Title 10. P2C10. সেমিকন্ডাক্টর ও ইলেক্ট্রনিক্স (With Solve).pdf ✅

†mwgKÛv±i I B‡jKUawb·  Engineering Practice Sheet .................................................................................................... 1 WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. 4.5 k 12 V Si 0.7 V Ge 0.3 V (i) eZ©bxi Zwor cÖevn KZ? (ii) Ge Wv‡qvW‡K Dwë‡q w`‡j †iv‡ai `yB cÖv‡šÍi wefe cv_©K ̈ KZ? [BUET 22-23] 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. 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 = 2 V (Ans) VBC = VBE – VCE = – 2.204 V (Ans.) 3. jwRK †MUwU n‡Z mZ ̈K mviwY •Zwi Ki| [BUET 21-22] A B A B 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 (Ans.) 4. bx‡Pi eZ©bxi Kv‡j±i wefe VC wbY©q Ki hLb UavbwR÷iwU active mode G wμqvkxj Av‡Q (A_©vr, †eBR-GwgUi Rvskb m¤§yLx †SuvK Ges Kv‡j±i-†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
2 ......................................................................................................................................  Physics 2nd Paper Chapter-10 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.) 5. 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 = 0.125 mA (Ans.)  IC = IE – IB = 8.8 – 0.125 = 8.675 mA (Ans.) wØZxq †ÿ‡Î, cÖevn jvf,  = IC IE = 8.675 8.8 = 0.986 (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 1. †Kv‡bv 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.) 2. †Kv‡bv Kgb †em UavbwR÷‡ii †eR Kv‡i›U I GwgUi 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.) 3. 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 (Ans.)   =  1 –  = 0.941 1 – 0.941    16 (Ans.) 4. †Kv‡bv 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 (Ans.)   =  1 –  = 57.82 (Ans.) 5. †Kv‡bv 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 (Ans.)   =  1 –  = 79 (Ans.) weMZ mv‡j BUTex-G Avmv cÖkœvejx 1. GKwU UavbwR÷‡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.) 2. (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| (Ans.) (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` transducer Gi D`vniY| (Ans.) 3. Rsk‡bi wefevšÍi 2.0-volt †_‡K evwo‡q 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.)
†mwgKÛv±i I B‡jKUawb·  Engineering Practice Sheet ................................................................................................... 3 MCQ weMZ mv‡j BUET-G Avmv cÖkœvejx 1. Kv‡i›U weea©K ̧Yv1⁄4  = ? [BUET Preli 22-23] IC IE IC IB IE IB †Kv‡bvwUB bq DËi: IC IE 2. Kgb †em UavbwR÷‡ii Collector current 47 A n‡Z 96 A Ges emitter current 100 A n‡Z 150 A n‡jv| Current 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 3. 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 4. †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 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. 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 7. p-n Rvsk‡bi 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 weMZ mv‡j CKRUET-G Avmv cÖkœvejx 1. 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 = (5 × 5) – (0.5 × 3) 0.5 × 3 = 15.67   =  –  = 6.67  cÖevn e„w× 6.67 2. wb‡Pi wP‡Î Si Ges Ge Wv‡qvW `ywUi bx-†fv‡ëR h_vμ‡g 0.7 V 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 = 2 mA Ge Wv‡qvWwU D‡ëv K‡i mshy3 Ki‡j I = 0  V = 0  R = 0 V
4 ......................................................................................................................................  Physics 2nd Paper Chapter-10 3. 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 4. GKwU UavbwR÷‡ii wb¤œwjwLZ gvb ̧‡jv cwigvc Kiv n‡jv| IC = 1.98 mA; IB = 20A; UavbwR÷‡ii ,  Ges IE Gi gvb †ei Ki| [CKRUET 21-22]  = 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 weMZ mv‡j KUET-G Avmv cÖkœvejx 1. 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 2. 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| 3. †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 4. GKwU UavbwR÷‡ii †ÿ‡Î  = 0.95 Ges IE = 0.9 mA n‡j  KZ n‡e? [KUET 13-14] 19 16 18 12 DËi: 19 e ̈vL ̈v:  =  1 –  = 0.95 1 – 0.95 = 19 5. 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 weMZ mv‡j RUET-G Avmv cÖkœvejx 1. GKwU mvaviY f~wg UavbwR÷‡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 2. GKwU UavbwR÷‡ii weea©b ̧Yv1⁄4 0.98 Ges GwgUi 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 e ̈vL ̈v:  = IC IE  IC = 0.98 × 1.5 = 1.47 mA