Inverter Circuit using IR2113, TLC555 and signal inverter IC

Discussion in 'Power Electronics' started by Dasun Umayanga M. A., Sep 10, 2019 at 4:56 AM.

  1. Dasun Umayanga M. A.

    Thread Starter New Member

    Aug 3, 2019
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    Inverter Circuit.jpg
    Input DC= 30V
    operating frequency = 10Khz
    problem =
    when I am at the no-load condition at the output, only the oscilloscope is connected, I am receiving proper output but when I connect a load, which is 10 Ohms resistor, the output voltage drops down to almost zero.

    I really need a bit of professional advice here
     
  2. Dodgydave

    AAC Fanatic!

    Jun 22, 2012
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    Firstly a 555 chip is the wrong choice, i would use a TL494,

    Ok what transformer have you got connected to terminals X3?
    And how are you regulating the output voltage??
     
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  3. Alec_t

    Expert

    Sep 17, 2013
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    Remove the 10k resistors R2 and R3. They may be preventing the bootstrap circuitry from operating correctly.
    Note that neither the IR2113 nor your way of generating Signal1 and Signal2 provide any dead-time, so you are likely to have significant shoot-through which could fry the FETs.
     
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  4. TeeKay6

    Member

    Apr 20, 2019
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    @Alec_t
    I assume that you meant to say "remove 10K resistors R8, R5, R7, R9." R2=R3=10Ω and likely have little effect.
     
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  5. TeeKay6

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    Apr 20, 2019
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    @Dasun
    With an oscilloscope you should be easily able to verify that Vgs of each MOSFET is about 10-12V when that MOSFET should be on. A lower voltage indicates a problem with the gate drive. Reference your 'scope to ground when viewing the gate signals.

    In your schematic, variable resistor R1 (near the TLC555) is not connected.
     
    Last edited: Sep 11, 2019 at 1:43 AM
    Dasun Umayanga M. A. likes this.
  6. Dasun Umayanga M. A.

    Thread Starter New Member

    Aug 3, 2019
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    I have not connected any transformer yet, simply i connect a oscillarscope to observe the output, then i received acceptable Square wave.

    Afterwards i connected a 100ohm resister across the x3 terminal then i connected to the oscillarscope. Then i am getting null output voltage.

    How this happens? Can you please give me an idea?
     
    Last edited: Sep 11, 2019 at 2:25 AM
  7. Dasun Umayanga M. A.

    Thread Starter New Member

    Aug 3, 2019
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    How could I prevent the FETs?
     
    Last edited: Sep 11, 2019 at 3:15 AM
  8. Dasun Umayanga M. A.

    Thread Starter New Member

    Aug 3, 2019
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    Currently,I am using the 74hc04 IC and function generator to generate Signal 1 and Singal 2 not the 555
     
  9. Dasun Umayanga M. A.

    Thread Starter New Member

    Aug 3, 2019
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    simply I connect an oscilloscope to observe the output at X3 terminal, then I received acceptable Square wave.

    Afterward, I connected a 100ohm resistor across the x3 terminal then I connected to the oscilloscope. Then I am getting null output voltage.

    How does this happen? Can you please give me an idea? why current is not passing and voltage drops?
     
  10. Alec_t

    Expert

    Sep 17, 2013
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    I meant remove the 10k resistors going from gate to drain on both high-side FETs. The post #1 pic is so blurred I probably mis-read the resistor designators. Dasun, have you removed those two resistors and confirmed that the bootstrap is working (i.e. the high-side FET gates go to a higher voltage than the source voltages?
     
  11. TeeKay6

    Member

    Apr 20, 2019
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    @Dasun Umayanga M. A.
    The first step to resolving this problem is to look at the MOSFET gate drive signals. For an N-channel enhancement mode MOSFET (such as IRLZ44N) to be "on", its gate voltage must be more positive than its source voltage by many volts. The IR2113 ought to provide 10-12V gate drive. For example, to turn Q1 "on", its gate voltage should be 10-12V more positive than its source voltage. After removing the 10K resistors per @Alec_t suggestion, verify with your 'scope that all gate signals are correct. Do this with your test load.

    If you review the circuit of the IR2113, you can see that HO and LO both provide pull-up and pull-down drive. Thus, none of the four 10K resistors you show are needed.

    In your post#1 you say the load is 10Ω but in post#6 you say it is 100Ω. Which is correct? How are you measuring the X3 voltage? This should be an AC voltage; when X3:1 (X3,pin1) is low (approximately 0V), X3:2 should be high(approx. VCC); when X3:1 is high, X3:2 should be low.

    Also verify that the timing of gate drives is correct. Q1 and Q2 should be "on" at the same time, and Q3 and Q4 should be "on" at the same time. Q1 and Q4 should never be "on" at the same time; Q3 and Q2 should never be "on" at the same time. Also, are the drive signals switching at the frequency you expected (and what is that frequency)?

    When measuring signals in this circuit, your 'scope ground should be connected to the circuit ground.

    @Alec_t mentioned the problem of "shoot through" (also called "punch through"). Using your schematic as an example, if Q1 and Q4 (or Q2 and Q3) are ever turned on at the same time, even for very brief periods, they present an extremely heavy load on the VCC-to-ground power supply; the supply is virtually short-circuited. Current is limited only by what the power supply can provide, the MOSFETs can carry, and wiring resistance and inductance. This can quickly overload the MOSFETs. During the both-"on" period, great power is wasted, reducing efficiency significantly for even tiny periods of overlap in "on" time. The solution is conceptually very simple: do not turn on Q4 until Q1 has had time to fully turn "off" and do not turn on Q1 until Q4 has had time to fully turn "off". (This means that Signal2 cannot be a simple inversion of Signal1 as in your circuit.) This delay is generally called "dead time." Note that the required dead time will vary for different MOSFET types, different loads, and different gate drivers. There are several schemes for providing the necessary dead time. If only a very small amount of delay is needed (e.g. <10ns), then the gate drive signal can sometimes be modified using R, C, and diode combinations. For the more usual case when larger delays are needed, they can be provided by the gate driver IC itself as a built-in function. When using drivers, such as the IR2113, that do not offer that function, the delay must be provided via the Signal1 & Signal2 timing. Various logic schemes can provide the delay and many modern microcontrollers provide the delay as a built-in function. You can find more info by searching AAC and the Web for "H-Bridge Driver".
     
    Last edited: Sep 11, 2019 at 11:01 PM
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