In a dynamic random access memory (DRAM)computer chip, each memory cell chiefly

Question

In a dynamic random access memory (DRAM)computer chip, each memory cell chiefly consists of a capacitor for charge storage. Each of these cells represents a single binary-bit value of 1 when its 35-fF capacitor (1 fF = 10^-15 F) is charged at 1.5 V, or 0 when uncharged at 0 V. When it is fully charged, how many excess electrons are on a cell capacitor's negative plate? Express your answer using two significant figures. After charge has been placed on a cell capacitor's plate, it slowly "leaks" off (through a variety of mechanisms) at a constant rate of 0.30 fC/s. How long does it take for the potential difference across this capacitor to decrease by 1.0% from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is "refreshed" many times per second.) Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

(a)
Q = C*V = (3.5*10^-14F)*(1.5V) = 5.25*10^-14 C
Q = n*e- --> n = Q/e- = (5.25*10^-14C)/(1.60*10^-19C) = 328125 excess electrons

(b)
1.0% decrease in Q = (5.25*10^-14C)*(0.01) = 5.25*10^-16C
Q = 5.25*10^-14C - 5.25*10^-16C = 5.1975*10^-14C

I = Q/t
--> t = Q/I = (5.1975*10^-14C)/(3.0*10^-16C/s) = 173.25 s

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