Question

The pathway for a binary electrical signal between gates in an integrated circuit can be modeled as an RCcircuit, as in Figure 3.13(b); the voltage source models the transmitting gate, and the capacitor models the receiving gate. Typically, the resistance is $\mathbf{100}\mathbf{\Omega }$and the capacitance is very small, say, ${\mathbf{10}}^{\mathbf{-}\mathbf{12}} \mathbf{F}$(1 picofarad, pF). If the capacitor is initially uncharged and the transmitting gate changes instantaneously from 0 to 5 V, how long will it take for the voltage at the receiving gate to reach (say) ? (This is the time it takes to transmit a logical “1.”)

Short answer

The time taken by the signal is $\mathbf{t}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{163}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{sec}$.

Step-by-step solution

Important formula.

The governing differential equation for RC circuit is$\frac{\mathbf{dq}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{q}}{\mathbf{CR}}\mathbf{=}\frac{\mathbf{E}}{\mathbf{R}}$

Evaluate the value of  Qt

Here resistance $\mathbf{\left(}\mathbf{R}\mathbf{\right)}\mathbf{=}\mathbf{100}\mathbf{\Omega }$, Capacitance $\mathbf{\left(}\mathbf{C}\mathbf{\right)}\mathbf{=}\mathbf{1}{\mathbf{0}}^{\mathbf{-}\mathbf{12}}\mathbf{F}$, initial charge ${\mathbf{Q}}_{\mathbf{o}}\mathbf{=}\mathbf{0}$, voltage supplied to circuit $\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\mathbf{5} \mathbf{V}$.

Now the differential equation of RC circuit is

$$\begin{gathered} \frac{\mathbf{Q}}{\mathbf{C}}\mathbf{+}\mathbf{IR}\mathbf{=}\mathbf{E} \\ \frac{\mathbf{dQ}}{\mathbf{dt}}\mathbf{+}\frac{\mathbf{Q}}{\mathbf{RC}}\mathbf{=}\frac{\mathbf{E}}{\mathbf{R}} \end{gathered}$$

Now the integrating factor is ${\mathbf{e}}^{\frac{\mathbf{t}}{\mathbf{RC}}}$.

The equation is

$$\begin{gathered} {\mathbf{Qe}}^{\frac{\mathbf{t}}{\mathbf{RC}}}\mathbf{=}\int {\mathbf{e}}^{\frac{\mathbf{t}}{\mathbf{RC}}}\frac{\mathbf{E}}{\mathbf{R}}\mathbf{dt} \\ {\mathbf{Qe}}^{\frac{\mathbf{t}}{\mathbf{RC}}}\mathbf{=}{\mathbf{ECe}}^{\mathbf{tkRC}} \\ \mathbf{Q}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{EC}\mathbf{+}\mathbf{k}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{RC}}} \end{gathered}$$

When $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{Q}\mathbf{=}\mathbf{0}$then $\mathbf{k}\mathbf{=}\mathbf{-}\mathbf{CE}$.

$\mathbf{Q}\mathbf{ }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{CE}\mathbf{ }\mathbf{ }\left(1-e^{-\frac{t}{\mathrm{RC}}}\right)$

Determine the value of  Qc

$$\begin{gathered} \mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{{\mathbf{v}}_{\mathbf{c}}} \\ {\mathbf{v}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{C}} \\ {\mathbf{v}}_{\mathbf{c}}\mathbf{=}\mathbf{E}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{RC}}}\mathbf{)} \end{gathered}$$

Find the value of time.

When solving for t and put the all required values then

$$\begin{gathered} \mathbf{t}\mathbf{=}\mathbf{-}\mathbf{RCln}\left(1-\frac{v_c}{E}\right) \\ \mathbf{t}\mathbf{=}\mathbf{-}\mathbf{\left(}\mathbf{100}\mathbf{\right)}\mathbf{(}\mathbf{1}{\mathbf{0}}^{\mathbf{-}\mathbf{12}}\mathbf{)}\mathbf{ln}\left(1-\frac{E(1-e^{\frac{-t}{\mathrm{RC}}})}{E}\right) \\ \mathbf{t}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{163}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{sec} \end{gathered}$$

Therefore, the time taken by the signal is$\mathbf{t}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{163}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\mathbf{sec}$.