The Voltage Divider, Potentiometer & Photoresistor

class: center, middle, inverse, title-slide

.title[
# The Voltage Divider, Potentiometer & Photoresistor
]
.subtitle[
## Programmable Electronics — Unit 1
]
.date[
### Class 6
]

---

# Learning Targets

.lt-box[
- Derive and apply the voltage divider equation.
- Explain how a potentiometer acts as an adjustable voltage divider — and, wired differently, as an adjustable resistor.
- Explain how a photoresistor's resistance changes with light, and use it in a voltage divider to sense brightness.
- Build a potentiometer-controlled LED dimmer and a photoresistor-driven light sensor.
]

.eu-label[Essential Understanding:] Two resistors in series don't just add up — the point *between* them sits at a predictable voltage. That single idea is how almost every sensor in this course turns a physical quantity into something a circuit can read.

---

# Agenda

.agenda-box[
1. The voltage divider: two resistors, one useful midpoint
2. Deriving the equation
3. The potentiometer: a divider you can turn — and a resistor you can turn
4. Quick exercises
5. Choosing resistor values for the dimmer
6. **Build 1 (15 min):** potentiometer LED dimmer
7. The photoresistor: resistance that changes with light
8. Photoresistor + voltage divider = a light sensor
9. Quick exercise
10. Choosing the fixed resistor for the photoresistor divider
11. **Build 2 (20 min):** photoresistor night-light preview
12. Homework
]

---

# Two resistors, one useful midpoint

.pull-left.w45[
<img src="assets/schematics/voltage-divider.svg" alt="Voltage divider: R1 and R2 in series, Vout tapped between them" style="max-height:340px;display:block;margin:0 auto;"/>
]
.pull-right.w50[
- Two resistors in series, from a supply voltage down to ground.
- The point *between* them sits at a voltage somewhere between `$V_{in}$` and 0 — exactly where depends on the ratio of the two resistors.
- That midpoint, `$V_{out}$`, is what you tap with a wire (or a multimeter probe) to use elsewhere in your circuit.
]

---

# Deriving the equation

Same current flows through both resistors — it's a series circuit, just like Class 4.

`$$I = \frac{V_{in}}{R_1 + R_2}$$`

`$V_{out}$` is just the voltage across `$R_2$` alone — apply Ohm's law to `$R_2$` using that same current:

`$$V_{out} = I \times R_2 = \frac{V_{in}}{R_1 + R_2} \times R_2$$`

`$$V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$$`

---

# Quick exercise

`$V_{in} = 6\,\text{V}$`, `$R_1 = 1\,\text{k}\Omega$`, `$R_2 = 2\,\text{k}\Omega$`.

1. Calculate `$V_{out}$`.
2. If you swapped `$R_1$` and `$R_2$`, would `$V_{out}$` go up or down? Predict before you calculate — then check.

Show your work in your notebook.

---

# The potentiometer

.pull-left.w40[
<img src="assets/schematics/potentiometer-divider.svg" alt="Potentiometer wired across Vin and ground, wiper tapped as Vout" style="max-height:320px;display:block;margin:0 auto;"/>
]
.pull-right.w55[
- A potentiometer ("pot") is a single resistor with a **third terminal** — a wiper that slides along it.
- The two ends are `$R_1 + R_2$` combined, fixed. The wiper's position sets how much is on each side — exactly like our two-resistor divider, but adjustable by hand.
- Turn the knob, and you're sliding the "midpoint" of the divider back and forth between the two ends.
]

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# The same pot, wired as a rheostat

.pull-left.w50[
<img src="assets/schematics/potentiometer-rheostat.svg" alt="Potentiometer using one end and the wiper, in series with a resistor and LED across the 6 V supply; third pin unused" style="width:100%;margin-top:1em;"/>
]
.pull-right.w45[
- Use only **two** terminals — one end, plus the wiper — and ignore the third end entirely.
- Now the pot behaves as a single **adjustable resistor**: turning the knob changes one resistance value directly, instead of splitting a voltage between two.
- This is called using the pot as a **rheostat**. It's the same physical part, wired differently, doing a different job — controlling current instead of tapping a voltage.
]

---

# Quick exercise

A potentiometer's two ends are connected across a 6 V supply, wiper output goes to `$V_{out}$`.

1. Where should the wiper sit for `$V_{out} = 6\,\text{V}$`?
2. Where should it sit for `$V_{out} = 0\,\text{V}$`?
3. Where should it sit for `$V_{out} = 3\,\text{V}$`?
4. Now imagine it's wired as a rheostat instead (end + wiper only), in series with an LED and a fixed resistor. At which of those same wiper positions would the LED be **brightest**? Why?

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# Choosing resistor values for the dimmer

Goal: the LED should visibly dim and brighten across the **whole** knob turn — bright at one end, still glowing (not dark) at the other — without ever exceeding a safe current.

As a rheostat, the pot's resistance *can* reach 0 Ω at one end — but the 220 Ω is always in series with it, so that end is never a bare short. It's the safety floor under every "brightest" number below.

`$$I = \frac{V_{in} - V_f}{R_{fixed} + R_{pot}}$$`

With `$V_{in} = 6\,\text{V}$`, `$V_f \approx 2\,\text{V}$` (red LED), `$R_{fixed} = 220\,\Omega$` — the same pair you've used since Class 2:

| `$R_{pot}$` | Brightest (`$R_{pot}=0$`) | Dimmest (`$R_{pot}=$` max) |
|---|---|---|
| 1 kΩ | 18.2 mA | 3.3 mA |
| 10 kΩ | 18.2 mA | 0.39 mA |
| 100 kΩ | 18.2 mA | 0.04 mA |

Only the **1 kΩ** pot keeps the LED visibly lit across its *entire* range — that's why Build 1 uses it.

**Quick exercise:** verify the 10 kΩ row's dimmest-current number yourself using the formula above, then explain in one sentence why that pot would make a disappointing dimmer.

---
class: center, middle

# Build 1 (15 min): potentiometer LED dimmer

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# Build 1 instructions

With your partner, using your breadboard, 6 V battery pack, multimeter, **your 1 kΩ potentiometer**, an LED, and a 220 Ω resistor:

1. Build a fixed voltage divider with two resistors of your choice. **Predict** `$V_{out}$` first, then measure it. Do they match?
2. Swap in the potentiometer in place of your two resistors — same two end-terminals across the supply, wiper to `$V_{out}$`. Measure `$V_{out}$` at three positions: fully one way, fully the other way, and roughly in the middle.
3. Now rewire it as a **rheostat**: one end terminal + the wiper only, in series with your 220 Ω resistor and the LED, across the 6 V supply. Leave the third terminal unconnected.
4. Slowly turn the knob end to end. The LED should dim and brighten smoothly — does it match your quick-exercise prediction?
5. In your notebook, sketch both circuits (3-terminal divider vs. 2-terminal rheostat) and label what each one is actually doing.

---

# The photoresistor

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.pull-right.w55[
- A photoresistor (LDR, "light-dependent resistor") is a resistor whose resistance changes with the light hitting it.
- **Dark → high resistance** (megaohms). **Bright → low resistance** (hundreds of ohms).
- It's a passive component — no polarity, no "forward voltage" to worry about, just a resistance that reacts to light.
]

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# Turning light into a voltage

.pull-left.w55[
A photoresistor alone doesn't give you a voltage you can read — it just changes resistance. Put it in the **voltage divider** you already know, with a fixed resistor, and the midpoint voltage now tracks light level.

`$$V_{out} = V_{in} \times \frac{R_{fixed}}{R_{LDR} + R_{fixed}}$$`

**Quick exercise:** if the LDR is on top (`$R_1$`) and the fixed resistor is on the bottom (`$R_2$`), what happens to `$V_{out}$` as the room gets darker (LDR resistance rises)? Predict, then check with the equation.
]
.pull-right.w40[
<img src="assets/schematics/photoresistor-divider.svg" alt="Fixed resistor R1 on top, photoresistor R2 on the bottom, Vout tapped between them" style="max-height:340px;display:block;margin:0 auto;"/>
]

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# Which side should the LDR go on?

Think ahead to Build 2 today — and to the Unit 1 milestone in a few classes: a nightlight that should get brighter as it gets **darker**.

That means you want `$V_{out}$` to go **up** as light goes **down** — which tells you exactly which position the LDR needs to sit in in the divider.

Work it out with your partner: top (`$R_1$`) or bottom (`$R_2$`)? Be ready to justify it with the equation, not just a guess.

---

# Choosing the fixed resistor

.pull-left.w50[
Every LDR is different, so there's no single "right" value — but there's a principle. We want `$V_{out}$` to swing *across* the LED's ~2 V threshold between light and dark, not sit stuck above or below it.

- Too **small** (close to `$R_{bright}$`) → `$V_{out}$` stays high even in bright light — LED never turns off.
- Too **large** (close to `$R_{dark}$`) → `$V_{out}$` stays low even in the dark — LED barely turns on.
- **Sweet spot** — the *geometric mean*:

`$$R_{fixed} \approx \sqrt{R_{bright} \times R_{dark}}$$`
]
.pull-right.w45[
**Worked example:** LDR reads 1 kΩ bright, 100 kΩ covered.

`$$\sqrt{1{,}000 \times 100{,}000} = 10\,\text{k}\Omega$$`

That's exactly your kit's 10 kΩ pot, left untouched — why Build 2's tip suggests trying it as the fixed resistor.
]

---
class: center, middle

# Build 2 (20 min): photoresistor night-light preview

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# Build 2 instructions

With your partner, using your breadboard, 6 V battery pack, multimeter, photoresistor, a fixed resistor, a second LED, and a second 220 Ω resistor:

1. With the photoresistor alone (not yet in the circuit), measure its resistance in normal room light, then covered by your hand. Calculate `$\sqrt{R_{bright} \times R_{dark}}$` — that's your target fixed resistor. Pick the closest value you have, including your 10 kΩ or 100 kΩ pot left untouched.
2. Build a voltage divider using the photoresistor and your chosen fixed resistor, with the LDR in the position you justified earlier.
3. Measure `$V_{out}$` under normal room light, then cover the photoresistor with your hand and measure again. Did it move the direction you predicted?
4. Now tap an LED (with its own 220 Ω resistor) from `$V_{out}$` to ground — same idea as the dimmer in Build 1, but now a sensor is turning the knob instead of your hand.
5. Cover and uncover the photoresistor. Watch the LED.
6. Try a different fixed-resistor value, further from your calculated target. Is the effect less dramatic, like the calculation predicts? .small-font[(If you used a pot as the fixed resistor, this is just a knob-twist.)]

.small-font[Notice: once the LED is connected, it's drawing current out of the divider too, which nudges Vout away from your clean equation. That's a real-world wrinkle — and exactly why, in a few classes, you'll use a transistor to let the sensor *control* the LED without being loaded down by it.]

---
class: center, middle

# Homework

.hw-box[
- In your notebook, explain in your own words why a potentiometer is "a voltage divider you can turn" — and separately, why the same part can also act as "a resistor you can turn."
- Describe what you saw in Build 2: which direction did the LED respond to darkness, and does that match what a real nightlight needs to do?
- Record any measurements from either build you didn't finish in class.
]