# How to Power a Tank Circuit with Mosfet without the Inductor Undesireably Triggering the Mosfet

BACKGROUND: I have this circuit that I asked about in a previous question:

Someone kindly explained that the problem I'm having with the circuit is that the mosfet is being triggered by negative voltage in the tank circuit causing the mosfet to stay on longer than I want it to. It's being caused by the initial field collapse of the inductor before the first oscillation begins. The result is the appearance of a delay between the end of the gate pulse and the start of the first oscillation. The result is the following messed up scope shot (green signal is mosfet gate, and yellow signal is mosfet source voltage):

MY NEW QUESTION: Can some folks give me direction on a better way to power the tank circuit without causing this behavior? I guessed that using a BJT would probably fix the problem, but I was wondering if there's a way to stick with mosfet. I'm finding that I'm running into the same problem in other configurations with inductors and mosfets.

• What waveform do you ideally expect to see? What amplitudes, what value L and C? What is the initial current pulse value? – Andy aka May 13 '13 at 9:33
• Please also specify the type of MOSFET used (N or P channel as a minimum) and what Vcc is applied to the 555 chip. – Michael Karas May 13 '13 at 9:41
• I want a freely oscillating circuit with a clean sin wave. If it wasn't for the negative voltage in the circuit biasing on the mosfet when I didn't want it, then you'd see a perfect oscillating sin wave. You'd also have resonant rise behavior. – JamesHoux May 13 '13 at 10:11
• As for the mosfet, it's an n-channel. The 555 is +12v. I want resonant rise in the tank circuit so the voltage on 555 shouldn't matter. The tank circuit will be high enough, that the gate-source voltage is going to be a problem regardless of the 555 being at 5v or 12v or even 18v. I need a way to isolate the negative voltages in the tank circuit from triggering the mosfet completely. – JamesHoux May 13 '13 at 10:14
• Rushing - quick thought. Try TWO MOSFETS in series - botyh identical N Channel. Connects source to source and gate to gate. Input and output are the two Drtains. Drive needs to be gates positive relative to source. Can use 555 to gates with high value resistor sources to ground to see what happens BUT you will need a flaoting drive in due course. The two back to back MOSFETS mean there is no body diode through path. MOSFET will turn on with either polarity Vds as loing as Vgs is positive for N channel. – Russell McMahon May 13 '13 at 10:33

## 3 Answers

One problem with your present circuit is that by putting the N-channel MOSFET on the "high side" of the tank circuit (between the power supply and the tank), it is dissipating far more power than necessary. You're essentially forcing the drain-source voltage to be equal to the gate threshold voltage (about 4V, based on your waveforms), rather than allowing it to be as low as possible.

One obvious solution would be to put the MOSFET on the low side of the circuit, between the tank and ground. Or you could stay with a high-side switch, but make it a P-channel MOSFET instead (which will require an inverted drive signal).

Either way, the MOSFET will stop conducting as soon as the gate pulse ends. However, this means that you may see some very high (or low) voltages at the drain of the MOSFET because of the inductive kick of the coil. You will want to add something to the circuit that limits the voltage to whatever the MOSFET can tolerate — perhaps a large-value zener diode.

Just to put some numbers to this, and assuming zero losses, the peak current in the coil will be

$$I_{peak} = \frac{V}{L}\cdot t_{ON}$$

And the peak voltage after the MOSFET cuts off will be

$$V_{peak} = I_{peak} \sqrt{\frac{L}{C}}$$

Which means that you can control Vpeak by either limiting the on time of the gate drive signal, or controlling the ratio of L to C, or a combination of both.

Using some numbers pulled from your scope traces, it looks like if your capacitor is 5 µF, your coil must be about 2.5 mH. Also, your tON looks to be about 1.6 ms.

Therefore, Ipeak is going to be about 7.68 A (!)

Vpeak will be about 172 V.

• The OP definitely needs to be using a much shorter time (like 5usec) for doing this and then the repeat pulses need to be exactly coincident with the same part of the cycle. – Andy aka May 13 '13 at 15:05

There are some things to be aware of with LC resonant circuits:

• Parallel LC circuits (like you have here) are best driven by current sources.
• Voltage sources (like you seem to be trying to do here) are good for driving series LC circuits.
• Any ringing in the circuit will decay due to resistive losses (dampening) in the circuit.

The easiest way to get sustained ringing will be to use a switched voltage source into a series LC, and apply pulses at a period of t = $2 \pi \sqrt{\text{LC}}$. You could use a half bridge or synchronous switched source like this.

Pulse width of the pulses will depend on the loss of the circuit. The less loss, the narrower the pulse width would need to be. For the example LC values shown you would want to apply a pulse every 200uSec. For example a one time pulse of 50uSec (or 1/4 the LC resonant period) width would give a maximum amplitude of $2 V_{\text{cc}}$. After that you would only want to apply maintenance pulses of narrower width to keep the ringing going.

HighDrive and LowDrive signals could be provided by a half bridge drive IC of your choice (like a IR2104, or LM5104 for example).

Based on what everyone put in, it appears to me that I really need to switch to BJT's. My purpose is to maximize voltage rise, and now I understand how it's extremely difficult to do this when using MOSFETs since any excessive voltages will impact the gate/source field and destroy it by putting too high of voltage difference between them. BJT's will be a lot more rugged and don't have the field stress problem.

• Changing to a BJT did not fix the unwanted delay after the square pulse turns off. Evidentally, the original explanation for the cause having to do with the mosfet's gate-source voltage was incorrect. Something else must be going on. I don't see why the same thing would be happening on a BJT. – JamesHoux May 14 '13 at 7:00
• I changed my mind. I believe the original problem with MOSFET is correct but it also happens on the BJT. The reason is probably that when the voltage goes low enough, the 555 starts sourcing current when it shouldn't be. That's the only explanation I can come up with. – JamesHoux May 14 '13 at 8:09
• I'm ditching all the 555 stuff and going to a completely different style of circuit so that I can get better control over this behavior. The 555 was only a temporary timing mechanism anyway. And I should be able to automatically drive the circuit at its own frequency with the right transistor combination and avoid the problem described above. – JamesHoux May 14 '13 at 8:10
• Just a follow up: I built a BJT oscillator without a 555 and it worked perfectly. I'm pretty certain the 555 itself was having the same issues as the mosfets. When voltage dropped into the negatives, it caused the 555 to source current that would bias on the BJT unexpectedly. – JamesHoux May 15 '13 at 9:52