Consider a p-n-p bipolar junction transistor (BJT). The emitter is uniformly dop
ID: 2082735 • Letter: C
Question
Consider a p-n-p bipolar junction transistor (BJT). The emitter is uniformly doped with N_AE = 2 times 10^19/cm^3, the base is uniformly doped with N_D = 2 times 10^17/cm^3 and the collector is also uniformly doped with N_AC = 5 times 10^15/cm^3. Assume the minority carrier mobility in the base region is 400 cm^2/Vsec, the minority carrier lifetime is 0.5 mu sec and the BJT is operated at 300K a. If the base width is 1.4 mu m, then calculate the transport factor beta^+. b. Given the above parameters estimate the emitter injection efficiency in the forward and inverse (reverse) directions. c. What are the common base, alpha, and common emitter, beta, current gains for the transistor. d. What is the V_CE (off-set) at I_C = 0 for this transistor. e. If the input base current is 10 mu A, then what is the output collector current? f. If the Early voltage is 25V, then what is the output conductance for Part (e)?Explanation / Answer
We kno emitter efficiency factor gama(y)=[1/(1+Nd/Nae)] [1/(1+2*10^17/2*10^19)]=0.99
Base transport factor (aT)=[1/cosh(Xb/Lb)].....eun (|)...=1/cosh(9.9*10^-3)=0.9967
As we know Xb=1.4m=1.4*10^-4cm;
Lb=(Db*Taub)...[minority carrier mobility in base region Db=400cm2/sec,minority carrier life time Taub=5*10^-7sec];
Putting the above two values we get
Lb=(400*5*10^-7)=0.01414cm;
Now the ration Xb/Lb=9.9*10^-3;
All of these value put in equn ( |)
again delta(d)=1/(1+jro/jso*(exp(-evbe/2kT)))..||
By putting all of the above value we get delta(d)=0.999667;
As we know alpha(a)=aT*d*y=0.9967*0.999667*0.99=0.9864
bita()=a/1-a=72.52
Given input base current Ib=10*10^-6;
Then output collector current Ic=*Ib=72.52*10*10^-6=725.2A
As we know collector current(Ic)=(Vcc-Vce)/Rc;
If Ic=0,then Vcommon collector=V collector emitter;
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