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# 500W 12v to 220v Inverter

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hie, i just joined the community and i was hoping that somebody,anyone pliz help by explaining the modified inverter that Ronnie.B.Tabanao showed.I could really use some help coz i can't understand it.
I

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Hi Doron,
Welcome to our forum. ;D
What don't you understand about the simple inverter?
1) It uses a CD4047 Cmos IC as its oscillator. It has a frequency divider inside to produce perfect square waves. It has two outputs and one is inverted so their voltage waveforms are opposing.
2) The opamp is used to amplify the low current from the Cmos oscillator to about 15mA.
3) The pre-driver transistors are darlington emitter followers and amplify the 15mA to about 1A.
4) The driver transistors are also darlington emitter followers and they take the 1A from the pre-driver transistors and amplify it to about 12A.
5) The paralleled output transistors are also darlington emitter followers and amplify the 12A to about 50A.
6) The 50A from each side is conducted through each half of the center-tapped winding of the transformer, alternating.
7) The transformer steps-up the voltage to 220VAC.

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audioguru thanks,what you said is what i had in mind but i wasn't sure about the calculations of the gains,for example,the LM358 OP AMP.
The 2SC1061 has a minimum gain of 35, the 2N3055 has a gain of 20 so aren't you supposed to use that in your calculations?

You know what?Maybe it would be better if you could show me how you got the output current form the chip(assuming that the input current to the chip is about 10mA) to 15mA to 1A then to 12A and finally to 50A.

My assumption is that you would be using the transistor's current gain but its not adding up.
Hey,initially how much current would be flowing when you switch on the inverter?I thought it was around 42Amps.

Once again thanks for the invaluable wisdom you dished out,you really helped me understand the circuit better.

Looking forward to hearing from you soon

Doron

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Hi Doron,
The inverter draws about 50A from a 12V battery when it has a 500W load. 41.7A for the load and the rest for heating itself. Without a load or when starting without a load, the current from the battery is probably only 2A if the transformer is good.

The gain of transistors depends on their collector current. I didn't look at the gain of the oriental 2SC1061 since if I built one I would use a local TIP31. At 1A, the min gain of a TIP31 is 25.
With a 500W load, each of the 4 output transistors for each side will divide the 41.7A so would have a collector current of about 9.7A (typical gain of 15 at 10A, and its 700mA base current is also powering the load). At 10A, the min gain of a 2N3055 is only 5!
A min 2N3055 saturates with 3V across it at 10A when it has a whopping 3.3A of base current!
Therefore the modified circuit was designed with extra gain and extra transistors so it will work with weak or strong transistors.
If all the transistors have high gain, the inverter won't draw more current. The transistors will just use less input current.

The typical input current of an LM358 opamp is only 45nA. Its max input current is only 250nA. Therefore the CD4047 Cmos oscillator IC will have nearly nothing for a load and will swing its outputs very close to 12V and ground. The LM358 current-limits its output at about 15mA, so the pre-driver transistors get plenty of base current.

You can multiply the current gains of the stages using min-spec transistors if you want.  Start calculating current at the output and work your way forward.
Just remember that the darlington connected transistors add their base current to the load. ;D

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My inverter uses MJ15015 from /\/\otorola which has a higher power compared to 2N3055 and more expensive too.. :P
I've been using this inverter for sometime and just perform flawlessly..

rhonn  ;) ;)

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Hi Rhonn,
How are ya doin'?
Its great to hear that your inverter operates flawlessly. ;D

The expensive Motorola transistors have exactly the same electrical spec's as the cheaper 2N3055 transistors. They use a better mechanical mounting of the chip to the case, or maybe the case is aluminum instead of steel, so they are able to transfer more heat to the heatsink. Therefore the heatsink can be smaller than if 2N3055 transistors are used.

To stay cool enough, 2N3055 transistors in this project should not have an insulator. Two heatsinks must be used and be insulated from each other and from a grounded case.
The expensive transistors might be cool enough with individual insulators and a single grounded heatsink.

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• 1 month later...

hey you guys how are you doing?It's been long hasn't it?Anyway i'm back and i got more  questions than answers.
First,does the ratio of the 0.1R,1/4W Resistor to the 0.1R,10W Resistor have any effect on the power output because i managed to get 4.7Ohms,10W and 4.7k,1/4W so i don't know if i can still get the desired output.I did my calculations and got about  122W!!This comes from
( V/R)=12/4.7
=2.553Amps
This value is then multiplied by 4 for the 4 transistors to get
2,553*4=10.212Amps
Now the Power available is then P=V*I
= 12*10.212
=122.553Watts

Secondly,the output on the Transformer has a frequency of 50Hz Square Wave.Does this work on ordinary household appliances because i was thinking of adding a  sine wave filter at the output of the transformer?Do you think it will work and do you have any idea if i can build one?
You guys are the best,so i hope you can really help me out on this one coz i could really use yor profound knowledge
Later

Doron

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First,does the ratio of the 0.1R,1/4W Resistor to the 0.1R,10W Resistor have any effect on the power output because i managed to get 4.7Ohms,10W and 4.7k,1/4W so i don't know if i can still get the desired output.I did my calculations and got about

Hi Staigen,

I

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Hi Doron,

This is what audioguru is talking about using two heat sinks (see pictures attached). These are the ones I have used however; I have not got it all together yet so let

Hi Staigen,

I
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audioguru,Cooperman how are you?The reason why i thought of using the 4.7Ohms was because i could not find any 0.1 Ohms,10W resistors in my town(Harare,Zimbabwe) so had not put into consideration that voltage drop across the resistors.However I thought that the current from the pre driver transistors would compensate for the losses incurred.
My question is can i not use any other value of resistance besides the 0.1 Ohms,10W resistance and the 100 Ohms,1/4 W to get the same power output of 500W using the same components i.e the 2N3055 & 2SC1061 TRANSISTORS?
I do know that when fully switched on(the 0.1 Ohm,10W) that is when i can get the 500w assuming that i do have a 500W load connected to the output of th transformer.
Do you have any idea where i can get the 12v-0-12v to 220v transformer from any old household appliance coz i can't seem to get it from anywhere?
Thanks in advance you guys and Cooperman would it not be easier if you mounted the 2N3055  transistors the other way coz then it would be easier for you to then connect the resistors to the emitter of each transistor?

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My question is can i not use any other value of resistance besides the 0.1 Ohms,10W resistance and the 100 Ohms,1/4 W to get the same power output of 500W using the same components i.e the 2N3055 & 2SC1061 TRANSISTORS?

Some people think that the 0.1 ohm resistors aren't required. I think that the strongest transistor will hog the load without them and burn out. Do you want to try the circuit without them?
The 100 ohm resistors' value is not critical. 33 ohms to 1k will work.

Do you have any idea where i can get the 12v-0-12v to 220v transformer from any old household appliance coz i can't seem to get it from anywhere?

I don't know.
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• 2 weeks later...

Audioguru,i think using resistors is a wise idea coz i'm positive the transistors will blow but i have decided to use the 0.2 Ohms,3W in place of the 0.1 Ohm,10W so does that have any effect on the circuit?Hope not.

I got the transformer so lets wait and see what happens when i connect the circuit.Can you fit 4 power transistors on the heatsink used by Cooperman coz i have 2 Heat sinks that are similar to the one used by him.I thought we needed  4 power Transistors for the circuit to work but i'm only seeing 3 on Cooperman's circuit,does that do the same work?

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Audioguru,i think using resistors is a wise idea coz i'm positive the transistors will blow but i have decided to use the 0.2 Ohms,3W in place of the 0.1 Ohm,10W so does that have any effect on the circuit?Hope not.

The 0.1 ohm/10W resistors are needed if the load is 500W. The circuit will draw about 600W from the 12V battery which is 50A. Each output transistor and its emitter resistor will share the 50A so each will have 50/4= 12.5A for half the total time, so your 0.2 ohm/3W emitter resistors will each dissipate 12.5 squared x 0.2= 31.25W for half the time and will burn out!

I got the transformer so lets wait and see what happens when i connect the circuit.Can you fit 4 power transistors on the heatsink used by Cooperman coz i have 2 Heat sinks that are similar to the one used by him.I thought we needed
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just how much power can you handle if you use 0.24 Ohms.3W on the emitter resistance and how do you find that out(mathematically i mean)?
How is one heatsink different from the other?I thought they were all the same.

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just how much power can you handle if you use 0.24 Ohms.3W on the emitter resistance and how do you find that out(mathematically i mean)?

The 500W inverter will draw 600W from a 12V battery. The transistors draw a total current of 50A so each of your emitter resistors will have 50/4= 12.5A through it and will dissipate (12.5 squared) x 0.24 ohms= 37.5W each for half the time. So the inverter's output will be reduced by 37.5W x 4= 150W.

How is one heatsink different from the other?I thought they were all the same.

The heatsinks are the same and must dissipate about 50W to 100W each.
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audioguru if  i use my circuit on a device that requires about 400w of power using the 0.24,3W emitter resistors,they should work shouldn't they?I'm asking this because the last time i spoke to you,you said they would blow

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Hi Doron,
I don't know if your resistors are rated at 0.3W or at 3W, but they certainly will blow.
Their value is 2.4 times too high, so they will get 2.4 times hotter than the required 0.1 ohm/10W resistor, but since your load is "only" 400W then each of your little resistors will dissipate 24W half the time which is 12W.

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just how much  load would the circuit be able to withstand if i use my 0.24 Ohms which dissipates about 3Watts on the emitter side of the resistors because they are the only ones that i can find close enough to the 0.1 Ohms,10W resistors?
I am asking this because i don't necessarily have to produce 500W as yet,i just want to test the circuit with the resources that i have.

just how much
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• 2 weeks later...

What type of loads/appliances/devices can be used with this inverter because i believe we can't use it on critical loads like  a computer or some other data capturing devices?

What is the purpose of the 0.1Ohm,10Watt resistor?Why can't we use a wire for short circuiting if that is what we want to achieve when the current is being drawn by  the load as 0.1 Ohm is really no resistance at all?

What is the purpose of the 4 2N3055NPN Transistors bacause I thought they were for current sharing because the 50Amps is too much for one transistor,eg,
50/4=12.5Amps.The transistors max rating is 15Amps so i thought they could handle this much current?

What type of power MOSFETS can i use in place of the transistors?

How do you connect the collector of the 2N3055NPN transistors when you have mounted them on the base of the heat sink?

How do you place the resistors  onto the heatsink?

I've found some 0.47 Ohms,5Watt resistors and I wanted to parallel them in 3s so i get an effective resistance of
(0.47*0.47)/(2*0.47)=0.235

Do you think this will work in place of the 0.1 Ohm,10Watt resistors?

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What type of loads/appliances/devices can be used with this inverter because i believe we can't use it on critical loads like
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On the schematic given for the 500W inverter,ther are values given as 100R,1/4W is this suppossed to be 100 Ohms,1/4W or I can use any value for R as long as it has 1/4w dissipated?

The darlington paired transistors;are they to be used for amplifyng the signal out of the CD4047 chip or  for some other purpose?

If you can't mount the emitter resistors on the heatsink,where are you going to place them?I saw the pictures by COOPERMAN and he had mounted them on the side of the heatsink(see his pictures attached)

You said that the 3 paralled 0.47 Ohms,5W will have the project's voltage output reduced,by how much will the output voltage be reduced and what are the effects on the load connected? ;)

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On the schematic given for the 500W inverter,ther are values given as 100R,1/4W is this suppossed to be 100 Ohms,1/4W or I can use any value for R as long as it has 1/4w dissipated?

100 ohms, 1/4W resistors are very common. Use them.
You should never use a resistor at its rated power dissipation, it will be too hot. Its rated power is its absolute max allowed.

The darlington paired transistors;are they to be used for amplifyng the signal out of the CD4047 chip or

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