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Sine Wave Inverter Design Help

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Is this circuit correct ? Can I build it ? MOSFET will be STE40NK90ZD.

I will be using Full Bridge Output ( 4 signals) from PIC to drive the IR2184 and MOSFETs.

See attached simulation. The attached image shows the signals.

I will be adding more MOSFETs in parallel to obtain more VA ratings for the Inverter. I will use another PIC to generate PWM for switching transformer. PWM frequency for switching transformer will be 10 KHz.

A 12V to 75V DC will be input to switching transformer from a 140 Ah APC Battery.
 

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  • To Edaboard.rar
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Please see the attached circuit and tell me how UF5408 can be used ?

Actually I will not be using UC3843. Instead I will use PIC to provide 50 KHz PWM signal to T1 Mosfet. I will use 100V
4A Logic gate Mosfet.

I am designing 5 KW Inverter. If current drawn by Inverter load is 40A then Tr1 secondary has to provide 40A from say a 48V 140Ah battery (12V batteries in series).

If 40A flows through UF5408 then how can it handle ? UF5408 Forward Current max is 3A.

In the circuit why is D2 and D3 used ? Should I use Tr2 also ? If yes, why ?

Edit:

My switching transformer calculation referring

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

https://tahmidmc.blogspot.in/2012/11/feedback-in-sine-wave-inverter-pic16f.html


Npri = (48 * 10^8) / (4 * 50000 * 1500 * 1.25)

Npri = 12.8 (approx 13)

Voltage ratio secondary:primary = 330:(48 * 0.98) = 7

Nsec = N * Npri = 7 * 13 = 91 turns

Na = Nsec/Naux = Vsec / (Vaux + vd)

two ultafast rectifiers used.

Na = 310V / (18 + 1.4) = 310 / 19.4 = 15.9 (approx 16)

Nsec / Naux = Na

Naux = Nsec / Na = 91 turns / 16 = 5.6875 = approx 6 turns

The Inverter will be designed for 5 KW max but max load will not exceed 2.5 KW. If 2.5 KW load is applied then to handle the load current what guage wire I have to use for the switching transformer secondary ?

P = VI

I = P / V = 2.5 KW / 230 = 10.8 approx 11 A rms

Vpk = 1.4142 * 11 = 15.55 = approx 16 A

16 A current will flow through secondary of switching transformer. Right ? What guage copper wire I have to use for the secondary ?

16A * 2 = 32 A = 14 AWG ?

https://www.powerstream.com/Wire_Size.htm
 

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I have generated SPWM signal for the Mosfet curcuit.
 

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  • psi.png
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Here is the updated circuit in pdf format. What should be the I(forward) of D1 and D9 in the circuit ?

I have referred this circuit partly for DC-DC converter circuit for my Inverter.

http://danyk.cz/menic230_6_en.html

Why is he using UF5408 (I(forward = 3A) at the output of switching transformer ?Actually He has designed a 150W Inverter.

So, P = VI, I = 150 / 230 = 0.65 A

Irms = 0.65 A

Ipk = 1.4142 * 0.65 = 0.922A (approx 1A)

So, UF5408 can handle this 1A.

But in my case

Ipk will be 31A and so I have to choose D1 and D9 which can handle say 40A.

Is my diode selection good ? I have used APT40DQ60B.


I still have to add fast charging circuit (battery charging).
 

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  • Pure Sinewave Inverter.PDF
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  • SPWM signals.png
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I will be using this fan for Mosfet cooling.

**broken link removed**

There will be 4 fans. One fan for 3 Mosfets on heatsink.

The current consumtion of one fan is 0.18A and so total current drawn by 4 fans will be 0.18A * 4 = 0.72A and so LM7812 (1.5A) will be sufficient for my needs. 12V supply is used only for fans. Fan has 3 pins. 12V, GND and PWM. I want to know can I drive the 4 PWM pins of the fans directly from PIC PWm out pin ?

This is the updated circuit. I still have to add current sense for overload protection and fast battery charging circuit.

As 12V is only used for fans can I go with LM7812 or should I go with LM2576-12 ?
 

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  • Pure Sinewave Inverter.PDF
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Discard all previous circuits in this thread. The circuit is getting finalized. Here is the latest circuit.

Use this circuit for future discussion.

My max current will be 31A.

With my knowledge of electronics I found that I have chosen proper diodes for D6 and D7 but the datasheet doesn't mention about max power it can handle ?

How to find max power it can dissipate ?

Vf(max) @ 40A is 2.4V

Pd = Vf * If = 2.4V * 31A = 74.4W

How to dissipate this much power ? The diode is APT40DQ60B (TO245 package, two terminal)

I know I have to mount it on heatsink but do I also need a fan to cool it ?


The attached image shows mosfet placement on heatsink and also fan. 140 mm High Static Pressure Corsair fan will be used.

There will be 4 heatsinks and each heatsink will have 3 mosfets and 1 fan.

Datasheets attached.
 

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  • PSI.PDF
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  • Mosfet Placement.png
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  • APT40DQ60B_S(G)_C.pdf
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  • en.CD00003691.pdf
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The power rating of the DC/DC converter doesn't seem to fit the output power. Single transistor forward converter is a bad idea, I think.
 

@FvM

What do you mean by that ? What modifications I have to do ?

I used parallel diodes to reduce power dissipation but out of these two circuits which one is better ?

I am taking into consideration what happens if one or more parallel diodes in the DC-DC converter or parallel Mosfets at the output fails. The other diodes and mosfets should be able to dissipate the power.

How to convert Power dissipated into temperature (degree C). Will the LM35DT be able to measure the temperatures produced in my circuit ?

Added one more LM35DT and fan for diode(s) cooling.

How about using this diode for DC-DC converter ?

STTH100W06C

This has low Vf (1.65V) and hence less power has to be dissipated in each parallel diodes. I am not able to find a better ultrfast recovery diode with low Vf. If somebody finds better diode then mention its part name.
 

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  • PSI Ver2.PDF
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Hi,

Btw: paralleling diodes doesn't reduce the dissipated power significantely. It just spreads the dissipated power to several diodes.

Klaus
 

@FvM

How ?

The current at the output of the switching transformer divides through the parallel diodes. Right ? Then why not power dissipated per diode reduce ? Please explain clearly. I don't want to see flying diodes and mosfets.

My max current is 31A. With 6 diodes in parallel it divides to 31/6 = 5.17A per diode path.

So, Pd(per diode) = Vf * If = 2.4 * 5.17A = 12.4W. With heatsink and a fan this will be kept cool.

Can somebody point to to a better ultrafast soft recovery disode (600 V 40A or more) from this link preferably in TO247 package.

**broken link removed**

Btw: paralleling diodes doesn't reduce the dissipated power significantely. It just spreads the dissipated power to several diodes.

Are you telling in terms of efficiency ? That is total power dissipated in parallel diode combination is same as that dissipated in single diode path ?

How to increase efficiency ?

I think this diode is better. Vf is less.

Please tell me which of the 3 diodes I have mentioned is suitable for my purpose ? DC-DC Converter PWM frequency is 100 KHz. Diodes should be able to switch at this speed.
 

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  • en.DM00063615.pdf
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I think the rectifier selection is a minor problem. But nobody makes a single-ended forward converter for 2 or even 5 kW.
 

Should I use a two transistor forward converter as shown in page 19 of attached document. Will this be enough ?

What about the components I have used in DC-DC converter (diodes) ? Are they able to switch at 100 KHz ?

How to provide PWm to two transistor forward converter ? Two PWms with inverted polarity (complementary PWMs) ?

Now I am reading this link and also attached documents. I need sometime to make changes to the DC-DC Converter circuit.

http://schmidt-walter-schaltnetzteile.de/smps_e/hdw_hilfe_e.html
 

Attachments

  • Two Switch Forward Converter.rar
    1.5 MB · Views: 148
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Two-switch forward converter might be an option, it still requires a primary switch transistor with higher voltage and possibly higher current rating than IRLU3110Z.

Input current ripple and transformer core utilization however suggests a bridge converter, either hard switching or resonant converter.
 

Hi,

Are you telling in terms of efficiency ? That is total power dissipated in parallel diode combination is same as that dissipated in single diode path ?

How to increase efficiency ?
You can increase efficiency by lowering voltage drop.
But paralleling doesn't reduce the voltage drop significantely, therefore the overall power dissipation is almost the same.

--> Use diodes with low voltage drop
Or use synchronous rectifiers. Where diodes are replaced with mosfets. Mosfets have low voltage drop when switched ON. You need extra logic to switch the Mosfet ON and OFF.

Klaus
 

I have to design a bridge converter (dc-dc converter) like the ones mentioned in the attached documents for better efficiency of the dc-dc converter stage ?

What should be the input DC voltage for the DC-DC-Converter (Bridge Converter) ?

I can't put more than four 12V batteries.

@KluasSt

I am still finding if there are any low Vf Ultrafast recovery diodes.

- - - Updated - - -

Edit:

I think this diode is suitable. What do you people say ? It has Vf = 1.5V and I will put 6 diodes in parallel.

Pd = Vf * If = 1.5 * 5.167 = 7.75W (with heatsink and fan).
 

Attachments

  • Bridge Converter for DC-DC Converter.rar
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  • vs-epu6006l-m3.pdf
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Can I provide 15V to Vcc pin of IR2184 ? Actually In my Inverter only IR2184 needs 12V. I am designing a 500W Inverter for someone and only one 12V battery will be used for DC_DC converter input. So, this battery output will vary from 10.4V to 15V. So, can I connect this output of battery directly to Vcc pin of IR2184. Will it work ? Datasheet is not clear. It mentions for Vcc (min -0.3V and max 25V). Can somebody explain this ?
 
How to calculate current of primary winding of DC-DC Converter switching transformer ?

I am designing 500W Inverter. I will be using Single Transistor forward converter.

My DC-DC converter PWM frequency will be 50 KHz.

For 500W Inverter

ETD39 ferrite core

Bmax = 1500

Ac = 1.25

Battery Voltage = 12V (10.4 to 15V)

Npri = (12 * 10^8) / (4 * 50000 * 1500 * 1.25)

Npri = 3.2 turns ~ 3 turns

98% duty

Voltage ratio secondary:primary = 330:(12 * 0.98) = 11.76

Nsec = N * Npri = 11.76 * 3 = 35.28 turns ~ 35 turns

Na = Nsec/Naux = Vsec / (Vaux + vd)

vd = 1.5 + 1.5 = 3.0V

Na = 310V / (18 + 3.0) = 310 / 21 = 14.76V

Nsec / Naux = Na

Naux = Nsec / Na = 35 turns / 14.76 = 2.37 turns ~ 3 turns

Isec(max) = (500W / 220V) * 1.4142 = 3.21A

How to calculate Ipri ?

Does the formula

Np/Ns = Vp/Vs = Is/Ip hold good for switching transformer current calculation ?

If yes, then

Ip = Ns/Np * Is = (35/3)*3.21A = 37.45A

Is this correct ?

Why this is not correct ?

Ip = Vs/Vp * Is = (325V/12V) * 3.12A = 84.5A

- - - Updated - - -

Edit:

The 5KW Inverter I am designing for home use. I am also designing 500W Inverter for someone else. It uses single transistor forward converter.

For 500W Inverter Single 12V Battery will be used.

I have seen many Inverter circuits which use bulky 12V to 230V transformer connected to H-Bridge. In such designs when mains is present the high side mosfets are turned OFF and Low side mosfets are made to Oscillate for battery charging but as I am design doesn't have bulky transformer I don't have that option for battery charging.

I need a fast battery charger for single 12V battery. I designed this circuit. Is it good ? A 230V Primary, 18V 15A secondary transformer is used with 35A rated bridge rectifier. PWM is used for Mosfet. 8A charging current will be used.

Is there any other method for fast battery charging without using transformer ?
 

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  • fast battery charger.png
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@FvM and other Power electronics experts

Is there a way I can do a fast charging of a 100Ah to 200 Ah 12V battery of a 500Watts UPS which uses DC-DC Converter without using a bulky transformer for battery charging ?

I designed this but it needs a separate bulky 230/12V 15A transformer.

In DC-DC Converter transformer please tell me how to calculate primary current.

primary voltage 12V
secondary voltage 325V (VBUS)
secondary current max = 500W/220 * 1.414 = 3.21A

- - - Updated - - -

Can I build a 220V AC input and 14.3V 15A DC Ouput SMPS and use it for battery charging ?

I am referring these.

https://ww1.microchip.com/downloads/en/DeviceDoc/dsPICSMPS AC_DC Users Guide.pdf

https://tahmidmc.blogspot.in/2013/02/ferrite-transformer-turns-calculation_22.html

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

https://tahmidmc.blogspot.in/2012/12/learning-smps-hard-way.html

https://tahmidmc.blogspot.in/2013/03/output-inductance-calculation-for-smps.html
 

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  • fast battery charger.png
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I finally decided to go with SMPS for battery charging because I don't want to use a bulky transformer.

The batteries that will be used will have Ah ratings from 100Ah to 200Ah. Batteries will be 12V. So, the charging currents will be 10A to 20A. I need to design a AC-DC SMPS Power supply for this. How should I proceed ? I am referring to the attached document. I will be using full bridge topology.

What should be my SMPS output for charging 12V batteriy with load (battery) connected ? 15V ?

For what current rating Iout I have to design the SMPS ? 30A ?

This SMPS will be a part of the Inverter. As both Inverter and SMPS will be using H-Bridge and FET drivers can I use the same mosfets and fet drivers (and also the PIC) for SMPS input side circuit and Inverter output side circuit ?
 

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  • 19.pdf
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I am referring this

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

for 500VA Inverter's DC-DC Converter.

He has calculated 3 turns for primary and then used 3 + 3 turns for primary. Why ? I will be using single mosfet forward converter. Should I use just 3 turns for the primary. How to calculate the primary and secondary currents ? For 500VA Inverter I will be using one 12V 140Ah battery.

He mentions that push-pull and full-bridge forward converters are used and push-pull type needs center tap in primary winding and no. of turns in primary will double. I will be using single transistor forward converter and so there is no need for center tap but what about the no. of turns ? Is it same as that for full-bridge forward converter ?
 

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