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I am just trying to understand why we use 2.2×Tau when calculate the rise time. I can't find a derivation anywhere, I don't understand where this 2.2 came from.

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For charging:

$$V(t) = V_0(1 - e^{-t/\tau})$$

Lets say rise time 10% to 90% of V0. So in this equation put: $$V(t) = 0.1 \cdot V_0$$ and find \$t_1\$. Similarly put: $$V(t) = 0.9 \cdot V_0$$ and find \$t_2\$. You'll get: $$rise time = t_2-t_1 = 2.2 \tau$$

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    \$\begingroup\$ Supplementing the above answer we have for the rise time (defined between 10% and 90% of the final value): Tr=tau*(ln0.9-ln0.1). \$\endgroup\$ – LvW May 4 '14 at 8:18
  • \$\begingroup\$ Thank you! that was the part i failed to do! (ln0.9-ln0.1) = 2.197 :) \$\endgroup\$ – user3474943 May 4 '14 at 8:46
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Using the standard formula of Vc=A+Be-(T/Tau) you solve for 90% and 10% and get T90%=-Tau*ln(0.1)=Tau*2.3026 & T10%=-Tau*ln(0.9)=0.1054 and subtracting T10% from T90% you get Trise=2.1972*Tau or about 2.2*Tau; where Tau=R*C.

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