# PWM inverter

1. Jan 1, 2014

### Physicist3

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. Jan 1, 2014

### Okefenokee

An inverter doesn't require any sine input for power. It uses DC. They're may be a AC-DC 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.

This is essentially correct. The part that's missing is that the PWM generates the on-off signal (chopping signal) for the IGBT's. The current being driven by the IGBT's is the actual sine output.

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.

Last edited: Jan 1, 2014
3. Jan 2, 2014

### Baluncore

For single phase the peak value of the rectified input voltage will be Sqrt(2) * the input RMS voltage.

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. Jan 2, 2014

### Physicist3

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. Jan 2, 2014

### Baluncore

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 H-bridge, now usually made from MOSFETs or IGBTs.

http://en.wikipedia.org/wiki/Inverter_(electrical)
http://en.wikipedia.org/wiki/Variable-frequency_drive

Last edited: Jan 2, 2014
6. Jan 2, 2014

### Physicist3

Hi, thanks for the info and the links. Much appreciated.

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. Jan 3, 2014

### Baluncore

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. Jan 3, 2014

### Physicist3

So, in very simple terms, when for instance the user sets the frequency of the drive output to be 10Hz, does the circuitry within the drive create a sine wave (green in the figure) of that frequency and with a max amplitude equal to the DC value and then compare that sine wave (modulating signal) with the saw tooth wave (carrier signal)? From this, are the IGBTs then switched on and off according to the points of intersection of the sine and saw tooth, hence causing the PWM output to change between on and off.

9. Jan 3, 2014

### Baluncore

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

10. Jan 3, 2014

### Physicist3

Ah right, I think I understand. But the Saw Tooth would still be compared against this numerical model sine wave having a frequency equal to the desired output in order to produce the on-off pulses for the pwm? Also, when analysing the PWM signal going to the motor, I noticed that the voltage amplitude was the same for all frequencies when looking on an oscilloscope, however, when using a multimeter to measure the voltage to the motor, it got lower as the frequency decreased. Is this because the change in frequency affects the on-off ratio of the PWM wave, meaning that the RMS voltage will change? The DC link voltage appeared to be equal to the peak value of the AC supply signal. Does this sound correct

Last edited: Jan 3, 2014
11. Jan 4, 2014

### Baluncore

In theory yes, but simplistic theory and understanding no longer describe the way it is now done numerically with a micro-controller.
Did your analysis simulate a hypothetical system? or did you measure a real system?

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.