Chapter 3: Q 3.5-3E (page 121)

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 $100 Ω$ and the capacitance is very small, say, $10^{- 12} F$ (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.”)

Expert verified

The time taken by the signal is $t = 9.163 \times 10^{- 11} \sec$ .

Step by step solution

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

Here resistance $(R) = 100 Ω$ , Capacitance $(C) = 1 0^{- 12} F$ , initial charge $Q_{o} = 0$ , voltage supplied to circuit $(E) = 5 V$ .

Now the differential equation of RC circuit is

$\frac{Q}{C} + IR = E \frac{dQ}{dt} + \frac{Q}{RC} = \frac{E}{R}$

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

The equation is

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

When $t = 0, Q = 0$ then $k = - CE$ .

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

$C = \frac{Q}{v_{c}} v_{c} = \frac{Q}{C} v_{c} = E (1 - e^{\frac{- t}{RC}})$

When solving for t and put the all required values then

$t = - RCln (1 - \frac{v_{c}}{E}) t = - (100) (1 0^{- 12}) \ln (1 - \frac{E (1 - e^{\frac{- t}{RC}})}{E}) t = 9.163 \times 10^{- 11} \sec$

Therefore, the time taken by the signal is $t = 9.163 \times 10^{- 11} \sec$ .

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