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The physical layout of a MOSFET is very similar to BJT transistor, if we ignore the gate. Normally a FET requires a gate voltage to turn on and allow current flow from Source to Drain or vice vera.

My question is: Can a very high \$ V_{ds} \$ force charge carries drift through the channel and reach the other side?

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    \$\begingroup\$ I have read an article the other day on MOSFETS that actually mentions a parasitic BJT and subsequently discusses cause and effect. If I only could recall where I read that. \$\endgroup\$ – jippie Mar 12 '16 at 7:02
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    \$\begingroup\$ The physical layout of the Himalaya is very similar to Antarctica, if you ignore the mountains... \$\endgroup\$ – PlasmaHH Mar 12 '16 at 7:55
  • \$\begingroup\$ This application not writes about the parasitic BJT I mentioned earlier: fairchildsemi.com/application-notes/AN/AN-9010.pdf Mentioned several times, but figure 21 is nice. \$\endgroup\$ – jippie Mar 12 '16 at 8:40
  • \$\begingroup\$ Anyways as you can read from the app note, it is not so much a high V(DS) that triggerst the BJT, but a high \$\dfrac{\text{d}v}{\text{d}t}\$. In other words high rate of change in voltage. \$\endgroup\$ – jippie Mar 12 '16 at 16:46
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As we increase the drain voltage the depletion region around the drain gets wider while the depletion region around the source does not change. If we keep on increasing it further, the depletion region around the drain will finally reach the source side.

This undesired behavior is known as punch-through and can be regarded as an extreme case of channel-length modulation.

For this reason the drain current will then strongly depend on the drain-source voltage.

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You must look at the fundamental equation of an NMOS:

\$ I_D \approx \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{TH}) V_{DS}. \$

Where:

\$ \mu_n \$ is charge mobility (depend on technology which NMOS is built upon)

\$ C_{ox}\$ is Oxide Capacitance and also depends on IC technology.

\$ W \$ is Width of transistor.

\$ L \$ is Length of transistor.

\$ V_{TH} \$ is the Gate voltage threshold.

You can easily see that no matter how high is \$ V_{DS} \$, if \$ V_{GS} = V_{TH} \$ then \$ I_D \$ will be zero, which means no current will flow through the channel.

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  • \$\begingroup\$ This is just a mathematical modeling of the underlying physics. I believe the physics under very strong electrical fields between the drain and source changes the equation above. \$\endgroup\$ – user45498 Mar 12 '16 at 7:59
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    \$\begingroup\$ Yes it DOES change the equation above. It blows up the transistor. \$\endgroup\$ – Ehsan Mar 12 '16 at 8:26
  • \$\begingroup\$ Maybe a very small current (like Is in BJTs) start to flow between D and S before the blow up? \$\endgroup\$ – user45498 Mar 12 '16 at 8:29
  • \$\begingroup\$ aka "you can get a lot of V/metre across a micron". \$\endgroup\$ – user_1818839 Mar 12 '16 at 11:20

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