更新时间2019-10-03 00:27:17
1、Ri = RB1//RB2//rbe = ( 3.9//20 )//1 = 780/1019 k ≈ 765 Ω;
Ro = Rc//RL = 2/3 k ≈ 667 Ω;
Ku = - βRo/rbe = -40 * (2/3) / 1 = -80/3 ≈ -26.7 倍。
2、RB1 开路,三极管无偏置,截止;
UB = Ue = 0v;Uc = Ucc = +12v;
RB2 开路,三极管饱和导通,饱和压降Uce 可视作 0.3v;
因为是饱和导通,Ic ≠ βIb,所以求电位要解方程。
Ue = ( Ib + Ic ) * RE = ( Ic + Ib ) * 1 = Ic + Ib;
Ic = ( Ucc - Uce - Ue )/Rc = ( 12 - 0.3 - Ue )/2
= 11.7/2 - Ue/2 = 11.7/2 - Ic/2 - Ib/2;
3Ic = 11.7 - Ib,3Ic + Ib = 11.7;①
Ib = ( Ucc - UBE - Ue )/RB1 = ( 12 - 0.7 - Ue )/20
= 11.3/20 - Ue/20 = 11.3/20 - Ic/20 - Ib/20;
21Ib = 11.3 - Ic,3Ic + 63Ib = 33.9;②
② - ①,62Ib = 22.2,Ib = 0.358 mA;
代入① ,3Ic = 11.7 - 0.358,Ic = 3.78 mA;
因为 Ib > Ic/β ,所以三极管饱和导通 。
Ue = Ic + Ib = 3.78 + 0.358 = 4.14v;
Ub = Ue + UBE = 4.14 + 0.7 = 4.84v;
Uc = Ue + Uce = 4.14 + 0.3 = 4.44v 。
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