Plot the Instantaneous Junction Temperature

The power loss of a device is shown in Figure 17.9. Plot the instantaneous junction temperature rise above the case. $P_2 = P_4 = P_6 = 0, P_1 = 800 W, P_3 = 1200 W$ , and $P_5 = 600 W$ . For
$t_1 = t_3 = t_5 = 1 ms$ , the data sheet gives
$Z(t=t_1)=Z_1=Z_3=Z_5=0.035^\circ C/W$
For $t_2 = t_4 = t_6 = 0.5 ms$ ,
$Z(t=t_2)=Z_2=Z_4=Z_6=0.025^\circ C/W$

Question

Plot the Instantaneous Junction Temperature

The power loss of a device is shown in Figure 17.9. Plot the instantaneous junction temperature rise above the case. [latex]P_2 = P_4 = P_6 = 0, P_1 = 800 W, P_3 = 1200 W [/latex], and [latex]P_5 = 600 W [/latex]. For
[latex]t_1 = t_3 = t_5 = 1 ms [/latex], the data sheet gives

[latex]Z(t=t_1)=Z_1=Z_3=Z_5=0.035^\circ C/W[/latex]

For [latex]t_2 = t_4 = t_6 = 0.5 ms[/latex],

[latex]Z(t=t_2)=Z_2=Z_4=Z_6=0.025^\circ C/W[/latex]

Accepted Answer

Equation (17.4)

[latex]T_J(t)=T_{J0} +\sum\limits_{n=1,3,. . .}^{m}{P_n(Z_n - Z_{n+1})}[/latex] (17.4)

can be applied directly to calculate the junction temperature rise.

[latex]\Delta T_J(t = 1 ms) = T_J(t = 1 ms) - T_{J0} = Z_1 P_1 = 0.035 imes 800 = 28°C[/latex]

[latex]\Delta T_J(t = 1.5 ms) = 28 - Z_2 P_1 = 28 - 0.025 imes 800 = 8°C[/latex]

[latex] \Delta T_J(t = 2.5 ms)= 8 + Z_3 P_3 = 8 + 0.035 imes 1200 = 50°C[/latex]

[latex] \Delta T_J(t = 3 ms) = 50 - Z_4 P_3 = 50 - 0.025 imes 1200 = 20°C[/latex]

[latex] \Delta T_J(t = 4 ms)= 20 + Z_5 P_5 = 20 + 0.035 imes 600 = 41°C[/latex]

[latex] \Delta T_J(t = 4.5 ms) = 41 - Z_6 P_5 = 41 - 0.025 imes 600 = 26°C[/latex]

The junction temperature rise above case is shown in Figure 17.10.

Question 17.1: Plot the Instantaneous Junction Temperature The power loss o...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions