A RAM (random access memory) integrated chip contains transistors, in which there are doped semiconductors, and capacitors - Leaving Cert Physics - Question b - 2019
Question b
A RAM (random access memory) integrated chip contains transistors, in which there are doped semiconductors, and capacitors.
What is a semiconductor? What is meant b... show full transcript
Worked Solution & Example Answer:A RAM (random access memory) integrated chip contains transistors, in which there are doped semiconductors, and capacitors - Leaving Cert Physics - Question b - 2019
Step 1
What is a semiconductor?
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Answer
A semiconductor is a material with a conductivity value intermediate between that of a conductor and an insulator. It can conduct electricity under certain conditions, making it essential for electronic devices.
Step 2
What is meant by doping a semiconductor?
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Doping a semiconductor involves adding an impurity to change its electrical properties. This procedure modifies the number of charge carriers available in the semiconductor material.
Step 3
How can a semiconductor be doped so that (i) its majority charge carriers are electrons?
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To create a n-type semiconductor with electrons as majority charge carriers, an element with more valence electrons than silicon (which has four) is added. For example, phosphorus, which has five valence electrons, donates one electron, thereby increasing the number of free electrons.
Step 4
How can a semiconductor be doped so that (ii) its majority charge carriers are holes?
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To make a p-type semiconductor with holes as majority charge carriers, an element with fewer valence electrons than silicon is introduced. For instance, adding boron, which has three valence electrons, creates 'holes' or vacancies in the semiconductor, enhancing hole concentration.
Step 5
Calculate (i) the energy stored in the capacitor when it is fully charged.
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The energy stored, E, in a capacitor can be calculated using the formula:
E=21CV2
Substituting the values:
Capacitance, C = 90 ff = 90 \times 10^{-15} F
Voltage, V = 1.2 V
Thus,
E=21×(90×10−15)×(1.2)2
Calculating gives:
E=6.48×10−14J
Step 6
Calculate (ii) the number of additional electrons that are on the negative plate of the capacitor as a result of it being fully charged.
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Using the formula for charge, Q:
Q=CV
Substituting in our values:
Q=(90×10−15)×(1.2)
We find:
Q=1.08×10−13C
To find the number of additional electrons, we divide the total charge by the charge of a single electron ($e = 1.6 \times 10^{-19} C):
numberofadditionalelectrons=1.6×10−191.08×10−13=675000
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