Electronics & Circuits · Study deck

IoT Electricity: Kirchhoff Analysis and Diagnostics

Series and parallel rules explain many circuits, but a real IoT board also has branches and loops.

Voltage Vera is your guide for this deck.

electricity
Voltage Vera, the module guide, in a scene from this chapter.
iotclass.org

After studying this chapter

Learning objectives

You will be able to:

  • Explain: The Sensor Squad was going to build their very first LED circuit. "First, we need a resistor," said the microcontroller, holding up a tiny striped cylinder. "Without it, Lila could get hurt!".
  • Explain: Understanding the relationship between voltage (electrical "pressure"), current (flow rate), and resistance (opposition to flow) through Ohm's Law (V = I × R) is essential for designing and troubleshooting IoT circuits.
  • Explain: These two laws catch problems that Ohm's Law alone cannot: KCL catches overcurrent on shared rails, and KVL catches voltage headroom failures in multi-component loops.
iotclass.org

Major section

Deep dive: Kirchhoff Laws in IoT

Current cannot appear or disappear -- it must flow somewhere.

  • Every volt the battery provides must be "used up" by components in the loop.
  • A standard 1/8 W (0.125 W) resistor is sufficient.
  • Problem found by KVL: The regulator cannot maintain 3.3V output when the battery drops below 4.3V.

Key terms

Actual runtime
Actual runtime is lower; for 85% overall power conversion efficiency, this simplified continuous-active estimate is about 6.0 hours.
iotclass.org

Major section

Deep dive: Kirchhoff Laws in IoT (continued)

Step 3: Apply KVL to Check Regulator Headroom: The LDO regulator needs a minimum dropout voltage of 1.0V (AMS1117 datasheet).

  • A single-cell LiPo reaches about 4.2 V when fully charged and falls below that during discharge.
  • With the stated 1.0 V dropout, this LDO cannot maintain 3.3 V from that battery at any normal charge level.
  • Longer runtimes require the MCU, sensor, and indicator loads to sleep too.
iotclass.org

Major section

Key Takeaway from Kirchhoff's Laws

KCL answered: "Can my regulator handle the total current?" (92.4 mA -- yes, AMS1117 supports up to 1A).

  • These two laws catch problems that Ohm's Law alone cannot: KCL catches overcurrent on shared rails, and KVL catches voltage headroom failures in multi-component loops.

KVL answered: "Does my LED get the right current?" (yes, 120 Ohm resistor) AND revealed a hidden problem: "Will the LDO maintain output as the battery discharges?" (no -- must switch to buck-boost).

iotclass.org

Major section

Key Takeaway

Passive components (resistors, capacitors, inductors) shape every IoT circuit: resistors limit current and create voltage dividers, capacitors filter noise and store energy, and inductors enable efficient power conversion.

  • The power budget calculation (average current = sum of duty-cycle-weighted component currents) is the single most important analysis for battery-powered IoT deployments.
iotclass.org

Major section

For Kids: Meet the Sensor Squad!

Temperature Terry and friends learn about building real circuits!: Temperature Terry was excited -- today was Build Day!

  • The Sensor Squad was going to build their very first LED circuit. "First, we need a resistor," said the microcontroller, holding up a tiny striped cylinder. "Without it, Lila could get hurt!".

How?" Max explained: "If too much electricity flows through you, you'll burn out -- like eating too much candy too fast gives you a tummy ache.

iotclass.org

Major section

For Kids: Meet the Sensor Squad! (continued)

The resistor is like a speed bump that slows the electricity down to a safe amount.".

  • the battery was ready to power everything. "I supply 9 volts!" she announced proudly.
  • Max did some quick math: "Lila needs about 2 volts, so the resistor needs to handle 7 volts.
  • Using Ohm's Law -- that's V divided by I -- we need a 350 ohm resistor to target 20 milliamps.".
iotclass.org

Major section

Chapter Summary

Electricity is the foundation of all IoT systems, driven by the flow of electrons through conductors.

  • Understanding the relationship between voltage (electrical "pressure"), current (flow rate), and resistance (opposition to flow) through Ohm's Law (V = I × R) is essential for designing and troubleshooting IoT circuits.

Key terms

Series configurations
Series configurations are useful for voltage division and cumulative resistance, while parallel configurations provide redundancy and current distribution.

Why it matters

Understanding these principles enables proper component selection and circuit design.

iotclass.org

Major section

Chapter Summary (continued)

Real-world circuits combine these components to condition sensor signals, filter noise, store energy, and protect sensitive electronics.

  • Circuit configurations significantly impact system behavior: series circuits share current but divide voltage, while parallel circuits share voltage but divide current.
  • Series configurations are useful for voltage division and cumulative resistance, while parallel configurations provide redundancy and current distribution.
  • Safety margins (typically 2×) prevent component failure and ensure long-term reliability in IoT deployments.
iotclass.org

Major section

LED Resistor Current Error

Many beginners size LED resistors for the maximum rated current (20 mA) because it is listed prominently in the datasheet.

  • At 20 mA, LEDs are near their thermal limits and consume unnecessary power.

Use 5-10 mA for indicator LEDs — they are still clearly visible and last orders of magnitude longer.

iotclass.org

Major section

2. Voltage Divider Loading Error

When a load (ADC input, voltmeter, next stage circuit) is connected to the voltage divider output, it appears in parallel with R2 and changes the voltage.

  • If the load resistance is comparable to R2, the output voltage drops significantly below the calculated value.

Use an operational amplifier buffer if the load impedance is too low to maintain divider accuracy.

iotclass.org

Major section

Battery Rated vs Usable Capacity

Devices with brownout reset voltages of 2.8-3.0 V stop working before the battery is fully discharged.

  • For an initial battery-life estimate, you may assume 80–90% of rated capacity is usable; then replace that assumption with the chosen cell’s discharge curve at the device cutoff voltage, load and temperature.

Battery rated capacity (mAh on the label) assumes discharge to a cutoff voltage (typically 2.5-3.0 V for Li-ion, 1.0 V for alkaline).

iotclass.org

Major section

Reverse Diode Voltage Drop

A 3.3 V system with a diode sees 2.6-3.0 V at the circuit.

  • For 3.3 V logic that requires VCC > 2.7 V minimum, this can bring the supply dangerously close to the minimum operating voltage.

A series protection diode drops 0.3-0.7 V off the supply voltage.

iotclass.org

Deck summary

Key takeaways

Current cannot appear or disappear -- it must flow somewhere.

  • Step 3: Apply KVL to Check Regulator Headroom: The LDO regulator needs a minimum dropout voltage of 1.0V (AMS1117 datasheet).
  • KCL answered: "Can my regulator handle the total current?" (92.4 mA -- yes, AMS1117 supports up to 1A).
  • Passive components (resistors, capacitors, inductors) shape every IoT circuit: resistors limit current and create voltage dividers, capacitors filter noise and store energy, and inductors enable efficient power conversion.
  • Temperature Terry and friends learn about building real circuits!: Temperature Terry was excited -- today was Build Day!
iotclass.org

Retrieval practice

Recall check 1 of 3

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

Q1A circuit has 12V voltage and 4Ω resistance. What is the current?

A3A
B48A
C8A
D16A
Show answer

Answer: A Answer: A) 3A I=VR=12V4Ω=3A \begin{aligned} I &= \frac{V}{R} \\ &= \frac{12V}{4\Omega} \\ &= 3A \end{aligned} I​=RV​=4Ω12V​=3A​.

Q2You need to limit current to 20mA for a 2V LED powered by a 5V supply. What resistor value?

A100Ω
B220Ω
C150Ω
D330Ω
Show answer

Answer: C Answer: B) 150Ω R=Vsupply−VLEDI=5V−2V0.02A=150Ω \begin{aligned} R &= \frac{V_{supply} - V_{LED}}{I} \\ &= \frac{5V - 2V}{0.02A} \\ &= 150\Omega \end{aligned} R​=IVsupply​−VLED​​=0.02A5V−2V​=150Ω​ In practice, use 150Ω or the nearest standard value (220Ω).

Q3Two 1kΩ resistors are connected in parallel. What is the total resistance?

A2kΩ
B1kΩ
C250Ω
D500Ω
Show answer

Answer: D see answers page

iotclass.org

Retrieval practice

Recall check 2 of 3

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

Q4A 220Ω resistor carries 50mA (0.05A) current. How much power does it dissipate?

A0.55 W dissipated by the resistor
B0.055 W dissipated by the resistor
C5.5 W dissipated by the resistor
D11 W dissipated by the resistor
Show answer

Answer: A Answer: B) 0.55W P=I2×R=(0.05A)2×220Ω=0.0025×220=0.55W \begin{aligned} P &= I^2 \times R \\ &= (0.05A)^2 \times 220\Omega \\ &= 0.0025 \times 220 \\ &= 0.55W \end{aligned} P​=I2×R=(0.05A)2×220Ω=0.0025×220=0.55W​ This requires at least a 1W rated resistor for safe operation (with safety margin).

iotclass.org

Retrieval practice

Recall check 3 of 3

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

Q5You need to select a resistor that will dissipate 0.3W of power. Which power rating should you choose for safe operation with a 2× safety margin?

AA 1/8 W rating, below the 0.3 W load
BA 1/4 W rating, below the 0.3 W load
CA 1 W rating, above the 0.6 W margin
DA 1/2 W rating, below the 0.6 W margin
Show answer

Answer: C see answers page

Q6What pair of resistors scales a voltage to a safe ADC input fraction?

ALED current limiting resistor
BPull-up resistor
CPower budget estimate
DVoltage divider
Show answer

Answer: D Two resistors scale an input voltage to a fraction that an ADC can measure safely.

iotclass.org

Print reference

Answers 1 of 4

Answer key.

  1. A · Answer: A) 3A I=VR=12V4Ω=3A \begin{aligned} I &= \frac{V}{R} \\ &= \frac{12V}{4\Omega} \\ &= 3A \end{aligned} I​=RV​=4Ω12V​=3A​.
  2. C · Answer: B) 150Ω R=Vsupply−VLEDI=5V−2V0.02A=150Ω \begin{aligned} R &= \frac{V_{supply} - V_{LED}}{I} \\ &= \frac{5V - 2V}{0.02A} \\ &= 150\Omega \end{aligned} R​=IVsupply​−VLED​​=0.02A5V−2V​=150Ω​ In practice, use 150Ω or the nearest standard value (220Ω).
iotclass.org

Print reference

Answers 2 of 4

Answer key.

  1. D · Answer: C) 500Ω For parallel resistors: 1Rtotal=1R1+1R2=11kΩ+11kΩ=21kΩ \begin{aligned} \frac{1}{R_{total}} &= \frac{1}{R_1} + \frac{1}{R_2} \\ &= \frac{1}{1k\Omega} + \frac{1}{1k\Omega} \\ &= \frac{2}{1k\Omega} \end{aligned} Rtotal​1​​=R1​1​+R2​1​=1kΩ1​+1kΩ1​=1kΩ2​​ Therefore: Rtotal=1kΩ2=500Ω \begin{aligned} R_{total} &= \frac{1k\Omega}{2} \\ &= 500\Omega \end{aligned} Rtotal​​=21kΩ​=500Ω​ For identical resistors in parallel: Rtotal=Rn \begin{aligned} R_{total} &= \frac{R}{n} \end{aligned} Rtotal​​=nR​​ where n is the number of resistors.
iotclass.org

Print reference

Answers 3 of 4

Answer key.

  1. A · Answer: B) 0.55W P=I2×R=(0.05A)2×220Ω=0.0025×220=0.55W \begin{aligned} P &= I^2 \times R \\ &= (0.05A)^2 \times 220\Omega \\ &= 0.0025 \times 220 \\ &= 0.55W \end{aligned} P​=I2×R=(0.05A)2×220Ω=0.0025×220=0.55W​ This requires at least a 1W rated resistor for safe operation (with safety margin).
iotclass.org

Print reference

Answers 4 of 4

Answer key.

  1. C · Answer: D) 1 W With 2× safety margin: 0.3W×2=0.6W0.3W \times 2 = 0.6W0.3W×2=0.6W The next standard power rating above 0.6W is 1W. Safety margins prevent overheating and ensure long-term reliability. Common practice is to use components rated for at least 2× the expected power dissipation.
  2. D · Two resistors scale an input voltage to a fraction that an ADC can measure safely.
iotclass.org