Collapse effe5ct of field-effect transistors in power applications 7

In the exported formula (3), it can also be applied to continuous forms of pulses. Formula (3) can be modified as follows to obtain its instantaneous thermal resistance:transistor

Zθjc(t)=[t1/t2+(1-t1/t2)xr(t1+t2)+r(t1)–r(t2)]Rθjc=t1/t2Rθjc(1uS)+(1-t1/t2)Zθjc(t1+t2)+Zθjc(t1)-Zθjc(t2)-----(4)

Where t1: Pulse width of continuous pulses t2: Total duration of continuous pulses

Formula (4) is suitable for the case of infinite pulse waves. When the number of shock waves is finite, their R θ jc (1uS) should be replaced by Z θ jc (T), where T is the duration of the finite shock wave.transistor

Assuming a situation where a set of switching power supply lines is undergoing a short-circuit test, the measured voltage across the Drain and Source terminals of the field-effect transistor exceeds its maximum rated voltage and continues for a period of time until the short-circuit protection is activated. The parameters under this condition are as follows: FQA9N90C. html "target=" -blank "title=" FQA9N90C ">FQA9N90C is its field-effect transistor, 100nStAV, 9.2uS cycle length, and 20mS delay time. Under these conditions, the instantaneous thermal resistance is calculated as follows

Zθjc(t)=0.01xZθjc(20mS)+(1-0.01)Zθjc(9.3uS)+Zθjc(100nS)-Zθjc(9.2uS)=0.00274 ℃/W

We assume that a power consumption of 5KW will be consumed on the field-effect transistor, and the resulting junction temperature will rise as follows

D T = 5000 x 0.00274 = 13.7℃