# Time taken to charge a capacitor

I was told by a lecturer to study the 555 timer in astable mode, I'm pretty confident I know just about everything now except for how the formula for the output high and low time was derived. Things I know:

• The high time is the time the capacitor takes to charge

• The low time is the time the capacitor takes to discharge

I tried to derive a general formula for both myself:

Q = It = CV = CIR t = RC But the formula is: t = 0.693RC

• There are many online resources for using the 555 timer. Have you searched for the answers to your questions? – Elliot Alderson Oct 5 '18 at 18:16
• Wikipedia en.wikipedia.org/wiki/RC_circuit. – Oldfart Oct 5 '18 at 18:16
• Do you know how to derive the charging rate for an RC circuit? – Andy aka Oct 5 '18 at 18:32
• @Andy aka, I don't know how – lekarane Oct 5 '18 at 18:36
• That's your starting point - not the 555 – Andy aka Oct 5 '18 at 18:37

Here are the schematic and naming conventions used :

simulate this circuit – Schematic created using CircuitLab

# Charge and discharge bounds

You probably know that already, but the theory of operation is as follows : When TRIGGER is pulled at $$\2/3 \cdot V_{CC}\$$, the output goes $$\\text{LOW}\$$, and the DISCHARGE pin becomes a short to ground. $$\C_1\$$ is then discharged through $$\R_2\$$ to ground. When the THRESHOLD's pin voltage (and $$\C_1\$$'s voltage reaches $$\1/3\cdot V_{CC}\$$, the output goes $$\\text{HIGH}\$$, the DISCHARGE pin becomes open-circuited, and $$\C_1\$$ charges to $$\V_{CC}\$$ through $$\R_1\$$ and $$\R_2\$$.

The $$\\text{HIGH}\$$ period is then the time the capacitor takes to charge from $$\1/3\$$ to $$\2/3\$$ of the supply through the equivalent resistance $$\R_{tot} = R_1 + R_2\$$.
The $$\\text{LOW}\$$ period is the time the capacitor takes to discharge from $$\2/3\$$ to $$\1/3\$$ of the supply through $$\R_2\$$.

# The series RC circuit charge-discharge formula

The charge-discharge formula of an RC circuit is : $$V_C(t) = \Delta V(1 - e^{-t/RC}) + V_I$$ Where

• $$\V_I\$$ is the initial voltage across the capacitor.
• $$\\Delta V\$$ is the difference between the initial voltage $$\V_I\$$ and the steady state voltage once the circuit will have stabilized at the limit.

Often, it is assumed $$\V_I = 0\$$ and the formula takes the more well-known form of : $$V_C(t) = V_{\text{terminal}}(1 - e^{-t/RC})$$

# Application to the 555 timer's astable mode circuit

Only the charging part will be discussed here, the discharging part of the period follows the same principles.

During the charging part of the cycle, the capacitor charges (if the 555 didn't exist) to $$\V_{CC}\$$ from $$\1/3 \cdot V_{CC}\$$. It is trivial to see that , $$\V_I = 1/3\cdot V_{CC}\$$ and $$\\Delta V = 2/3 \cdot V_{CC}\$$. For $$\t = 0\$$ the moment the capacitor begins to charge, the $$\\text{HIGH}\$$ period is $$\t\$$ such that $$\V_C(t) = 2/3 \cdot V_{CC}\$$, as when this is satisfied, the capacitor begins its discharge, and the output becomes $$\\text{LOW}\$$.

\begin{align} 2/3 \cdot V_{CC}(1 - e^{-t/R_{tot}C}) + 1/3\cdot V_{CC} = 2/3\cdot V_{CC} & \Rightarrow 2/3(1 - e^{-t/R_{tot}C}) + 1/3 = 2/3 \\ & \Rightarrow 1 - e^{-t/R_{tot}C} + 1/3 \cdot 3/2 = 1 \\ & \Rightarrow -e^{-t/R_{tot}C} = 1 - 1 - 1/2 = -1/2 \\ & \Rightarrow -t/(R_{tot}C) = \ln(1/2) \\ & \Rightarrow t = -\ln(1/2) \cdot (R_{tot}C) \approx 0.693 \cdot R_{tot}C \end{align}

Changing what needs to be, the discharge part of the cycle follows the same line of reasoning.

I think you did great for deriving the general formula t=RC for a simple RC circuit.

But the problem is that with the 555 circuit, the capacitor voltage varies between 1/3 Vcc and 2/3 Vcc and you did not consider that.

You must introduce a formula containing an exponential part and solve that for time if you want to get the right result.