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PWM inverterby Physicist3
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#1
Jan114, 08:20 PM

P: 72

Hi, could someone confirm whether I am right in thinking that for a PWM inverter, the input sine voltage is converted to DC using a bridge rectifier? Does this DC voltage then have a value which is the same as the PEAK value of the sine input? Is the DC voltage then 'chopped' using IGBT's to produce a new 'sine' wave, which is then compared with a triangle wave to get the PWM output??
Thanks 


#2
Jan114, 10:17 PM

P: 242

An inverter doesn't require any sine input for power. It uses DC. They're may be a ACDC converter that creates the DC. The upper limit for the voltage of a DC rectifier is going to be the RMS value of the mains voltage.
The sawtooth that you are talking about is compared to a reference signal of the desired frequency. The reference signal could come from a VCO (voltage controlled oscillator). The ouput of the comparison between the reference and sawtooth signal becomes the PWM signal which then controls the IBGT's. 


#3
Jan214, 02:41 AM

Sci Advisor
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P: 1,897

With an inverter used for a variable speed drive, a single or three phase AC input is first rectified to DC. The rectified DC is then PWM to produce the output AC at the required frequency and in the required phase sequence to control the speed and direction of the motor. 


#4
Jan214, 05:05 PM

P: 72

PWM inverter
I think I may be a little confused as I always thought that for something such as an inverter drive used for motors etc, the value of the DC link voltage was sqrt(2) * RMS input value, and that this was then turned into an artificial sine wave using fast switching IGBT's? I was then under the impression that this sine wave was compared with a triangle (sawtooth) wave and in the sections where the sine wave had a higher amplitude than the triangle wave, This is where the pwm output wave to the motor was 'high' or 'on'?



#5
Jan214, 05:53 PM

Sci Advisor
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P: 1,897

In a sine wave inverter the PWM duty cycle is varied continuously to adjust the output phase waveforms to the required output sine wave amplitude and frequency. The output of the comparison/error amplifier may be compared with a low amplitude, say 50 kHz, triangle wave to generate the digital PWM switch on/off timing.
With single phase to three phase VFDs the maximum voltages needed for the output phases is the same as the maximum rectified input voltage available, so while an output phase is at an extreme voltage the appropriate PWM switch can stay on for a short period. The input circuit is often a bridge rectifier with several DC energy storage capacitors. The output circuit is a three phase Hbridge, now usually made from MOSFETs or IGBTs. http://en.wikipedia.org/wiki/Inverter_%28electrical%29 http://en.wikipedia.org/wiki/Variablefrequency_drive 


#6
Jan214, 11:34 PM

P: 72

Just to clarify, on the second wiki link you supplied about VFDs, looking at figure 1, it shows a green sine wave with a blue sawtooth wave, and then a pink pwm wave below. Am I correct in thinking that the pink wave is the wave which goes to the motor stator? The bit that is confusing me is the bit with the sine wave and the sawtooth. Is that sine wave (green wave in fig 1 on wiki) produced using the igbt's from the DC bus before being compared with the blue sawtooth wave to produce the pink pwm wave, which is 'high' when the sine wave is above the sawtooth and 'low' when the triangle is above the sine wave?? Thank you again :) 


#7
Jan314, 01:14 AM

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P: 1,897

Yes, the pink wave could go to the stator. But that picture is simplistic in that it does not show the pink to have a very much higher switching frequency than the sine wave being generated. Some form of inductance is needed in the load to average the current, such as a motor winding or an LC network.
The green sine wave is a reference sine wave. To generate digital PWM switch control, the reference sine and a saw or triangle wave are compared. If that pink signal was low pass filtered it should look like the green sine wave. In reality, current regulation and load variation demands more than a simple comparator. To be robust it needs a current monitor and the computation of an output error voltage. Remember that since v = L * di/dt the voltage applied to the inductive load causes the current to gradually rise or fall in the motor inductance. So the motor voltage is a PWM wave, but the phase current is a slowly varying current with a small triangle wave current impressed on it. 


#8
Jan314, 03:08 PM

P: 72




#9
Jan314, 03:42 PM

Sci Advisor
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P: 1,897

Yes, but now the sine wave is not usually generated physically. The external load voltage and current are read by AD converters, then compared with a numerical model sine wave for each phase inside the microcontroller.



#10
Jan314, 04:57 PM

P: 72




#11
Jan414, 05:56 PM

Sci Advisor
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P: 1,897

In theory yes, but simplistic theory and understanding no longer describe the way it is now done numerically with a microcontroller.
With a real motor, voltage will be expected to fall as frequency falls. The VFD will regulate motor torque which is proportional to current, not to voltage. At low RPM the motor will not generate a back EMF to overcome. That may explain the reduction in voltage with lower drive frequency while the torque and current are both rising at lower speed and frequency. Your multimeter will be sensitive to wave shape and if it is a simple average or a true RMS meter? Do not use a multimeter to measure digital switching signals if you want a useful result. It will lead you astray when you least expect it. 


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