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| -rw-r--r-- | 2459/CH3/EX3.4/Ex3_4.PNG | bin | 0 -> 10113 bytes | |||
| -rw-r--r-- | 2459/CH3/EX3.4/Ex3_4.sce | 33 |
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diff --git a/2459/CH3/EX3.4/Ex3_4.PNG b/2459/CH3/EX3.4/Ex3_4.PNG Binary files differnew file mode 100644 index 000000000..085bcbd1b --- /dev/null +++ b/2459/CH3/EX3.4/Ex3_4.PNG diff --git a/2459/CH3/EX3.4/Ex3_4.sce b/2459/CH3/EX3.4/Ex3_4.sce new file mode 100644 index 000000000..2368bf5b8 --- /dev/null +++ b/2459/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,33 @@ +//chapter3 +//example3.4 +//page50 + +Eb=250 // V +Ec=-3 // V + +// given that Ib=0.003*(Eb+30*Ec)^1.5 mA +// differentiating w.r.t Ec with Eb=constant, we get +gm=0.003*1.5*(Eb+30*Ec)^0.5*30*10^-3 +mutual_inductance_micro=gm*10^6 + +printf("mutual conductance = %f mho or %.3f micro mho \n",gm,mutual_inductance_micro) + +// differentiating given equation w.r.t Ec with Ib=constant, we get +// 0=0.003*10^-3*1.5*(Eb+Ec)^1.5*(mu+30) where mu is equal to ratio of changes in Eb and Ec i.e. amplification factor +// thus mu+30=0 hence we get +mu=-30 + printf("here negative sign of amplification factor indicates that Eb and Ec are in opposite direction. \n \n") +// here we need not worry as to if mu may be positive because the equation given in problem statement will always give mu+30=0 i.e. mu=-30 + +printf("amplification factor = %.3f \n",mu) + +rp=mu/gm +if rp<0 // rp can not be negative + rp=-rp +end + +printf("plate resistance = %.3f ohm \n",rp) + +//in book, the answers are less accurate. The accurate answers are +// gm=1707.630 micro mho +// plate resistance=17568.209 ohm |
