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output waveform push pull inverter using toroidal transformer not sine wave

rizqiuhuy

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hi, i'm new in here, i have project with 48v inverter using toroidal transformer with topology is push pull. I use spwm with frequency 15kHz. I use mosfet irfb4332 3x each channel connect to drain directly to 48v toroidal transformer, on the secondary output transformer use 1.2uF capacitor. but the output transformer wave like on attachment (yellow signal is drain & blue signal is output transformer).
IMG-20210331-WA0018.jpg
how to fix this to get pure sine wave ? thanks.
 

KlausST

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Hi,
You don't show the negative half of the blue signal.
Is there a reasin why?
Please adjust vertical setting.

Klaus
 

rizqiuhuy

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thanks, for all reply
Please explain the inverter topology and switching pattern.
the schematics like this, but on the board i use 3 mosfet irfb4332, for switching pattern i use the code from GitHub - Irev-Dev/Arduino-Atmel-sPWM: Implementation of an sPWM signal, but i implement to stm32f103 use timer interrupt from PeriodElapsedCallback to switch signal.
capture.jpg

Hi,
You don't show the negative half of the blue signal.
Is there a reasin why?
Please adjust vertical setting.

Klaus
sorry for negative half wave, i don't get the capture because oscillscope limit. the signal like show on the oscilloscope but like inverted

rizqi
 

KlausST

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Hi,

There are a lot of issues in the schematic.
At least we miss the transformer connection and the output filter. (Besides decoupling capacitors.,)
--> show transformer and filter schematic

You say it is "SPWM" but there is nothing about SPWM in the scope picture.
--> Show the driver signals.

--> why do you use optocouplers when you connect input and output side (GND)?

--> to measure the output voltage use a self made resistive divider 100k/1k. Should work for low frequency signals.

Klaus
 

FvM

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I guess you are operating the two low side switches with a center tapped transformer? One problem of this topology is that current commutation goes through the transformer leakage inductance. Energy stored in the leakage inductance can't be recovered and has to be dissipated, either in clamping zener diodes or by avalanch breakdown of the switch MOSFETs. The avalance energy rating is probably exceeded if the inverter is operated with larger output current.

To achieve sine waveform without resistive load, a synchronous switching topology (half or full bridge with respective drive pattern) is required. Your asynchronous push-pull topology with "on-off" switch pattern can't work.
 

Easy peasy

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you appear to have a 50Hz push-pull, slapping a cap on the o/p will achieve very little filtering, you will need big bulky LC filters, 2 stage if you want to get a sine wave, total volume larger than your transformer by quite some margin, avoid filter resonances near 50Hz, you must be below this ...
--- Updated ---

@FvM, you can do it with a push-pull, ideally with low leakage on the pri wdgs, and low energy stored in the Tx core, but it can be done, just need a bit more filtering and pwm control that can go near zero. A few watts of dummy load on the o/p doesn't hurt either ...
 

dick_freebird

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I've seen schematics of simple push-pull inverters which
purport to be on a 50/60Hz frequency but no mention of
output waveshape although there's an obvious stab at filtering.

PushPullInverter.jpg

A "modified sine wave" inverter could be had by additional
windings and additional switches (2, 4, 6, ...) depending on
how "modified" or how "sine" you want. Maybe a better
candidate for digital control than analog, trying to keep
all the leading and trailing edge phase relations right.
 

Easy peasy

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I've seen the above ckt on linked-in recently, the chap stated sine wave out, I checked up on the Co. he works for Pwrtrnx.com - seems they have been around since 1994 doing power electronics - possibly they know what they are doing ... ?
 

FvM

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Remember that the original post intends to achieve sine waveform with pwm, not a 50 Hz filter.

The problem with an asynchronous switcher is that the output voltage is only proportional to pwm duty cycle if the current through transistor switch and diode is continuous which requires a sufficient real load. In discontinuous mode, the duty cycle need to be pre-distorted to achieve the intended waveform.
 

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@FvM - correct - that's what a feedback loop is for - to shape the output to a sine wave,

if one is striving for feed-forward only ( no volt feedback ) - then one can only use a direct action type converter, e.g. buck type - with a clean drive i.e. a drive voltage or a short on the Tx pri ( e.g. phase shift )

We did make a push pull like this once where the Tx pri was shorted in the dead time, this gives the same output voltage as a phase shift, or MPW, i.e. a Tx short in any dead time period.

Alternatively one can use a 2 level approach where 50% pwm = 0v output ... ( classic S_PWM )
 

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