Not generally. But you're most likely able to damage the device by considerable over currents. This is mainly a matter of control algorithm and can't be seen from your circuit. You may want to think about a latching over current sense immediately shutting down the IGBT driver. Furthermore, possible problems of circuit layout should be considered.Could it be that the IGBT I'm using is not powerful enough for this application?
//Written for MikroC 8.2
void main()
{
//Power-up timer
Delay_ms(100);
TRISA = 0xFF;
TRISC = 0xF0;
//Enable analog inputs
ADCON1 = 0x07;
//Initialize the PWM
PWM_Init(10000); //10kHz
PWM_Change_Duty(179); //70%
//Turn working LED off
PORTC.F3 = 0;
while(1)
{
//Stall until the charge button is pressed
while(1)
{
if(PORTC.F7 == 1)
{ break; }
}
//Continually charge until 200V is reached
while(1)
{
if(ADC_Read(0) < 410)
{
PWM_Start();
PORTC.F3 = 1;
}
else
{
PWM_Stop();
PORTC.F3 = 0;
break;
}
}
}
}
//Pseudo-code for the control algorithm
while true
//2V is the output of the voltage divider at 200V
if ADC_Read(0) is less than 2V
PWM_Start();
else
PWM_Stop();
break;
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?