In the circuit i use optocouplers to separate the control part from the power part, and to generate the pulses for the ir2113 i use a 18V power supply but reference to the ground of the control part, and my question is, instead of use a 18 v power supply reference to the ground of the control part do i need to use a 18V power supply reference to the ground of the 4 IGBT's circuit, and finally can anyone tell me if there is errors in the schematic, many thanks in advance.
The whole point of the high side drive is it is meant to wap up and down and the chip level shifts the input to where it wants to be. Reference to inputs to the lowest voltage in your circuit and level shift the inputs down to where they need to be.
As I write this I am forced to listen to someone slapping his chops about some food.
What is your advice for separate the control part from the power part?, the retardation that the optocoupler introduces is not a problem due to the inverter is supposed to work in very low frequencies less than 5 Hz.
#Answer for Genome
So, you mean that i put as reference for the optocoupler the lowest voltage in teh circuit? i got three power supply, one for +/-5 V, the
+/- 18V and a DC 300 V for the inverter, so i think that i should put as reference the +/- 5 v supply because is the power supply with the lowest value.
The 18V power supply to the IR2113's should be reference to power ground at the emitters of the two low-side IGBT's, otherwise the opto-couplers are redundant.
Also I don't think you can rely on the internal pull-down resistors in the IR2113 to hold the inputs LOW when the opto's are switched off. You should take pin 4 of the opto's directly to +18V and then connect a resistor of say 10k from pin 3 down to power ground.
Of course this help, thank so much Mark, so the scheme should be something like that,
formatting link
Now i reference the 18v to the power ground, to do this i have donde this
formatting link
to do the VCC of the inverter i use a transformer, diode bridge and some capacitors, and to do the 18V supply use the same source than the VCC, using the same source the grounds of the 18V and the ground of the Vcc will be equal?.
For the optocouplers i put directly the pin4 to the power supply, and i connect the pin 3 with a resistor to the power ground, i put the resistor between pin4 and the 18v to limite the current, in the datasheet i read that the maximum collector current is 80mA without the resistor the current will be lower than the maximud permited it?.
The frequency of the current in the load will be less than 5 Hz, but to commute the IGBT's i use SPWM and the commute frequency is about 250 Hz, so the 1uF cap will be enough?.
ANd finally for the diode i put the diode that you say, i put the
1N4376 because i still havent chosen a diode.
Again thanks so much, and you help me a lot :), thanks.
Not exactly sure what you mean? If you mean how much time does it takes to charge up the 1uF caps then these are refreshed every time the low-side igbt is switched on. Assuming that pin 5 (Vs) on the IR2113's is connected to the emitter of the relevent high-side igbt (not shown in your schematic).
If you mean how long will the 1uF caps hold enough charge to keep the igbt's switched on then this depends on the quiescent current of the floating supply, gate charge required to turn on the igbt and leakage currents in the gate and cap.
Just considering the quiescent current- from the IR2113 datasheet, figure 17B gives a quiescent current of
300uA max @ 18V. If we assume we can allow a voltage droop of around 1V and that the quiescent current remains fairly constant then using I=C*dv/dt dt=C*dV/I
then ton_max = 1e-6 * 1 / 300e-6 so ton_max= 3.3mS
you might be able to allow more droop if you want, just depends on the igbts characteristics. This gives a minimum operating frequency (assuming 50% duty cycle) of around 150Hz. once you include the effects of gate charge and leakage currents then you would have to increase the minimum operating frequency.
Thanks for your response, i apologise for the time in responding. In my question i mean how much time will take the cap to store enough charge to turn on the igbt, i mean to turn on the igbt the capacitor must store some charge, and therefore the on time of the low side switch (or the off time of the high side switch) must be sufficient to ensure that the charge drawn from the cap by the high side driver, can be fully replenished, can you tell me how can i caluclate the minimum off time of the high side driver?, i hope you understand my question now.
Therefore the on time of the low side switch (or the off time of the high side switch for a high side driver) must be sufficient to ensure that the charge drawn from the Cbs capacitor by the high side driver, can be fully replenished.
And finally can you tell me about the questions that i did in the post previous to this "One question Mark, can you tell me how much time can take to the capacitor to store the charge necessary to turn on the high side IGBT?". Mark many thanks again, you are helping me a lot, can i do anything for you?.
Sorry if my english sometimes sound a little abrupt that's because i'm spanish.
If your control circuitry is operating at a different potential to the low-side igbt emitters (Power Ground), then you need to have an isolated 18V supply for the gate drivers which should connected to power ground. That way you can have the control circuitry at a safe voltage, this also keeps all the noisy power circuit away from you control.
With the 10k resistors the current will be less than 2mA.
1uF should be okay at 250Hz
Thinking about it again you should probably us an ultra-fast diode like a uf4005 or uf4006, the byv95c is a bit slow.
In your circuit Cbs will probably be charged during the turn-on transition of the low-side igbt with a large spike of current though the diode. The time this will take depend on the igbt's and the gate drive resistors but probably around 1uS or less. All the circuits ive seen running at around this frequency have had a resistor in series with the diode to limit the peak current.
Ok, I will make some calculus and i will use a faster diode,.
My load, is a 15 ohms, and 250 mH, and i'm using an irg4pc50fd. I have dome more simulations in orcad and i get with this signal control for T1 a good output,
formatting link
the frecuency is
50 Hz, so i will require a bigger cap than 1uF, i have donde the calculus for the bootstrap cap and i think i will require a 10 uF cap, but my problem is, if there is enough time for the cap to refresh its charge to turn on the igbt and to hold on the igbt during 20ms, what do you think?. And about the resistor in series with the diode to limit the peak current, what is the resistor value?.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.