GPIO — input - output, pull-up - pull-down, interrupt on pin change: Exercises

5.5.2 · D4 · Coding › Embedded Systems & Real-Time Software › GPIO — input - output, pull-up - pull-down, interrupt on pin

Do symbols baar baar aate hain, toh unhe ek baar plain words mein pin kar lete hain:

Definition Wo do quantities jo hum poore page compute karte hain

$V_{D D}$ = supply voltage — wo "full HIGH" level jis par chip run karta hai, volts (V) mein measure hota hai. Ise ek tank ke top par paani ka pressure samjho.
$I$ = electric current — ek wire se per second kitna charge flow karta hai, amperes (A) mein measure hota hai. $1 mA = 1 0^{- 3} A$ , $1 μ A = 1 0^{- 6} A$ . Ise litres of water per second samjho.
$R$ = resistance — ek component current se kitna fight karta hai, ohms ( $Ω$ ) mein; $1 k Ω = 1000 Ω$ . Socho: ek narrow pipe.

Yeh Ohm's law se tied hain, wo ek equation jis par yeh poora page lean karta hai: $V = I \cdot R ⟺ I = \frac{V}{R} ⟺ R = \frac{V}{I}$ Kuch fancier kyun nahi? Kyunki ek pull resistor sirf ek resistor hai, aur hum jo ek hi sawaal poochte hain woh hai "given {voltage across it, current through it, its resistance} mein se do, teesra nikalo." Ohm's law exactly yahi answer karta hai.

Figure — GPIO — input - output, pull-up - pull-down, interrupt on pin change

Upar ki figure dekho: yeh almost har problem ke liye reference circuit hai — ek internal pull-up resistor $R$ pin se $V_{D D}$ tak, ek button pin aur $GN D$ ke beech, aur CPU ka sense buffer pin par "sunne" wala. Red element wahi hai jo har problem mein poochha jaata hai.

Level 1 — Recognition

L1.1 Ek pin is tarah configure ki gayi hai ki jab kuch connected ya pressed nahi hota, CPU logic 1 (HIGH) read karta hai. Kaun sa resistor enable hai — pull-up ya pull-down?

Recall Solution

Pull-up. Definition ke hisaab se pull-up pin ko $V_{D D}$ se ek weak resistor ke through tie karta hai, isliye idle (undriven) read HIGH = logic 1 hoti hai. Pull-down use karein toh woh $GN D$ se tie hota aur idle LOW = logic 0 hota.

L1.2 Ek button active-LOW wire kiya gaya hai. CPU pin ko 0 read karta hai. Kya button pressed hai ya released?

Recall Solution

Pressed. "Active-LOW" ka matlab hai ki active (pressed) state LOW = 0 produce karta hai. Released → pull-up → HIGH = 1. Toh 0 ⇒ pressed.

L1.3 Almost har MCU par ek GPIO pin use karne ke liye aap jinhe touch karte hain, un teen registers ke naam batao, aur batao har ek kya control karta hai.

Recall Solution

Direction register (DDRx / MODER / TRIS) — input vs output set karta hai.
Output-data register (PORTx / ODR) — woh HIGH/LOW jo aap output hone par drive karte ho.
Input-data register (PINx / IDR) — woh level jo aap input hone par read karte ho.

Level 2 — Application

L2.1 $V_{D D} = 3.3 V$ . Ek internal pull-up of $R = 47 k Ω$ use kiya jaata hai. Jab button pressed hota hai (pin GND se button ke through short ho jaata hai), pull-up se kitna current flow karta hai? Apna jawab $μ A$ mein do.

Recall Solution

Jab pressed hota hai toh pin $\approx 0 V$ par hoti hai, toh poora $V_{D D}$ , $R$ ke across drop hota hai. Poora $V_{D D}$ kyun? Kyunki button short $R$ ke ek end ko GND par hold karta hai aur $V_{D D}$ doosre end ko, isliye resistor ke across voltage $V_{D D} - 0 = V_{D D}$ hoti hai. $I = \frac{V _{D D}}{R} = \frac{3.3}{47 000} = 7.02 \times 1 0^{- 5} A \approx 70.2 μ A$

L2.2 Aapko $V_{D D} = 5 V$ ke saath pressed-state current $50 μ A$ ya us se neeche rakhni hai. Minimum pull-up resistance kya hai?

Recall Solution

Chhota $R$ ⇒ bada current, toh ek maximum current ek minimum $R$ deta hai. $R$ ke liye Ohm's law rearrange karo: $R \geq \frac{V _{D D}}{I} = \frac{5}{50 \times 1 0 ^{- 6}} = 100000 Ω = 100 k Ω$ Toh ≥ 100 kΩ use karo. Isse chhota kuch bhi button hold hone par allowed $50 μ A$ se zyada waste karega.

L2.3 Ek real output transistor pin ko LOW drive karta hai on-resistance $R_{d r i v er} = 8 Ω$ ke saath, ek pull-up $R_{p u l l} = 33 k Ω$ se $V_{D D} = 3.3 V$ tak fight karta hua. Pin actually kis voltage par settle hoti hai? Kya yeh valid LOW hai?

Recall Solution

Pull-up aur driver ek voltage divider form karte hain: $V_{D D}$ $R_{p u l l}$ se neeche push karta hai, driver GND ki taraf $R_{d r i v er}$ se pull karta hai. Divider kyun? Kyunki dono resistors $V_{D D}$ se GND tak series mein hain pin as the middle tap ke saath, aur pin voltage woh fraction of $V_{D D}$ hai jo bottom resistor $R_{d r i v er}$ ke across land hoti hai: $V_{p in} = V_{D D} \cdot \frac{R _{d r i v er}}{R _{d r i v er} + R _{p u l l}} = 3.3 \cdot \frac{8}{8 + 33 000} = 8.0 \times 1 0^{- 4} V \approx 0.8 mV$ Yeh essentially $0 V$ hai — ek rock-solid LOW. Yahi poora reason hai ki "weak" pull-ups kaam karte hain: strong driver unhe overwhelm kar deta hai.

Level 3 — Analysis

L3.1 Ek input pin mein koi pull resistor enable nahi hai aur kuch bhi drive nahi kar raha. Ek colleague kehta hai "yeh bas 0 read karta hai, yeh theek hai." Circuit terms mein explain karo ki read actually unreliable kyun hai, aur stray voltage physically kahan store hoti hai.

Recall Solution

Input ek high-impedance sense buffer hai — yeh almost koi current draw nahi karta, isliye yeh pin ko kahin nahi pull kar sakta. Pin aur nearby traces ek tiny capacitor (charge store) form karte hain. Kuch bhi voltage fix karne ke liye nahi hai, toh woh capacitor mains hum, nearby switching wires, yahan tak ki aapki ungli ke field se charge pick up karta hai. Toh read wahi hoti hai jo noise ne node ko last charge kiya tha — kabhi 0, kabhi 1, randomly. Ek weak pull resistor $V_{D D}$ ya $GN D$ ke liye ek defined path supply karta hai, isliye capacitor hamesha ek known level par relax karta hai.

L3.2 Ek active-LOW button ke liye aapko interrupt ka trigger choose karna hai. Rising, falling, ya level? Justify karo ki baki do kyun galat hain.

Recall Solution

Falling edge. Active-LOW idle HIGH rehta hai aur press par LOW ho jaata hai, yaani ek $1 \to 0$ transition = falling edge. Yeh exactly ek baar per press fire karta hai — woh event jis ki hume parwah hai.

Rising edge release ( $0 \to 1$ ) par fire karega, press par nahi — galat moment.
Level (LOW) continuously re-fire karta rahega jab tak button held hai, CPU ko flood kar dega. Ek edge ek one-shot event hai; ek level ek sustained state hai.

L3.3 Ek pressed button ek burst of ~14 edge interrupts 6 ms mein generate karta hai jabki aapne sirf ek baar press kiya. Is phenomenon ka naam batao, physical cause explain karo, aur ek concrete time window ke saath software fix do.

Recall Solution

Yeh switch bounce hai. Metal contacts physically slam karte hain, spring apart hote hain, aur ~1–10 ms mein settle hone se pehle kai baar re-touch karte hain, toh har micro-contact hardware ke liye ek real edge hai. Fix = debounce: pehle edge par, ek timestamp record karo aur aage ke edges ko ignore karo jab tak, maano, 20 ms elapsed nahi ho jaata (20 ms ~6 ms bounce window se comfortably zyada hai). Alternatively line ko ek RC network se filter karo taaki fast wiggles pin tak kabhi pahunche hi nahi. Switch Debouncing dekho.

Level 4 — Synthesis

L4.1 Ek keypad line ke liye pull-up design karo. Constraints: $V_{D D} = 3.3 V$ ; pressed-state current $60 μ A$ se neeche rehni chahiye; pull-up itna strong hona chahiye ki pin tak $1 μ A$ tak ka leakage LOW ko $0.4 V$ se upar na utha sake. Ek valid resistance range do.

Recall Solution

Upper bound (current cap). Bada $R$ = kam current, toh current limit ek minimum $R$ set karta hai: $R_{min} = \frac{V _{D D}}{I _{ma x}} = \frac{3.3}{60 \times 1 0 ^{- 6}} = 55000 Ω = 55 k Ω$ Lower bound... ruko, leakage ek maximum $R$ set karta hai. Jab button LOW pull karta hai, ek leakage current $I_{l e ak} = 1 μ A$ pull-up se bhi flow karta hai aur pin ko lift karta hai. GND ke upar pin voltage $V = I_{l e ak} \cdot R$ hai (pull-up ke across Ohm's law, $I_{l e ak}$ as current ke saath). $V \leq 0.4 V$ rakhne ke liye: $R_{ma x} = \frac{V}{I _{l e ak}} = \frac{0.4}{1 \times 1 0 ^{- 6}} = 400000 Ω = 400 k Ω$ Toh $55 k Ω \leq R \leq 400 k Ω$ mein koi bhi $R$ kaam karega. Ek classic 100 kΩ us range mein safely baith jaata hai. (Related sizing logic: Pull-up Resistor Sizing & Ohm's Law.)

L4.2 Pin 3 par ek active-LOW button ke liye ek correct interrupt-driven skeleton (pseudo-C) likho jo (a) internal pull-up use kare, (b) per press ek baar trigger kare, (c) 20 ms se debounce kiya gaya ho, aur (d) -O2 optimization ke under safe ho. Batao kyun har safety choice wahan hai.

Recall Solution

volatile bool pressed = false;     // shared ISR<->main => volatile
volatile uint32_t last_edge_ms = 0;
 
void ISR_PIN3() {                  // hardware jumps here on falling edge
    uint32_t now = millis();
    if (now - last_edge_ms >= 20)  // debounce: ignore bounce burst
        pressed = true;
    last_edge_ms = now;
    clear_interrupt_flag(PIN3);    // else it re-fires immediately
}
 
int main() {
    pin_mode(3, INPUT_PULLUP);         // idle = HIGH (active-LOW)
    attach_interrupt(3, FALLING, ISR_PIN3);  // once per press
    while (1) {
        if (pressed) { handle_press(); pressed = false; }
    }
}

pressed aur last_edge_ms par volatile: compiler otherwise unhe ek register mein cache kar leta hai aur main loop kabhi ISR ki writes nahi dekhta — exactly woh bug jo "debug mein kaam karta hai, -O2 par break karta hai."
Falling trigger: active-LOW ke liye per press ek ISR.
clear_interrupt_flag: hardware ek flag latch karta hai; clear nahi kiya toh ISR forever re-enter karta hai.
20 ms debounce window: bounce burst swallow karta hai. Tiny ISR body: ek flag set karo, heavy kaam main mein karo. Interrupts & ISR Design aur volatile & Memory-Mapped Registers dekho.

Level 5 — Mastery

L5.1 Ek student pin 5 ko ek push-pull output ke roop mein set karta hai 1 drive karte hue, phir usi loop mein pin 5 read karta hai ek button sense karne ki umeed mein jo us se wired hai. Read hamesha 1 hoti hai button se independent. Poori tarah diagnose karo: read stuck kyun hai, button current kya karta hai, aur correct fix kya hai.

Recall Solution

Ek push-pull output line ko actively drive karta hai: 1 drive karna top transistor on karta hai, pin ko $V_{D D}$ se kuch ohms ke through hard-connect karta hai. Input register tab woh level return karta hai jo output force kar raha hai — 1 — button nahi. Isse bhi bura: agar button pressed hai toh yeh $V_{D D} \to$ button $\to GN D$ ko us low-ohm transistor ke through short karta hai, ek near-short jo wastefully large current sink kar sakta hai ya pin ko damage kar sakta hai. Fix: ek button sense karne ke liye aapko pin direction = INPUT pull-up ke saath set karni hai (INPUT_PULLUP), jo drive transistors disconnect karta hai aur world ko level set karne deta hai. Output aur input physically alag paths hain (Push-Pull vs Open-Drain Outputs aur Polling vs Interrupt-Driven I/O dekho).

L5.2 Ek shared line par do devices, ek open-drain output ek external pull-up $R = 4.7 k Ω$ ke saath $V_{D D} = 3.3 V$ tak. Jab ek device line ko LOW pull karta hai ( $V_{p in} \approx 0$ ), bus current compute karo. Phir explain karo ki aisi shared bus par open-drain kyun mandatory hai, push-pull kyun nahi.

Recall Solution

Jab ek device line ko LOW hold karta hai, poora $V_{D D}$ pull-up ke across drop hota hai: $I = \frac{V _{D D}}{R} = \frac{3.3}{4700} = 7.02 \times 1 0^{- 4} A \approx 0.70 mA$ Open-drain mandatory kyun hai: har device sirf LOW pull ya release kar sakta hai (Hi-Z ho jaao); shared pull-up HIGH provide karta hai. Toh agar do devices ek saath act karein, worst case dono LOW pull karein — harmless. Push-pull ke saath, ek device HIGH drive kare aur doosra LOW drive kare toh $V_{D D}$ directly $GN D$ se do low-ohm transistors ke through connect ho jaata: ek destructive short. Open-drain wired-AND bus ko safe banata hai. Exactly aisa Open-drain & I²C Bus mein hota hai.

Recall Har formula ka ek-line recap jo use hua

Current through a pull-up when pressed: $I = V_{D D} / R$ .
Minimum $R$ from a current cap: $R_{min} = V_{D D} / I_{ma x}$ .
Maximum $R$ from a leakage/voltage limit: $R_{ma x} = V / I_{l e ak}$ .
LOW voltage under a real driver (divider): $V_{p in} = V_{D D} R_{d r i v er} / (R_{d r i v er} + R_{p u l l})$ .

Parent GPIO note par wapas jaao.