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I want to replace a MOSFET in a legacy product. The MOSFET is a P-channel Power MOSFET that is used for PWM switching purposes (VGS @ -12V, PWM is 20% @ 200us). The FET is switching in to a 200uH, 10ohm load (see image, 5 and 6 are a coil of 5ohms 200uH @ 10KHz). The FET has quite a large heatsink attached due to it getting hot. I'm wondering if the RDS(on) of 1.5 ohms may be the culprit as this is fairly high resistance in saturation mode? Would it be worth looking at alternatives with similar characteristics but with lower RDS(on) (in the milli-ohm range)? Or is the issue potentially more complicated / caused by the load? Any help would be great.

enter image description here

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    \$\begingroup\$ The "thermal characteristics" are given by the MOSFET, if both MOSFETs have the same thermal resistance then the "thermal characteristics" are the same. What you mean is the power dissipation which will indeed decrease with a lower \$R_{DS,on}\$ assuming everything else stays roughly the same (if the low \$R_{DS,on}\$ switches much slower, things are not the same). A lower power dissipation means that less heat needs to be dissipated and that can result in lower temperatures. \$\endgroup\$
    – Bimpelrekkie
    Feb 10, 2021 at 15:15
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    \$\begingroup\$ I think you're assuming that a low RDSon does not come without a cost. But it does; increased gate capacitance. This makes each switching cycle possibly longer, which could cause more heat dissipation. You'll need to analyze this very thoroughly. \$\endgroup\$
    – rdtsc
    Feb 10, 2021 at 15:47
  • \$\begingroup\$ Looking at the thermal specifications, the proposed substitute device has a lower Rth(j-c) but a higher thermal derating coefficient. \$\endgroup\$
    – vir
    Feb 10, 2021 at 16:55
  • \$\begingroup\$ @rdtsc you're correct I was assuming that as this is a particularly old product - I was wrong to do so. It seems that the cost of increased gate capacitance tends to mean a slower turn on in the range of 10-100ns but this can be at a reduced RDSon of just a few 100 milliohms. As the switching frequency is fairly "slow" at 5KHz I'm wondering how much of a difference the slower switch on will make, I don't think it matter a lot. Time to experiment. \$\endgroup\$
    – ChrisD91
    Feb 12, 2021 at 8:23
  • \$\begingroup\$ You could try to simulate the two using LTspice for a rough estimation. \$\endgroup\$
    – rdtsc
    Feb 12, 2021 at 15:57

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Lower RDson often means higher gate charge/capacitance which slows down switching and increases switches losses. Things could get hotter depending on the balance between conduction and switching losses. So pay attention to both total gate charge and RDson when choosing.

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  • \$\begingroup\$ Also, if the load is inductive, it may generate significant back-emf. I see no freewheeling diode, so the mosfet may be exposed to high voltages; OP needs to take that into account when choosing a replacement. \$\endgroup\$
    – marcelm
    Feb 10, 2021 at 16:26
  • \$\begingroup\$ It should work just fine, use a device with an avalanche rating and you have no worries about adding a diode. Your gate drive is going to be restricted by R22. Be sure you pick a device with enough gate voltage to work properly. The legacy device statement clues me as a few years back P-MOS devices were expensive in comparison to N-Channel devices and low RDSon parts were not readily available, heatsinks were cheaper. \$\endgroup\$
    – Gil
    Feb 10, 2021 at 20:22
  • \$\begingroup\$ Looking at comparable devices, although you're 100% correct in saying that lower RDSon means slower switching, it tends to be in the 10s of ns while the RDSon can be significantly smaller (100s of milliohms vs 1.5 ohms). If the device is switching at a frequency of 5KHz I'm wondering how much of a real difference those few nano seconds make. I guess the answer is to experiment. \$\endgroup\$
    – ChrisD91
    Feb 12, 2021 at 8:19
  • \$\begingroup\$ @ChristopherDyer It really depends on the operating conditions. At low currents and high frequency. R losses are already low relative to switching losses so going 10x lower RDson may still make things hotter because most losses are already switching losses and you're already past the point of diminishing returns on low RDson. Also, note the times shown in datasheet are not the switching times in your circuit. That is dependent on gate charge and your gate drive circuit. \$\endgroup\$
    – DKNguyen
    Feb 12, 2021 at 14:55
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    \$\begingroup\$ @ChristopherDyer Ideally, you want switching losses = conducting losses. So you can set up non-switching R loss only circuit separately at your operating current and measure temp rise over ambient, then in circuit if you get double the temperature rise over ambient then your two losses should be approximately equal. If you can pin that down you will have better idea of the gate charge vs RDson tradeoff in your circuit by comparing those values against other transistors. \$\endgroup\$
    – DKNguyen
    Feb 12, 2021 at 15:00

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