Electronic – LC oscillator versus Kirchhoff’s law for meshes

Can anyone explain how does the current amplitude go to infinity, at resonance?

The brief answer is that the AC steady state current isn't infinite but, rather, the circuit has no AC steady state solution.

Recall that one of the assumptions justifying AC (phasor) analysis is that the circuit is in AC steady state, i.e., that all transients have decayed.

For the circuit given, the time domain solution for the current is proportional to

$$i(t) \propto t \cos\left( \frac{t}{\sqrt{LC}}\right),\, t \ge 0$$

The amplitude of the current starts at zero and grows linearly with time once the switch is closed but for any value of time \$t\$, the current is finite, i.e., the current is never infinite.

Note that this solution has no sinusoidal steady state - the amplitude does not approach a constant as \$t \rightarrow \infty\$ so this solution has no phasor representation and, thus, we should not be surprised that applying phasor analysis to this problem produces an undefined division by zero result.

Given the solution for the current, one can solve for the voltages across the inductor and capacitor as well as the energy stored in each as a function of time.

Different initial conditions will have different initial energies but will not affect the main result that the amplitude of the current will grow without bound once the switch is closed.

Can you please list the steps that lead to the derivation of the expression i(t)∝ tcos(t/√LC),t≥0 ?

By KVL, we have

$$v_S = v_L + v_C = L\frac{di}{dt} + \frac{1}{C}\int_0^ti(\tau)d\tau$$

(we assume zero initial voltage across the capacitor).

Differentiating both sides with respect to time and dividing through by \$L\$ yields

$$\frac{d^2i}{dt^2} + \frac{1}{LC}i = \frac{1}{L}\frac{dv_S}{dt}$$

Assuming \$v_S = V\cos\omega_0 t\$ yields the following non-homogeneous 2nd order ODE:

$$\frac{d^2i}{dt^2} + \frac{1}{LC}i = -\frac{\omega_0 V}{L}\sin \omega_0 t $$

where

$$\omega_0 = \frac{1}{\sqrt{LC}} $$

Assume a solution of the form

$$i(t) = t\left(A \cos \omega_0 t + B \sin \omega_0 t \right) $$

Substitute this \$i(t)\$ into the ODE to find

$$A = \frac{V}{2L}, B = 0 $$

First, look at the circuit with the capacitor removed. You will see that when power is applied, Q1 is turned on by the base resistors R3 and R4. The Q1 collector current goes into the Q2 base, turning on Q2. The Q2 collector current turns on Q1 even harder, which in turn drives Q1. This is a positive feedback loop. These two transistors will always be conducting, and the stable state is for both of them to be in saturation. They are locked up. Yes, R1 and R2 do provide some negative feedback, but not nearly enough to cancel the strong positive feedback of the cross-coupled collector to base connections.

Now add the capacitor back to the circuit. When Q1 turns on, the emitter voltage goes more negative. This transition is coupled to the Q2 emitter, driving it more negative, which causes Q2 to turn on as well. This is also positive feedback. Yes, over time current through the capacitor will diminish, but since the transistors are stable in saturation there will be no oscillation.

You need to provide negative DC feedback to stabilize the transistors in a linear operating region. Only then will the positive feedback through the capacitor cause oscillation. Look at your collector-to-base connections. These are the source of your latch-up problem.

If you redraw your schematic you might find it easier to understand:

^{simulate this circuit – Schematic created using CircuitLab}