Q95GP In the DRAM computer chip of Pro... [FREE SOLUTION]

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Chapter 17: Q95GP (page 473)

In the DRAM computer chip of Problem 94, suppose the two parallel plates of one cell鈥檚 35-fF capacitor are separated by a 2.0-nm-thick insulating material with dielectric constant K= 25.
(a) Determine the area A\(\left( {{\bf{\mu }}{{\bf{m}}^{\bf{2}}}} \right)\)of the cell capacitor鈥檚 plates.
(b) If the plate area A accounts for half of the area of each cell, estimate how many megabytes of memory can be placed on a\({\bf{3}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\)silicon wafer.\(\left( {{\bf{1}}\;{\bf{byte = 8}}\;{\bf{bits}}{\bf{.}}} \right)\)

Short Answer

Expert verified

(a) The area of the cell capacitor鈥檚 plates is \(0.32\;\mu {{\rm{m}}^2}\).

(b) The number of megabytes of memory is \(58\;{\rm{Mbytes}}\).

Step by step solution

Understanding the capacitance

The capacitance of a capacitor relies on the area of the call鈥檚 capacitor鈥檚 plates and their separation.

The capacitor is given by,

\(C = K\frac{{{\varepsilon _0}A}}{d}\) 鈥� (i)

Here, K is the dielectric constant,\({\varepsilon _0}\)is the permittivity of free space, A is the area and d is the plate separation.

Given Data

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

The thickness is,\(d = 2\;{\rm{nm}}\).

The dielectric constant is,\(K = 25\)

The area is, \({A_{\rm{s}}} = 3\;{\rm{c}}{{\rm{m}}^2}\).

Determination of the area of each cell

The relation of area of each cell is given by,

\(A = \frac{{C \times d}}{{K{\varepsilon _0}}}\)

Here, \({\varepsilon _0}\)is the permittivity of free space and Ais the area of each cell.

Substitute the values in the above expression.

\(\begin{array}{l}A = \left[ {\frac{{\left( {35\;{\rm{fF}} \times \frac{{{{10}^{ - 15}}\;{\rm{F}}}}{{1\;{\rm{fF}}}}} \right) \times \left( {2\;{\rm{nm}} \times \frac{{{{10}^{ - 9}}\;{\rm{m}}}}{{1\;{\rm{nm}}}}} \right)}}{{\left( {25} \right)\left( {8.85 \times {{10}^{ - 12}}\;{\rm{F/m}}} \right)}}} \right]\\A = \left( {3.16 \times {{10}^{ - 13}}\;{{\rm{m}}^2} \times {{\left( {\frac{{{{10}^6}\;\mu {\rm{m}}}}{{1\;{\rm{m}}}}} \right)}^2}} \right)\\A = 0.32\;\mu {{\rm{m}}^2}\end{array}\)

Thus, the area of the cell capacitor鈥檚 plates is \(0.32\;\mu {{\rm{m}}^2}\).

Calculation of the number of megabytes

Since half of the area cell is utilized for capacitance, the available area is,

\(\begin{array}{c}{A_{\rm{s}}} = \frac{{3\;{\rm{c}}{{\rm{m}}^2}}}{2}\\{A_{\rm{s}}} = 1.5\;{\rm{c}}{{\rm{m}}^2}\end{array}\)

The relation to find the number of megabytes of memory is given by,

\(\begin{array}{l}n = \left( {1.5\;{\rm{c}}{{\rm{m}}^2}} \right){\left( {\frac{{{{10}^6}\;\mu {\rm{m}}}}{{{{10}^2}\;{\rm{cm}}}}} \right)^2}\left( {\frac{{1\;{\rm{bit}}}}{{0.32\;\mu {{\rm{m}}^2}}}} \right)\left( {\frac{{1\;{\rm{byte}}}}{{8\;{\rm{bits}}}}} \right)\\n = \left( {5.8 \times {{10}^7}\;{\rm{bytes}} \times \frac{{1\;{\rm{Mbytes}}}}{{{{10}^6}\;{\rm{bytes}}}}} \right)\\n = 58\;{\rm{Mbytes}}\end{array}\)

Thus, thenumber of megabytes of memory is \(58\;{\rm{Mbytes}}\).