Electronics & Circuits · Study deck

Semiconductors: Fabrication and Switching Devices

An integrated circuit puts many components on one chip.

Voltage Vera is your guide for this deck.

dopingdiodes
Voltage Vera, the module guide, in a scene from this chapter.
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After studying this chapter

Learning objectives

You will be able to:

  • Outline the main stages of integrated-circuit fabrication.
  • Compare MOSFET and BJT switching limits and losses.
  • Explain: The second option is an integrated circuit (IC), and the printing process is a direct extension of the doping this chapter just covered.
  • Explain: The Misconception: Many beginners assume any MOSFET can be driven directly from microcontroller GPIO pins (3.3V or 5V), leading to circuits that "sort of work" but run hot, waste power, and fail prematurely.
This is the TO-220 package Option B in the worked example above ships in -- three leads for gate, drain, and source, plus a metal tab that would bolt to a heatsink if losses climbed past the 0.0055W the IRLZ44N dissipates driving the 12V fan at 0.5A. Photo: Suyash Dwivedi, CC BY-SA 4.0
This is the TO-220 package Option B in the worked example above ships in -- three leads for gate, drain, and source, plus a metal tab that would bolt to a heatsink if losses climbed past the 0.0055W the IRLZ44N dissipates driving the 12V fan at 0.5A. Photo: Suyash Dwivedi, CC BY-SA 4.0
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Major section

How ICs Are Actually Made

The second option is an integrated circuit (IC), and the printing process is a direct extension of the doping this chapter just covered.

  • Lithography -- the circuit design is drawn onto a photomask, then projected through it onto a photoresist-coated silicon wafer, exposing the pattern in UV light.
  • Etching -- the exposed pattern is "carved" into the wafer, opening windows down to the silicon where each component will form.
  • Deposition -- dopants and metal are deposited into those windows, forming the p-type and n-type regions for diodes and transistors and the metal traces that wire them together.
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Major section

GPIO Voltage MOSFET Limits

The Misconception: Many beginners assume any MOSFET can be driven directly from microcontroller GPIO pins (3.3V or 5V), leading to circuits that "sort of work" but run hot, waste power, and fail prematurely.

  • The Reality with Real Numbers: Standard MOSFETs require Vgs = 10-12V to fully turn ON and achieve their datasheet specifications.
  • Component damage: excess junction temperature can shorten service life; compare the selected MOSFET’s thermal and reliability limits with the operating profile.
  • Debugging time: checking RDS(on) at the actual gate voltage early can avoid chasing a hot-switch fault later.
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Major section

For Kids: Meet the Sensor Squad!

the microcontroller smiled. "Don't worry -- we have a special protector called a DIODE.

  • Electricity can flow through in one direction, but if it tries to go backwards, the door slams shut!".
  • "How does that work?" asked the LED.
  • One side has extra electrons, the other side has holes where electrons are missing.
  • "Wait," said Lila, "my name has LED in it -- Light Emitting Diode!
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Major section

Flyback Diode for 12V Relay

Scenario: You are building an ESP32-controlled smart home hub that switches a 12V electromagnetic relay (coil: 120mA, inductance: 150mH).

  • The relay controls a 120V AC lamp.
  • Step 1: Calculate the inductive voltage spike (worst case): When the relay coil current is cut, the inductor generates: V = -L × (di/dt).

Key terms

Average power
Average power is negligible for low-frequency relay switching.
The 1N4007 chosen above for this 12V relay is exactly this part class -- a cheap, axial-leaded general-purpose rectifier -- installed in reverse bias across the coil, where its slow ~30 microsecond recovery time never matters because the relay switches under 1 Hz. Photo: oomlout, CC BY-SA 2.0
The 1N4007 chosen above for this 12V relay is exactly this part class -- a cheap, axial-leaded general-purpose rectifier -- installed in reverse bias across the coil, where its slow ~30 microsecond recovery time never matters because the relay switches under 1 Hz. Photo: oomlout, CC BY-SA 2.0
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Major section

Flyback Diode for 12V Relay (continued)

Step 2: Select diode voltage rating: The flyback diode must withstand at least the supply voltage in reverse bias: V_reverse > 12V.

  • Worst-case scenario: transistor switches off in 100ns (modern MOSFETs are fast!).
  • All three diodes exceed this (300mA to 1A rating).
  • Average power is negligible for low-frequency relay switching.
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Major section

Flyback Diode for 12V Relay (continued)

V_spike = -(150 × 10⁻³ H) × (0.12A / 100 × 10⁻⁹ s) = -180,000V: Without protection, this idealized spike would instantly destroy the transistor (rated for 60V max).

  • In a real circuit, parasitic capacitance and arcing limit the spike to hundreds of volts rather than 180kV — still far beyond the transistor's rating, which is why the diode is non-negotiable.
  • Step 4: Check reverse recovery time: When the relay switches ON again, the diode must stop conducting quickly to avoid shorting the power supply.
  • For a relay switching at <1 Hz, even the slow 1N4007 is adequate.
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Deck summary

Key takeaways

The second option is an integrated circuit (IC), and the printing process is a direct extension of the doping this chapter just covered.

  • The Misconception: Many beginners assume any MOSFET can be driven directly from microcontroller GPIO pins (3.3V or 5V), leading to circuits that "sort of work" but run hot, waste power, and fail prematurely.
  • the microcontroller smiled. "Don't worry -- we have a special protector called a DIODE.
  • Scenario: You are building an ESP32-controlled smart home hub that switches a 12V electromagnetic relay (coil: 120mA, inductance: 150mH).
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Retrieval practice

Recall check 1 of 6

Voltage Vera says: answer from memory, then check your reasoning.

Q1You're reverse-engineering a commercial IoT smart plug. The teardown reveals the main switching device is a triac, not a MOSFET. The product switches 120V AC mains at 15A. Why use a triac?

ABecause triacs cost less than MOSFETs at every current and voltage rating.
BBecause suitably rated MOSFETs cannot handle 120 V mains voltage.
CBecause a triac itself isolates the control circuit from the mains load.
DBecause a rated triac switches both AC half-cycles of the mains load.
Show answer

Answer: D A triac conducts in both directions when triggered, making it useful for AC load switching.

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Retrieval practice

Recall check 2 of 6

Voltage Vera says: answer from memory, then check your reasoning.

Q2A 2N2222 switching design assumes a typical current gain β = 100 exactly, but measured devices range from β = 75 to 250. What design risk must be checked?

AUsing typical β alone risks incomplete saturation across units.
BCircuit will fail completely - must use transistors with exact β=100
CHigher β transistors will burn out due to excessive current
DNo impact - β variation doesn't affect switching circuits
Show answer

Answer: A Measured β from 75 to 250 shows that a typical value is not a guaranteed switching value.

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Retrieval practice

Recall check 3 of 6

Voltage Vera says: answer from memory, then check your reasoning.

Q3A red LED is connected directly between a 3.3 V ESP32 GPIO output and ground, with no series current-limiting resistor. What is the electrical risk?

AThe LED self-limits at full brightness without a series resistor.
BThe GPIO or LED can exceed current limits without a series resistor.
CThe red LED cannot light from 3.3 V because its drop is higher.
DThe ESP32 GPIO automatically limits LED current to a safe value.
Show answer

Answer: B see answers page

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Retrieval practice

Recall check 4 of 6

Voltage Vera says: answer from memory, then check your reasoning.

Q4A 12 V, 500 mA solenoid is switched by a logic-level N-channel MOSFET from a 3.3 V ESP32 GPIO. During an ON command the gate measures 3.3 V to ground, but the source measures 1.5 V above ground through an unintended resistive return path; Vds measures 2 V. Which terminal connection should be corrected for low-side switching?

AGate connection to 12 V, removing GPIO control rather than fixing return.
BSource connection directly to ground to remove the resistive return rise.
CDrain connection to ground instead of the solenoid's switched end.
DAll gate, drain and source terminals tied together to switch the load.
Show answer

Answer: B see answers page

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Retrieval practice

Recall check 5 of 6

Voltage Vera says: answer from memory, then check your reasoning.

Q5An H-bridge motor works in both directions, but supply current spikes on direction changes. A scope shows high-side and low-side MOSFETs in the same bridge leg conducting together during each spike, even with the motor disconnected. What fault does this identify?

AShoot-through during high/low-side MOSFET overlap.
BMotor is defective - internal short circuit
CH-bridge MOSFETs have insufficient current rating
DPower supply voltage too high for motor
Show answer

Answer: A The scope establishes shoot-through during transitions: both MOSFETs in one bridge leg conduct at once, causing a supply-current spike without the motor.

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Retrieval practice

Recall check 6 of 6

Voltage Vera says: answer from memory, then check your reasoning.

Q6What gate-controlled channel switches load current in the circuit?

AN-type semiconductor
BP-type semiconductor
CPN junction diode
DMOSFET switch
Show answer

Answer: D Its gate field controls a channel that can switch the load current.

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Print reference

Answers 1 of 4

Answer key.

  1. D · A triac conducts in both directions when triggered, making it useful for AC load switching.
  2. A · Measured β from 75 to 250 shows that a typical value is not a guaranteed switching value.
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Print reference

Answers 2 of 4

Answer key.

  1. B · A 3.3 V output can forward-bias a red LED. With no series current-limiting element, current depends on the LED I-V curve and the GPIO output characteristics; it is not established that the parts immediately fail, but damage is a real risk. Size a resistor from the supply, expected LED forward voltage and target current, and check the GPIO current rating.
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Print reference

Answers 3 of 4

Answer key.

  1. B · The source’s measured 1.5 V rise reduces actual Vgs to 1.8 V even though the GPIO reads 3.3 V to ground. A resistive source return therefore can leave the MOSFET inadequately enhanced and raise Vds. Correct the low-side source-to-ground path, then confirm on-resistance at the actual gate drive and fit the required solenoid flyback protection. A source tied to 12 V would not explain the given partial-on measurements.
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Print reference

Answers 4 of 4

Answer key.

  1. A · The scope establishes shoot-through during transitions: both MOSFETs in one bridge leg conduct at once, causing a supply-current spike without the motor.
  2. D · Its gate field controls a channel that can switch the load current.
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