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Stiff Voltage divider bias with centered Q point:-
Given:-
Vcc=10V , Vce=Vcc/2 , Ic=10mA , βdc=100-300(100 to 300)
Solution:-
BJT ANALYSIS:-
Ic=10mA
Since Vcc<20V, We can't use ideal or approximations. We need to be more accurate.
(So, Vbe≈0.8V)
Ie=(β+1) /β *Ic (take the lowest beta value)
Ie=101/100 *10mA=10.1mA
Since Vce=Vcc/2=0.5Vcc, let's takeVc=0.4Vcc and Ve=0.1Vcc
Vc=0.4Vcc=0.4*10V=4V
Ve=0.1Vcc=0.1*10V=1V
Re=Ve/Ie=1/10.1mA≈99.1Ω(≈100Ω)
Rc=Vc/Ic=4V/10mA=400Ω
Voltage Divider analysis:-
The main condition:
R2≤0.01βdc*Re
R2≤0.01*100*99.1Ω
R2=99.1Ω
R1=V1/V2 * R2
By KVL,
V2=Vbb=Ve+0.8V=1V+0.8V=1.8V
V1=Vcc-V2=10V-1.8V=8.2V
So,
R1=8.2V/1.8V * 99.1Ω=451.45Ω≈452Ω
Disclaimer:-
You can use Vbe=0.7V, when Vbb or The voltage given to the base is greater than 20V
Since Vcc is given to Base through a resistor, Vbb≤Vcc
So, Vbb can't exceed Vcc, and Vcc is not even close to 20V, you need to use 0.7V.
The calculations are done as accurately as possible to avoid errors.
The Voltage divider bias (VDB) will be used in BJT based amplifiers
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