Diode Capacitor Circuit – How Negative Input Voltage Affects Output

No, there is no circuit being completed, hence no means for the "charge" to flow into the battery.

It appears that you expect a Coulomb static charge behavior to occur, such as in a Van De Graaf generator. That's static electricity, which albeit analogous to battery type "Direct Current" electronics, is not quite the same thing.

For one thing, batteries are not Coulombic charge storage devices: They are, at the simplest, reversible chemical reaction storage of energy resulting from ion flow caused by a potential difference. "No actual electrons are harmed in the battery process".

Also, though capacitors are technically charge storage devices, they would require a discharge path towards a lower absolute static charge potential point, for what you have asked, to occur. That's not the case here.

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.