4-9 10K R 10K 2.1K 解:(1)L EE e3 C3 R C3 RB E3 R+0.7 +3K 3K Ⅰ=2L,=2×=k R R Cl -15V (2)R2=3+(1+B)Rg EE 1/(Rc‖R BR c 3 2 RB+rel 1be3+(1+ B)REb e E C B b e C i u i b e E CR C R E E C E E E C r R R R r R R A R r R RU I I U I R RU I IC C ( 1 ) ( || ) 21 ( 2 ) ( 1 ) 2 2 0.7 1 3 3 1 2 2 3 1 1 3 3 3 3 1 1     + + −  + = − = + + = =  = + 解：（ ）  = 4 - 9 R C