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In this circuit and explanation, it all makes sense except I'm having a hard time picturing Q2 not turning on until C1 has discharged first.

When C1 is discharging through Q1 why doesn't Q2 turn on as it's discharging since it's base is higher than 0.6v through R2?

Also is C1 discharging only by the + plate having a path to ground rather than both + and - plates of C1 being connected together?12

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  • \$\begingroup\$ Draw the waveforms. \$\endgroup\$ Aug 14, 2017 at 16:25

1 Answer 1

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Suppose C1 is charged with 4 Volts with the positive voltage at side marked +.

Now suppose that Q1 switches on pulling the collector of Q1 to 0 V (ground). Since C1 is still charged with 4 V, the base of Q2 will be at - 4 V. So Q2 is forced to be off as Vbe < 0.6 V. It is actually negative !

The base of Q2 needs to be above 0.6 V before it can conduct so C1 has to discharge to at least -0.6 V for that to happen (still assuming Q1 to be on so C1's + plate is 0 V).

So in reality C1 first discharges completely (to 0 Volt) and then charges with opposite polarity to at least 0.6 V.

For discharging a capacitor it is not really needed to short the capacitor plates. As long as the voltage on both plates is the same then the capacitor is discharged. It does not matter how this is achieved (making the voltages the same).

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  • \$\begingroup\$ Thank you. If you could help me wrap my head around why the base of Q2 sees -4v when C1 is still charged? Why is the voltage there negative if R2 connects the base directly to Vcc? Is it because when Q1 is on it's actually pulling Vcc down as well? \$\endgroup\$
    – user140123
    Aug 14, 2017 at 14:32
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    \$\begingroup\$ C1 capacitor previously charged to 4V. The voltage polarity across the capacitor is: left plate of C1 is at + and the right plate is at "-". So when Q1 turns on he connected the left plate of a C1 capacitor to gnd, so the right plate becomes -4V and this is why we have a negative voltage at the Q2 base. \$\endgroup\$
    – G36
    Aug 14, 2017 at 14:40
  • \$\begingroup\$ What G36 writes. The Vcc is not pulled down, it is the voltage at the collector of Q1 which is pulled down. When C1 is charged you can see it as a 4 V battery with the + at the collector of Q1. Then Q1 closes, collector becomes 0 V so the - side of "battery" C1 becomes - 4 V. \$\endgroup\$ Aug 14, 2017 at 14:43
  • \$\begingroup\$ @Bimpelrekkie Oh so a capacitor can have a negative charge as well on the negative plate? Even electrolytics? \$\endgroup\$
    – user140123
    Aug 14, 2017 at 15:00
  • \$\begingroup\$ You mean that it can have a negative charge on the positive plate. Yes any capacitor can be charged either way. However on polarized capacitors (for example electrolytic caps) there is a limit on the reverse voltage. Here that will be less than 1 V. Normal electrolytic caps do not have a problem with that. You can charge a 25 V rated electrolytic cap to -25 V but it will not like that, it will start getting warm and eventually it will blow off pressure or even blow itself up. After that it must be replaced. \$\endgroup\$ Aug 14, 2017 at 15:14

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