Hi, I made the inverter circuit as shown. But I used irfz44n Mosfet and a 7.6 - 0 7.6v transformer. First to make sure the circuit was working I tried with a 9v battery and then got the output of 121v. Then I connected to a used hybrid battery pack of around 15.8v. Suddenly a spark came from the circuit I think now the MOSFETs are blown. Can u guys figure out what's wrong? I am waiting for a quick reply!! Thank You!!

 

With a wiring diagram instead of a circuit schematic diagram it is quite a challenge to see what is happening in the inverter. There are a lot of places where a spark could have happened with a circuit board trace fusing. so examine the circuit board and let us know where in the circuit a conductor failed. And certainly the failure was a conductor burning open.

 

There is nothing like that in the circuit

 

You should always have a current limiting power supply or low current fuse when testing circuits like that. Using a 15.8 battery with no circuit protection can be a recipe for trouble as you found out.

https://electronics-project-hub.com/modified-sine-wave-inverter-circuit/

 

"A Spark" will either be the result of a current carrying conductor suddenly failing open, or else a breakdown in insulation, possibly an air gap, as the voltage exceeds the withstanding ability of the gap. So the spark might have been between the source and drain connection on one of the transistors, because that is the only point in the circuit where there could be a high voltage, and there is nothing in what I can see of the circuit that is intended to reduce the inductive spikes from very fast switching.
And this is not a "modified sine wave" inverter circuit, it is a stepped square wave circuit. A bit of an improvement, certainly, but still in need of some spike reduction effort. That oscilloscope used to display the waveform is probably not fast enough to see narrow spikes.
Most well designed inverter circuits include some sort of provision to reduce those spikes. And many articles recommend that a load be connected to the output before power is applied to the circuit.

And once again, a schematic circuit drawing will be far more useful than a wiring diagram when evaluating the function of any circuit with more than 3 components.
I did visit that linked website and unfortunately they misrepresent their wiring diagram as a circuit diagram. Many other threads in this forum show circuit schematic diagrams, and you can see the difference immediately.

 

"A Spark" will either be the result of a current carrying conductor suddenly failing open, or else a breakdown in insulation, possibly an air gap, as the voltage exceeds the withstanding ability of the gap. So the spark might have been between the source and drain connection on one of the transistors, because that is the only point in the circuit where there could be a high voltage, and there is nothing in what I can see of the circuit that is intended to reduce the inductive spikes from very fast switching.
And this is not a "modified sine wave" inverter circuit, it is a stepped square wave circuit. A bit of an improvement, certainly, but still in need of some spike reduction effort. That oscilloscope used to display the waveform is probably not fast enough to see narrow spikes.
Most well designed inverter circuits include some sort of provision to reduce those spikes. And many articles recommend that a load be connected to the output before power is applied to the circuit.

And once again, a schematic circuit drawing will be far more useful than a wiring diagram when evaluating the function of any circuit with more than 3 components.
I did visit that linked website and unfortunately they misrepresent their wiring diagram as a circuit diagram. Many other threads in this forum show circuit schematic diagrams, and you can see the difference immediately.

Thank You So much for the detailed reply. I will try provide u a schematic diagram

 

Looks to me like a 555 oscillator driving a 4017 as a divide-by-2 counter.
There's quite a lot wrong with that circuit. . .
1. Transformer voltage is too low. To make a squarewave inverter you need a higher voltage transformer than 12V. A 12V transformer expects a voltage no higher than 12.√2.2/π, because the average DC voltage and the rms voltage of a sinewave half-cycle are not the same. You can use a 12V 50Hz transformer if you increase the frequency to 60Hz.
2. There is no dead-time, so there is a very good chance that one MOSFET will have turned on before the other has turned off. That will blow it up.
3. A 4017 really doesn't have enough output to drive a big MOSFET, even at 50Hz. Use a low-side MOSFET driver.
Better still, throw the whole lot away and use an SG3525, which will do the oscillator, dead-time and MOSFET drive all in one package.

 

Looks to me like a 555 oscillator driving a 4017 as a divide-by-2 counter.
There's quite a lot wrong with that circuit. . .
1. Transformer voltage is too low. To make a squarewave inverter you need a higher voltage transformer than 12V. A 12V transformer expects a voltage no higher than 12.√2.2/π, because the average DC voltage and the rms voltage of a sinewave half-cycle are not the same. You can use a 12V 50Hz transformer if you increase the frequency to 60Hz.
2. There is no dead-time, so there is a very good chance that one MOSFET will have turned on before the other has turned off. That will blow it up.
3. A 4017 really doesn't have enough output to drive a big MOSFET, even at 50Hz. Use a low-side MOSFET driver.
Better still, throw the whole lot away and use an SG3525, which will do the oscillator, dead-time and MOSFET drive all in one package.

If you don't mind can u provide me a schematic diagram for the one you suggest that will work for 7.6v - 0 - 7.6v transformer and 15.8v battery pack. I don't know much deep in electronics so this will be a very good help

 

If you don't mind can u provide me a schematic diagram for the one you suggest that will work for 7.6v - 0 - 7.6v transformer and 15.8v battery pack. I don't know much deep in electronics so this will be a very good help

There isn't one.

 

There isn't one.

So is there a way that I can modify the existing one?

 

If you look at the claimed waveform

Looks to me like a 555 oscillator driving a 4017 as a divide-by-2 counter.
There's quite a lot wrong with that circuit. . .
1. Transformer voltage is too low. To make a squarewave inverter you need a higher voltage transformer than 12V. A 12V transformer expects a voltage no higher than 12.√2.2/π, because the average DC voltage and the rms voltage of a sinewave half-cycle are not the same. You can use a 12V 50Hz transformer if you increase the frequency to 60Hz.
2. There is no dead-time, so there is a very good chance that one MOSFET will have turned on before the other has turned off. That will blow it up.
3. A 4017 really doesn't have enough output to drive a big MOSFET, even at 50Hz. Use a low-side MOSFET driver.
Better still, throw the whole lot away and use an SG3525, which will do the oscillator, dead-time and MOSFET drive all in one package.

The circuit shown in that link, which claims to use the same circuit, there is very wide dead time, as the gates are connected to Q0 (pin#3), and Q2 (pin#4). So there would be one whole pulse period of dead time. The output high level current at 13.5volts is listed as just 3.5mA, while the output low current is listed as 8mA. So the turn on drive may not have been adequate.
But we have no information about the transformer resistance or inductance, and the claimed scope traces that I see in that link are from a scope that I would not trust to show spikes.

 

If you look at the claimed waveform

The circuit shown in that link, which claims to use the same circuit, there is very wide dead time, as the gates are connected to Q0 (pin#3), and Q2 (pin#4). So there would be one whole pulse period of dead time. The output high level current at 13.5volts is listed as just 3.5mA, while the output low current is listed as 8mA. So the turn on drive may not have been adequate.
But we have no information about the transformer resistance or inductance, and the claimed scope traces that I see in that link are from a scope that I would not trust to show spikes.

What can I do to reduce spikes? What if I use a Zenner diode between the gate and source

 

For starters, the instructions for some simple inverters say to connect the load before switching on the power. An incandescent light bulb makes a good load.

But before anything else, check the data sheet of the transistors that you used and compare them with the data sheet for the specified transistors. It might be that the transistors failed because they were not rated for the voltage that high.

 

• Small signal N-Channel MOSFET
• Continuous Drain Current (ID) is 49A at 25°C
• Pulsed Drain Current (ID-peak) is 160A
• Minimum Gate threshold voltage (VGS-th) is 2V
• Maximum Gate threshold voltage (VGS-th) is 4V
• Gate-Source Voltage is (VGS) is ±20V (max)
• Maximum Drain-Source Voltage (VDS) is 55V
• Rise time and fall time is about 60ns and 45ns respectively.
• It is commonly used with Arduino, due to its low threshold current.
• Available in To-220 package

• For starters, the instructions for some simple inverters say to connect the load before switching on the power. An incandescent light bulb makes a good load.
  
  But before anything else, check the data sheet of the transistors that you used and compare them with the data sheet for the specified transistors. It might be that the transistors failed because they were not rated for the voltage that high.

 

• Small signal N-Channel MOSFET
• Continuous Drain Current (ID) is 49A at 25°C
• Pulsed Drain Current (ID-peak) is 160A
• Minimum Gate threshold voltage (VGS-th) is 2V
• Maximum Gate threshold voltage (VGS-th) is 4V
• Gate-Source Voltage is (VGS) is ±20V (max)
• Maximum Drain-Source Voltage (VDS) is 55V
• Rise time and fall time is about 60ns and 45ns respectively.
• It is commonly used with Arduino, due to its low threshold current.
• Available in To-220 package

this is what it says

 

Hi! I just found a video that is about 4 years old on how to make an inverter using a 555 and 4017 IC . I want to know whether this would work with an input of around 16v and a 7.6 - 0 - 7.6v transformer. And can u say me why there is a 220-ohm resistor in the gate of the MOSFET? Here is the video link -

Here is the transformer I have :

These are the batteries I use - http://www.hybrids.co.nz/battery-specs/
I waiting for a quick reply.

 

It will probably work but from my limited experience driving transformers backwards you might be disappointed with the overall performance.

 

With that 15+ volt battery pack the Gate to source max could have been approached , or exceeded by even a small spike of just a few volts.. And with no load on the transformer you have no idea what was happening.

 

The gate-source voltage of the MOSFET is around 20v so I guess it would never go above that. Because the battery can charge up to a maximum of 16.8v and what if I add a 220ohm resistor to the gate of the MOSFET and a 10k resistor to the gate and source of the MOSFET?

 

Consider that when one side of the transformer primary is pulled up to Vsupply, the other side is pulled down to Minus Vsupply. So the gate to drain voltage can be twice the supply voltage. My point being that voltages in an active circuit can be higher than the supply voltage.
Have you checked to see if the two transistors are damaged, or not? It seems that the 4017, being directly connected, might have been damaged.
Looking back at post #1, no hint as to where the spark happened, or any mention of any part looking damaged. The spark might even have been across the transformer secondary terminals.
We have no hint as to the actual construction of the circuit assembly. Is it a circuit board (PCB)?, or a white breadboard? Or just wires in air?? Are those transistors mounted on heat sinks?? A whole lot of variables that could assist answers.