Need help for Pure Sine Wave Inverter

[Tahmid]

I don't really see much of a problem. If you use a MOSFET rated for the required current, ie, you use one single MOSFET, it will be very expensive and it is not as widely available. Also, the larger MOSFETs have larger Ciss. Either way, paralleling or using a single large MOSFET, Ciss for such MOSFET(s) handling this large current will naturally be very high. That is why you must use drivers to drive them. These drivers must be able to handle the required gate current.

But by paralleling common MOSFETs such as the IRF3205, you get a cost-effective solution with commonly available parts, whilst getting other benefits such as reduced RDS(on), greater current handling capacity, etc. MOSFETs can also be easily paralleled as they have a negative temperature coefficient, so there is less chance of encountering thermal runaway.

In all inverters in Bangladesh, India and Pakistan, for large capacities, common MOSFETs are paralleled as mentioned. And they yield no problems and run for years.

In exchange for the higher gate drive current requirement, you gain other benefits.

Hope this helps.
Tahmid.

---------- Post added at 00:08 ---------- Previous post was at 00:03 ----------

For example, let's take the IXTK250N10 and the IRF3205 MOSFETs. The IXTK250N10 (rated for 250A, RDSon of 5mΩ) has a Ciss of about 13nF; IRF3205 (rated for 110A, RDSon of 8mΩ) has a Ciss of about 3nF. If we parallel four IRF3205s, the total Ciss is about 12nF (still less than that of the IXTK250N10). The RDS(on) is 2mΩ (less than half that of IXTK250N10). Max current capacity is about 440A (little less than double that of the IXTK250N10). So, we have less than half the RDS(on), almost double the max current capacity and yet less Ciss than the single MOSFET. The only downside is that we're using 4 MOSFETs, instead of 1. But the IXTK250N10 is much more difficult to find than the IRF3205 and probably more expensive.

Hope this helps.
Tahmid.

 

[goldsmith]

Hi again Dear Tahmid , i used a simple IRFP450LC at the frequency up to 10 MHZ , without any problem . and i don't know how much are the price of these mosfets in your country , but here in my country those are cheap .
Regards
Goldsmith

 

[Tahmid]

I'm sorry, but I'm not getting what you're saying. I've told him to use the IRFP460 for the DC-AC conversion. It has Ciss much higher than IRFP450, but with greater current.

I was talking about the DC-DC part, where with 11V input, the current may be 114A (assuming 80% efficiency). That's where I was talking of paralleling MOSFETs. If IRFP450LC (a 500V MOSFET) were to be used here, (for 114A peak maybe more than 220A), which would require at least 16 MOSFETs if each was to be stressed to its maximum capacity. But for IRF3205, even 4 may be used (more for lower losses). In that case, even the advantage of reduced Ciss is lost, since 16 IRFP450LC will have Ciss of about 35nF compared to about 13nF for the 4 IRF3205s. So, I don't get what you're trying to say about using one MOSFET (IRFP450LC).

The one MOSFET may be used in the DC-AC part, but that wasn't where I was talking of paralleling MOSFETs - there, too, I talked of using one MOSFET.

Hope this helps.
Tahmid.

 

[goldsmith]

Tahmid ,
My mean was , that if , he can buy IRFP 450LC or IRFP 460 LC , it will be better . but i don't think at this low frequency there is any problem . ( as he mentioned 64 KHZ ) . and it is a low frequency really .
Respect
Goldsmith

 

[ROCKET SCIENTIST]

Such a nice thread.. Very helpful too.I've been following it.. Well, then, what will be the best gate driver IC and MOSFET to use?? I'm trying to make a 650VA 230V 50Hz inverter..

 

[goldsmith]

Dear ROCKET SCIENTIST
Hi
But how much is your carrier frequency ? do you want to use iron core transformer or ferrite core ?
Best Wishes
Goldsmith

 

[babatundeawe]

i guess at 50hz osscilator frequency an iron core transformer will do well

 

[goldsmith]

Dear babatundeawe
Hi
The iron core transformers aren't economical .
Best Wishes
Goldsmith

 

[babatundeawe]

goldsmith actually i meant the laminated steel core u know the e and i stuff. i think they are ok at such frequency. but if not please do enlighten me more. thanks

 

[goldsmith]

Dear babatundeawe
Hi
you can use SPWM method , and thus you'll need HF transformer with ferrite material for first stage and for latest stage a ferrite core inductor , just . and you can achieve very high powers with this method , simply , and with lowest price .
Best Wishes
Goldsmith

 

[babatundeawe]

thanks goldsmith. ok can u strike the difference between a ferrite torroid and a wound laminated steel torroid.

thanks

 

[goldsmith]

ferrite transformers are smaller lighter weight , and cheaper . and you'll be able to get higher powers with an small ferrite core .
( naturally if you increase the frequency ) .
But iron cores ( however EI or EE or L or T or the other types of that ) will be heavier and more expensive and at las they aren't economical . and if you increase the frequency of current through an iron core inductor or transformer it will be hot !
My mean is that the temperature of that will increase , ( did you hear about inductive melt systems ? ) so iron cores aren't appropriate choices for high frequencies .
Best Wishes
Goldsmith

 

[sam781]

@Goldsmith: Could you please describe a bit detail about ferrite core transformer and iron core transformer? e.g, core size, primary/secondary turns etc calculation w.r.t desired current ratings.

 

[goldsmith]

Dear Sam
Hi
The iron core transformer and inductor , will be heavier . and iron is a conductor but ferrite isn't conductor . and according to the magnetic physics , magnetic flux , at center of each insulator will be higher than each conductor ( about *Pi )
so , if we use ferrite core we can use from all sides of it because flux can move entire of it , easily . but it isn't possible about iron core .
So , iron cores are heavier and more expensive and larger and more dissipative . but ferrite transformers are more suitable . and you can trust them .
Best Wishes
Goldsmith

 

[samEEEf]

Let's take ROCKET SCIENTIST's requirement (650VA, 230V, 50Hz) as an example to learn more about designing an efficient Inverter.

For more clarification, I'm describing high level steps to make an Inverter after reading the valuable posts (specially Tahmid & Goldsmith) of this thread.

1. DC 12V to DC *400v(?) converter
2. Filtering the DC voltage to be used as input of the inverter
3. DC 400v to AC 230v Inverter
4. Filtering the inverter output voltage to get pure(!) 50Hz 230v AC

Stage-1: DC 12V to DC 400v converter
Let's chose full bridge converter with 50kHz modulation frequency and Ferrite core transformer
Major semiconductor devices -
a) IC SG3525
b) MOSFET IRF3205

Taking minimum voltage of battery as 11v, max. current drawn from the battery = 650/11=60Amp,
Now, adding extra 25% current to this value considering inverter efficiency as 80% the final value reaches 75Amp.

So, the transformer's primary current ratting should be 75Amps. Hence, the transformer ratting should be - primary side 12v, 75amp; secondary side 400v (?), 2.25Amp
*Q1: What should be the secondary voltage (here we used 400v) of the transformer and how can we calculate this value to get desired output of 230v AC?

We'll get 400v unfiltered DC here.

Stage-2: Filter
Please comment to make appropriate filter circuit with calculation...


Stage-3: DC 400v to AC 230v Inverter
Let's chose H-Bridge SPWM Inverter with 48kHz modulation frequency
Major semiconductor devices -
a) Any suitable Microcontroller chip capable of generating gate pulse for the MOSFETs.
b) MOSFET IRFP460

At the end of this stage we'll get 230v AC (unfiltered)

Stage-4: Filter
Please comment to make appropriate filter circuit with calculation to get 50Hz pure sine wave AC voltage...


Question to all experts: Is this okay? Please provide your valuable feedback for each and every important stages to complete it.

 

[masud58]

Simple design.

 

[ROCKET SCIENTIST]

thanx to all... This post is so helpful.. Wanna learn more details...

 

[samEEEf]

Expecting comments on my previous post (#35) specially from Mr. Tahmid & Mr. Goldsmith

 

[Tahmid]

Let's take ROCKET SCIENTIST's requirement (650VA, 230V, 50Hz) as an example to learn more about designing an efficient Inverter.

For more clarification, I'm describing high level steps to make an Inverter after reading the valuable posts (specially Tahmid & Goldsmith) of this thread.

1. DC 12V to DC *400v(?) converter
2. Filtering the DC voltage to be used as input of the inverter
3. DC 400v to AC 230v Inverter
4. Filtering the inverter output voltage to get pure(!) 50Hz 230v AC

Stage-1: DC 12V to DC 400v converter
Let's chose full bridge converter with 50kHz modulation frequency and Ferrite core transformer
Major semiconductor devices -
a) IC SG3525
b) MOSFET IRF3205

If you plan to use H-bridge, you need hi-lo side drivers for the H-bridge. So, add 2 drivers to the list of major semiconductor devices. The IR2110/IR2113 are very popular drivers. You can also use drivers such as L6385E, L6387, etc.

Taking minimum voltage of battery as 11v, max. current drawn from the battery = 650/11=60Amp,
Now, adding extra 25% current to this value considering inverter efficiency as 80% the final value reaches 75Amp.

So, the transformer's primary current ratting should be 75Amps. Hence, the transformer ratting should be - primary side 12v, 75amp; secondary side 400v (?), 2.25Amp
*Q1: What should be the secondary voltage (here we used 400v) of the transformer and how can we calculate this value to get desired output of 230v AC?
We'll get 400v unfiltered DC here.

The transformer's primary current rating should be 75A continuous average. Do keep in mind that the peak current is higher - about 2.4 times for full-bridge and push-pull converters. Keep this in mind when selecting MOSFETs.

We're making a sine wave inverter. So, for 230VAC RMS, we need a peak voltage of approximately 230* √2 ≈ 325VDC. Assuming 80% efficiency, we'd need a voltage of 325/0.8 ≈ 406V. This isn't going to be the exact voltage as the voltage will be varied according to the output voltage and load as well as battery voltage by the feedback section. This is for reference. So, at least 406VDC must be "available". So, 406VDC must be "available" when battery voltage drops to the minimum voltage, which you have taken as 11V. For 12V, then, it'd correspond to ≈ 443V. So, the transformer should be rated at 12V primary, ≈ 440V secondary. 400V may (or may not) be enough depending on the efficiency of the system. In reality, the actual voltage at the output will vary according to the PWM duty cycle set through the feedback section in order to maintain the 230VAC output.

After rectification through a bridge rectifier, you'll get the unfiltered high voltage DC.

Stage-2: Filter
Please comment to make appropriate filter circuit with calculation...

Use inductor-capacitor (LC) filter. One calculator is here:

Details:
**broken link removed**

You can get all required formulae in a book. Grab a book and read it. You can find a list here:

Stage-3: DC 400v to AC 230v Inverter
Let's chose H-Bridge SPWM Inverter with 48kHz modulation frequency
Major semiconductor devices -
a) Any suitable Microcontroller chip capable of generating gate pulse for the MOSFETs.
b) MOSFET IRFP460

At the end of this stage we'll get 230v AC (unfiltered)

Add 2 drivers to the list of major semiconductor devices. You'll need 2 hi-lo side drivers (eg IR2110/IR2112, L6385E/L6387, etc).

Stage-4: Filter
Please comment to make appropriate filter circuit with calculation to get 50Hz pure sine wave AC voltage...

You can find the formulae online. You can also use trial and error for getting it right. Either way, you'll need to use an LC (inductor-capacitor) filter.

Question to all experts: Is this okay? Please provide your valuable feedback for each and every important stages to complete it.

I've put forward my comments above. I hope they can now help give you a clearer picture on the topic.

Hope this helps.
Tahmid.

 

[samEEEf]

For rectifier, which diode need to be used here in 50kHz?