Reading Schematics & Breadboard Anatomy

class: center, middle, inverse, title-slide

.title[
# Reading Schematics & Breadboard Anatomy
]
.subtitle[
## Programmable Electronics — Unit 1
]
.date[
### Class 5
]

---

# Learning Targets

.lt-box[
- Recalculate total resistance, current, and voltage in series and parallel resistor networks (review).
- Calculate the power dissipated by a resistor and use its wattage rating to diagnose an overheating failure.
- Identify standard schematic symbols for the components in your kit.
- Explain how a breadboard's rows and rails connect internally.
- Build a circuit directly from a schematic, without step-by-step wiring instructions.
]

.eu-label[Essential Understanding:] A schematic isn't a picture of a circuit — it's a map of *connections*. Once you can read that map and know how a breadboard actually connects underneath, you don't need step-by-step instructions anymore. And before we move on: a resistor that runs hot is telling you something specific about current and power — today we learn to read that signal too.

---

# Agenda

.agenda-box[
1. Review: series & parallel resistance
2. Wattage rating, revisited — the 10 Ω incident
3. Why does a smaller resistor run hotter?
4. Why schematics, not photos
5. Schematic symbols you already know
6. Breadboard anatomy: rows and rails
7. Quick check: schematic → breadboard
8. **Build (25–30 min):** build from a schematic alone
9. Homework
]

---
class: center, middle

# Review: Series & Parallel Resistance

Before we move on to schematics, let's make sure Class 4's math is solid — we're going to need it again today.

---

# Review exercises

**Series:** 100 Ω, 220 Ω, and 470 Ω in series, off a 9 V supply.

- `$R_{total} = ?$`
- `$I = ?$` (same through all three)
- Voltage drop across each resistor? (they should add up to 9 V)

**Parallel:** the same three resistors — 100 Ω, 220 Ω, 470 Ω — in parallel, off 9 V.

- `$R_{total} = ?$`
- Current through *each* branch?
- Which branch carries the most current, and why?

Work both in your notebook before we check answers.

---

# Review exercises — solutions

**Series:**

`$$R_{total} = 100 + 220 + 470 = 790\,\Omega \qquad I = \frac{9\,\text{V}}{790\,\Omega} \approx 11.4\,\text{mA}$$`

`$$V_{100} \approx 1.14\,\text{V} \qquad V_{220} \approx 2.51\,\text{V} \qquad V_{470} \approx 5.35\,\text{V} \qquad (\text{sum} \approx 9\,\text{V})$$`

---

# Review exercises — solutions

**Parallel:**

`$$\frac{1}{R_{total}} = \frac{1}{100} + \frac{1}{220} + \frac{1}{470} \implies R_{total} \approx 60\,\Omega$$`

`$$I_{100} = \frac{9}{100} = 90\,\text{mA} \qquad I_{220} \approx 40.9\,\text{mA} \qquad I_{470} \approx 19.2\,\text{mA}$$`

The smallest resistor (100 Ω) always carries the most current — same voltage, least resistance to it.

---

# Wattage rating, revisited

Three equivalent ways to write the same power equation — pick whichever one you already know two variables for:

`$$P = IV \qquad P = I^2R \qquad P = \frac{V^2}{R}$$`

All three come from the same place: substitute Ohm's law (`$V = IR$`) into `$P = IV$`, and you can derive either of the other two yourself.

Every resistor is rated for a **maximum power** it can dissipate as heat — usually ¼ W (0.25 W) for the resistors in your kit. Cross that line, and the resistor doesn't fail gracefully — it overheats.

---
class: center, middle

# What happened to the 10 Ω resistor?

A pair in this class built their LED circuit with a **10 Ω** resistor instead of 220 Ω. It got very hot, and the resistor discolored — turned visibly brown.

Let's find out why, with the numbers.

---

# Work it out first

Same circuit as always: 6 V supply, red LED (`$V_f \approx 2.0\,\text{V}$`) — but this time, `$R = 10\,\Omega$` instead of 220 Ω.

1. What's the voltage across *just the resistor*? (Remember: the LED takes its share first.)
2. What current flows through the circuit?
3. How much power does the resistor dissipate?
4. How does that compare to a ¼ W (0.25 W) rating?

Work through it in your notebook before the next slide.

---

# The reveal

`$$V_R = V_{supply} - V_f = 6\,\text{V} - 2.0\,\text{V} = 4\,\text{V}$$`

`$$I = \frac{V_R}{R} = \frac{4\,\text{V}}{10\,\Omega} = 0.4\,\text{A} = 400\,\text{mA}$$`

`$$P = I \times V_R = 0.4\,\text{A} \times 4\,\text{V} = 1.6\,\text{W}$$`

`$1.6\,\text{W}$` against a `$0.25\,\text{W}$` rating is **6.4× over budget**. That's not a close call — that's a resistor being asked to dissipate more than six times the heat it was built for. The browning is the resistor's material literally starting to burn.

That same 400 mA was also flowing through the LED — twenty times its ~20 mA rating.

---

# Why does a smaller resistor run hotter?

The LED's forward voltage pins `$V_R$` to roughly the same ~4 V, almost no matter which resistor you choose. So use the form of the power equation written in terms of a *fixed* voltage:

`$$P = \frac{V_R^2}{R}$$`

`$V_R$` stays about the same — but `$R$` is in the *denominator*. Shrink `$R$`, and `$P$` goes up. A 22× smaller resistor (10 Ω vs. 220 Ω) doesn't just carry more current — because power depends on `$R$` this way, it dissipates dramatically more heat in a much smaller part.

---

# Quick check: the other direction

Same circuit, but now try `$R = 1\,\text{k}\Omega$` instead of 220 Ω.

1. Calculate `$I$` and `$P$`.
2. Is this resistor in any danger of overheating?
3. What trade-off do you give up by going this high?

This is exactly why Class 3 told you to **round up**, not down, when your calculated resistor value falls between two standard sizes — the failure modes are not symmetric.

---

# Why schematics, not photos?

- A photo of a circuit shows you *where things sit*. A schematic shows you *what connects to what* — which is the only thing that actually matters electrically.
- The same schematic can be built a dozen different physical ways on a breadboard, and they're all "correct" if the connections match.
- Every build day from here on will hand you a schematic, not a wiring diagram. Today is about learning to trust that map.

---

# Schematic symbols

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<div>battery</div>
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<div>switch (open)</div>
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<div>ground</div>
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---

# A full schematic, read left to right

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Battery → switch → resistor → LED → back to battery. Nothing about *where* these sit on a breadboard — only what connects to what, in what order.

---

# Breadboard anatomy

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- The **power rails** (red +, blue –) run the full length of the board — any hole along a rail is connected to every other hole on that same rail.
- The **terminal strip** is grouped in short columns of five holes. Every hole in a highlighted column is *one single electrical point* — plugging two legs into the same column connects them, no wire needed.
- The columns on one side of the center gap are **not** connected to the columns on the other side.

---
class: small-font

# Quick check: schematic → breadboard

Look back at the schematic two slides ago.

1. If the LED's cathode (the bar side) and the resistor's right leg both need to connect to the same point, where would you plug them on the breadboard?
2. Does the switch need its own column, or could it share one with something else?

Talk it over with your partner — we'll build it next.

---
class: center, middle

# Build (25–30 min): build from a schematic alone

No wiring diagram this time — just the schematic and your own reasoning.

---

# The challenge: a 4-LED array

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One switch, two parallel branches, **two LEDs in series** per branch. Branch 1 uses a single resistor; branch 2 splits the same job across **two** resistors in series — three resistors, four LEDs, one schematic.

---

# Sizing R1, R2a, and R2b

Each branch has to account for **two** LED forward-voltage drops in series before you even get to the resistor(s):

`$$R_{target} = \frac{V_{supply} - (V_f + V_f)}{I_{LED}}$$`

For two red LEDs (`$V_f \approx 2.0\,\text{V}$` each) off the 6 V supply, at a 15–20 mA target — work out `$R_{target}$` with your partner first.

- **Branch 1:** pick a single resistor from your kit close to `$R_{target}$` (round up, same rule as always).
- **Branch 2:** you don't have to match it with one resistor — pick **two** resistors from your kit whose values *add together* to roughly `$R_{target}$`. Remember: resistors in series just add, same as this morning's review.

---

# Build instructions

With your partner, using your breadboard, kit, and both the 6 V battery pack and the 9 V battery:

1. **Warm-up:** build the single-LED schematic from earlier in this deck. Choose your own resistor value and justify it in your notebook. Test it.
2. **The challenge:** calculate `$R_{target}$`, then choose `$R_1$` for branch 1 and a pair `$R_{2a}$`, `$R_{2b}$` for branch 2 that sum to roughly the same value. Build both branches.
3. Test it — confirm all four LEDs light, and that the switch controls all of them together.
4. **Predict, then check:** before you power it on, predict whether branch 1's LEDs and branch 2's LEDs will be equally bright. Explain why in your notebook — then confirm by eye.
5. **Swap the 6 V pack for the 9 V battery.** Two LEDs in series drop ~4 V, so on 6 V the resistor only had ~2 V to work with — recompute `$R_{target}$` for a 9 V supply and note how much more headroom you now have. Rebuild with the new values. Did the branches get easier to balance?
6. **If time allows:** design your own variation — a third parallel branch, a different LED color in one branch (its `$V_f$` is different, so its resistor needs recalculating), or your own visual arrangement of the LEDs. Sketch the schematic before you touch the breadboard.
7. Trade schematics with another group and build theirs, without asking them how they built it.

---
class: center, middle

# Homework

.hw-box[
- In your notebook, redraw tonight, from memory, the five schematic symbols from today.
- Sketch the schematic for one circuit you've already built this unit (your choice).
]