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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)