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Sine Wave Inverter Design Help

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Why do we actually need a High-Side and Low-Side Driver for driving N-Mosfet H-Bridge circuit ? Whay cant we just generate two SPWM signals which are complementary to each other and input SPWM1 to Q1 & Q3 (diagonally opposite mosfets) and SPWM2 to the other two diagonally opposite mosfets (Q2 & Q4) ?
 

1. The high side transistors need floating gate drive voltage, on top of the bridge center voltage
2. SPWM preferably uses three-level pwm scheme which includes other switch states than only diagonally on. The scheme is discussed in "pwm techniques final" linked in your previous post.
 

Let me ask one or two questions at a time because I am not getting answers. Please quote the question while answering.

This is what I want to design.

A 230 V AC 50 Hz 5.25 KW Transformerless Pure Sine Wave Inverter using two PIC Microcontrollers.

PIC18F26K22 will generate SPWM signals for the Inverter and PIC18F46K22 will generate PWM for the DC-DC Converter.

I will use N-Chall MOSFETs based H-Bridge Circuit for the output.

The Quality of Inverter should be very good.

Now what PWM frequencies can I use for the SPWM and PWM ?

For PWM what duty I can use ?

SPWM duty will vary from 0 to x. What should be x so that it eliminates dead time ?
 

@FvM

I am designing a 2 KW Inverter now.

P = VI

Current = P/V = 2000 / 230 = 8.69 approx 9 A

I will double it it is 18 A

So, Can I use 600V 30A Bridge Rectifier to get the 325 V DC VBUS from 230 V AC mains ?
 

Attachments

  • Pure Sine Wave Inverter.rar
    25.1 KB · Views: 171

This is the calculation for the code in post #26. I am not getting sine wave. Please tell me what is wrong in the calculation.

The calculation for the ISR code is like this (according to thamid's method)

My PWM frequency is 10 KHz and hence Timer2 Interrupt is for 100 us.

I have 100 entries for 180 degree of Sine Wave in the Sine table.

100 us * 100 (no. of sine table entries) = 10 ms

1/(10ms * 2) = 50 Hz

SET_FREQ = 65536 / (100 * 1) = 655.0 (approx)

If 655 is called 100 times then time will be 10 ms and TBL_POINTER_NEW will be 65500 and on 101th interrupt it will overflow and PWM direction will change.
 

This is the calculation for the code in post #26. I am not getting sine wave. Please tell me what is wrong in the calculation.

The calculation for the ISR code is like this (according to thamid's method)

My PWM frequency is 10 KHz and hence Timer2 Interrupt is for 100 us.

I have 100 entries for 180 degree of Sine Wave in the Sine table.

100 us * 100 (no. of sine table entries) = 10 ms

1/(10ms * 2) = 50 Hz

SET_FREQ = 65536 / (100 * 1) = 655.0 (approx)

If 655 is called 100 times then time will be 10 ms and TBL_POINTER_NEW will be 65500 and on 101th interrupt it will overflow and PWM direction will change.

This much appears Ok.

100 points for half-cycle is a lot. 16 or 32 points would have been sufficient. The load would have smoothened any rough edges. I cannot imagine a case where some minor jaggedness would have mattered.

But what is not working?
 

In PIC18F46K22 which PWMs are CCP and which PWMs are ECCP ?

There are 5 PWM channels. PWM1 - PWM5

Does that mean PWM1 and PWM2 are CCP and PWM3 - PWM5 are ECCP ?

PIC18F46K22 has 2 CCP and 3 ECCP modules.

- - - Updated - - -

I have generated Sine wave using CCP module. Now I have a question.

I am referring this.

https://tahmidmc.blogspot.in/2012/12/ferrite-transformer-turns-calculation.html

My max load is 2000 W and so P = VI and I = P/V

I = 2000 W / 230 V = 8.7 A

I double it so it is approx 20A.

So, should I get a SMPS transformer manufactured which can handle 20A in its secondary. The secondary will be giving 325V DC.
 

My max load is 2000 W and so P = VI and I = P/V

I = 2000 W / 230 V = 8.7 A

I double it so it is approx 20A.

So, should I get a SMPS transformer manufactured which can handle 20A in its secondary. The secondary will be giving 325V DC.

So finally you land up with 20A at 325V that is close to 6 kW. Do not overrate randomly; If you need 2 kW output, target a margin of 10-20% and compute accordingly.

When you will come to selection of available cores (commercially available), just select the one with the next size (that will give enough margin for power).
 

No, 20A will never flow through the transformer secondary. 8.7 A is the max current that will flow but to make the devices work at 50% of the max rating I made current as approx 20A and if I choose guage of secondary winding copper wire to handle 20A, I want to know will it work ?

So, 8.7A approx 10A will flow through the Inverter when on full load. It will be never run at full load. There will be MCB and RCBO to trip the Inverter if current is more than 10A.

I just want to know should I choose the copper guage of the wire for secondary to handle 20A ?

- - - Updated - - -

No, 20A will never flow through the transformer secondary. 8.7 A is the max current that will flow but to make the devices work at 50% of the max rating I made current as approx 20A and if I choose guage of secondary winding copper wire to handle 20A, I want to know will it work ?

So, 8.7A approx 10A will flow through the Inverter when on full load. It will be never run at full load. There will be MCB and RCBO to trip the Inverter if current is more than 10A.

I just want to know should I choose the copper guage of the wire for secondary to handle 20A ?

- - - Updated - - -

I am referring this.

https://tahmidmc.blogspot.in/2013/02/demystifying-use-of-table-pointer-in.html

I am not getting Sine Wave. What is the problem ? I am attaching the project files and also Proteus Simulation.

I also tried sine table with peak value 1023 (10 bit duty) using your sine table generator and also this.

https://www.daycounter.com/Calculators/Sine-Generator-Calculator2.phtml

but I am not getting proper sine waves.
 

Attachments

  • Pure Sine Wave Inverter.rar
    36.1 KB · Views: 110
  • Schematic and Simulation Result.png
    Schematic and Simulation Result.png
    120.2 KB · Views: 133
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For selecting wire sizes, please look up AWG sizes and current capacity from standard tables. These values are for free air and you may consider upgrading one size. I see that comes to about size 9 (which is a fat wire) but you may consider using litz wire that may be easier to wind. It will also help with skin effect.

I do not know why you are not getting sine wave: I cannot debug your data but first thing first: just cut and paste the values in a worksheet column and plot and see whether the plot looks like a sine wave. That will confirm the first step: your numbers are right.
 

Here is the final code that I have made. It gives sine wave. I don't know why but the sine table has to be 8 bit (pwm duty) instead of 10 bit (pwm duty).

ECCP supports 10 bit pwm duty but it is not working and gives incorrect sine wave.

The code inside ISR which assigns duty is for 10 bit duty but if 10 bit duty is assigned (commented sine table) then wrong since wave appears and if 8 bit duty is assigned with the same code then (which is actually incorrect) then I get sine wave.

Page No. 202 of Rev F datasheet of PIC18F46/26K22 tells how to assign 10 bit duty.
 

Attachments

  • Pure Sine Wave Inverter - PIC18F26K22.rar
    36.7 KB · Views: 124
  • Schematic and Simulation Result.png
    Schematic and Simulation Result.png
    119.8 KB · Views: 129

Finally I am making a 2KW inverter. I still have to do the calculation for the SMPS Ferrite Transformer.

I am referring this.

https://tahmidmc.blogspot.in/2013/01/using-high-low-side-driver-ir2110-with.html

I will be using tantalum Capacitor for the bootstrap circuit. I will be using IRS2186 for High-side and Low-side driver and STY60NM60 N-Channel Mosfets for the Full-Bridge.

For bootstrap diode I will be using UF4007.

My VBUS = 325V DC

Output required is 230 V AC Constant 50 Hz, 1 phase.

I need help with these.

Please refer to MQP_D_1_2.pdf and tell me if these are correct.

Qgs = 266nC (STYM60NM60)
Qls = 5nC (given in an-978 application note)
VCC = 12V (from Battery to VCC pin of IRS2186)
Vf = 1.7V (UF4007 Fairchild)

Where to find these ?

Vls = ?
f = ?
Iqbs = ?
Icbs(leak) = ?

In the equation in page 28 of the attached document what is VMin ?

My PWM frequency is 12.821 KHz and I am using PIC18F45K22 to get the Full-Bridge Outputs P1A, P1B, P1C and P1D for the Full-Bridge Mosfet Circuits driven by 2x IRS2186.


In the MQP_D_1_2.pdf page no. 29

Vls = Voltage drop across low side FET = 1.5V for his Mosfet which is IRFB20N50KPBf (datasheet attached)

Is it Body Diode Voltage (1.5V mentioned in page no. 2 of datasheet) ?

In STY60NM60 datasheet page no 3 it is mentioned Forward On Voltage = 1.5V. Should I take it ?
 

Attachments

  • Datasheets and Application Notes.rar
    6.7 MB · Views: 121
Last edited:

@FvM

I am referring an-978 Rev D PDF page no. 5.

Q g = Gate charge of high-side FET
f = frequency of operation
I Cbs (leak) = bootstrap capacitor leakage current
I qbs (max) = Maximum V BS quiescent current
V CC = Logic section voltage source
V f = Forward voltage drop across the bootstrap diode
V LS = Voltage drop across the low-side FET or load
V Min = Minimum voltage between V B and V S .
Q ls = level shift charge required per cycle (typically 5 nC for 500 V/600 V MGDs and 20 nC for
1200 V MGDs)


Q ls = level shift charge required per cycle (typically 5 nC for 500 V/600 V MGDs and 20 nC for
1200 V MGDs)

For IRS2186 High Side Floating Absolute Voltage is 620V max. So, I should take Q ls as 5 nC ?

I Cbs (leak) = bootstrap capacitor leakage current (it is 0 as I am using Tantalum Capacitor)


Q g = Gate charge of high-side FET (mentioned in Previous post)

f = ? frequency of PWM ?

Should I take f as 12.821 KHz as it is my PWM frequency ?

V f = 1.7 for UF4007

How to fine V Min ?
 

@FvM

Ok. I am reading the datasheet of IRS2186. I have a few more questions. How to find these

I qbs (max) = Maximum V BS quiescent current
V LS = Voltage drop across the low-side FET or load
Q ls = level shift charge required per cycle (typically 5 nC for 500 V/600 V MGDs and 20 nC for
1200 V MGDs)

As I am using IRS2186 how to know whether it is 500/600V MGD or 1200V MGD ?

Where to find V ls (Voltage Drop across low side FET) ?

Where to find I qbs.

I read the an-978.pdf and also the datasheets but not able to find these.

Mine is a 2 KW 230V 50 Hz 1 phase Inverter. All components will work at less than 50% of their max ratings.
VBUS is 325 V DC.

How to calculate the value of capacitor connected across the secondary winding of switching transformer ? Is there any equation or can I blindly use 1000 uF 400V electrolytic capacitor ?
 

If you take the operating frequency as 16K (more accurate 16384=2**14), the hardware and software would be happy. This is the only parameter you can freely choose in the beginning.
 

@ c_mitra

I didn't understand. How the hardware and software would be happy ?

Where can I get information about ferrite switching transformer ? I want to use a easily available switching transformer for 2 KW inverter. I will then change the PWM frequency according to the transformer.
 


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