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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惟and the capacitance is very small, say, 10-12鈥�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 鈥�1.鈥�)

Short Answer

Expert verified

The time taken by the signal is t=9.163脳10-11sec.

Step by step solution

01

Important formula.

The governing differential equation for RC circuit isdqdt+qCR=ER

02

Evaluate the value of  Qt

Here resistance (R)=100惟, Capacitance (C)=10-12F, initial charge Qo=0, voltage supplied to circuit (E)=5鈥�V.

Now the differential equation of RC circuit is

QC+IR=EdQdt+QRC=ER

Now the integrating factor is etRC.

The equation is

QetRC=鈭�etRCERdtQetRC=ECetkRCQ(t)=EC+ke-tRC

When t=0,鈥�Q=0then k=-CE.

Q鈥�(t)=CE鈥�鈥�(1-e-tRC)

03

Determine the value of  Qc

C=Qvcvc=QCvc=E(1-e-tRC)

04

Find the value of time.

When solving for t and put the all required values then

t=-RCln(1-vcE)t=-(100)(10-12)ln(1-E(1-e-tRC)E)t=9.163脳10-11sec

Therefore, the time taken by the signal ist=9.163脳10-11sec.

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