ET-210 Quiz 1

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CUNY Queensborough Community College *

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210

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Electrical Engineering

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Apr 3, 2024

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ET 210 Quiz 1 1) Define the term Semiconductor . A semiconductor is a material with electrical conductivity between that of a conductor and an insulator. Its conductivity can be altered by introducing impurities or applying external factors. Silicon is a commonly used semiconductor, but other materials like germanium are also used. They enable controlled electron flow, crucial for the operation of modern technology, including computers, smartphones, and various electronic systems. 2) What is a Valence electron ? A valence electron is an electron located in the outermost energy level or shell of an atom. These electrons are involved in chemical reactions and bonding with other atoms. The number of valence electrons largely influences an element's chemical properties and its ability to form bonds. Elements in the same group or column of the periodic table typically have similar valence electron configurations, contributing to similarities in their chemical behavior. Understanding the arrangement and interactions of valence electrons is essential in predicting and explaining the formation of chemical compounds. 3) How many Valence electrons do semiconductors typically have? In the case of silicon and germanium, two commonly used semiconductor materials, they both belong to Group 14 of the periodic table. Elements in Group 14 typically have four valence electrons. 4) Sketch the electron configuration of a Silicon atom. 5) Sketch the electron configuration of a Germanium atom (see a textbook or do an internet search) 6) Find the Energy (Work) required to move a charge of 8 mC between two points when the voltage is 40 V W = (8 x 10 − 3 C) x (40V) = W = 0.32 J 7) Is the charge of an electron positive, negative, or zero? The charge of an electron is negative. Electrons are negatively charged particles, and they contribute to the negative charge of an atom. 8) What are holes in semiconductors? In semiconductors, holes are conceptualized as the absence of an electron in the crystal lattice of a material, typically silicon or germanium. When an electron is excited from its normal position in the lattice, it leaves behind a vacancy, or hole. Although holes are not physical entities with positive charge, they are treated as carriers of positive charge for theoretical and practical purposes. Holes play a crucial role in semiconductor physics, particularly in the operation of electronic devices such as transistors. 9) Is the charge of a hole considered to be positive, negative, or zero?

The charge of a hole is positive. 10) How are N type semiconductors created? N-type semiconductors are created by introducing impurities into a pure semiconductor crystal, typically made of silicon or germanium. The impurity atoms used are known as donor atoms, and they have more valence electrons than the host semiconductor atoms. Common donor atoms include phosphorus or arsenic. When these donor atoms are added to the crystal lattice, they contribute extra electrons to the semiconductor material, creating an excess of negatively charged carriers. These additional electrons become the majority charge carriers in the N-type semiconductor. The term "N-type" comes from the excess of negative charge carriers, indicating that electrons are the dominant charge carriers in the material. 11) The majority (that is many, many in number) carriers in N type material are (choose from the following ): Electrons 12) How are P type semiconductors created? P-type semiconductors are created by introducing acceptor impurities, such as boron or aluminum, into a pure semiconductor crystal, leading to the formation of holes as the majority charge carriers in the material. 13) The majority (that is many, many in number) carriers in P type material are (choose from the following ): Holes 14) The current in N type material consists mostly of the movement of _____________ (choose from the following ): Electrons 15) The current in P type material consists mostly of the movement (flow) of _____________ (choose from the following ): Holes 16) Draw each of the following for a Diode (Semiconductor PN junction diode) and label the terminals : Internal construction of diode Physical package of diode Schematic symbol of diode

17) Draw the schematic symbol of a diode. Show the Voltage (V D ) polarity reference label with a plus and minus sign pair, and show the current (I D ) direction reference label with an arrow, on your schematic symbol diagram. In addition, write the Shockley diode math equation with appropriate variables and parameters. The quiz is continued on the next set of pages. 18) Sketch the “actual” shape of the plot of diode current I D versus diode voltage V D . Label all axes with appropriate variables and units. Label the Forward Bias region and label the Reverse Bias region. State the mathematical signs (positive or negative) of the current (I D ) and voltage (V D ) variable quantities in both the Forward Bias and Reverse Bias regions. 19) Draw the Approximate Diode model when a diode is Forward biased . Label the actual physical voltage polarity and label the actual physical current direction on your Forward Biased diode diagram. For the Forward biased diode, also indicate the approximate numerical values of the voltage V D for a Silicon (Si) diode and for a Germanium (Ge) diode.

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20) Draw the Approximate Diode model when a diode is Reverse biased . Label the actual physical voltage polarity and label the actual physical current direction on your Reverse Biased diode diagram. For the Reverse biased diode, also indicate the approximate numerical value of the current I D 21) What is the rough approximate value of the voltage across a Forward biased LED (Light Emitting Diode) ? 0.6 V- 0.75 V 22) Given a diode with the parameters : n = 1.3, I S = 3.0 nA, V T = 26 mV Use the Shockley Diode equation to : a) Find the diode current (I D ) when the diode voltage is V D = 0.61 V Vd = 0.61V ID = IS (e VD/nvt -1) ID = 3x 10^-9 ( e 0.61/1.3 x 26 x 10^-3 – I ) ID = 3 x 10 ^-9 (e^18 -1) ID = 1.96mA b) Find the diode current (I D ) when the diode voltage is V D = – 5 V ( minus 5 V) ID = 3 x 10^-9 (e -5/1.3 x 26 x 10^-3 – I) ID = -3nA c) Find the diode current (I D ) when the diode voltage is V D = 0.51 V ID = IS ( VD /NVT – I) ID = 3 x 10^-9 ( e 0.51 x 10^3/33.0 -1) ID = 3 x 10 ^-9 (e^15.45 -1 ) ID = 5.62mA d) Find the diode current (I D ) when the diode voltage is V D = 0.81 V ID = 3 x 10^-9 (e 0.81 x 10^5 /33 -1) ID = 3 x 10^-9 (e ^24.54 -1 )

ID = 45.45 Amp 23) Given a diode with the parameters : n = 1.3, I S = 3.0 nA, V T = 26 mV Find the diode voltage (V D ) when the diode current is I D = 100 mA Hint: First, use algebra to solve the Shockley diode equation for V D and then substitute in the given values. The natural logarithm (Ln) function will be involved in your solution for V D. VD = VT * ln(ID / IS ) VD = 26mV * ln(100mA / 3nA) VD = 0.7V Diode voltage = 0.7V Diode current = 100mA The quiz is continued on the next set of pages. Part TWO: Problems Related to the Previous Electrical Circuit Analysis Course (These problems must also be completed for this quiz) Refer to the generic circuit drawing below when completing problems 1, 2, 3, and 4 : 1) Find the current I if the voltage Vs = 36 V and the resistance R = 4 kΩ 36V / 4 kΩ = 9 mA 2) Find the current I if the voltage Vs = 40 V and the resistance R = 8 MΩ 40V / 8 MΩ = 5 μ A 3) Find the voltage Vs if the current I = 4.26 µA and the resistance R = 39 kΩ 4.26 µA x 39 kΩ = 166.14 mV

4) Find the power P if the voltage Vs = 15 V and the current I = 29 mA 15 V x 29 mA = 435 mW 5 ) Use Kirchhoff’s Voltage Law (KVL) to find the voltage V X in the circuit shown directly below: 12 V 15 V Vx 6 V + _ + _ + _ + _ 6 ) Use Kirchhoff’s Voltage Law (KVL) to find the voltage V y in the circuit shown directly below: 7 V 10 V Vy 20 V + _ + _ + _ + _ The quiz is continued on the next page. 7) Use Kirchhoff's Current Law (KCL) to find the current I X in the diagram shown directly below: 8) Use Kirchhoff's Current Law (KCL) to find the current I y in the diagram shown directly below:

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9) Given the series-parallel circuit drawn below (note: prefix k is kilo in the resistances): Es 560 V R1 80kΩ R2 150kΩ R3 30kΩ R4 120kΩ R5 168kΩ + _ a) Find the equivalent (total) resistance R eq = R T of the entire circuit. 14.97k Ω b) Find the current through the source (battery). That is, find I S 37.4 mA c) Find the currents through each of the resistors R1, R2, R3, R4, and R5. That is, find I 1 , I 2 , I 3 , I 4 , and I 5 . I 1 = 560 V / 80k Ω = 7 mA, I 2 = 560 V / 150k Ω = 3.73 mA, I 3 = 560 V / 30k Ω = 18.66 mA, I 4 = 560 V / 120k Ω = 4.66 mA, I 5 = 560 V / 168k Ω = 3.33 mA d) Find the voltages across each of the resistors R1, R2, R3, R4, and R5. That is, find V 1 , V 2 , V 3 , V 4 , and V 5 V 1 = 3.92 W / 7 mA = 560 V, V2 = 2.09 W / 3.73 mA = 560 V, V3 = 10.4 W / 18.66 mA = 557 V, V4 = 2.61 W / 4.66 mA = 560 V, V5 = 1.86 W / 3.33 mA = 558 V e) Find the power that the source delivers to the circuit. That is, find P S . 560V x 37.4 mA = 20.94 W f) Find the powers dissipated in resistors R1, R2, R3, R4, R5. That is, find P 1 , P 2 , P 3 , P 4 , and P 5 . P 1 = 560V^2 / 80k Ω = 3.92 W, P 2 = 560V^2 / 150k Ω = 2.09 W, P 3 = 560V^2 / 30k Ω = 10.4 W, P 4 = 560V^2 / 120k Ω = 2.61 W, P 5 = 560V^2 / 168k Ω = 1.86 W g) Verify your power results by using Conservation of Energy (or powers) with the powers found in part e and part f .