MOSFET INVERTER CIRCUIT PLZ HELP heating problem

Discussion in 'General Electronics Chat' started by Emil sonu sam, Aug 12, 2015.

  1. Emil sonu sam

    Thread Starter New Member

    Jan 31, 2013
    4
    0
    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..
     
  2. #12

    Expert

    Nov 30, 2010
    16,298
    6,810
    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.
     
    Roderick Young likes this.
  3. Emil sonu sam

    Thread Starter New Member

    Jan 31, 2013
    4
    0
    TANX..
    the pwm for 2 signals phase shifted was actually by programing arduino ..

    bit-banging Pulse Width Modulation
    You can "manually" implement PWM on any pin by repeatedly turning the pin on and off for the desired times. e.g.
    https://www.arduino.cc/en/Tutorial/SecretsOfArduinoPWM
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,540
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    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
     
    alfacliff likes this.
  5. Emil sonu sam

    Thread Starter New Member

    Jan 31, 2013
    4
    0
    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!
     
  6. Emil sonu sam

    Thread Starter New Member

    Jan 31, 2013
    4
    0
    Pull ups are of 10k .
    Is it necessary to provide a dealt btwn on and off of upper and lower MOSFETs.


    Any suggestions
     
  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,540
    1,251
    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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