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The voltage across a discharging capacitor in an RC network is

$$V_c = V_s (e^{-\frac{t}{RC}})$$

Rearranging this equation for t gives

$$t = -RC \ln \left( \frac{V_c}{V_s} \right)$$

Say I have a capacitor charged such that the voltage across it is 3.6V and I want to find the time it takes to reach 0.9V. I remove the voltage source and the capacitor begins to discharge. Since I have removed the source, what is the value of \$V_s\$ in this equation, and how do I use it to find the time it takes the capacitor voltage to reach 0.9V?

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The more general form of the first equation is

$$v_C(t) = v_C(0)\cdot e^{\frac{-t}{RC}}$$

This must be so since, for \$t=0\$

$$e^{\frac{-0}{RC}} = 1$$

So, assuming the capacitor has charged up to the source voltage before you disconnect the source at \$t=0\$, we have

$$v_C(0) = V_S$$

and then

$$t(v_C) = - RC \ln \frac{v_C}{V_S} = RC \ln \frac{V_S}{v_C} $$

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  • \$\begingroup\$ Some extra information: This is for a schmitt trigger relaxation oscillator (which I'm looking in to). C = 1mF, R = 10k. Using your last equation gives t to be 13.9 seconds. However, using the oscillator period equation T = 0.5RC, I get 5 seconds. What's the right way? \$\endgroup\$ – imulsion Jan 29 '15 at 19:31
  • \$\begingroup\$ @imulsion, please post a schematic or a link to a schematic of the oscillator. \$\endgroup\$ – Alfred Centauri Jan 29 '15 at 23:03
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Vs is best described in this equation as "the voltage at time zero" not "the source voltage." Considering that V_c is "the voltage across the capacitor" it's actually a time dependent variable, so expressing it as such might help you understand it.

$$V_c(t) = V_c(0) * e^{-\frac{t}{RC}}$$

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  • \$\begingroup\$ So then (in my example) Vs would be 3.6V and Vc would be 0.9V? \$\endgroup\$ – imulsion Jan 29 '15 at 19:15
  • \$\begingroup\$ yes I think that's correct. Think of it like this - the time it takes to reach a certain voltage from a starting voltage is related to the log of the percentage of decay from that initial state. \$\endgroup\$ – vicatcu Jan 29 '15 at 19:21
  • \$\begingroup\$ Excellent, thank you \$\endgroup\$ – imulsion Jan 29 '15 at 19:22
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"However, using the oscillator period equation T = 0.5RC, I get 5 seconds. What's the right way?"

This depends on the oscillator design. You need to know the threshold voltages at which it flips over and use these values in your equation. In the design below this is set by R1,R2. Different designs have different threshold settings though.

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