audioguru2
- Apr 6, 2004
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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.
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.
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
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
audioguru2
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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
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
K
Kasamiko
- Jan 1, 1970
- 0
My inverter uses MJ15015 from /\/\otorola which has a higher power compared to 2N3055 and more expensive too.. 
I've been using this inverter for sometime and just perform flawlessly..
rhonn
I've been using this inverter for sometime and just perform flawlessly..
rhonn
audioguru2
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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.
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.
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
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
audioguru2
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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
View attachment 38673
View attachment 38674
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
View attachment 38673
View attachment 38674
audioguru2
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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?
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?
audioguru2
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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?doron said:
The 100 ohm resistors' value is not critical. 33 ohms to 1k will work.
I don't know.
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?
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?
audioguru2
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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!doron said:
audioguru2
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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.doron said:
The heatsinks are the same and must dissipate about 50W to 100W each.
audioguru2
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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.
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.
