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if I stick a MOSFET that should be going (S-D) to positive in backwards, and make it (D-S) to positive, all the current drops across the diode instead. This diode then approaches an impedance of 0.

Technically the current wouldn't go through the higher impedance S-D capacitance, and wants to travel through the diode instead.

Can I ignore the capacitance of source-drain while the voltage drops across the diode?

(I know this is not a ~practical~ thing, but I'm wondering more about the diode versus S-D capacitance than the practicality of backwards FETs)

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Consider the case of a diode. Diodes have a low "resistive" impedance (i.e., current is in phase with voltage), and they also have capacitance. At low frequencies most of the current flows through the diode resistively, and at high frequencies most current flows through the capacitance.

When the S-D of a MOSFET is biased backwards, it's exactly the same thing. It's a diode.

Be mindful that under forward bias, a diode has more capacitance than under reverse bias. Namely, the capacitance (called a "diffusion capacitance") scales linearly with the DC current. Therefore, the frequency at which capacitive current dominates resistive current remains roughly constant regardless of DC current level.

To answer your question of whether you can ignore the source-drain capacitance: It depends on the frequency of interest.

In the synchronous FET of a switching converter (which is where you'll usually see a FET with reverse S-D bias), you do care about this capacitance. In this context though, because the behavior is large-signal and the frequencies are so high to cause non-quasistatic behavior, it's usually called something different: reverse recovery charge. This emphasizes that in this context, thinking of the diode as a simple conductance or capacitance is incomplete.

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  • \$\begingroup\$ So if I put a couple back to back, and ran AC through the two MOSFETs (so basically an SSR), since they both are reverse/forward biased at either end of the cycle, I can just use the datasheet specific value of c charge at V(sd)? aosmd.com/res/data_sheets/AOK42S60.pdf (120kHz) \$\endgroup\$ – ARMATAV May 22 '15 at 20:54
  • \$\begingroup\$ I think I'm still kind of confused. The datasheet specifies a a value of 100pF S-D at 1Mhz, which is WAY higher than my 120kHz, so I could just use the constant capacitance of 100pF (Figure 9) across S-D, right? Why bother with the reverse recovery charge if at 600V the S-D capacitance is still relatively constant at 1MHz. \$\endgroup\$ – ARMATAV May 22 '15 at 21:03
  • \$\begingroup\$ Also, if I'm operating at 120kHz, my MOSFET SSR will have an impedance of around 13,000; will it be able to stop, or dampen, the AC if I turn it off? Rds on is 0.099, and Rds off is simply the 13000 from the capacitor. \$\endgroup\$ – ARMATAV May 22 '15 at 21:21
  • \$\begingroup\$ I don't think you got my point. 100pF is the reverse-biased capacitance. Its forward-biased capacitance manifests itself as a reverse-recovery charge (\$Q_{RR}\$ -- it's in the electrical characteristics table as 10μC). As for the impedance of a 100pF capacitor being 13kΩ, it's not meaningful to think of things that way. Not for a switching converter with a switching frequency of 120kHz anyway. \$\endgroup\$ – Zulu May 22 '15 at 21:37
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    \$\begingroup\$ I see. I was presuming you cared about switching converters, since 120kHz is a pretty typical switching frequency. In this case, yes, ~100pF is the capacitance you care about. \$\endgroup\$ – Zulu May 22 '15 at 21:46

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