MOSFET INVERTER CIRCUIT PLZ HELP heating problem

Thread Starter

Emil sonu sam

Joined Jan 31, 2013
4
I am working to develop a 12v dc/ac inverter with 4 mosfets 2 pchannel irf9540 (High side and 2 nchannel 55nf06(Low side) as in circuit.
driver circuit - 4 independent drivers input from arduino isolated by tlp250 and amplifies by a tottem pole BJT circuit as shown in figure.

TLP 250 supply is provided feom a 230/12v transformer.

2 pulses with phase shift of 180 degree and amplitude about 6v were generated to gates. gate pulses to Q1 AND Q3 are identical and Q2 and Q4 are identical.

frequency of pulses 50Hz Ton = 10ms.

I am facing some problem , while connecting the circuit to 12v Vcc THE 2 pchannel mosfets are highly heating and circuit is not working.

employed 10k bleeding resistors.

PLZ help..
 

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#12

Joined Nov 30, 2010
18,224
Consider the word, "shoot through". I don't see any method of delaying the time between one pair turning off and the next pair turning on.

You could also do better with a logic level mosfet that is seriously "on" with 6 volts on the gate.
 

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AnalogKid

Joined Aug 1, 2013
12,142
That's not the timing issue #12 is referring to. The two P-channel MOSFETS are turned off by nothing more than a pullup resistor of unknown value discharging the gate capacitance, while the two N-channel fets are being turned on by direct drive from the driver circuit. There is an excellent chance that the upper FETs still are partly on and turning off slowly when the lower FETs come on. Since the upper FETs are not saturated during a slow turn-off time, the power dissipation increases.

ak
 

Thread Starter

Emil sonu sam

Joined Jan 31, 2013
4
Thanx for d reply..
Pull up resistors - 10 k.
So is it necessary to make a delay between turn off and on of upper and lower MOSFET?

any otr suggestion!
 

AnalogKid

Joined Aug 1, 2013
12,142
If both the upper and lower FETs are even partially on at the same time, current runs directly through them without going through the load. This is called cross-conduction or shoot-through. The solution is more complex gate drive circuits that produce a very short time when both FETs are guaranteed off. This is called a deadtime or deadband circuit. This can be done with some timing parts, or by reprogramming the arduino with timing delays, or by using bridge driver chips from LT, Maxim, or Unitrode.

ak
 
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