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sinewave inverter under load

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freeman3020

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Hi all

I have finished of design and implementation for pure sine wave inverter , first stage use push-pull to convert 12v dc to 310v dc with feedback , so no problem with load in dc side range measure between 35 to 315

the problem in full bridge mosfet using pic uC as signal , under 100w load , I get pure sinewave form and ac side 220v aprox. , but when add more load the signal convert to triangle not sine.

is sinewave inverter have feedback , all tutorial on internet explain how to make sine wave signal from lookup table , is there secret , no full detail can u find for puresine wave inveter
 

Is the DC voltage still steady 310-315 Vdc when you increase the load? Is there a filter on the full bridge output?
 

Is the DC voltage still steady 310-315 Vdc when you increase the load? Is there a filter on the full bridge output?
yes dc voltage steady, I'm using two indictor in both side of dc in series with 1.mf cap in parallel, also used 33nf cap from both ac output to ground of 320

on 100w load signal form same as mains ac

- - - Updated - - -

sorry inductors used in AC side
 

Can you post a schematic and images of the sinusoidal and triangle like signals? Is it possible that sine wave reference or carrier changes when the load changes? If you update the sine wave reference on timer interrupt (without modification to sample value) and carrier is made with a pwm module, then modulation is independent of load. If you update the reference on main loop it is possible that reference is not regularly updated anymore. If amplitude of the reference for some reason is bigger than amplitude of the carrier, resulting waveform can start to look more like triangle wave than a sine wave. Based on simulation there should be a flat top so not exactly a triangle wave.
 

thanks tsan500

I have not feedback or reference to generated spwm signal, so I asked if sinewave inverter need feedback from H-bridge to recalculate signal value according to feedback voltage?

I used smart sine lookup table generation that created by Tahmid and use PIC uC with static values(160 value to generate 16khz signal)

as u said signal "in load" look more like triangle wave ? is this normal

Filter circuit attached
 

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  • Capture.JPG
    Capture.JPG
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Last edited:

thanks tsan500

I have not feedback or reference to generated spwm signal, so I asked if sinewave inverter need feedback from H-bridge to recalculate signal value according to feedback voltage?

There is no need for feedback to have sinusoidal output with higher load, although I simulated only with resistive load. I attached LTspice simulation file for 50 Hz output "sine wave inverter_bipolar_edaboard.txt" that I tried to make according to your configuration. It doesn't matter if the load is 100 ohm or 1000 ohm based on simulation. This means 487 w and 49 w loads. It is necessary to change file extension to .asc to use with Ltspice. Put file "160_samples_for_16kHz.txt" to the same directory with the simulation file. It contains the sine wave reference samples. Keep the .txt extension on the sample file. RMS voltage change is also small even when there is no feedback. RMS voltage changes 1,5 volt between 100 ohm and 1000 ohm load.

Perhaps your output/filter inductors are saturating. Can you test with inductors having bigger current rating?
 

Attachments

  • 160_samples_for_16kHz.txt
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  • sine wave inverter_bipolar_edaboard.txt
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I think simulation not always reflex the reality, after a lot of search I found PID feedback , need more info about it
 

As you can see in video after changing SPWM pulse in software , I got this sinewave

https://www.youtube.com/watch?v=N6gpbmJ0XZw

the problem is AC voltage is low you can see it's 204v although DC voltage 330v

how can increase amplitude , the duty cycle in degree 90 of since wave about 95% ,no change to increase it
 

What do you mean duty cycle in 90 degree? Usually modulation index is used to describe the ratio of amplitude of sine reference vs. amplitude of carrier. On your case, peak of the sine reference should equal peak of the carrier. It is possible, that now amplitude of the sine ref samples are too low, therefore producing lower output voltage.
 

Hi,

Duty cycle:
If dury cycle is less than 100% then you never can get full peak output voltage = bus voltage.
If duty cycle is 100% then your driving circuit (depending on what circuit you use) may fail because of bootstrap circuit failure.

There will be voltage drop....in capacitors, traces, switching circuit, filters....

Klaus
 

What do you mean duty cycle in 90 degree? Usually modulation index is used to describe the ratio of amplitude of sine reference vs. amplitude of carrier. On your case, peak of the sine reference should equal peak of the carrier. It is possible, that now amplitude of the sine ref samples are too low, therefore producing lower output voltage.

I used this lookup table 0, 13, 26, 39, 52, 64, 77, 89, 101, 112, 124, 135, 145, 155, 165, 174, 183, 191, 199, 206, 212, 218, 223, 227, 231, 234, 237, 239, 240

it's from 0 to degree 90 of sine signal , each value will repeat 3 times

as you see last value is 240 , and timer in uC totally overflow in 255 ,I can't increase this value because uC need processing time,

it's about 94% of duty cycle of the highest value, so I think there isn't problem in software
 

I used this lookup table 0, 13, 26, 39, 52, 64, 77, 89, 101, 112, 124, 135, 145, 155, 165, 174, 183, 191, 199, 206, 212, 218, 223, 227, 231, 234, 237, 239, 240

it's from 0 to degree 90 of sine signal , each value will repeat 3 times

as you see last value is 240 , and timer in uC totally overflow in 255 ,I can't increase this value because uC need processing time,

it's about 94% of duty cycle of the highest value, so I think there isn't problem in software
I calculated quickly on excel and got close the same values for sine ref table. Last value of 240 means that amplitude of the generated wave is 240/255 of the amplitude that would result when the last value is 255. You can compensate for it by increasing the DC voltage. Also other losses can be compensated by increasing DC voltage. If losses are too high, then change the components or circuit.

Hi,

Duty cycle:
If dury cycle is less than 100% then you never can get full peak output voltage = bus voltage.
If duty cycle is 100% then your driving circuit (depending on what circuit you use) may fail because of bootstrap circuit failure.

There will be voltage drop....in capacitors, traces, switching circuit, filters....

Klaus
To avoid bootstrap circuit failure, bootstrap capacitor can be made bigger than basic calculation using carrier frequency gives. The reference is sine wave so it will be on maximum only limited time. Limiting duty cycle and having higher DC voltage seems better option, though.

Voltage drop on filter will be low with low load. On simulation I attached earlier, voltage drop was 1,5 V with 500W load. It is also possible to calculate voltage drop on the inductor using basic formula XL=2*pi*f*L, 50 Hz frequency and current according to load. I got similar result (1,36 V) as on simulation when using 2mH inductor (2x1 mH given earlier on schematic) and 500W load. If BJT's or IGBT's are used on the H-bridge, then there can be for example 4V voltage drop on no load too.
 

Hi,

I calculated quickly on excel and got close the same values for sine ref table.
Me too.
I found out (or at least it seems so) your last item is not 90°, but 86.9°.
With 29 items per 90°, so step sizte is about 3.10°.
I wonder why the 90° is missing. For 29 items per 90° you need 30 items in total.

But maybe I´m wrong.

Klaus
 

Hi,


Me too.
I found out (or at least it seems so) your last item is not 90°, but 86.9°.
With 29 items per 90°, so step sizte is about 3.10°.
I wonder why the 90° is missing. For 29 items per 90° you need 30 items in total.

But maybe I´m wrong.

Klaus
You are correct. I merely compared amplitude values. I checked the numbers and plotted the samples over half cycle but the problem does not appear with half cycle. The issue comes when only 90 degrees of the cycle are used. I have used on my project full cycle which made table indexing really easy but perhaps here there is only very limited memory. I made a picture showing principle of 90 degrees sampling I think would work. It is only for 5 samples for 90 degrees to show the principle. The 90 degrees sample is used only once per half cycle.

sine wave reference sample example for 90 degrees.png

On the youtube video given, the scope shows 50,76 Hz frequency. There is most likely 90 degrees samples missing from the sequence, so there is only 116 samples but there should be 118 samples. 50*118/116=50,86 which agrees well with scope reading.
 

Thanks KlausST and tsan500

I modified uC software with new lookup table and increased degree 90 to value 249

so from 0 degree to 90 will be

0, 13, 25, 38, 50, 62, 75, 86, 98, 110, 121, 132, 142, 152, 162, 172, 180, 189, 197, 204, 211, 218, 224, 229, 234, 238, 241, 244, 246, 248, 249,249
then will reverse 249,248,268,246,.......

there will be three 249 value in center

I will try on PCB today and give you feedback
 

Here is an image showing situation where there are sine wave reference samples for full cycle. This requires more memory for the table but code on the timer interrupt is easy. On some micro controllers it is possible to put look up tables to program memory. If that is possible then look up table doesn't take ram.

sine wave reference 17 samples per cycle.png
 

now I got 220 volt at 150w load with dc bus 320

So I will increase dc bus to steady 340volt dc , then use feedback for AC to change duty cycle to be steady in 220v A
C

note: I saved in memory 60 table for different cycle amplitude ,and will use one depending on AC feedback Value

Is I'm in correct way?
 

I think I'm in finishing of 80% of my inverter, the problem here is with no load I got square wave form , is this normal?
 

I think I'm in finishing of 80% of my inverter, the problem here is with no load I got square wave form , is this normal?

It is not normal. Is it a 50 Hz square wave? You have not shown a schematic so I can only assume what you possible have, and I assume that your full bridge is a typical H-bridge and you use bipolar modulation. Perhaps something wrong with circuit connections. If you increase the load from zero, at what load the output waveform becomes sinusoidal? Is the load resistive?
 

yes it's 50hz square ,
it's typical H-Bridge with unipolar PWM,if I increase the load by resistive or inductive it's convert to sinewave
 

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