Tiêu đề 10. P2C10. সেমিকন্ডাক্টর ও ইলেক্ট্রনিক্স Merge OK_Sha 28.4.24_Mahee -Ok.pdf ✅

†mwgKÛv±i I B‡jKUawb·  Practice Content 1 `kg Aa ̈vq †mwgKÛv±i I B‡jKUawb· Semiconductor and Electronics weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv wjwLZ cÖkœmg~n 1| A B A B jwRK †MUwU n‡Z mZ ̈K mviwY •Zwi Ki| [BUET 21-22] mgvavb: – A – B .  AB – = – AB + A – B = A  B A B – AB A – B A  B = – AB + A – B 0 0 0 0 0 0 1 1 0 1 1 0 0 1 1 1 1 0 0 0 2| IB, IE, VBC, VBE, VCE wbY©q Ki| [ = 49, IC = 5 mA] [BUET 21-22] Ic IB 2.2 k 2 k Vcc = 13 V mgvavb:  = IC IB  49 = 5 IB  IB = 0.102 mA (Ans)  IE = IB + IC = 5.102 mA (Ans) VBE = – 2  103  0.102  10–3 = – 0.204 V (Ans) VCE = Vcc – 2.2  103  5  10–3 = 2V (Ans) VBC = VBE – VCE = – 2.204 V (Ans.) 3| bx‡Pi eZ©bxi Kv‡jKUi wefe VC wbY©q Ki hLb UavbwR÷iwU active mode G wμqvkxj Av‡Q (A_©vr, †eBR-GwgUvi Rvskb m¤§yLx †SuvK Ges Kv‡jKUi-†eBR Rvskb wegyLx †SuvK) †`Iqv Av‡Q, VBE = 0.7 V Ges  = 100. [BUET 19-20] 10V RC=4.7 k 4V B E C RE=3.3 k mgvavb: 10V 4.7 k 4V B E C 3.3 k IB IC 10V 0V IE cÖ_g jy‡c, 4 = VBE + 3.3 × 103 × IE  IE = 1 mA   = IC IB = IC IE – IC  100 = IC 1 – IC  IC = 0.99 mA GLb, 10 – VC = 4.7 × 103 IC  VC = 5.347 V (Ans.) 4| GKwU mvaviY wbtmviK weea©‡Ki cÖevn jvf 70 nq| hw` wbtmviK cÖevn 8.8 mA nq, Zvn‡j msMÖvnK Ges cxV cÖev‡ni gvb wbY©q Ki| UavbwR÷iwU hLb mvaviY cxV weea©K wnmv‡e KvR K‡i, ZLb cÖevn jvf KZ? [BUET 16-17] mgvavb: cÖ_g †ÿ‡Î, cÖevn jvf,  = IC IB = IE – IB IB  70 = 8.8 – IB IB  IB = 0.124 mA (Ans.)  IC = IE – IB = 8.8 – 0.124 = 8.676 mA (Ans.) wØZxq †ÿ‡Î, cÖevn jvf,  = IC IE = 8.676 8.8 = 0.986 (Ans.)
2  Physics 2nd Paper Chapter-10 5| †Kvb UavbwR÷‡i 8.0 mA wbtmviK cÖevn cwieZ©‡bi Rb ̈ 7.9 mA msMÖvnK cÖev‡ni cwieZ©b NUj| cÖevn weea©K ̧YK  Ges Kv‡i›U †MBb  †ei Ki| [KUET 03-04] mgvavb:  = IC IE = 7.9 8 = 0.9875   =  1 –  = 79 (Ans.) 6| †Kvb UavbwR÷‡ii Kgb †em mvwK©‡U GwgUi Kv‡i›U 100 A †_‡K 150 A G DbœxZ Kivq Kv‡j±i Kv‡i›U 98 A †_‡K 147 A DbœxZ nj| Kv‡i›U A ̈vgwcøwd‡Kkb d ̈v±i Ges Kv‡i›U †MBb wbY©q Ki| [RUET 18-19] mgvavb: Kv‡i›U A ̈vgwcøwd‡Kkb d ̈v±i, = IC IE = 147 – 98 150 – 100 = 0.98 (Ans.) Kv‡i›U †MBb,  =  1 –  = 49 (Ans.) 7| †Kvb Kgb †em UavbwR÷‡ii †eR Kv‡i›U I GwgUvi Kv‡i›U h_vμ‡g 5 × 10–4 amp I 1 × 10–3 amp| Kv‡j±i Kv‡i›U Ges Kv‡i›U †MBb d ̈v±i  wbY©q Ki| [RUET 08-09] mgvavb: IC = IE – IB = 10–3 – 5 × 10–4 = 5 × 10–4 A (Ans.)  = IC IE = 5 × 10–4 10–3 = 0.5 (Ans.) 8| GKwU Kgb wbtmviK UavbwR÷i ms‡hv‡M wbtmviK cÖevn 0.85 mA Ges †em cÖevn 0.05 mA|  I  Gi gvb wbY©q Ki| [RUET 07-08] mgvavb:  = IC IE = IE – IB IE   = 0.85 – 0.05 0.85 = 0.941   =  1 –  = 0.941 1 – 0.941    16 (Ans.) 9| †Kvb UavbwR÷‡ii msMÖvnK cÖevn 0.95 mA Ges wbtmviK cÖevn 0.966 mA| UavbwR÷iwUi  Ges  wbY©q Ki|[RUET 04-05, 03-04] mgvavb:  = IC IE = 0.95 0.966 = 0.983   =  1 –  = 57.82 (Ans.) 10| GKwU Kgb BwgUv‡i Ggwcødvqvi eZ©bxi †jvW †iva 4000 | evqvwms †fv‡ëR 20 mV n‡j cxV cÖev‡ni 40 A cwieZ©b N‡U Ges msMÖvnK cÖev‡ni 3 mA cwieZ©b cvIqv hvq| eZ©bxi †fv‡ëR †MBb Ges cvIqvi †MBb wbY©q K‡iv| [BUTex 23-24] mgvavb: BbcyU †iva, Rin = Vin IB = 20  10–3 40  10–6 = 500  †jvW †iva, RL = 4000   †fv‡ëR †MBb, AV =   RL Rin  AV = IC IB  RL Rin  AV = 3 40  10–3 4000 500 = 600 cvIqvi †MBb, AP =  2  RL Rin =     3 40  10–3 2  4000 500  AP = 45000 11| GKwU Ua ̈vbwR÷‡ii Rb ̈  = 0.95 Ges IE = 1 mA n‡j IB †ei Ki| GLv‡b cÖZxK ̧‡jv cÖPwjZ A‡_© e ̈eüZ| [BUTex 21-22; CKRUET 20-21] mgvavb:  = IC IE = IE – IB IE  0.95 = 1 – IB 1  IB = 0.05 mA (Ans.) 12| (K) n-UvBc †mwgKÛv±i wK? [BUTex 10-11] (L) UavÝwWDmvi wK? mgvavb: (K) †h me Aa©cwievnx ev †mwgKÛv±‡ii mv‡_ †fRvj wn‡m‡e c‡hvRx †gŠj mvgvb ̈ cwigv‡Y †gkv‡bv nq Zv‡`i‡K n-UvBc †mwgKÛv±i e‡j| (L) Transducer nj GK ai‡bi iƒcvšÍiK hv GK kw3‡K Ab ̈ kw3‡Z iƒcvšÍi K‡i| †hgbÑ gvB‡μv‡dvb, TV Antena BZ ̈vw` transducerGi D`vniY| 13| GKwU p-n Rsk‡bi wefevšÍi 2.0-volt †_‡K evwo‡Z 2.2-volt Kiv nj| G‡Z Gi Zwor cÖevn 400 mA †_‡K †e‡o 800 mA nj| MZxq †iva KZ? [BUTex 01-02] mgvavb: R = V I = 2.2 – 2 (800 – 400) × 10–3  R = 0.5  (Ans.) weMZ eQ‡i BwÄwbqvwis fvwm©wU‡Z Avmv eûwbe©vPbx cÖkœmg~n 1. [BUET 22-23] 4.5k 12 V Si 0.7 Ge 0.3 (i) eZ©bxi Zwor cÖevn KZ? (ii) Ge Wv‡qvW‡K Dwë‡q w`‡j †iv‡ai `yB cÖv‡šÍi wefe cv_©K ̈ KZ? mgvavb: (i) Zwor cÖevn, I = 12 – 0.7 – 0.3 4.5 × 103  I = 2.44 × 10–3 A (Ans.) (ii) Ge Wv‡qvW‡K Dwë‡q w`‡j Zwor cÖevn = 0  VR = 0  V = 0 V (Ans.) 2. Kv‡i›U weea©K ̧Yv1⁄4  = ? [BUET Preli 22-23]
†mwgKÛv±i I B‡jKUawb·  Practice Content 3 IC IE IC IB IE IB †Kv‡bvwUB bq DËi: IC IE 3. Kgb †em UavbwR÷‡ii Collector current 47 A n‡Z 96 A Ges emitter current 100 A n‡Z 150 A n‡jv| Kv‡i›U Amplification factor = ? [BUET Preli 22-23] 0.98 0.99 0.49 0.48 DËi: 0.98 e ̈vL ̈v: = IC IE = 96 – 47 150 – 100 = 0.98 4. Transistor Kx wn‡m‡e KvR Ki‡Z cv‡i? [BUET Preli 21-22] Amplifier, Rectifier Amplifier, Switch Rectifier, Switch Amplifier, Rectifier, Switch DËi: Amplifier, Switch 5. GwU †Kvb †MBU? [BUET Preli 21-22] X Y Output 1 1 0 1 0 0 0 1 0 0 0 1 OR Gate AND Gate NAND Gate NOR Gate DËi: NOR Gate 6. †Kvb UavbwR÷‡i BbcyU I AvDUcyU wmMb ̈v‡ji g‡a ̈ `kv cv_©K ̈ KZ? [BUET Preli 21-22] 0 180 90 – 90 DËi: 180 7. wb‡Pi †KvbwU‡K †Wv‡c›U wnmv‡e e ̈envi Ki‡j p-UvBc Aa©cwievnxi ag© cvIqv hv‡e bv? [BUET 11-12] A ̈vjywgwbqvg Gw›Ugwb M ̈vwjqvg BwÛqvg DËi: Gw›Ugwb 8. p-n Rvskb ms‡hvM ̄’‡j wW‡cøkb ͇̄ii m„wói KviY njÑ [BUET 10-11] †nv‡jb Zvob Avavb evn‡Ki e ̈vcb B‡jKUa‡bi Zvob Ac`ae ̈ Avqb-Gi ̄’vbvšÍi DËi: Avavb evn‡Ki e ̈vcb 9. wb‡Pi wP‡Î Si Ges Ge Wv‡qvW `ywUi wb-f‡ëR h_vμ‡g 0.7 Ges 0.3 V| 5.5 k †iv‡ai ga ̈ w`‡q KZ we`y ̈r cÖevwnZ n‡e? Ge Wv‡qvWwU D‡ëv K‡i ms‡hvM w`‡j †ivawUi `yB cÖv‡šÍi wefe cv_©K ̈ KZ n‡e? [CKRUET 23-24] +12 V Si Ge 5.5 k 2.0 mA and 0 V 2.0 mA and 0.7 V 2.0 A and 0 V 2.0 mA and 0.3 V 2.0 mA and 11.3 V DËi: 2.0 mA and 0 V e ̈vL ̈v: I = 12 – 0.7 – 0.3 5.5  103 = 2mA Ge Wv‡qvWwU D‡ëv K‡i mshy3 Ki‡j I = 0  V = 0  R = 0 10. GKwU Kgb †em ms‡hv‡M _vKv UavbwR÷‡ii wbtmviK I †em cÖevn h_vμ‡g 5 mA Ges 0.5 mA| wbtmviK Gi cÖevn 5 ̧Y I †em cÖevn 3 ̧Y Kiv n‡j, UavbwR÷iwUi cÖevn e„w× KZ? [CKRUET 23-24] 0 6.67 9 15.67 24.67 DËi: Blank e ̈vL ̈v: Kgb †em ms‡hv‡M cÖevn jvf =  = IC IE = 5 – 0.5 5 = 0.9  = IC IE = IE – IB IE = (5 × 5) – (3 × 0.5) 5 × 5 = 0.94   = 0.04 myZivs Kgb †em Kv‡bKk‡b cÖevn jvf | †m‡ÿ‡Î DËi †bB| cÖevn jvf =  ai‡j,  = IC IB = IE – IB IB = 5 – 0.5 0.5 = 9  = IC IB = IE – IB IE
4  Physics 2nd Paper Chapter-10 = (5 × 5) – (0.5 × 3) 0.5 × 3 = 15.67   =  –  = 6.67  cÖevn e„w× 6.67 11. Kgb G ̈vwgUvi eZ©bx‡Z GKwU n-p-n UavbwR÷‡ii msMÖvn‡K 8 V e ̈vUvix mshy3 Av‡Q| msMÖvnK eZ©bxi RC †iv‡ai `yB cÖv‡šÍ wefe cZb 0.5 V n‡j †em Kv‡i›U wbY©q K‡iv| [†`Iqv Av‡Q, RC = 800  Ges  = 0.96] [CKRUET 22-23] 0.625 mA 0.000625 mA 0.974 mA 0.026 mA 0.000026 mA DËi: 0.026 mA e ̈vL ̈v:IC = VC RC = 0.5 800 = 6.25  10–4 A IB = IC  = 6.25  10–4 0.96  (1 – 0.96) A = 2.6  10–5 A = 0.026 mA 12. GKwU UavbwR÷‡ii wb¤œwjwLZ gvb ̧‡jv cwigvc Kiv n‡jv| [CKRUET 21-22] IC = 1.98 mA; IB = 20A; UavbwR÷‡ii ,  Ges IE Gi gvb †ei Ki|  = 0.99;  = 99 and IE = 5.02 mA  = 99;  = 0.99 and IE = 5.02 mA  = 0.99;  = 99 and IE = 7.00 mA  = 99;  = 0.99 and IE = 5.02 mA  = 0.99;  = 0.99 and IE = 5.02 A DËi: Blank e ̈vL ̈v:  = IC IE = IC IC + IB = 1.98 1.98 + 20 × 10–3 = 0.99  =  1 –  = 99 IE = IB + IC = 1.98 + 20 × 10–3 = 2 mA 13. wbtmviK cÖev‡ni 11.6 mA cwieZ©b msMÖvnK cÖev‡ni 10.92 mA cwieZ©b NUvq|  Gi gvb KZ? [KUET 18-19] 19 18 16 13 15 DËi: 16 e ̈vL ̈v:  = IC IB = IC IE – IC = 10.92 11.6 – 10.92    16 14. 98 †K evBbvix msL ̈vq iƒcvšÍi Ki| [KUET 15-16] 1100010 0100011 1000011 100011 110001 DËi: 1100010 e ̈vL ̈v: Calculator e ̈envi K‡i| 15. †Kvb UavbwR÷i mvaviY cxV ms‡hv‡M mshy3| Gi wbtmviK cÖevn 0.88 mA Ges cxV cÖevn 0.065 mA| cÖevn weea©b ̧YK KZ? [KUET 14-15; CUET 13-14; RUET 09-10] 0.942 0.93 0.95 0.96 0.926 DËi: 0.926 e ̈vL ̈v:  = IC IE = IE – IB IE = 0.88 – 0.065 0.88   = 0.926 16. GKwU UavbwR÷‡ii †ÿ‡Î  = 0.95 Ges IE = 0.9 mA n‡j  KZ n‡e? [KUET 13-14] 19 16 18 12 10 DËi: 19 e ̈vL ̈v:  =  1 –  = 0.95 1 – 0.95 = 19 17. 0.02 A wbtmviK cÖev‡ni d‡j GKwU UavbwR÷‡i 18 mA msMÖvnK cÖevn cvIqv †Mj| UavbwR÷‡ii f~wg cÖev‡ni gvb KZ? [KUET 12-13] 38 mA 2 A 2 mA 0.2 A 0.38 A DËi: 2 mA e ̈vL ̈v: IB = IE – IC = 0.02 – 18 × 10–3  IB = 2 mA 18. GKwU mvaviY f~wg UavbwR÷v‡i msMÖvnK cÖevn 0.85 A Ges f~wg cÖevn 0.05 mA| cÖevn weea©K ̧YK KZ? [RUET 13-14] 0.99994 1.99994 0.49999 4.9999 None DËi: 0.99994 e ̈vL ̈v:  = IC IE = IC IC + IB = 0.85 0.85 + 0.05 × 10–3   = 0.99994 19. GKwU UavbwR÷‡ii weea©b ̧Yv1⁄4 0.98 Ges A ̈vwgUvi Kv‡i›U 1.5 mA n‡j Kv‡j±i Kv‡i›U KZ? [RUET 12-13] 7.47 mA 4.74 mA 4.17 mA 1.74 mA 1.47 mA DËi: 1.47 mA