Hello,
I m back again for understanding. I can't understand how Irms is different of Iavg. Ok there is a high peak current. If Iavg = 1 A and if the diode are conducting between t1 and T with T the period. Ipeak = Iavg*T/(T-t1) = Iavg/(1-t1/T). If t1/T is equal to 0.8. So Ipeak is 5 A. However Irms is equal to the following expression for a periodic signal :
If you resolve it you will have something like this sqrt(1/T*Ipeak²*(T-t1), then sqrt[Iavg² *T/²(T-t1)²*(T-t1)] and so sqrt(Iavg²/(1-t1/T)) and the limit of this tends to Iavg. It seems to be normal to me that Iavg is equal to Irms ? Did i do a mistake ?
So diodes have no reason to heat more than they "should" do ?
Thanks,
I m back again for understanding. I can't understand how Irms is different of Iavg. Ok there is a high peak current. If Iavg = 1 A and if the diode are conducting between t1 and T with T the period. Ipeak = Iavg*T/(T-t1) = Iavg/(1-t1/T). If t1/T is equal to 0.8. So Ipeak is 5 A. However Irms is equal to the following expression for a periodic signal :
If you resolve it you will have something like this sqrt(1/T*Ipeak²*(T-t1), then sqrt[Iavg² *T/²(T-t1)²*(T-t1)] and so sqrt(Iavg²/(1-t1/T)) and the limit of this tends to Iavg. It seems to be normal to me that Iavg is equal to Irms ? Did i do a mistake ?
So diodes have no reason to heat more than they "should" do ?
Thanks,