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Physics: Electricity and Magnetism - Semiconductors, Electronic Devices, and Smart Grids

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Semiconductors are materials whose electrical conductivity lies between that of conductors and insulators. Their conductivity can be altered by temperature or the addition of impurities, known as doping.

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The Energy Band Theory explains that in semiconductors, there is a small energy gap EgE_g between the valence band (filled with electrons) and the conduction band (empty at 0 K0 \text{ K}).

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Intrinsic semiconductors are pure (e.g., Silicon or Germanium), while Extrinsic semiconductors are doped. nn-type doping adds donor atoms (Group 15) providing extra electrons (e−e^-). pp-type doping adds acceptor atoms (Group 13) creating 'holes' (h+h^+).

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A P-N Junction Diode is formed by joining pp-type and nn-type materials. It allows current to flow in Forward Bias (Positive to pp, Negative to nn) and blocks current in Reverse Bias.

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Light Dependent Resistors (LDRs) and Thermistors are sensing devices. For an LDR, resistance RR decreases as light intensity increases. For a Negative Temperature Coefficient (NTC) thermistor, RR decreases as temperature TT increases.

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Transistors (like Bipolar Junction Transistors) consist of three layers (Emitter, Base, Collector) and can act as an electronic switch or an amplifier where a small base current IBI_B controls a larger collector current ICI_C.

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A Smart Grid is a modernized electrical grid that uses digital communication technology and sensors to monitor and manage the transport of electricity. It improves efficiency η\eta and integrates renewable energy sources.

📐Formulae

V=I×RV = I \times R

P=V×IP = V \times I

η=(PoutPin)×100%\eta = \left( \frac{P_{out}}{P_{in}} \right) \times 100\%

E=P×tE = P \times t

IE=IB+ICI_E = I_B + I_C

💡Examples

Problem 1:

A smart grid monitoring system measures a power input of 1200 W1200 \text{ W} at a local substation. Due to transmission losses in the smart cables, the power output delivered to the neighborhood is 1152 W1152 \text{ W}. Calculate the efficiency η\eta of this section of the grid.

Solution:

Using the efficiency formula: η=(11521200)×100%\eta = \left( \frac{1152}{1200} \right) \times 100\% η=0.96×100%\eta = 0.96 \times 100\% η=96%\eta = 96\%

Explanation:

Efficiency represents the ratio of useful energy output to total energy input. A smart grid aims to keep this percentage as high as possible by reducing heat losses.

Problem 2:

An NTC thermistor is connected to a 9 V9 \text{ V} battery. At room temperature, its resistance is R1=3000 ΩR_1 = 3000 \, \Omega. After being placed near a heat source, its resistance drops to R2=600 ΩR_2 = 600 \, \Omega. Calculate the change in current ΔI\Delta I flowing through the circuit.

Solution:

Initial current I1I_1: I1=VR1=93000=0.003 AI_1 = \frac{V}{R_1} = \frac{9}{3000} = 0.003 \text{ A} Final current I2I_2: I2=VR2=9600=0.015 AI_2 = \frac{V}{R_2} = \frac{9}{600} = 0.015 \text{ A} Change in current ΔI\Delta I: 0.015−0.0030.012\begin{array}{r} 0.015 \\ - 0.003 \\ \hline 0.012 \end{array} ΔI=0.012 A\Delta I = 0.012 \text{ A}

Explanation:

In an NTC thermistor, resistance decreases as temperature increases. According to Ohm's Law V=IRV = IR, if voltage remains constant and resistance decreases, the current must increase.