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let's assume I have this transistor (IRFR640): https://www.infineon.com/dgdl/Infineon-IRFR6215-DS-v01_02-EN.pdf?fileId=5546d462533600a40153563595592114

$$R_{DS(ON)} =0,295 \Omega $$ $$V_{max} = -150V$$ $$I_{max} =-13A $$ $$I_{load)} =260mA $$ $$T_{a} =50°C \Omega $$ $$R_{\theta ja} =50°C/W$$ We apply on the drain 100V (\$V_d = 100V\$)

What is the good calculation?

1) $$T_{PMOS} =T_a + R_{\theta ja}*R_{DS(ON)}*I_{load}^2 $$ So $$T_{PMOS} =50 + 110*0,295*(260*10^{-3})^2 = 52,2°C $$

2)$$T_{PMOS} =T_a + R_{\theta ja}*V_{d}*I_{load} = 894°C $$

What is the good answer? And why? I use same power dissipation Tthanks

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    \$\begingroup\$ When Vd = 100 V and Rdson = 0.295 ohm then 338A will flow, that is unrealistic, that would exceed Id_max and P_max so calculation 2 is unrealistic. \$\endgroup\$ – Bimpelrekkie Jan 22 at 11:07
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    \$\begingroup\$ Equation 2 is more correct, but it is not Vd. It is Vds, the voltage from drain to source of the transistor. I guess that would be 100V - Vload, assuming there is some load connected between 100V and drain of transistor. Really, to get a good answer, you should draw a circuit schematic. \$\endgroup\$ – mkeith Jan 22 at 11:21
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    \$\begingroup\$ Use the built-in schematic editor to draw your schematic, don't make us click through a link. \$\endgroup\$ – Elliot Alderson Jan 22 at 13:21
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    \$\begingroup\$ What is your load? And why P-Channel MOSFET? \$\endgroup\$ – G36 Jan 22 at 14:04
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    \$\begingroup\$ OK, so you want to switch ON/OFF the 100V supply rail that needs 260mA max? \$\endgroup\$ – G36 Jan 22 at 14:16
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Your assumptions must be invalid. It is not possible that \$V_{DS} = 100\mathrm{V}\$, \$I_{DS}=260\mathrm{mA}\$, and \$R_{DS}=0.295\Omega\$. Before you can calculate the transistor's die temperature you must analyze the circuit and determine the actual operating conditions.

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  • \$\begingroup\$ I didn't understand. Because the transistor can handle these voltages and current. So why is there something wrong? \$\endgroup\$ – Tack Jan 22 at 13:57
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    \$\begingroup\$ @Tack Ohms law, if you have 100 V and 0.295 ohms you have more than 260 mA. \$\endgroup\$ – Colin Jan 22 at 14:08
  • \$\begingroup\$ I'm sorry, I didn't understand. 100V with 0,295 ohms => 344A. And that's more than what the transistor could give (13A). \$\endgroup\$ – Tack Jan 22 at 14:11
  • \$\begingroup\$ I think I found the answer We assume that: $$I_d = 260mA$$ $$V_S = Vd - R_{DS(ON)}*I_d = 100-0,295*260*0,001 = 99,9233V$$ And so $$V_{DS} = V_d-Vs = 0,0767$$ $$T_{PMOS} =T_a + R_{\theta ja}*V_{d}*I_{load} $$ So T_{PMOS} =50 + 0,295*260*10^{-3} = 52,19°C Same results. :) \$\endgroup\$ – Tack Jan 22 at 14:32
  • \$\begingroup\$ I can't edit my post, sorry. $$T_{PMOS} =T_a + R_{\theta ja}*V_{ds}*I_{load}$$ $$T_{PMOS} =50 + 0,295*260*10^{-3}*0,0767 = 52,19°C$$ Expected: 52,2°C Thank you all. :) \$\endgroup\$ – Tack Jan 22 at 14:39
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Your circuit should look something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

And the static power dissipation

\$P = 0.56\Omega \cdot 260mA \cdot 260mA \approx 38mW \$

Hence the static power dissipation is not a problem (if the ON/OFF period is long <1KHz).

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  • \$\begingroup\$ Thank you very much for your circuit. I just have questions, why do we need R1? And what is the 0,56 ohms? \$\endgroup\$ – Tack Jan 22 at 14:46
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    \$\begingroup\$ R1 is needed to pull-up the P-MOS gate to 100V rail and to discharge P-MOS Cgs capacitance to turn OFF the P_MOS when Q2 is OFF. And 0.56ohms is a Rds_max. \$\endgroup\$ – G36 Jan 22 at 14:54
  • \$\begingroup\$ And so why R2? Did you added Q2 because of Vghth of Q1? \$\endgroup\$ – Tack Jan 22 at 15:23
  • \$\begingroup\$ In this case, I add R2 because your MOSFET Vgs_max is equal to +/-20V. Hence R2 resistor limits the Vgs value to 10V. R2 also limits the Cgs charging current when P-MOS turn-ON. And Q2 is needed to control ON/OFF of a P-MOS from a 3.3V signal referenced to ground. Because to turn-OFF the P-MOS, the Voltage at the gate should be equal to about 96V, and 3.3V cannot do this. And to fully turn it ON about 90 volts is needed( but cannot be lower than 80V). \$\endgroup\$ – G36 Jan 22 at 16:03
  • \$\begingroup\$ Basically R2 is needed to limit the Zener diode (D1) current when Q2 turns on. Without R2, the Zener current could be very high. \$\endgroup\$ – mkeith Jan 22 at 22:45

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