Electrical – What makes the charge on a capacitor change

The instant that transistor, TR1 switches "ON", plate "A" of the capacitor immediately falls to 0.6 volts.

In an NPN BJT, the collector (c) plate will normally never drop below the base (b) plate. This is because a BJT consists of two diodes, one from the base to the emitter, the other from the base to the collector. The BJT diagram actually shows the diode between base and emitter. Thus, current flowing from collector to emitter must first pass the base.

Thus, the collector can never drop below the base, as then there no longer is a voltage that drives current from collector to base. And if the current can not reach the base, it can never reach the emitter. Current would no longer flow down the collector, and resistor R1 would pull the collector voltage up. However, transistor in this circuit operate in deep saturation, so voltages will differ from the 0.6 volts normal for silicon BJT's.

So how am I suppose to read this negative voltage on the capacitor ?

• plate A drops from 6V -> 0,6V (a 5,4V drop)
• plate B also needs to drop 5,4V ? from 0V to -5,4V ?

Exactly. The voltage over a capacitor can not change instantaneously, so if one plate drops, the other must drop likewise, even if this results in a negative voltage outside the supply rails. A current must flow for some time to change the voltage over a capacitor. This same trick is used in charge pumps to generate very high voltages. Coils work exactly the opposite: there, the current can not change instantaneously.

After the swing, transistor TR2 blocks, meaning no current flows from the base to the emitter. Thus all current flowing through R3 will flow into the capacitor, raising the voltage of plate B. At some point, its voltage will raise above the point where TR2 starts to conduct again, and the circuit switches to its other state. This happens very quickly: as TR2 starts to conduct, output 2 starts to drop. This drop is fed to the base of TR1 through C2, causing TR1 to conduct less, which causes output 1 to rise, which causes B to rise, which causes TR2 to conduct even more, etc. This is an example of positive feedback.

It's like the "paradox" of the immovable object meeting the irresistible force. In reality, neither can exist.

A real power source will have some impedance. A real capacitor will have some impedance. Real wires connecting them have resistance and inductance.

So in reality when you slap a fairly 'stiff' power source across a fairly good real capacitor there's a spark and the capacitor charges through those series resistances with some ringing and stuff due to the inductances. Ignoring the inductances, the voltage difference would simply divide in ratio to the internal resistance of the capacitor and the internal resistance of the power supply and the wire resistance.

To answer your specific question: If the capacitor and voltage source and wires are ideal, you have a mathematical problem, like division by zero. It's of no consequence in the real world- it just illustrates that the ideal models of the power source and the capacitor and wires are insufficiently accurate to describe their real-world behavior. Modelling any one of those as a real part with real resistance (and inductance) will make the mathematical problem go away, but it won't likely give you an accurate indication of what is actually happening.

For example, if the wire (or the capacitor or the power supply) had 10m\$\Omega\$ resistance, and there is a 10V difference, you could predict you'd see 1000A (which is very high, but not infinity) and the capacitor would charge very quickly. In reality that isn't likely going to happen because of other non-ideal factors. If the 10m\$\Omega\$ was modeled as in the power supply, the power supply voltage would drop. If the 10m\$\Omega\$ was modeled as in the capacitor, the voltage would suddenly appear across the capacitor terminals. If the 10m\$\Omega\$ was modeled as in the wire, the voltage would appear across the wire. But none of those is very realistic.

If you modeled as a circuit with no resistance at all and a tiny bit of inductance (even superconducting wires of any length have inductance) then a simple mathematical model would predict ringing that would persist forever, energy sloshing back and forth between the inductance and capacitance at an angular frequency of \$\omega_0 = {1 \over \sqrt{LC}}\$.