Diode logic gate basics — Hinglish notes

2.3.9 · Hardware › Diodes & Applications

Ek diode logic kyun karta hai?

Yahan diode KYA kar raha hai? Ek diode tab conduct karta hai (ON hota hai) jab uska anode, uske cathode se $0.7 V$ (silicon ke liye) zyada positive ho. Warna woh block karta hai. ON hone par, woh roughly ek battery ki tarah behave karta hai: woh apne upar voltage ko $\approx V_{D} \approx 0.7 V$ tak clamp kar deta hai.

YEH logic kyun deta hai: Logic bas yahi hai — "kya yeh node HIGH hai ya LOW?" Ek conducting diode ek shared node ko kisi ek input ki taraf pull kar sakta hai. Agar hum kai diodes ko aise arrange karein ki woh sab ek hi node ko pull karne ki koshish karein, toh jo diode sabse zyada hardest conduct karta hai woh node voltage ko dominate karta hai. Diode ki orientation choose karna decide karta hai ki HIGH-inputs jeetenge (OR) ya LOW-inputs jeetenge (AND).

DIODE OR gate ko scratch se derive karna

Setup (derive karo, memorise mat karo): Do inputs $A$ aur $B$ rakho. Har input se ek common node $Y$ tak ek diode connect karo, anode input par, cathode $Y$ par (arrow $Y$ ki taraf point karta hai). $Y$ se ground tak ek resistor $R$ connect karo.

Yeh arrangement kyun? Anode-at-input ka matlab hai ek diode sirf current ko ek HIGH input se $Y$ mein push kar sakta hai. Toh koi bhi HIGH input $Y$ ko upar utha sakta hai.

Chalo case by case reason karte hain ( $V_{H} = 5 V$ , $V_{D} = 0.7 V$ ):

$A = 0, B = 0$ : Koi bhi input apne diode ko forward-bias nahi kar sakta (dono anodes 0 V par hain, current ke liye node ko 0 V se neeche jaana padega). Koi diode conduct nahi karta $\Rightarrow$ $R$ , $Y$ ko ground tak pull karta hai $\Rightarrow Y = 0$ . ✔ (OR of 0,0 = 0)
$A = 5, B = 0$ : $A$ ka diode conduct karta hai. Current flow karta hai $A \to$ diode $\to Y \to R \to$ gnd. $Y$ tab tak rise karta hai jab tak diode ON nahi ho jaata, clamping: $Y = V_{A} - V_{D} = 5 - 0.7 = 4.3 V (HIGH) .$ ✔
$A = 0, B = 5$ : Symmetric hai, $Y = 4.3 V$ . ✔
$A = 5, B = 5$ : Dono conduct karte hain, $Y = 4.3 V$ . ✔

DIODE AND gate ko scratch se derive karna

Sab kuch flip karo. Ab diodes cathode input par, anode node $Y$ par rakho (arrows $Y$ se door, inputs ki taraf point karte hain). Resistor $R$ ko $Y$ se upar $+ V_{C C}$ (supply) tak connect karo.

Flip kyun? Cathode-at-input ka matlab hai ek LOW input $Y$ se current pull out kar sakta hai, $Y$ ko neeche kheench sakta hai. Pull-up resistor $Y$ ko high rakhne ki koshish karta hai; koi bhi grounded input use defeat kar deta hai.

Case analysis ( $V_{C C} = 5 V$ ):

$A = 0, B = 0$ : Dono diodes conduct karte hain, $Y$ ko neeche kheenchte hain. $Y = V_{A} + V_{D} = 0 + 0.7 = 0.7 V (LOW)$ . ✔
$A = 5, B = 0$ : $B$ ka diode conduct karta hai (path: $V_{C C} \to R \to Y \to$ diode $\to B$ =0). $Y = 0 + 0.7 = 0.7 V (LOW)$ . $A$ ka diode reverse biased hai (uska anode $Y = 0.7$ V < cathode $A = 5$ V) → OFF. ✔
$A = 0, B = 5$ : Symmetric hai, $Y = 0.7 V$ . ✔
$A = 5, B = 5$ : Koi diode conduct nahi kar sakta (dono cathodes 5 V par hain, node unse itna upar nahi ja sakta). $R$ , $Y$ ko $V_{C C} = 5 V (HIGH)$ tak pull karta hai. ✔

Worked Examples

Common Mistakes (Steel-man + fix)

Recall Feynman: 12-saal ke bachche ko samjhao

Ek paani ki bucket (output node) imagine karo jisme ek chhota drain hole hai (resistor). OR gate: kai pipes bucket mein andar point karti hain, har ek mein ek one-way flap hai toh paani sirf andar jaata hai. Agar koi bhi pipe on ho, bucket bhar jaati hai — output HIGH. Sirf tab jab saari pipes off hoon, bucket drain hokar khaali hoti hai. "Koi bhi ON → full." AND gate: ab bucket ko top par ek tap se bhara jaata hai, aur pipes paani bahar drain karti hain agar unka end open (LOW) ho. Agar koi bhi pipe ka end open ho, paani bahar nikalta hai → bucket khaali → LOW. Sirf tab jab har pipe plugged ho (sab HIGH) bucket bhari rehti hai. Tiny $0.7$ V waisa hai jaise bucket ke kinare par ek chhota sa lip hai jise paani poori tarah cross nahi kar pata — isliye yeh kabhi perfectly full ya empty nahi hoti.

Active-Recall Flashcards

Diode OR gate output formula

Y = max (V_{A}, V_{B}) - V_{D}

Diode AND gate output formula

Y = min (V_{A}, V_{B}) + V_{D}

Diode OR gate mein resistor kahan connect hota hai?

Ground se (pull-down); diodes anode→node point karte hain.

Diode AND gate mein resistor kahan connect hota hai?

V_{C C}

se (pull-up); diodes cathode→input point karte hain.

Diode-OR HIGH kabhi full 5 V kyun nahi hota?

Conducting diode ke across hamesha ek diode drop (

\approx 0.7

V) loss hota hai.

Sirf diodes se NOT gate kyun nahi bana sakte?

Diodes passive hain — woh invert ya amplify nahi kar sakte; transistor chahiye.

OR gate mein node ko kaun sa diode dominate karta hai?

Jo highest input se connected hai; woh baaki sab ko reverse-bias kar deta hai.

AND gate ka LOW output voltage (Si) kya hai?

\approx 0.7

V, 0 V nahi.

OR vs AND ke liye mnemonic kya hai?

"OR pulls the flOOR (max), AND pulls the ceiling (min)."

Diode logic chain karne ki main limitation kya hai?

Har stage ~0.7 V lose karti hai, isliye HIGH levels degrade hote hain aur deeply cascade nahi ho sakti.

Connections

PN Junction Diode — forward-drop $V_{D} \approx 0.7$ V jo hamare saare offsets set karta hai.
Diode as a Switch — ON/OFF model jo voting behaviour ke neeche kaam karta hai.
DTL Diode-Transistor Logic — invert + restore levels ke liye transistor add karta hai.
Logic Levels and Noise Margin — level degradation kyun khatarnaak hai.
Boolean Algebra basics — $max$ =OR, $min$ =AND correspondence.
Pull-up and Pull-down Resistors — resistor jo default node state set karta hai.