Semiconductors and Semiconductor devices

Flashcards for topic Semiconductors and Semiconductor devices

Intermediate136 cardsphysics

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Card 1

Front

Compare and contrast the characteristics of donor and acceptor impurities in semiconductors, including their energy levels relative to conduction and valence bands.

Back

Donor impurities: • Elements: Group V (P, As, Sb) in Si or Ge • Valence electrons: 5 (one more than host) • Energy level: Slightly below conduction band

  • For P in Si: 0.045 eV below conduction band • Carrier contribution: Donates electrons to conduction band • Result: n-type semiconductor (electrons are majority carriers) • Representation: Extra electron orbits donor atom with large radius (~30Å)

Acceptor impurities: • Elements: Group III (B, Al, Ga) in Si or Ge • Valence electrons: 3 (one less than host) • Energy level: Slightly above valence band

  • For Al in Si: 0.057 eV above valence band • Carrier contribution: Accepts electrons from valence band, creating holes • Result: p-type semiconductor (holes are majority carriers) • Representation: Missing electron creates incomplete bond

Key difference: Donor impurities create energy levels near conduction band, while acceptor impurities create levels near valence band.

Card 2

Front

How does hole movement occur in a semiconductor crystal lattice?

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Back

Hole movement in a semiconductor occurs through a sequence of electron shifts:

  1. When a covalent bond is broken in the lattice, an electron gains enough energy to enter the conduction band, creating a vacancy (hole)
  2. An adjacent bonded electron can then shift to fill this vacancy
  3. This electron movement appears as the hole moving in the opposite direction
  4. The process requires minimal energy since electrons are only moving between bonds of similar energy levels

Example: In silicon, when an electron from the bond between atoms A and B breaks free, it creates a hole. When an electron from the A-C bond moves to fill this hole, the vacancy appears to have moved from the A-B position to the A-C position.

Note: This apparent movement of positive charge (hole) forms the basis of p-type semiconductor conduction.

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Card 3

Front

What happens when silicon is doped with a pentavalent impurity like phosphorus, and how does this affect semiconductor behavior?

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Back

When silicon is doped with phosphorus:

  • The phosphorus atom substitutes for a silicon atom in the crystal lattice
  • Four of phosphorus's five valence electrons form covalent bonds with neighboring silicon atoms
  • The fifth electron becomes a "free" conduction electron that requires minimal energy (≈0.045 eV) to enter the conduction band
  • This creates an n-type semiconductor where:
    • Electrons are the majority charge carriers
    • Phosphorus atoms act as "donor impurities"
    • Conductivity increases by orders of magnitude (can be 10^6 times higher)
    • Impurity energy levels exist just below the conduction band

Example: Adding just 1 part per million of phosphorus to silicon can increase its conductivity by a factor of a million.

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Card 4

Front

Explain the function and operation principle of a self-sustaining LC oscillator circuit using a transistor.

Back

  • Function: Produces alternating voltage of specific frequency using only DC power sources

  • Core components:

    • Transistor amplifier section (provides gain)
    • LC network (determines resonant frequency)
    • Feedback path (sustains oscillation)
  • Operation principle:

    1. LC network resonates at frequency f0=12πLCf_0 = \frac{1}{2π\sqrt{LC}}
    2. Small oscillations are amplified by transistor
    3. Portion of output fed back to input in correct phase
    4. Components with frequency f₀ get reinforced through positive feedback
    5. Oscillation amplitude stabilizes due to nonlinear transistor characteristics
  • Self-sustaining mechanism: Energy transfers between inductor (magnetic) and capacitor (electric), with losses compensated by transistor amplification

Card 5

Front

How does a p-n junction diode function as a half-wave rectifier in an AC circuit?

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Back

A p-n junction diode creates half-wave rectification by:

• Allowing current in only one direction (forward-biased) • During positive half-cycles: diode conducts, current flows through load resistance • During negative half-cycles: diode is reverse-biased, blocking current flow • Result: unidirectional current with approximately this behavior:

  • When forward-biased: i ≈ (V₀/R)sin(ωt + φ)
  • When reverse-biased: i ≈ 0

This creates a pulsating DC output that flows only during alternate half-cycles of the input AC signal.

Note: The rectification efficiency depends on the forward resistance of the diode compared to load resistance.

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Card 6

Front

Compare and contrast the truth tables for OR and AND logic gates as implemented with switches in electrical circuits. When would you use each configuration in practical applications?

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Back

OR Gate (parallel switches):

  • Truth table:
    • A=0, B=0 → C=0
    • A=0, B=1 → C=1
    • A=1, B=0 → C=1
    • A=1, B=1 → C=1
  • Implementation: Switches in parallel
  • Output is 1 if ANY input is 1

AND Gate (series switches):

  • Truth table:
    • A=0, B=0 → C=0
    • A=0, B=1 → C=0
    • A=1, B=0 → C=0
    • A=1, B=1 → C=1
  • Implementation: Switches in series
  • Output is 1 only if ALL inputs are 1

Practical Applications:

  • OR gates (parallel):

    • Safety systems where any trigger should activate the system
    • Lighting controlled from multiple locations
    • Alarm systems with multiple sensors
    • Redundant systems where backup is needed
  • AND gates (series):

    • Security systems requiring multiple conditions
    • Machine safety interlocks requiring multiple checks
    • Authorization systems requiring multiple approvals
    • Control systems where all conditions must be met
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Card 7

Front

How does a diode-based OR gate circuit implement Boolean logic function X = A OR B?

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Back

The diode-based OR gate works through these mechanisms:

  • Two diodes (D₁ and D₂) connect inputs A and B to the output node X
  • When either input is HIGH (logic 1/+5V):
    • The corresponding diode becomes forward-biased
    • Current flows through the diode to the output
    • The output X becomes HIGH (+5V)
  • When both inputs are LOW (logic 0/0V):
    • Neither diode conducts
    • No current flows
    • The output X remains LOW (0V)
  • The resistor R connects to ground, ensuring the output is pulled LOW when no input is active

Truth table verification:

  • A=0, B=0 → X=0 (neither diode conducts)
  • A=1, B=0 → X=1 (D₁ conducts)
  • A=0, B=1 → X=1 (D₂ conducts)
  • A=1, B=1 → X=1 (both diodes conduct)
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Card 8

Front

Why does an OR gate implementation require pull-down resistors while AND gate implementation requires pull-up resistors?

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Back

The difference arises from the default output state and how the diodes control it:

OR gate with pull-down resistor:

  • Default state: Output pulled to ground (logic 0) through resistor R
  • Active operation: Any HIGH input forward-biases its diode, pulling output HIGH
  • The resistor creates a path to ground when all inputs are LOW
  • Without the pull-down resistor, the output would float when all inputs are LOW

AND gate with pull-up resistor:

  • Default state: Output pulled to +V (logic 1) through pull-up resistor
  • Active operation: Any LOW input forward-biases its diode, pulling output LOW
  • The resistor creates a path to +V when all inputs are HIGH
  • Without the pull-up resistor, the output would float when all inputs are HIGH

This complementary design reflects the logical duality between OR and AND functions, where OR activates with any HIGH input, while AND deactivates with any LOW input.

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Card 9

Front

If you need to implement a NOT gate using only NOR gates, how would you do it, and what Boolean identities does this demonstrate?

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Back

To implement a NOT gate using only NOR gates:

  • Connect both inputs of a NOR gate to the same input signal A
  • This creates NOT(A OR A) = NOT(A) = A'

Mathematical proof:

  • A NOR A = A + A = A + A = A'
  • Since A OR A = A, then NOT(A OR A) = NOT(A)

Boolean identity demonstrated:

  • Idempotent law: A + A = A
  • Double negation: (A')' = A
  • De Morgan's law: (A + B)' = A' · B'

This demonstrates that NOR is a universal gate, capable of implementing any logical function when used in appropriate combinations.

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Card 10

Front

Explain the complete operation of a transistor-based NOT gate (inverter) circuit, including how it processes both logical 0 and logical 1 inputs to produce inverted outputs.

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Back

A transistor-based NOT gate (inverter) functions by exploiting the switching behavior of a bipolar junction transistor (BJT) to invert the input signal:

When input is logical 0 (0V):

  • The base-emitter junction remains unbiased (insufficient voltage to turn on)
  • Transistor remains in cutoff state (acts like an open switch)
  • No current flows through the collector-emitter path
  • No voltage drop occurs across the collector resistor (RC)
  • Output terminal X remains at VCC (+5V)
  • Therefore, output = logical 1 (5V)

When input is logical 1 (5V):

  • The base-emitter junction becomes forward-biased
  • Transistor enters saturation state (acts like a closed switch)
  • Current flows from collector to emitter through RC
  • Voltage drop across RC pulls the output terminal down
  • Output terminal X drops to approximately 0V
  • Therefore, output = logical 0 (0V)

Key principle: The transistor acts as a voltage-controlled switch that implements the logical NOT operation: when input = 0, output = 1; when input = 1, output = 0.

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