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I'm a beginner and now I'm learning how this circuit works.

This circuit is transistor tester. enter image description here

I wonder how capacitor C2 discharges, because it needs to go to the other side of capacitor, but it can't because the BE junction blocks it.

In my book they said C2 discharges through ground. Is this true? Does that mean it flows through R5>>ground>>9V>>R3>>R1>> and the other side of the capacitor?

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  • \$\begingroup\$ See the arrow labelled (1)? The BE junction is forward biased, so it doesn't block the current. T1 is the source of that current. The reverse current is shown by arrow (2), which follows the path you described. R3 is the source of that current. \$\endgroup\$
    – Dave Tweed
    Commented Nov 1, 2022 at 11:07
  • \$\begingroup\$ Does capacitor C2 discharge ? In this circuit C2 is non polarized capacitor I add positive sign to show that now it charged this side . \$\endgroup\$
    – Heroz
    Commented Nov 1, 2022 at 11:08
  • \$\begingroup\$ Yes. Once C2 approaches full charge, the current through it drops low enough that "NPN (TOT)" cuts off. This causes T1 to cut off as well, and R5 pulls the positive end of C2 to ground. This means that the other end of C2 becomes negative with respect to ground, reinforcing the fact that both transistors are off. Now the only path for current is through R3 and R1 -- arrow (2). Since this current is the reverse of the charging current, C2 now discharges -- relatively slowly, since R1 is so large relative to R3 and R5. \$\endgroup\$
    – Dave Tweed
    Commented Nov 1, 2022 at 11:26
  • \$\begingroup\$ As it does so, the negative voltage at the base of "NPN (TOT)" rises, and once it becomes positive by enough to turn the transistor on, the cycle repeats. This is a form of "relaxation oscillator". Don't be confused by the little sine wave notations in the schematic -- the actual waveform is quite different. \$\endgroup\$
    – Dave Tweed
    Commented Nov 1, 2022 at 11:30
  • \$\begingroup\$ @DaveTweed..... Thank you very much . So it discharge through R5>>ground>>9V>>R3>>R1>> and the other side of the capacitor path. \$\endgroup\$
    – Heroz
    Commented Nov 1, 2022 at 12:15

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