from ground and connect it to positive via a 10k resistor.

take a BC547 transistor and connect its colector to pin9, emitter to ground and base to pin3 of IC2 (555) via a 10k resistor...diodes are no longer required.

now your mosfet section is free so that you can connect as many of them in parallel.

make sure the gate of each mosfet has its own 100 ohm resistor and also put a 1k resistor across each mosfets gate and source.
daniel adusei said…
ok sir

please im confused sir cos you said should connect to positive i don't understand the word via it is (with) or what? 10k resistor please can you update it for me co what you said confuted me
Swagatam said…
via means "through", meaning the mentioned connections must not be joined directly to the respective points, rather they must be connected in series with the mentioned resistors.
if possible i'll make the diagram and update it
daniel adusei said…
ok sir
i did it but it didn't work i don't know maybe wrong connection,please make the diagram and update it for me cos i want to make a bigger to see if this inverter is good for my need thank you sir hope to hear from you son.
Swagatam said…
daniel, I 'll try to update it soon....may be tomorrow.
daniel adusei said…
ok sir

im waiting for it thank you
Swagatam said…
done! please check it out.
daniel adusei said…
thank you sir
it st going to produce same wave form?
Swagatam said…
I hope so, it will need to be confirmed practically
daniel adusei said…
ok sir thank you for your kindness
daniel adusei said…
hi sir
please i finally build this circuit but still big problem i used four fets two at one side and two at other side so total four,if i power the circuit the fets got very hots if i set the frequency to 50HZ it got hot and if set up to 60HZ the same problem i change all the 555 and cd4047 please i now i need your assist thank you.
Swagatam said…
Hi Daniel in that case you can revert to the previous concept and use the PWMs on the mosfet gates.

for adding more mosfets you can use 4049 NOT gates and terminate their outputs with mosfets.

I'll try to update the idea soon...
daniel adusei said…
ok sir
im still waiting for the update thank you
Swagatam said…
daniel you can do the following steps:

we have 6 NOT gates inside a 4049 package

connect the inputs of three NOT gates in parallel meaning join them together, connect this joint to one of the outputs of the 4047 IC VIA A 1K RESISTOR.

Now connect the individual outputs of these gates to three separate mosfets via 100 ohm resistors.

join all the drains of these mosfets to one of the outer taps of the trafo.

Repeat the above procedure for the other three NOT gates and configure them with the other output of the 4047 IC and the trafo tap.

After completing the above you can integrate the PWM input at the junctions of the NOT gates and 1K from the 4047 outputs.
Swagatam said…
don't forget to connect the 4049 supply pins to the supply from the battery otherwise the IC will not work
daniel adusei said…
sir
please as im talking to im on it but it didn't work maybe the connections are wrong connected sir i tried to send you the diagram but we are having a problem with our network so please can you try to update it for me?thank you.
daniel adusei said…
sir please i did as you said but it didn't work i think the corrections are wrong connected so can you please update it thank you looking forward to hear from you .
Swagatam said…
daniel, please draw the schematic as per my previous instructions and send it to me, I'll check it for you and confirm its accuracy or make the necessary corrections in there are faults.
daniel adusei said…
sir please im still waiting for the update please sir you promised to update we are still waiting for it thank you.
Swagatam said…
daniel, please draw the schematic as per my previous instructions and send it to me, I'll check it for you and confirm its accuracy or make the necessary corrections in there are faults.

here are the instructions:

we have 6 NOT gates inside a 4049 package

connect the inputs of three NOT gates in parallel meaning join them together, connect this joint to one of the outputs of the 4047 IC VIA A 1K RESISTOR.

Now connect the individual outputs of these gates to three separate mosfets via 100 ohm resistors.

join all the drains of these mosfets to one of the outer taps of the trafo.

Repeat the above procedure for the other three NOT gates and configure them with the other output of the 4047 IC and the trafo tap.

After completing the above you can integrate the PWM input at the junctions of the NOT gates and 1K from the 4047 outputs.
daniel adusei said…
sir please you did not upload the circuit im still waiting for it
Swagatam said…
daniel, please draw the schematic as per my previous instructions and send it to me, I'll check it for you and confirm its accuracy or make the necessary corrections in there are faults.
Odon said…
Dear Mr. Swagatam Majumdar,

Thank your for the circuit design schematic & breadboard pictures. Although difficult (for me), I enjoyed putting this together. I put together the last schematic on this page, with the BC547 included for parallel mosfets. It works nicely for me.

I have not yet added parallel mosfets.

I think my previous attempt to reply did not post because I had not logged in, so I will try this again.

For the benefit of others, I wanted to mention that I had a few problems, mostly regarding having the wrong parts, and selecting the wrong values per the schematics. It took me a while to get it completed correctly.

Let me see, I would guess that my biggest problem was with the capacitor and resistor at C1 and R1 of the 4047. I had to study the comments from the breadboard pictures in order to get that correctly. I had missed the period from other schematics, and had guessed at 22 uF (for C1). That was not correct, as it gave me a pulsed flash in about every second.

For C1, I have tried .01uF and .22uF, both of which work for me. I am currently runing from .22uF, rated at 630 volts, per the comments regarding the successful breadboard trials.

For R1, I am currently running with 22K Ohms, without P1 on the 4047. This works for me, but I suspect that I need to adjust the frequency in some systematic fashion.

I do not have a oscilliscope, and I am in the U.S., so I have looked at the equations for frequency from some of your other posts. I suspect that I will buy an oscilliscope in order to check the frequency.

I had ordered 2.7 volt Zener diodes, but when they arrived, I found that I had ordered 1/2 watt zeners, and not 1/4 watt zeners. The 1/2 watt zener diodes gave some slight flickering, so for the moment, I am using 1N4148 diodes. I will order the zener diodes again.

I am currently running from three NE555's. I had bought 7555's from Radio Shack, but removed them to trouble shoot my problems.

Finally, I have some small concern regarding the breadboard comments about the burned out mosfet, due to no protecting diodes. My front mosfet gets hot quickly, and I do not see the protective diodes surrounding the mosfets in the schematic at the bottom (with the 547 addition).

I am running with IRFP150 mosfests, as I am having difficulty finding IRF150 mosfets. I also have IRF3105 mosfets, but am not currently using the 3205's.

I think that the capacitor at C1 was my biggest problem there.

Thank you very much. Although difficult (for me), I got a thrill from seeing it work!

- Dave Hamm
Swagatam said…
Thanks Dave, for the great explanation!
Yes an oscilloscope will be quite handy for checking the overall response of the system, and would also allow you to study the output waveform.

In the meantime you could try varying the lower 10k preset and see if the output voltage also varies correspondingly, this will prove that the PWM section is working correctly as proposed in the article.
Odon said…
Thanks for the response.

Yes, I did look at the 10K preset. I had done the calculations from your web page on "IC 4047 Datasheet, Pinouts, Application Notes" and calculated that P1+R1 should equal about 55 MHz when C1 is 0.22uF

However, I have very little confidence in my understanding of your equations.

I noticed a clue about the frequency when I had C1=0.22uF and R1+P1 = 22K. My little test 1 watt night light would give a humming sound that was quite noticeable.

So, yes, I did place a 10K resistor in for P1+R1, leaving C1 at 0.22uF, and the humming sound of the night light went away as soon as the 10K (preset) was in place.

Also, while using the 10K resistor in place for R1+P1, I did a test of a light load, a 15 Watt incandescent and let it run for about 45 minutes. During that time, the front mosfet (the lower one on your schematic) did not heat up hardly at all.

Therefore, at first glance, it would appear that the 10K preset is much better than the 22K which I was initially using (while using C1 is at 0.22uF). I will leave it at the 10K until the oscilloscope arrives.

On a side note, it also appears to be very energy efficient, which came as a pleasant surprise to me. I don't know the figures for efficiency, but it appears to have less drain on the batteries than I would have normally expected.

Just wonderful to see it working.

Thank you
Swagatam said…
Thanks Dave, actually I was referring to the lower 10K preset associated wit the first 555 IC from left.

It's included for facilitating RMS adjustment or the average output voltage of the circuit.

Regards
Micky A. said…
Hello sir, i have tested this circuit on bread board. this is awesome. Can you give me its pcb layout please
Swagatam said…
Hello Micky, thanks very much, however due to time constraint making the layout won't be possible at the moment for me...
Phil Ko said…
Hello, sir:

I would like to have it output 5Vac and 10 Vac. How do I make that happen?

Thank you!
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