R 16V 习题3.5.3 RbI 3kQ 60kQ 射极偏置电路和BJT的 输出特性曲线如图题3.53.1b 所示,已知B=60。 (1)分别用估算公式和。20 6kQ Re 图解法求Q点; 2kQ (2)求输入电阻r (3)用小信号模型分析电压增益A4y; (4)求最大不失真幅度; (5)若电路其他参数不变,如果要使cE=4V, 问上偏流电阻为多大?
习题3.5.3 + – vi iB i 1 Rb1 b c iC Rc VCC + – vo RL Re Rb2 Cb1 + 60k 20k 2k 6k 3k + + 16V 射极偏置电路和BJT的 输出特性曲线如图题3.5.3 所示,已知=60。 (1)分别用估算公式和 图解法求Q点; (2)求输入电阻rbe; (3)用小信号模型分析电压增益 ; AV (4)求最大不失真幅度; (5)若电路其他参数不变,如果要使VCE = 4V, 问上偏流电阻为多大?
直流(静态)分析 R 16V (1)估算法求Q点: rbl 3kQ 60kQ2 分压式射VB= R bI 极偏置: 20k 16=4V 60k+20k 6kQ b2 20kQ2 Je KVL:Ic≈lg=-B R R 2kQ 4-0.7 1.65mA lc165 T放大:IB =27.54 B60 Jc KVL: VCE =VcC-Ic(R+Re =16-1.65×(2+3)=775
+ – vi iB i 1 Rb1 b c iC Rc VCC + – vo RL Re Rb2 Cb1 + 60k 20k 2k 6k 3k + + 16V (1)估算法求Q点: 直流(静态)分析 + – vi iB i 1 Rb1 b c iC Rc VCC + – vo RL Re Rb2 Cb1 + 60k 20k 2k 6k 3k + + 16V V V k k k V R R R V CC b b b B 16 4 60 20 20 1 2 2 = + = + = mA R V V I I e B BE C E 1.65 2 4 0.7 = − = − = A I I C B 27.5 60 1.65 = = = V VCE VCC I C Rc Re 16 1.65 (2 3) 7.75 ( ) = − + = = − + 分压式射 极偏置: Je KVL: T放大: Jc KVL:
直流(静态)分析 (1)估算法求Q点 (5)VcE=4V→Rb1=? 射极偏置:WB1+=Rn=C、vnBB2,(c-VB) R B 20k b2 60k+20k 16-55 20x 38.2AQ Je KVL:Ic≈lg B BE 5.5 R 4-0.7 Vn≈lcke+BE 1.65mA =2.4×2+0.7=5.5 lc165 T放大:IB =27.54 B60 Jc KVL: VCE=VCC -Ic(R+Re) CC CE 16-4 =16-1.65×(2+3) CRtr 2+3s2.4m4
V V k k k V R R R V CC b b b B 16 4 60 20 20 1 2 2 = + = + = V VCE VCC I C Rc Re 16 1.65 (2 3) 7.75 ( ) = − + = = − + 直流(静态)分析 mA R V V I I e B BE C E 1.65 2 4 0.7 = − = − = A I I C B 27.5 60 1.65 = = = 射极偏置: Je KVL: T放大: Jc KVL: (1)估算法求Q点: (5)VCE = 4V →Rb1=? mA R R V V I c e CC CE C 2.4 2 3 16 4 = + − = + − = V VB I C Re VBE = 2.42 + 0.7 = 5.5 + = − = = − − = k k V V V R R V V V R CC B B b b B CC B b 38.2 5.5 16 5.5 20 ( ) 2 2 1
idmA (1)图解法求Q: 100 直流负载线( JC KVL方程) CE CC Ic(R+R C ere tRe 40 2点:lC=0 lC=1.65 ==== CEV CC 05101520ucE/V CC =3.2m4 (Rc+Re) CE 做一水平线(I=165mA)与负载线相交为Q点
(1)图解法求Q: 0 5 10 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 iB=20A ( ) CE CC C Rc Re V =V − I + 直流负载线(Jc KVL 方程): 2点: IC=0 VCE= VCC mA R R V I c e CC CS 3.2 ( ) = + = VCE= 0 0 5 10 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 iB=20A Rc +Re VCC VCC 0 5 10 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 iB=20A Rc +Re VCC VCC 0 5 10 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 Q iB=20A Rc +Re VCC VCC IC=1.65 做一水平线(IC=1.65mA)与负载线相交为Q点
ic/mA 斜率 RQ∥R1 (4)求最大不失真幅度 画交流负载线,关键M点坐标 解三角形即可 Re +Re 0 斜率:k=ta lC=1.65 R,∥R L 三角形;如、I M’1520 又有:B=qg(x-a)=-ga Rc∥R 所以:x CO a0-cx(R∥/R1)=1.65×2=33
(4)求最大不失真幅度 0 5 10 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 Q iB=20A Rc +Re VCC VCC IC=1.65 画交流负载线,关键M’点坐标 解三角形即可。 0 5 10 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 Q iB=20A Rc +Re VCC VCC 斜率 1 Rc // RL IC=1.65 M’ 斜率: Rc RL k tg // 1 = = − 三角形: 又有: 所以 : Rc RL tg tg tg // 1 = ( −) = − = x I tg CQ = I R R V tg I x CQ c L CQ = = ( // ) = 1.65 2 = 3.3 0 5 x 15 20 2 4 6 8 100 iC/mA CE/V 80 60 40 Q iB=20A Rc +Re VCC VCC 斜率 1 Rc // RL IC=1.65 M’
交流分析 R 16V Rbl 3ks2l 60kQ2 R RL 6ks Re Rbi 20kg2 R 2ko (2)求输入电阻rm 26mk 26 Tbe = bb+(1+B 200+61 12 g(m4) 165 (3)分析电压增益 3×6 B(R∥R) 60x3+6 100 12
交流分析 = + + = + = k I mA mV r r E b e b b 1.2 1.65 26 200 61 ( ) 26 (1 ) ' + – e Vi Rb1 rbe b I b Ic Rc + – RL Vo c I b Rb2 Rb Re + – vi iB i 1 Rb1 b c iC Rc VCC + – vo RL Re Rb2 Cb1 + 60k 20k 2k 6k 3k + + 16V (2)求输入电阻rbe; 100 1.2 3 6 3 6 60 ( // ) = − + = − − = = b e c L i o V r R R V V A (3)分析电压增益;
习题3.54静态同 已知B=100试求:353 共射 (1)Q点 (2)电压增益Av1=V/和 V2 CC 共集 Rc 10V (3)输入电阻R; R 2kQ (4)输出电阻R1和R2 20k Vo Ro Ro2较难 Rs R v 15kQ R e 2kQ R 源电压增益
习题3.5.4 已知 =100。试求: (1)Q点; (2)电压增益AV1 =Vo1 /Vs和 AV2 =Vo2 /Vs; (3)输入电阻Ri; (4)输出电阻Ro1和Ro2。 + – vs Rb1 b c Rc VCC vo1 Rs Re Rb2 Cb1 + 20k 15k 2k 2k 2k + + 10V vo2 e Ro2 Ro1 静态同 3.5.3 共射 共集 源电压增益 Ro2较难
直流(静态)分析 R 10V (1)估算法求Q点: Rb 2kQ 20kQ2 分压式射vB R l Ch 极偏置: Rb1+ rb2 Rs k iRo 15k 10=4.3J2kg 15k+20k rb Vo2 15kQ VsD Re Je KVL:Ic≈IE= B一BE 2k2 R02 R 4.3-0.7 1.8mA R 16V Rbi 3kQ T放大:1=4c=18=18m4 boko iB Jc KVL: VCE=VCC -IC(R+Re) 10-18×(2+2)=28 Rbi 20k9 R 2kQ
+ – vi iB i 1 Rb1 b c iC Rc VCC + – vo RL Re Rb2 Cb1 + 60k 20k 2k 6k 3k + + 16V (1)估算法求Q点: 直流(静态)分析 V V k k k V R R R V CC b b b B 10 4.3 15 20 15 1 2 2 = + = + = mA R V V I I e B BE C E 1.8 2 4.3 0.7 = − = − = A I I C B 18 100 1.8 = = = V VCE VCC I C Rc Re 10 1.8 (2 2) 2.8 ( ) = − + = = − + Je KVL: T放大: Jc KVL: + – vs Rb1 b c Rc VCC vo1 Rs Re Rb2 Cb1 + 20k 15k 2k 2k 2k + + 10V vo2 e Ro2 Ro1 分压式射 极偏置:
交流分析 R 10V 2kQ B 20kQ R R Rol R Re R (1+)b 15kQ v R 2kQ Rb (3)求输入电阻 26ml 26 The=rbb+(1+B 200+101 166Q TE(mA) 18 RFRb//rb/( e+(1+BRe8.2kQ2
(3)求输入电阻 = + + = + = k I mA mV r r E b e b b 1.66 1.8 26 200 101 ( ) 26 (1 ) ' 交流分析 + – vs Rb1 b c Rc VCC vo1 Rs Re Rb2 Cb1 + 20k 15k 2k 2k 2k + + 10V vo2 e Ro2 Ro1 + – e Vi Rb1 rbe b I b Ic Rc + – Rs c I b Rb2 (1+)I b Rb Re Vo2 Vo1 + – Vs – + Ri=Rb1//Rb2//(rbe+(1+)Re=8.2k
(2)电压增益 BR R 1 ;Vbe+(1+β)ReR+R 100×2 8.2 0.8 1.66+101×22+8.2 02 (1+B)R Ri 2 1Ve+(1+β)R2R+Ra 101×2 8.2 166+101×22+820.8 R,=Pmbe+Rh1∥/Rh2∥/R ≈31g (1+B) Blb RoIR 2kQ2 (4输出电阻Rn和Rn Re v (1+β) Rb
(2)电压增益 0.8 2 8.2 8.2 1.66 101 2 100 2 (1 ) 1 1 1 = − + + − = + + + − = = = s i i b e e c s i i o s o V R R R r R R V V V V V V A 0.8 2 8.2 8.2 1.66 101 2 101 2 (1 ) (1 ) 2 2 2 = + + = + + + + = = = s i i b e e e s i i o s o V R R R r R R V V V V V V A + – e Vi Rb1 rbe b I b Ic Rc + – Rs c I b Rb2 (1+)I b Rb Re Vo2 Vo1 + – Vs – + Ro1=Rc=2k + + = 31 (1 ) // // // 1 2 2 be b b s o e r R R R R R (4)输出电阻Ro1和Ro2