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Why is the electric field the greatest on the drain side of the channel?

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  • \$\begingroup\$ Have you looked at a 4-terminal device and then how a 3-terminal device compares to it? \$\endgroup\$
    – jonk
    Dec 1, 2021 at 3:01
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    \$\begingroup\$ So are you saying that the source side of a MOSFET cannot have breakdown? \$\endgroup\$ Dec 1, 2021 at 8:59
  • \$\begingroup\$ Which type of transistor? I know of at least one type where breakdown (assumed: high-drain-voltage avalanche breakdown) occurs in the nominal-substrate region, which is shorted to source, so would seem to contradict the assumption in the title. \$\endgroup\$ Sep 8, 2023 at 16:06

2 Answers 2

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Why is the electric field the greatest on the drain side of the channel?

The voltage at the drain (with respect to the source) affects the width of the "conducting" channel between drain and source along its whole length. The effect is to compress the channel dimensions the closer you are to the drain (aka pinch-off): -

enter image description here

Image from here. In effect, and due to the effect of "pinch-off", the voltage stress at the drain is higher than that seen at the source (where the channel is much wider/thicker).

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  • \$\begingroup\$ Is it related that MOSFETs seem to have generally much stronger heatsinking capability (i.e. more pins/pads) for their drain connection than for their source connection ? I mean does the narrow channel mean, that the main resistance is in the drain, hence heat is dissipated close to the drain contact ? \$\endgroup\$
    – tobalt
    Dec 1, 2021 at 13:06
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    \$\begingroup\$ @tobalt exactly - MOSFET heatsinks are generally on the drain connection because there is more local heat generated there. And, also exactly again, the relative resistance of the channel per distance is greater at the drain so, at any particular current, the volt-drop per distance along the channel is greatest near the drain and hence the power dissipated is also greater closer to the drain. \$\endgroup\$
    – Andy aka
    Dec 1, 2021 at 13:52
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As there are many types of MOSFET devices, there exist many mechanisms of breakdown in any of these devices. To reconcile the title question, Why does the breakdown always occur on the drain site for MOSFET? and the text body question, Why is the electric field the greatest on the drain side of the channel?, I can only think about the hot carrier injection in lateral MOSFETs. The hot carrier injection causes multiple effects, as threshold voltage degradation and leakage current increase. Avalanching can also take place: hot carriers produce electron-hole pairs amplifying the carrier density and finally causing avalanche breakdown to happen.

To understand why avalanching, which can accompany the hot carrier injection, occurs in the drain region, see the first take of my answer, where I used the graduate channel approximation to show the electric field behavior in the channel with the maximum reached at the pinch-off region (see a detailed derivation in my answer to the question Why does the Pinch-off Point in the MOSFET move towards the source as we increase Vds?).

Avalanche breakdown can also happen in vertical MOSFETs, but the mechanism is different, see the second take of my answer here: avalanche develops when the MOSFET channel is OFF, but the parasitic BJT of the MOSFET structure is ON.

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  • \$\begingroup\$ This answer seems it should've been comments asking for clarification. \$\endgroup\$ Sep 8, 2023 at 16:04
  • \$\begingroup\$ @TimWilliams: There are two major basic structural types of MOSFET devices (as of Dec 8, 2021, these are LDMOS and VDMOS), and the answer is not just a comment asking for clarification. In fact, for both types the breakdown tends to occur on the drain side, although different regions can be shorted. I explained the mechanisms for both, see edit one (for LDMOS, electronics.stackexchange.com/revisions/597338/1) and edit two (for VDMOS, electronics.stackexchange.com/revisions/597338/2). \$\endgroup\$
    – V.V.T
    Sep 10, 2023 at 16:41
  • \$\begingroup\$ I was waiting for OP's clarification and, if required, I was going to merge both edits into the answer body, but the question was closed and I forgot about it. You can see both takes -- are you thinking, if the question be re-opened, it would make sense to complete my answer? \$\endgroup\$
    – V.V.T
    Sep 10, 2023 at 16:41
  • \$\begingroup\$ The thought occurred to me that there are some types where the premise isn't even true (substrate depletion dominates; see TI NexFET for example), hence the reaction; for what it's worth, I agree with your answer in the scope you've noted. \$\endgroup\$ Sep 10, 2023 at 17:15

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