Ambiguity regarding RMS output of inverter circuit...

Discussion in 'General Electronics Chat' started by Anirban Raha, Oct 12, 2015.

  1. Anirban Raha

    Thread Starter Member

    Dec 4, 2013
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    Please take a look at the following circuit and tell me whether the output will be 220V RMS AC as indicated.
    My question is that since the peak to peak input via PWM on the primary side is 12V...then shouldn't 8.486 (12/root(2)) be used to determine the RMS voltage on the secondary?
    inverter.jpg
     
  2. bertus

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  3. MrAl

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    Jun 17, 2014
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    Hi there Raha,

    You can not know that without knowing the switching pattern developed by the controller.

    What you are calculating is the RMS voltage of a sine wave with a 12v peak. But the switching pattern plays a role in the RMS voltage by keeping the switch 'on' cycle shorter if necessary.
    For example, if you calculate a switching pattern for a sine wave with a peak of 12v, then make every 'on' period 1/2 of what it was, then your output is 1/2 of what it was before so the RMS is much lower.

    That's not to say they are really doing this though, so you have to check the code for the controller to find out for sure, but that's the way it should be done or the voltage will be too high for the primary of the transformer which is only 6v as per schematic. 8.5v RMS is significantly higher than 6.0v RMS so it will most likely saturate the transformer unless the switching pattern was cut back to account for the lower voltage rating of the transformer primary. It is however possible to cut back the pattern to create just about any lower voltage level like 6vrms, 5vrms, 4vrms, 1vrms, 0vrms, etc.
     
    Last edited: Oct 12, 2015
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  4. Anirban Raha

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    Dec 4, 2013
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    The PIC is generating a Sinusoidal PWM signal. I understand the basic concept behind SPWM. However:
    1. If indeed an SPWM signal is being fed to the transistors to produce a sinusoidal voltage at the primary of the step-up transformer, can this voltage be treated as 12V RMS as per the circuit?
    2. If it is not 12V RMS but 12V peak voltage instead, then is it possible for the output to be 220V RMS at the secondary?
     
  5. AnalogKid

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    Aug 1, 2013
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    No. I understand your thinking, but you missed something.

    First, let's be clear on terminology. When I say "primary", I mean the 12 V center-tapped winding. This is the secondary winding for the transformer as originally designed, so some clarity is needed.

    It would be if a transformer primary were being driven without the centertap. But by using the transformer input center tap, the circuit is essentially the reverse of a full-wave centertap power supply input. Input current is flowing through a 6 Vrms winding only. (6 Vrms with respect to the 220 Vrms secondary). The fact that the other half cycle uses a different winding doesn't matter because that winding also is rated for 6 Vrms. So the 12 V power rail is more than enough to create an energy waveform at the transformer input that is equivalent to 6 Vrms.

    ak
     
  6. Anirban Raha

    Thread Starter Member

    Dec 4, 2013
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    I don't know if we are actually talking about the same thing...but let me rephrase my question to be sure. Since this microcontroller is switching a 6V supply, then the peak voltage that one can obtain would be 6V on either of the two coils on the LV side. Am I correct? If yes, then how can the LV side of the transformer have 12V rms voltage? Shouldn't it have 12/root(2) RMS voltage and produce and output of 220/root(2) rms voltage on the HV side?
     
  7. crutschow

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    Mar 14, 2008
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    Post deleted.
     
    Last edited: Oct 12, 2015
  8. tsan

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    Sep 6, 2014
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    For me it looks like 12 V is connected to center tap. If it's so, then there is 12 V on either of the two coils.
     
  9. alfacliff

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    Dec 13, 2013
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    why arent you using the turns ratio of the transformer?
     
  10. AnalogKid

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    No, it isn't. Each microcontroller output drives a 3-stage darlington transistor arrangement. The transformer primary center tap is connected to 12 V, and the two primary winding ends are pulled to ground through saturated switches Q5 and Q6. This puts over 11 V peak across each "6 V" winding. Follow the "thick wirings" paths in your schematic.

    ak
     
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  11. MrAl

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    Jun 17, 2014
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    Hello again,

    IF the controller is putting a 6vrms signal on each leg of the primary then the output should be 220vrms. It cant be 12vrms or it would probably blow out the transistors, unless the transformer was much under rated by the manufacturer.

    However, we dont know if this was designed right or not. If it was designed right, then it should put out 6vrms into the primary as indicated on the schematic, and that should work provided the balance between driver transistors does not create too much of an offset at the primary.
    If it was designed to put out 12vrms into the primary then it's probably going to either blow the transistors or get them very hot and make a lot of noise in the transformer, if you are lucky.

    Good converter circuits have current sense mechanisms in place to sense current in the primary or each transistor. They can shut the converter down if a problem comes up.
     
  12. panic mode

    Senior Member

    Oct 10, 2011
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    if this was designed right it would not use 12V source to power LM7812
     
  13. Anirban Raha

    Thread Starter Member

    Dec 4, 2013
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    That means in order to have an RMS voltage of 220V on the secondary, I need to have a peak voltage of 6*root(2) on each of the two primary winding? Am I correct?
     
  14. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    One other thing is strange in the circuit.
    Q1 and Q2 are used as emitor followers, so the input voltage would not need to be 12 Volts.

    Bertus
     
  15. RamaD

    Active Member

    Dec 4, 2009
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    Yes. But that is the peak voltage of the sine wave at 50Hz frequency. The PWM frequency would much higher.
    The pulses applied to the primary of the transformer would always be 12V peak ideally, that is not taking into account the voltage drops on the transistors. AK had done this and got about 11V. So the pwm width at the sine wave (50Hz) pk should be 6*root 2 / 11V = 0.77 of the total PWM period.
     
  16. AnalogKid

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    Aug 1, 2013
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    Not quite. The difference between an analog voltage and a PWM signal that represents that analog voltage can be confusing. You are correct about the peak voltage value, but only if you are driving the transformer with an analog signal. With PWM, the voltage swing always is the full power supply value. It is the duty cycle of the PWM waveform that creates the equivalent of other analog voltages. You want a pulse width ratio that produces an average value of [6*root(2)] when multiplied by the power source (12 V in your case, minus the saturation voltage of the switching transistors). RamaD did the math.

    ak
     
  17. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello,

    If you want to see 6vrms or 6*sqrt(2) volts peak sine at the transformer primary then you probably have to filter the PWM signal and apply any necessary gain adjustment factor for the filter. The voltage will go from 0v to +24v (pulsing) referenced to ground unless the output filter couples well enough to the primary and then you might see a sine wave.

    BTW a 12v regulator may work ok at 13.8v which is the normal car voltage when the engine is running.
     
  18. Anirban Raha

    Thread Starter Member

    Dec 4, 2013
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    That means...if I neglect transistor drops ( since I'm using N-channel MOSFETs)...then to produce an RMS voltage of 6V on each of the windings I need to switch a 6*root(2) volts source using SPWM (then at peak voltage the PWM duty would be 100%)...right?
     
  19. AnalogKid

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    No. You still don't seem to understand the fundamental nature of PWM, or why it is employed instead of a true analog voltage in some situations. PWM is a digital signal used to create an analog signal. There is no automatic relationship between the peak value of the PWM square wave and the RMS value of the created analog waveform. It is not *required* that the peak value of the PWM waveform equal the peak value of the analog waveform. As an extreme example, it is entirely possible for a 100 V peak=to-peak PWM waveform to create a 1 Vrms sinewave. More on that later.

    But first:
    1. You never can neglect transistor drops.
    2. MOSFETs are very different from BJTs in their drive requirements. They cannot be dropped into the schematic you posted without other circuit changes.
    2. There are no MOSFETs in the schematic you posted. Please post a corrected schematic.

    ak
     
  20. Anirban Raha

    Thread Starter Member

    Dec 4, 2013
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    @AnalogKid
    I am not just replacing the transistors with MOSFETs...I'm using MOSFET driving circuitry as well...but what if I eliminate the centre tapped transformer and use a full bridge configuration (with N-channel and P-channel MOSFETs)? How would I relate PWM RMS with RMS of the sinusoidal analog waveform?
     
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