Question

In the dynamic random access memory (DRAM) of a computer, each memory cell contains a capacitor for charge storage. Each of these cells represents a single binary bit value of “1” when its 35-fF capacitor \(\left( {{\bf{1}}\;{\bf{fF = 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{F}}} \right)\) is charged at 1.5 V, or “0” when uncharged at 0 V.

(a) When fully charged, how many excess electrons are on a cell capacitor’s negative plate?

(b) After charge has been placed on a cell capacitor’s plate, it slowly “leaks” off at a rate of about \({\bf{0}}{\bf{.30}}\;{\bf{fC/s}}\). How long does it take for the potential difference across this capacitor to decrease by 2.0% from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is “refreshed” many times per second.) Note: A DRAM cell is shown in Fig. 21–29.

Short answer

(a) The number of excess electrons on the capacitor’s plate is \(3.28 \times {10^5}\;{\rm{electrons}}\).

(b) The time taken for the potential difference across this capacitor to decrease is \(3.5\;{\rm{s}}\).

Step-by-step solution

Understanding the potential difference and charge

In this problem, use the relation of absolute charge on the parallel plate to find the number of electrons. The charge relies on the potential difference, which means that the charge will reduce with a reduction in potential difference.

Given Data

The capacitance is,\(C = 35\;{\rm{fF}}\).

The voltage is,\(V = 1.5\;{\rm{V}}\).

The leakage rate is,\(\frac{{\Delta q}}{{\Delta t}} = 0.30\;{\rm{fC/s}}\).

The percent decrease is \(P = 2.0\% \).

(a) Determination of the number of electrons

The number of electrons is given by,

\(\begin{array}{l}N = \frac{Q}{e}\\N = \frac{{CV}}{e}\end{array}\)

Here, Qis the charge and eis the charge of the electron.

Substitute the values in the above expression.

\(\begin{array}{l}N = \left[ {\frac{{\left( {35\;{\rm{fF}} \times \frac{{{{10}^{ - 15}}\;{\rm{F}}}}{{1\;{\rm{fF}}}}} \right)\left( {1.5\;{\rm{V}}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}} \right]\\N = 3.28 \times {10^5}\;{\rm{electrons}}\end{array}\)

Thus, the number of excess electrons on the capacitor’s plate is \(3.28 \times {10^5}\;{\rm{electrons}}\).

(b) Determination of the time taken for the potential difference to decrease

Since the potential difference and charge are directly proportional to each other. Therefore, the charge will decrease by 2%.

The relation to find time is given by,

\(\Delta t = \frac{{\Delta Q}}{{\left( {\frac{{\Delta q}}{{\Delta t}}} \right)}}\)

Here, \(\Delta Q\) is the decrease in charge.

Substitute the values in the above expression.

\(\begin{array}{l}\Delta t = \frac{{0.02q}}{{\left( {0.30\;{\rm{fC/s}}} \right)}}\\\Delta t = \frac{{0.02\left( {CV} \right)}}{{\left( {0.30\;{\rm{fC/s}}} \right)}}\\\Delta t = \frac{{0.02\left( {35\;{\rm{fF}} \times \frac{{{{10}^{ - 15}}\;{\rm{F}}}}{{1\;{\rm{fF}}}}} \right)\left( {1.5\;{\rm{V}}} \right)}}{{\left( {0.30\;{\rm{fC/s}} \times \frac{{{{10}^{ - 15}}\;{\rm{C/s}}}}{{1\;{\rm{fC/s}}}}} \right)}}\\\Delta t = 3.5\;{\rm{s}}\end{array}\)

Thus, thetime taken for the potential difference across this capacitor to decrease is \(3.5\;{\rm{s}}\).