diodespassive-networksresonance
Initially when the switch is closed ,at that instant the inductor is a short and capacitor is open ,so the diode cathode is floating ,so how does the diode conduct?
Does the LC circuit resonate?
My hit at the solution:
under steady state the cap should charge to 100v and inductor voltage approaches 0v.
so answer should be 100volts
the ambiguity regarding resonance came in because of the natural response had sinusoids in it..
The answer is 200 volts ,but i don't understand it..
Since this question essentially became about a [LT]spice simulation, here's the answer to that "mystery".
The diode model in SPICE does have an ohmic resistance parameter (RS in the figure below) to simulate bonds/wires/contacts. At DC it looks like:
(Source: http://www3.imperial.ac.uk/pls/portallive/docs/1/7292572.PDF)
Since this circuit used an 1N4148 model (not the ideal diode model), the ohmic resistance (Rs) was set to a non-zero value; actually to 0.568ohms.
If you put a single 1N4148 (forward biased) in series with a 12V DC source, you'll get about 19.3A through it in LTspice (so with two you'll get about half that). A trivial calculation/approximation with Ohm's law (ignoring the voltage drop over the current source || GMIN) shows that 12V through 0.568ohm will give a 21.12A current limit.
Of course, as correctly emphasized by several others on this page, in real life, the 1N4148 diode catches fire (bonds vaporize and what not) long before reaching those levels of current.
Also, it seems there's a large variation between SPICE-based simulators' databases on diode RS values. For example TINA-TI has only 2miliohms for 1N4148, so in that sim you get 5.3kA through it.
Of course it would be nice if SPICE-based simulators implemented power dissipation limits for component packages and flashed a newbie-friendly warning when it's exceeded... but alas they typically don't have that feature because it's not standard in the academic & free SPICE codebase they're based on. (If someone knows of a counterexample, please leave a comment.) So in SPICE-based simulators it's normally left to the user to figure out that the current or power dissipation limit is exceeded for a component and that it catches fire in real life.
By placing a zener diode (or similar) between the drain and gate, the gate voltage will be forced to rise if the drain exceeds the breakdown voltage of the diode. You are correct in thinking that this will turn the MOSFET on. This is how that protection method works. To see why you have to understand MOSFET avalanche. The body diode of the MOSFET will break down at some voltage like a zener diode (or avalanche diode to be strictly correct). A power MOSFET is made up of many cells in parallel with each other. Depending of the processing of the transistor, these cells may or may not share the avalanche energy well. Also the place within the cell where the avalanche energy is dissipated is not quite the same as where normal conduction dissipation occurs. For this reason some MOSFETs have a limited amount of energy that they can safely absorb in a single pulse of avalanche. This energy may be much less then they could take in a pulse of normal conduction. It may be zero. On the other hand, some MOSFETs have avalanche energy limited only by thermal considerations, in other words they can take any amount of avalanche energy as long as the die does not overheat.
By turning the MOSFET on before the drain-source voltage exceeds the avalanche point, the energy is dissipated by the normal conduction mechanism. This allows a MOSFET which cannot take high avalanche energy to safely dissipate a transient, for example from and unclamped inductive load. You still have to ensure that in your design the MOSFET cannot be exposed to a transient that is too large from a thermal point of view, the energy will heat the die up just the same and you must not exceed Tj max.
You have to be careful adding diodes to the gate terminal. In some circumstances diodes connected directly to the gate can result in very high frequency oscillation of the MOSFET. This results from parasitic inductance and capacitance in the circuit causing a feedback path at VHF. A diode can rectify this and pump up the DC level on the gate resulting in a stable oscillation maintaining the device in a partially on state when trying to turn it off. This is mitigated by adding a suitable gate resistor.
If your circuit topology ensures that the MOSFETs will never see excessive voltage, or if they can withstand any transients by avalanche then you may well not need to provide any protection components. This is a good solution because there can be more problems introduced by the protection devices (too big a subject to cover here). Decoupling the supply rails near to the output devices in power stages helps. Ensuring that devices in bridge configurations cannot cross-conduct is also very important.
Best Answer
200 V is indeed the correct answer, assuming ideal components.
Think about this in the time domain. When the switch is initially closed, all the 100 V of the battery is applied across the inductor. Voltage across a inductor causes current thru it to rise linearly, with the slope proportional to the voltage. Immediately after turnon, the inductor current will therefore rise linearly.
However, the capacitor voltage builds up with the integral of this current. After a little while, the capacitor will have charged up a little, and the voltage across the inductor therefore reduced. This reduces the rate of current increase, but note that the current is still increasing. This current causes more voltage to build up across the cap, which decreases the voltage on the inductor, which decreases the rate of rise of the current.
Eventually the cap voltage builds up to the same as the supply voltage. At that point the voltage on the inductor is zero. However, that only means the current stops increasing, not that it stops. In fact, this is the point with the largest current.
Since the current keeps flowing, voltage on the cap keeps building up, which is now so high that the voltage across the inductor is negative, and the current starts going down. Eventually, this negative voltage on the inductor brings the current to zero.
However, at that point, the cap has charged up to twice the supply voltage. If the diode weren't there, the reverse voltage on the inductor would continue decreasing the current, now making it negative. This would eventually discharge the cap until it is at zero. That causes the current to increase, and the whole thing happens again. With ideal components, both the current across the inductor and voltage across the capacitor are sines, with the system oscillating continually ad infinitum.
In your case, the diode prevents the current from going negative. The system stops changing when the zero current point is reached, then stays that way forever (again, with ideal components). At that point the cap has 200 V on it.