I feel like a total newb here, because I've built resonant tank circuits plenty of times before, but somehow I'm stumped on this one. I've run into a bizarre problem I can't figure out.

See scope shot: enter image description here

The top channel is my signal to the gate of a mosfet. The mosfet is driving a tank circuit that contains a large coil wrapped on ferrite with a 5uF capacitor. The coil is large enough that the frequency is near 100Hz.

The bottom channel shows the positive side of the tank capacitor where the mosfet source is connected to drive the circuit. What I'm confused as heck about is the low voltage delay after the input turns off and before the initial oscillation begins. I don't recall ever running into something like this before. I've traced my circuit leads and double checked everything and I just don't know what's going on. Anyone else ever seen something like this in a tank circuit?

Ok here's a circuit diagram. I should have posted before, but I doubt this is going to help: enter image description here

  • \$\begingroup\$ Maybe I shouldn't have even posted this. I think I know part of what's going on. I've never used such a large coil compared to such a small capacitor before. The DC pulse instantly charges the capacitor and then saturates the coil. After the pulse ends, since the coil is already saturated, the capacitor instantly discharges through it. The circuit arrangement I've got may be too primitive for the smallness of capacitor and largeness of coil. \$\endgroup\$
    – JamesHoux
    Commented May 12, 2013 at 13:39
  • \$\begingroup\$ try a shorter pulse \$\endgroup\$
    – Andy aka
    Commented May 12, 2013 at 14:06
  • \$\begingroup\$ Well. I'm still perplexed because that doesn't explain why there's a delay after the pulse ends. I can't seem to get my head wrapped around this one. \$\endgroup\$
    – JamesHoux
    Commented May 12, 2013 at 14:14

1 Answer 1


Since you didn't post a circuit diagram, most of what follows is speculation, but your waveform is very similar to what you see in any switching regulator that's operating in discontinuous mode.

When you turn on your MOSFET, you are "charging" the coil with a certain amount of current. At the moment when you turn off the MOSFET, this current still needs to flow. Since the current can't flow through the (smallish) capacitor, it is forced to flow through the "body" diode of the MOSFET (or some other diode you have in the circuit?), at least until it drops to a value that allows the voltage of that node to rise above zero volts. At that point, the diode stops conducting. Now the MOSFET is truly "open circuit" and you see the oscillations you expect.

OK, now that you've posted a diagram, I can see what's going on. When the gate signal goes high, this enables current to flow through the MOSFET, because the source is essentially tied to ground through the diode and the coil. Current builds up in the coil as described above, because it has a positive bias across it. When the gate signal goes low, the inductance of the coil pulls the source terminal low enough (a negative voltage) so that the MOSFET is still on, but now the magnitude of the current is decreasing instead of increasing, because the coil now has a negative bias across it. Once the current decays sufficiently to allow the MOSFET to cut off, the remaining residual current drives the oscillation you see in the latter part of each cycle.

  • \$\begingroup\$ Dave, you're the best man! That makes perfect sense, and I never would have realized what was going on myself. I didn't have the right mindset of how to think about transistor biasing. Thank you so much! Accepted! \$\endgroup\$
    – JamesHoux
    Commented May 13, 2013 at 7:32
  • \$\begingroup\$ I opened up an entirely new question related to this issue since I felt it was merited: electronics.stackexchange.com/questions/69079/… \$\endgroup\$
    – JamesHoux
    Commented May 13, 2013 at 9:29

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