Using the formula for RC circuits: V*e^(-t/RC) Show steps, thank you! I. Suppose
ID: 1778293 • Letter: U
Question
Using the formula for RC circuits: V*e^(-t/RC)
Show steps, thank you!
I. Suppose that C = 30 fF (that is femto Farads look it up!). Suppose that once the switch is closed we want to take 200,sec for the voltage to drop to 99% of its starting value. (In other words, after the switch has been closed for 200,sec the voltage across the capacitor should drop from V to 0.991). Determine the value required for R (the sense amplifier's input resistance) to make this happen. 2. If you instead used a value of R that is double the R value you found in #1, what would that do to the time to takes for the voltage to drop? State a simple rule that covers this relationship between R and the time to drop to 99%. Explain your answer!Explanation / Answer
Given,
C = 30 fF ; t = 200 us
V = 99% of V0 = 0.99 V0
We know that,
V = V0 e^-t/RC
0.99 V0 = V0 e^-t/RC
0.99 = e^-t/RC
taking natural log both sides
ln(0.99) = -t/RC
-0.0101 = -t/RC
R = t/0.0101 C = 200 x 10^-6 /0.0101 x 30 x 10^-15 = 6.6 x 10^11 Ohm
Hence, R = 6.6 x 10^11 Ohm
2)R' = 2R = 13.2 x 10^11
Again using the same relation
V = V0 e^-t/RC
0.99 V0 = V0 e^-t/RC
taking natural log both sides
ln(0.99) = -t/RC
-0.0101 = -t/RC
t = 0.0101 x RC = 0.0101 x 13.2 x 10^11 x 30 x 10^-15 = 0.0004 = 400 uS
Hence, t = 400 us
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