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To calculate capacitor discharge time the formula is:

But because the current being sink from the capacitor is constant from highest voltage to zero volt, I think the capacitor should discharge faster as the capacitor voltage drops. right?

schematic

simulate this circuit – Schematic created using CircuitLab

  • How can I calculate how fast the capacitor will discharge to 0V when SW1 opens?
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  • \$\begingroup\$ What constant current sink? V2 is upside down? \$\endgroup\$
    – winny
    Commented Jul 1, 2022 at 12:24
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    \$\begingroup\$ The formula you posted is correct. That's just how capacitors work. \$\endgroup\$
    – Elliot Alderson
    Commented Jul 1, 2022 at 12:24
  • \$\begingroup\$ If you intend to keep your schematic then an analytic formula will be much more than you would like (you hav a transistor in there, nonlinear). If you had a constant current source (or sink) then yes, the formula will hold. You can deduce it, yourself, from the well-know formula, and by considering that the current is constant. \$\endgroup\$
    – a concerned citizen
    Commented Jul 1, 2022 at 12:48
  • \$\begingroup\$ @winny Yes, Q1,R1 and V2 make a constant current sink. \$\endgroup\$
    – ElectronSurf
    Commented Jul 1, 2022 at 14:07
  • \$\begingroup\$ @ElliotAlderson I was a bit confused about the formula, it's clear now with the simulation result in the answer to my question. \$\endgroup\$
    – ElectronSurf
    Commented Jul 1, 2022 at 14:11

1 Answer 1

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The only missing information in the correct equation you posted is the value of the ''fixed'' current I. You could roughly estimate it by assuming a fixed \$V_{BE}\$, which will be correct as long as the voltage range allows the BJT to stay in the active region. The smaller the magnitude of the negative voltage, the larger the error in the calculation due to \$V_{BE}\$ variations.

For example, using these values (the switch and input voltage were replaced by the initial voltage):

enter image description here

\$I=\dfrac{5V-0.6V}{10k\Omega} = 440\mu A\$

\$\dfrac{10\mu F \times (10V-0V)}{I} = 227ms\$

Which is quite close to the simulation results:

enter image description here

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