• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

### Select your language

Suggested languages for you:

Americas

Europe

Q 3.5-3E

Expert-verified
Found in: Page 121

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

### Answers without the blur.

Just sign up for free and you're in.

# The pathway for a binary electrical signal between gates in an integrated circuit can be modeled as an RC circuit, 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.”)

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

See the step by step solution

## Step 1: Important formula.

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

## Step 2: 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

$\frac{\mathbf{Q}}{\mathbf{C}}{\mathbf{+}}{\mathbf{IR}}{\mathbf{=}}{\mathbf{E}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dQ}}{\mathbf{dt}}{\mathbf{+}}\frac{\mathbf{Q}}{\mathbf{RC}}{\mathbf{=}}\frac{\mathbf{E}}{\mathbf{R}}\phantom{\rule{0ex}{0ex}}$

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

The equation is

${{\mathbf{Qe}}}^{\frac{\mathbf{t}}{\mathbf{RC}}}{\mathbf{=}}{\int }{{\mathbf{e}}}^{\frac{\mathbf{t}}{\mathbf{RC}}}\frac{\mathbf{E}}{\mathbf{R}}{\mathbf{dt}}\phantom{\rule{0ex}{0ex}}{{\mathbf{Qe}}}^{\frac{\mathbf{t}}{\mathbf{RC}}}{\mathbf{=}}{{\mathbf{ECe}}}^{{\mathbf{tkRC}}}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{EC}\mathbf{+}\mathbf{k}{{\mathbf{e}}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{RC}}}$

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

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

## Step 3: Determine the value of  Qc

${\mathbf{C}}{\mathbf{=}}\frac{\mathbf{Q}}{{\mathbf{v}}_{\mathbf{c}}}\phantom{\rule{0ex}{0ex}}{{\mathbf{v}}}_{{\mathbf{c}}}{\mathbf{=}}\frac{\mathbf{Q}}{\mathbf{C}}\phantom{\rule{0ex}{0ex}}{{\mathbf{v}}}_{{\mathbf{c}}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{1}\mathbf{-}{{\mathbf{e}}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{RC}}}{\mathbf{\right)}}$

## Step 4: Find the value of time.

When solving for t and put the all required values then

${\mathbf{t}}{\mathbf{=}}{\mathbf{-}}{\mathbf{RCln}}\left(1-\frac{{v}_{c}}{E}\right)\phantom{\rule{0ex}{0ex}}\mathbf{t}\mathbf{=}\mathbf{-}\mathbf{\left(}\mathbf{100}\mathbf{\right)}\mathbf{\left(}\mathbf{1}{{\mathbf{0}}}^{\mathbf{-}\mathbf{12}}\mathbf{\right)}\mathbf{ln}\left(1-\frac{E\left(1-{e}^{\frac{-t}{\mathrm{RC}}}\right)}{E}\right)\phantom{\rule{0ex}{0ex}}{\mathbf{t}}{\mathbf{=}}{\mathbf{9}}{\mathbf{.}}{\mathbf{163}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{11}}{\mathbf{sec}}\phantom{\rule{0ex}{0ex}}$

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

### Want to see more solutions like these?

Sign up for free to discover our expert answers

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.