No. I understand your thinking, but you missed something.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?
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?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.
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.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?
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.Since this microcontroller is switching a 6V supply,
Hello again,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?
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?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.
Yes. But that is the peak voltage of the sine wave at 50Hz frequency. The PWM frequency would much higher.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?
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.I need to have a peak voltage of 6*root(2) on each of the two primary winding? Am I correct?
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?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.
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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.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?
by Duane Benson
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by Duane Benson