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+//chapter12

+//example12.23

+//page258

+

+gain_beta=40

+I_C1=2 // mA

+t1=25 // degrees

+t2=55 // degrees

+I_CBO1=5d-3 // mA

+

+// for I_CBO=5 micro ampere at 25 degrees

+I_CEO1=(1+gain_beta)*I_CBO1

+

+I_CBO2=I_CBO1*2^((t2-t1)/10) // since it doubles every 10 degrees. So for t2-t1, it becomes 2^((t2-t1)/10) times.

+I_CEO2=(1+gain_beta)*I_CBO2

+I_C2=I_CEO2+I_C1

+I_C_change=100*(I_C2-I_C1)/I_C1

+

+// for I_CBO=0.1 micro ampere at 25 degrees

+t1_dash=25 // degrees

+t2_dash=55 // degrees

+I_CBO1_dash=0.1d-3 // mA

+I_C1_dash=2 // mA

+

+I_CBO2_dash=I_CBO1_dash*2^((t2-t1)/10) // since it doubles every 10 degrees. So for t2-t1, it becomes 2^((t2-t1)/10) times.

+I_CEO2_dash=(1+gain_beta)*I_CBO2_dash

+I_C2_dash=I_CEO2_dash+I_C1_dash

+I_C_change_dash=100*(I_C2_dash-I_C1_dash)/I_C1_dash

+

+printf("collector cutoff current = %.3f mA \n \n",I_CEO1)

+printf("percent change in zero signal current given that \nI_CBO=5 micro ampere at 25 degree is = %.3f percent \n \n",I_C_change)

+printf("percent change in zero signal current given that \nI_CBO=0.01 micro ampere at 25 degree is = %.3f percent \n",I_C_change_dash)