Lecture 29: BJT Design (II)
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1 NCN C606: Solid State Devices Lecture 9: JT Design () Muhammad Ashraful Alam Alam C 606 S09 1
2 Outline 1) Problems of classical transistor ) Poly Si emitter 3) Short base transport 4) High frequency response 5) Conclusions RF: SDF, Chapter 11 and 1 Alam C 606 S09
3 Topic Map Diode quilibrium DC Small Large Circuits signal Signal Schottky JT/HT MOS Alam C 606 S09 3
4 Doping for Gain β dc D W n W D n N n p i, h ih i, N mitter doping: As high as possible without band gap narrowing ase doping: As low as possible, without current crowding, arly effect N N N C Collector doping: Lower than base doping without Kirk ffect ase Width: As thin as possible without punch through Alam C 606 S09 4
5 How to make better Transistor β n n i, N p i, D W W D n N Graded ase transport Polysilicon mitter Classical Shockley Transistor Heterojunction ipolar Transistor Alam C 606 S09 5
6 Poly silicon mitter P ase mitter Collector N+ N+ N P+ N+ N SiGe intrinsicbase Dielectric i trench emitter N + P N Poly silicon Alam C 606 S09 6
7 Poly silicon mitter n n n e = qdn i, qv n, W N poly β ( ) N + P v s p p 1 W 1 p,, pol y = q Dp = q υs p W = qυ p = qp p,, poy l s 1 ( Dp ) = q W p p,, si 1 p p p,, poly υs = psi,, Dp W + υs p D W = p p D W υ 1 p s υs D D W + Question: Why does poly onlysuppress the hole current, not electron current? p p W + υ Alam C 606 S09 7 s
8 Gain in Poly silicon Transistor υ D W = qp = s p p,, poly 1 p,, poly Dp W + υs ( Dp ) = q W p p,, si 1 = p,, poly s, poly D υ W + υ ps,, i p s, si β poly C C = = si, Dn W n i, N D p W + υs W D ni, N p υs, polyy, si, polyp y D n n i, N 1 ( υ ) s << Dp W W n N υ i, s Poly suppresses base current, increases gain Alam C 606 S09 8
9 Outline 1) Problems of classical transistor ) Poly Si emitter 3) Short base transport 4) High frequency response 5) Conclusions Alam C 606 S09 9
10 How to make better Transistor D W n N D n N β W D n N W n N υ n i, n i, 1 p i, i, s Polysilicon mitter Gradedase transport Heterojunction bipolar transistor Classical Shockley transistor Alam C 606 S09 10
11 Short base Quasi ballistic Transistor n n = qd = qυ n W 1 n, n th W n D n W n,, ballistic υ th = = n1 Dn W + υ th ns,, i Dn W + υth Δn n 1 n υ th N + P Alam C 606 S09 11
12 Gain in short base Poly silicon Transistor υ = D W + υ p,, poly s, poly ps,, i p s, si υth = D W + υ n,, ballistic ns,, i n th β poly, ballistic C, ballistic C, ballistic C, si, si = =, poly C, si, si, poly υ th Dn W n i, N D p W + υs Dn W + υth W Dp ni, N υs n n N υ i, N υth i, s Quasi-allistic transport in very short base limits the gain 1
13 Outline 1) Problems of classical transistor ) Poly Si emitter 3) Short base transport 4) High frequency response 5) Conclusions Alam C 606 S09 13
14 Topic Map Diode quilibrium DC Small Large Circuits signal Signal Schottky JT/HT MOS Alam C 606 S09 14
15 Small Signal Response log 10 β C V (in) P+ N P C VC (out) ( ) β DC ω β ω ω β log 10 f 1 f β f T 1 W W C kt = + + C + C jc, j, π f T Dn sat q υ C Alam C 606 S09 15
16 Small Signal Response (Common mitter) C μ C Cμ V (in) P+ N P C V C (out) ( α ) 1 F F C π R α α F F R R r π C π g m V e F qv / kt ( ) ( 1 α ) 1 d d F F q = = = r dv dv k T = F0 1 d( α F F ) gm = = dv π δ ( α ) q k C T = g δv = g υ F F m m = β 1 qc DC kt Alam C 606 S09 16
17 Short Circuit Current Gain C μ C β ( f ) i C = = i g υ + jωc υ m μ C 1 υ + jωcπυ + jωcμυ rπ C r π C π g υ m β ( f ) T gm jωt Cμ gm 1 = 1 jωt C + C + jωt Cπ + jωc μ rπ ( π μ) 1 1 Cπ + Cμ kt kt ω π f = g = q + + q + T T ( C ) j, C Cj, ( Cd, C Cd, ) m C C kt C dq d, C Cd, C = = C C C q d dv d Alam C 606 S09 17
18 ase Transit Time Δn n 1 Ref. Charge control model N + P 1 dq Q q nw 1 = W = = d C n C 1 q D n W Alam C 606 S09 18
19 Collector Transit Time N + P W C τ = υ sat? i τ t τ eff, C q τ WC = = = i υ sat 1 i τ = q Alam C 606 S09 19
20 Putting the Terms Together log10 f T Kirk Current Collector transit time (slide 19) ase transit ittime (slide 18) 1 W W C = + + π ft Dn υsat K log 10 C kt C q C + C j, C j, Junction charging time (slide 17) Do you see the motivation to reduce W and W C as much as possible? What problem would you face if you push this too far? Alam C 606 S09 0
21 High Frequency Metrics (current gain cutoff frequency, f T ) 1 W WC kt q W W k T q ( C ) j+ C jc + ( Rex R c ) C cb τ =,, π f = D + υ + + T n sat C (power gain cutoff fffrequency, f max ) f max = f T 8π R bb C cbi Alam C 606 S09 1
22 Summary We have discussed various modifications of the classical JTs and explained why improvement of performance has become so difficult in recent years. The small signal analysis illustrates the importance of reduced junction capacitance, resistances, and transit times. Classical homojunctions JTs can only go so far, further improvementis is possiblewith heterojunction bipolar transistors. Alam C 606 S09
23 Aside: On ase Collector reakdown Vlt Voltages Alam C 606 S09 3
24 ssence of Current Gain C V (in) P+ N P C VC (out) V V C nput Response nput qd n W p i, q N qd ( V 1) e β n i, ( qv β 1 ) e W Response n N Alam C 606 S09 4
25 Collector reakdown (Common ase, Fixed ) P V + N P V (in) V C (out) Common ase ( fixed, variable) Alam C 606 S09 5
26 V Collector reakdown (Common mitter) V (in) P+ N P C C V C (out) Common mitter ( variable, ibl fixed) Common emitter breakdown voltage is smaller than common base breakdown voltage. Alam C 606 S09 6
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