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Dead times are necessary to mitigate any risk of current shoot through between the switching power FETs. Integrated FET drivers often have a feedback based self timed FET switching sequence to ensure the smallest possible dead times while avoiding any shoot through current.do you still need dead time bertus?
/*
This code was based on Swagatam SPWM code with changes made to remove errors. Use this code as you would use any other Swagatam’s works.
Atton Risk 2017
*/
const int sPWMArray[] = {500,500,750,500,1250,500,2000,500,1250,500,750,500,500}; // This is the array with the SPWM values change them at will
const int sPWMArrayValues = 13; // You need this since C doesn’t give you the length of an Array
// The pins
const int sPWMpin1 = 10;
const int sPWMpin2 = 9;
// The pin switches
bool sPWMpin1Status = true;
bool sPWMpin2Status = true;
void setup()
{
pinMode(sPWMpin1, OUTPUT);
pinMode(sPWMpin2, OUTPUT);
}
void loop()
{
// Loop for pin 1
for(int i(0); i != sPWMArrayValues; i++)
{
if(sPWMpin1Status)
{
digitalWrite(sPWMpin1, HIGH);
delayMicroseconds(sPWMArray[i]);
sPWMpin1Status = false;
}
else
{
digitalWrite(sPWMpin1, LOW);
delayMicroseconds(sPWMArray[i]);
sPWMpin1Status = true;
}
}
// Loop for pin 2
for(int i(0); i != sPWMArrayValues; i++)
{
if(sPWMpin2Status)
{
digitalWrite(sPWMpin2, HIGH);
delayMicroseconds(sPWMArray[i]);
sPWMpin2Status = false;
}
else
{
digitalWrite(sPWMpin2, LOW);
delayMicroseconds(sPWMArray[i]);
sPWMpin2Status = true;
}
}
}
what kind of quiz!?Dead times are necessary to mitigate any risk of current shoot through between the switching power FETs. Integrated FET drivers often have a feedback based self timed FET switching sequence to ensure the smallest possible dead times while avoiding any shoot through current.
Which brings up another question?
What is shoot through current?
that no two IGBT in your inverter or for that matter your H-bridge in the same limb (power buss),turns ON at the same time. An ideal power electronic switch during ON state behaves like a conductor and thus if two switches in the same limb turn ON simultaneously, they will cause the DC bus to short-circuit, This is called a shoot-through fault. because I am an Electronic Technician I have troubleshooted various advanced microcontroller used for gate-driving, there are various causes which may lead to shoot-through fault to appear thus rupturing the costly switches. Few reasons are Electromagnetic Interference due to HF (high-frequency) operation, faulty snubber circuit design etc.
Class dismissed...
There will be a quiz tomorrow be ready for it...![]()
Hello,
There are delays of 500 us between the pulses.
Bertus
Apologies mosfets not igbt.that no two IGBT in your inverter or for that matter your H-bridge in the same limb (power buss),turns ON at the same time
thanks doodApologies mosfets not igbt.
No quiz. I was Just kidding.
You doing a great job you're almost there.