pic based modified sine wave inverter

hi am trying to make a pic based inverter circuit.
the circuit uses 2 p-channel mosfets(f9540n) with a centre tapped 12-012/230v transformer.
i programmed the pic to turn on the mosfets inorder to create modified sine wave.

i assembled circuit with o/p of transfrmr open circuited.i switched on the 12v dc supply but within 5s my breadboard got heated up and started to melt.
how did this happen???
pls help

A schematic is needed to answer that question but I can make a guess at the reason - if you are directly driving the gates of the MOSFETS from the PIC they never turn off, you would be applying the wrong polarity signals to the gates. I assume you have the tap in the transformer grounded and the MOSFET source pins connected to a +12V supply so the best you could manage with a conventianally wired PIC would be VGS between -7V and -12V (12V minus PIC output voltages) when the worst case specification is they turn on with -4V.

Brian.

laxman289 said:

hi am trying to make a pic based inverter circuit.
the circuit uses 2 p-channel mosfets(f9540n) with a centre tapped 12-012/230v transformer.
i programmed the pic to turn on the mosfets inorder to create modified sine wave.

i assembled circuit with o/p of transfrmr open circuited.i switched on the 12v dc supply but within 5s my breadboard got heated up and started to melt.
how did this happen???
pls help

Click to expand...

If I correctly understand you pass high current over protoboard (breadboard) ? :shock:

Breadboard are not for higher currents, its for signals and smaller currents, and for higher currents you should use appropriate wires outside of breadboard.

here is a rough schematic

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here is a rough schematic

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but how come the current is so high..i havnt loaded the transformer

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betwixt said:

A schematic is needed to answer that question but I can make a guess at the reason - if you are directly driving the gates of the MOSFETS from the PIC they never turn off, you would be applying the wrong polarity signals to the gates. I assume you have the tap in the transformer grounded and the MOSFET source pins connected to a +12V supply so the best you could manage with a conventianally wired PIC would be VGS between -7V and -12V (12V minus PIC output voltages) when the worst case specification is they turn on with -4V.

Brian.

Click to expand...

but i checked the operation of mosfet by using led.i turns off with 5v and turns on with 0v(p-channel)

Where is the PIC ground (VSS pin) connected? Remember that to switch the MOSFET on and off you have to apply or remove the voltage between the gate and source. That's difficult to do when the source is connected to the transformer like that. Please post the full schematic so we can check.

As tpetar points out - breadboard construction isn't really suitable when you are switching many Amps.

Brian.

You switch MOSfet sides.

Example with P-MOSfets




Example with N-MOSfets :





MOSFets pins :





This design gives you square wave shape, and its not good for inductive loads. Search EDABoard threads for modified or pure sinewave inverters.



;-)

Wouldn't "modified sine wave" require a different pulse generation scheme with < 50% duty cycle?

It certainly SHOULD be less than 50% although that design doesn't do it. The MOSFETs Laxman289 is using are P-channel though, so the transformer center tap would have to be ground and the source pins conneced to +12V. That should work equally well (OK - reasonably well!) as long as the drive signals are +12V for 'off' and < 8V for 'on' and I doubt the PIC circuit is producing that.

Brian.

betwixt said:

It certainly SHOULD be less than 50% although that design doesn't do it. The MOSFETs Laxman289 is using are P-channel though, so the transformer center tap would have to be ground and the source pins conneced to +12V. That should work equally well (OK - reasonably well!) as long as the drive signals are +12V for 'off' and < 8V for 'on' and I doubt the PIC circuit is producing that.

Brian.

Click to expand...

bt i have checked the mosfets using a led..its switching correctly.the led is flickering.also i seperately used a pic high as well as low output pin to check mosfets switching..i dont think the ampere drawn during no load condition of transformer is high enough to melt bread board and connecting wires.

Please try this then:

1. remove the PIC completely then connect both the gate connections to the PIC Vss (ground) pin. This SHOULD turn both the MOSFETS off and no current should be drawn at all. Please tell me what actually happens.

2. With the PIC still removed, connect both the gate connections to the PIC VDD (supply) pin. Again let me know what happens.

What I'm getting you to do is simulate a condition where the MOSFETs are both conducting and both not-conducting to see if the drive levels can possibly be adequate to switch them on and off. I think I know what will happen but I need you to confirm it.

Brian.

U r right the mosfet is aways on.how can i turn off p mosfet while using 12v supply.

The only way to turn the MOSFET off is to raise the gate voltage to meet the source voltage, in other words 12V. You confirmed what I suspected, any voltage lower than about 8V would turn the MOSFET on and the PIC is only capable of going up to 5V so it is impossible to it to turn the MOSFET off.

You have two solutions, one is to replace the MOSFETS with N-Chanel types so the PIC and source pins can both be grounded or, you can add transistor driver stages between the PIC and MOSFET. The transistor driver is probably the easiest solution, just wire NPN signal transistors with emitter grounded, base to the PIC via 1K resistor and collector to 12V through a 470 Ohm resistor. The gate of the MOSFET goes to the transistor collector. Now the PIC can drive the NPN transistor hard on and off and it's collector will go from 12V down to near zero so the MOSFET gets the drive voltage it needs.

You should note that adding the transistors will invert the PIC signals (PIC pin = 0 means the MOSFET conducts) but as the drive is symetrical anyway it shouldn't make any difference in this design.

Brian.