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H-Bridge power supply shorting

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I just noticed the waveforms in post #14. They are simply unsuitable for a bootstrap driver. And you can't operate a high side switch without a complementary low side switch. So the circuit is inappropriate as well.
 

FvM he is just using the high side of the left IR2113 and the low side of right IR2113, how can the bootstrap capacitor of the left chip be charged in that case?

Alex

---------- Post added at 10:30 ---------- Previous post was at 10:27 ----------

when you say the low side switch, is it the low side switch on the same side of the highside IGBT or do you mean the lowside pair of the highside igbt?

The left side IR2113 has to have a low side device so that the bootstrap capacitor gets charged in order to drive the upper left side device correctly.

Alex
 

how can the bootstrap capacitor of the left chip be charged in that case?
Of course it can't. It would be so much easier to operate a single half bridge with an AC coupled load for test.
 

I just noticed the waveforms in post #14. They are simply unsuitable for a bootstrap driver.

Can you please explain what is wrong with these PWM?
It is my understanding then he used a high rate PWM (to control the load current) and a lower rate switching between HIN-LIN (to control polarity), for example 20KHz PWM and changes the active side HIN-LIN with 500Hz rate (just an example).

Alex
 

Can you please explain what is wrong with these PWM?
I was mainly referring to the fact, that the low side driver has to be activated before the high side driver will ever work. But strictly spoken, we should ask, which hardware is drivem by the waveforn. If you refer to post #16, there won't be an output current, unless both are switched on simultaneously, apart from the bootstrap precharge problem.

Generally, the discussion suffers form the problem, that the intended PWM scheme was never mentioned. Your idea of superimposing a high rate PWM with low rate polarity switching can basically work. But it involves utilizing the MOSFET substrate diodes for the buck conversion which should be avoided if efficiency is an objective. Professional inverters would use an unipolar PWM scheme instead. It requires four separate control signals, but achieves a better leveling of switching losses or alternatively a doubled output PWM frequency, and synchronous MOSFET operation.
 

FvM can you please provide a link, article of schematic where I can see how this unipolar PWM scheme works.
Is it a schematic like the one above using two IR2113 and four independent PWM signals?
I can't think of a different way to control the current and polarity at the same time apart from the one I have described.
The only other solution I can think of is a pwm driven mosfet controlling the positive supply that goes to the high side mosfets of the bridge.

Alex
 

Here's a diagram of unipolar PWM operation. The first line shows an analog pwm signal generation with triangle voltage. uAN and uBN are the resulting output voltages of both halfbridges, u2 is the difference voltage.


P.S.: You need four individual pwm signals for software deadtime control. If you have halfbridge drivers with internal deadtime generation or equivalent external hardware, two pwm signals are sufficient.
 
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So uAN is the output of one half of the bridge (say left side) and uBN is the right side.
In uAN/uBN waves when the pulse is high it means that the high side switch is conducting and when it is low that the low side switch is conducting, as a result when both uAN/uBN are high or low the load gets the same voltage from both sides which essentially doesn't let current flow since there is no voltage difference.
The voltage difference that the load actually sees is the wave shown in u2 and depending on the difference of uAN/uBN you get different polarities.
Cool!!!!

Thank you for sharing that
Alex
 

Hi, here is an update...

we tried the exact circuit of alexan_e, the PWM that we still used was the one in the post #14
https://obrazki.elektroda.pl/98_1306999454.jpg

schematic here:


the problem of our supply shorting was gone but the output wave that we obtained on the resistor load, was almost the same as in post #16
https://obrazki.elektroda.pl/32_1306996803.jpg

do you have an idea as to why there is still a straight dc voltage before the square waves that we produced?

we were expecting a continuous square wave output since if its already a full bridge...

we read the discussion of FvM and alexan_e and our intended output waveform was the one that FvM showed in post #27 u2 waveform.

Does the problem of the output waveform now, is due to the PWM that we are using? we have no idea with regards to the unipolar PWM. we are getting our PWM from our pic microcontroller.

Another question, does the resistance in the gate igbts matter? we are using 1kohms at the moment. and also the C1 and C2 in the images, what specific value should we be using?
 

we are getting our PWM from our pic microcontroller.
To understand about the available PWM option, we need to know the PIC type and connected output pins.

By chance, I got to know, that you are using mikroC PWM library. Unfortunately, I'm not using the compiler and don't know which PIC PWM generation features it supports. Regarding previously mentioned unipolar PWM, it's only available with dsPIC processors as far as I know.

the PWM that we still used was the one in the post #14
As said, you need an alternating activation of high- and lowside drivers to make the bootstrap circuit work. You have a 500 ms delay between both, the bootstrap capacitors won't keep charged for this long amount of time.

we were expecting a continuous square wave output since if its already a full bridge...
The PWM waveform isn't right. Why don't you refer to a Microchip PWM application note?

Another question, does the resistance in the gate igbts matter? we are using 1kohms at the moment.
Yes, it matters. Where did you get the 1k number? The datasheet is suggesting something like 10 ohm, if I remember right.
 

alright, we'll use a lower resistor.

Another question... if we use a resistor as our "LOAD" is that wrong? because the waveform is still distorted... does it affect the waveform since the load is a resistor? should we use an "inductive load" like an inductor?
 

No a resistor as a load is not the problem , it is perfectly fine to use it.

Lowering the gate resistor should give you sharper transition and faster square wave edges, this is because the gate is charged at a higher rate which speeds up the mosfet/IGBT switching.

Another note is why do you use IGBT instead of mosfets, is there any particular reason?
Take a look at **broken link removed**

Alex
 
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we've already lowered the resistor, we used 22ohms first. but the waveform is still the same... what possible solution could be done?

we used an IGBT since we wanted to do very fast switching.
 

I'm not sure you are getting a faster switching compared to a mosfet device like IRL2203
What you will get for sure is a voltage drop of about 1v to 1.5v in each IGBT because of the saturation voltage, so you loose efficiency and increase the heat because of the power consumption on these devices.
To have a similar level voltage drop across the mosfet you would need a current of over 200A for a device having an Rds-on of 7m ohm (like the mosfet above)

Alex
 
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