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<p>in an avalanche-like manner until finally Q1 switches off and Q2 switches on. The forward-biased Q2 base-emitter junction fixes the voltage of C1 right-hand plate at 0.6 V and does not allow it to continue rising toward +V.</p>

<p><b>State 2 (Q1 is switched off, Q2 is switched on)</b></p>

<p>Now, the capacitor C2 is fully charged (in the previous State 1) to the power supply voltage V with the polarity shown in Figure 1. Q2 is on and connects the right-hand positive plate of C2 to ground. As its left-hand negative plate is connected to Q1 base,</p><p>
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