# Why parallel LC oscillator producing very high voltage?

I was trying to generate 1 MHZ frequency using LC parallel circuit using 2.53 mH inductor and 0.01 nF capacitor and I successfully did so. But its amplitude in KVs is surprising me. Any idea why? • Because you don't have any losses! Try simulating real components. – Leon Heller Dec 27 '18 at 12:15
• @LeonHeller How is the voltage amplified? – Muhammad Naufil Dec 27 '18 at 12:19
• Add 1 ohm series resistor, or 50 ohms as the Rout of a function generator. Then Re-sim. – analogsystemsrf Dec 27 '18 at 19:35

Imagine, if you can, that prior to opening the switch, there is (say) 10 amps flowing from V1 (the 12 volt supply) into inductor L1 (2.53 mH). The energy stored (W) in an inductor is: -

$$W = \dfrac{1}{2}\cdot L\cdot I^2$$

This works out at about 0.127 joules. When the switch opens, that energy is released by the inductor and will "fill-up" the capacitor C1 (10 pF). The energy equation for a capacitor is: -

$$W = \dfrac{1}{2}\cdot C\cdot V^2$$

And, if you calculate the peak voltage across the capacitor, that comes to 159 kV. That voltage will rise and fall sinusoidally as energy swishes back and forth between capacitor and inductor.

Maybe if you limited the current to something like 10 mA you would get a more realizable circuit.

When you close key A, you might think you're charging the capacitor to 12v. You're also charging the inductor to a large current, which will store a lot of energy.

How large a current? With the circuit components you have there, 12v and zero resistance, the current ought to be infinite. However, as the simulator has come up with some finite numbers, it looks like it imposes its own lowish value for infinite things, or maybe uses a finite resistance in voltage sources or inductors to limit the current.

In real life, the inductor or battery would have some resistance. However, you would still expect a significant voltage rise from energy stored in the inductor.

• Good old-fashioned SPICE would have barfed at a zero-resistance path from a voltage source to ground. I'm not sure what that simulator is, but I suspect that it's trying to use "realistic" resistances in either the coil, the switch, or the "battery". – TimWescott Dec 27 '18 at 15:45

I added the switch B in resonance circuit.

First let the capacitor charge via closed switch A and opened switch B, so that no current passes through inductor because inductor opposes the change of current. When it was fully charged, I opened switch A and closed switch B to complete the loop[! Now it is working as it should be.