2.4.1
Ek Bipolar Junction Transistor (BJT) ek teen doped semiconductor regions ka sandwich hota hai jo is tarah stack kiya jaata hai ki beech wale region mein ek bahut chhota current, poore device mein flow hone wale ek bade current ko control karta hai. Bipolar ka matlab hai electrons aur holes dono current carry karte hain (ek FET ke unlike, jo sirf ek carrier type use karta hai). Saari magic is baat se aati hai ki beech wali layer ko purposely bahut thin aur lightly doped banaya jaata hai — taaki charge usse cross kar sake instead of usme trap ho jaane ke.
Hume ek aisa device chahiye jo amplify kare — jahan ek chhota signal ek bade ko control kare. Iske liye hume ek aisi physical structure chahiye jahan ek chhota input current ek bahut bade output current ko "steer" kar sake. BJT yeh do PN junctions ko back-to-back rakh ke achieve karta hai, jo ek common middle region share karte hain.
Ek BJT ek three-terminal device hai jisme teen semiconductor regions N-P-N ya P-N-P order mein hote hain, jo do PN junctions banate hain. Teen terminals hain Emitter (E) , Base (B) , aur Collector (C) .
Region
Doping
Size
Kaam (WHY)
Emitter heavily doped (n⁺ ya p⁺)medium
Base mein majority carriers emit (inject) karta hai. Heavy doping → inject karne ke liye bahut saare carriers.
Base lightly doped bahut thin Flow ko control karta hai. Thin + light doping → zyaadatar carriers recombine hone ki jagah through nikal jaate hain.
Collector moderately doped
sabse bada Carriers ko collect karta hai. Sabse bada taaki heat dissipate kar sake; collection ke liye area bada hota hai.
Intuition Base KYUN thin aur lightly doped hona chahiyeAgar base mota ya heavily doped hota, toh inject kiye gaye carriers bahut saare opposite carriers se milte aur collector tak pahunchne se pehle recombine ho jaate — koi current through nahi jaata, koi amplification nahi. Base ko thin (∼ 1 μ m recombine hone se pehle diffuse karke cross ho jaate hain . Yeh BJT ka sabse important structural fact hai.
Kyunki do PN junctions hain, har ek ko ek naam milta hai:
Emitter–Base junction (EBJ) → input junction.Collector–Base junction (CBJ) → output junction.
Normal amplification (active mode ) ke liye hum inhe opposite bias karte hain:
Emitter (n⁺) – Base (p) – Collector (n). Electrons inject kiye jaane wale carriers hain. Conventional current collector aur base mein andar flow karta hai, emitter se bahar .
Emitter (p⁺) – Base (n) – Collector (p). Holes inject kiye jaane wale carriers hain. Conventional current collector aur base se bahar flow karta hai, emitter mein andar .
Mnemonic Arrow rule (schematic symbol par)Arrow hamesha Emitter par hota hai.
NPN = "N ot P ointing iN " → arrow base se BAHAR point karta hai.
PNP = "P ointing iN P roudly" → arrow base ki taraf ANDAR point karta hai.
Arrow conventional current ki direction dikhata hai emitter par.
Maano I E I B I C
Step 1 — Charge conservation (transistor par ek node ki tarah KCL):
Jo kuch bhi andar jaata hai woh bahar aana chahiye.
I E = I B + I C Yeh step kyun? Transistor ek closed node hai; koi charge create ya store nahi hota (steady state mein).
Step 2 — Sirf ek fraction base se bachta hai. Emitter current I E I B Common-base current gain define karo:
α = I E I C , α ≈ 0.95 – 0.99 1 ke itna kareeb kyun? Kyunki base thin/lightly doped hai → almost kuch recombine nahi hota → almost saara I E I C
Step 3 — Base-referenced gain β I C = α I E I B = I E − I C = ( 1 − α ) I E β = I B I C = ( 1 − α ) I E α I E = 1 − α α Yeh kyun matter karta hai: Agar α = 0.99 β = 0.99/0.01 = 99 99× bade collector current ko control karta hai — yahi amplification hai, purely geometry se derive kiya gaya (thin base ⇒ α → 1 β
Worked example Example 1 — Currents seβ I C = 4.95 mA I B = 0.05 mA β = I B I C = 0.05 4.95 = 99 I E = I C + I B = 5.00 mA α = I C / I E = 4.95/5.00 = 0.99 Yeh step kyun? I E α
Worked example Example 2 — Pehle forecast karo phir verify karoDiya hai α = 0.98 β forecast karo: kyunki α β bada hona chahiye.
β = 1 − 0.98 0.98 = 0.02 0.98 = 49 Backward verify karo: α = β / ( β + 1 ) = 49/50 = 0.98 β
Worked example Example 3 — NPN carrier directionActive mode mein NPN mein, electrons Emitter → Base → Collector flow karte hain.
Conventional current electrons ke opposite hai, isliye conventional I C , emitter se exit karta hai. Isliye NPN mein emitter arrow BAHAR point karta hai.
Yeh step kyun? Carrier type (electrons) current direction fix karta hai, jo symbol ka arrow fix karta hai.
Common mistake "Emitter aur Collector interchangeable hain kyunki dono N hain (NPN mein)."Kyun sahi lagta hai: Dono end regions same doping type ke hain, isliye symbol symmetric lagta hai.
Fix: Ye doping level aur size mein alag hain — emitter heavily doped aur chhota hai; collector bada aur moderately doped hai. Inhe swap karne par ("reverse active" mode) bahut chhota gain milta hai. Structure ≠ symmetry.
Common mistake "Base mein hi saara current flow karta hai, kyunki yeh beech mein hai."Kyun sahi lagta hai: Base control terminal hai, isliye yeh central lagta hai.
Fix: Base sabse chhota current carry karta hai (I B control mein hai, carrying mein nahi — chhota I B I C
Common mistake "PNP mein current directions NPN jaisi hi hain."Kyun sahi lagta hai: Same structure, bas relabeled.
Fix: Carriers holes hain aur biasing polarities reverse hain, isliye saari current directions reverse ho jaati hain . PNP emitter arrow ANDAR point karta hai.
BJT ke teen regions order mein kya hain? Emitter – Base – Collector (N-P-N ya P-N-P)
Base ko thin aur lightly doped kyun banate hain? Taaki inject kiye gaye carriers recombine hone se pehle diffuse karke cross ho jaayein, aur zyaadatar collector tak pahunch sakein (high α).
Kaun sa region sabse heavily doped hai aur kyun? Emitter — heavy doping bahut saare carriers deta hai base mein inject karne ke liye.
Kaun sa region sabse bada hai aur kyun? Collector — carrier collection aur heat dissipation ke liye bada area.
Fundamental current equation batao. I E = I B + I C
α (common-base gain) define karo. α = I C / I E
β define karo aur α se relate karo. β = I C / I B = α / ( 1 − α )
Active mode mein do junctions kaise biased hote hain? EBJ forward biased, CBJ reverse biased.
NPN vs PNP mein kaun se carriers conduct karte hain? NPN: electrons; PNP: holes.
Symbol par emitter arrow NPN vs PNP ke liye kahan point karta hai? NPN: base se bahar (Not Pointing iN); PNP: base ke andar.
BJT mein "bipolar" ka matlab kya hai? Electrons aur holes dono conduction mein participate karte hain.
Agar α = 0.99 ho toh β kya hai? 99.
Recall Feynman: 12 saal ke bacche ko explain karoEk paani ka gate imagine karo. Emitter ek paani se bhari badi tank hai, Collector woh jagah hai jahan paani pahunchna chahiye, aur Base beech mein ek patli diwar hai jisme ek chhota valve hai. Jab tum chhota valve dheemare ghumaate ho (ek tiny base current), tab tank se collector tak paani ki ek BADI baadh aa jaati hai. Diwar ko super thin banaya gaya hai taaki paani seedha cross kar sake instead of wahan atak jaane ke. Ek chhoti koshish se bada flow control hota hai — isi tarah ek transistor amplify karta hai!
Tables cover karo. Kya tum bata sakte ho: kaun sa region sabse thin hai, kaun sa sabse zyaada doped hai, active mode mein do junctions kaise biased hote hain, aur β = α / ( 1 − α )
Amplification: small controls large
Bipolar: electrons and holes
Collector largest moderately doped
