From FDA75N28 datasheet:

What I am having difficulty understanding are the reasons behind overshoot and ringing in V_DS.

Now, to what I already know:

  • The current that passes in an inductor cannot change abruptly, but has to change gradually.

  • Initially current is flowing through the Driver FET, but then this transistor turns off.

  • The current has to keep going somewhere, and the easiest way is to flow through the freewheeling diode of DUT transistor. This makes V_DS negative, as the drain pin has a higher potential than the source pin, thus forward biasing the body diode.

  • Before all the excess current is dissipated in whatever parasitic resistances the circuit has, the driver transistor turns on again.

(I need help undestanding what happens between these two steps and am not 100% sure of the next)

  • The body diode of DUT transistor becomes reverse biased right after it was forward biased. Because the depletion region takes some time to reestablish, current starts flowing in opposite direction to what is normal in the diode. From the datasheet, this phenomenon does not take too much long, 320ns which is the reverse recovery time. The current passing in the inductor keeps flowing in the same direction but this time through the driver transistor instead of through the body diode of the DUT transistor.

Additional remark:

As far as I can tell, the potential of the positive probe, V_D, never changes, so for an overshoot to happen, the negative probe has to go below the negative pin of the power supply. I cannot see what can cause this.

Why the ringing after overshoot? I have talked with a teacher that only has a general knowledge in electronics and he suggested that it could be a parasitic capacitance somewhere that was auto-oscillating with the inductance. I would like to know what exactly causes this.

  • \$\begingroup\$ A ringing system has to be a second order system, at least. So yes, it hints to a stray capacitance interacting with the inductance. Any second order system that also includes resistance can have a ringing response to a step input. \$\endgroup\$ Apr 8, 2016 at 10:14

2 Answers 2


What I am having difficulty understanding are the reasons behind overshoot and ringing in V_DS.

The ringing is due to the parasitic capacitance of the MOSFET reacting with the inductor as an underdamped 2nd order system. This capacitance can be in the range of 100 pF to 1nF. Read the MOSFET data sheet - it looks like it will be about 900 pF.

  • \$\begingroup\$ Any other question I didn't spot? \$\endgroup\$
    – Andy aka
    Apr 9, 2016 at 21:18
  • \$\begingroup\$ Ah sorry for not elaborating, I don't understand why the overshoot in the first place, but that might still be related to ringing correct? \$\endgroup\$
    – rmarques
    Apr 10, 2016 at 9:24
  • \$\begingroup\$ Yers it's all down to the transient response of the 2nd order filter formed by L and mosfet C. \$\endgroup\$
    – Andy aka
    Apr 10, 2016 at 9:37

I think your explanation is a good shot in the right direction !

Regarding the diode changing from forward mode to reverse mode, there's an article from Fairchild, perhaps you've seen it already. An interesting picture is this one:

enter image description here

The minority carriers changing position takes some time. The time it takes to change from the conducting stage to the blocking stage is the reverse recovery time.

The negative voltage can be explained by realizing that the inductance of the load behaves as a current source when charged. When it is the discharging itself through the (forward biased) body diode of the DUT MOSFET it will simply create the forward voltage across that diode. The diode will behave as it's own voltage source so to say. So that's why the voltage can be below supply or ground, it simply behaves as a separate voltage source in series with the supply.

This is similar to how DCDC boost converters can make an output voltage which is higher than their own supply voltage. An inductor is charged and that charged inductor allows you to dump the charge almost anywhere and achieve a higher voltage.

Andy's answer explains the ringing. It is charge resonating back and forth between the capacitors (of the diodes in the MOSFET) and the inductance of the load.

Note that the body diode will not conduct current in this situation. The voltage is below the forward voltage. If the diode would conduct it would "dissipate" (part of) the energy in that moving charge and the ringing would die out more quickly.

  • \$\begingroup\$ It's not the gate capacitance that is the culprit - it is the output capacitance. The gate capacitance (although 5nF) is reduced thru the reverse transfer capacitance (90 pF) to a minor player IMHO. \$\endgroup\$
    – Andy aka
    Apr 8, 2016 at 11:46
  • \$\begingroup\$ Uh, of course :-) \$\endgroup\$ Apr 8, 2016 at 12:51

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