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The small signal amplifier used here is the low power audio amplifier. Everything is calculated finely. You can check more in my Everycircuit page. Please visit everything. Let's dive into the calculations.
Given:-
Vcc=12V Ic=1mA β=100 Av=60
R(load), Rl=10kΩ Frequencyf=100 - 10kHz
Formulae used:-
Av=-(Rc||Rl) /r'e, (i used α=-(Rc||Rl), so by doing common algebra, Rc=(αRl) /(Rl- α) )
Vce=Vcc/2
C=1/(2πfXc)
Calculations:-
First we need to find Rc:-
Rc=(αRl) /(Rl- α)
In this we need to find α,
So α=Av*r'e=60*25mV/1mA=1500Ω
Now we can find Rc
Rc=(αRl) /(Rl- α)=(1500*10kΩ)/(10kΩ-1500) Rc=1.76kΩ
Vc=Vcc-IcRc=12V-(1mA*1.76kΩ) =10.24V
Ve=Vc-Vce=Vc-Vcc/2=10.24V-6V=4.24V
Ie=Ic((β+1) /β)=1mA*(100+1) /100=1.01mA
Re=Ve/Ie=4.24V/1.01mA=4.19kΩ≈4.2kΩ
Stiff voltage condition for VDB:-
R2≤0.01*β*Re→R2≤0.01*100*4.2kΩ
R2≤4.2kΩ→R2=4kΩ
By voltage divider ratio:
R1=V1/V2 * R2, we need to know what are V1 and V2 are,
V2=Ve+Vbe=4.24V+0.8V=5.04V
V1=Vcc-V2=12V-5.04V=6.96V
Then,
R1=V1/V2 *R2=6.96V/5.04V * 4kΩ=5.52kΩ
Coupling and bypass capacitor calculations:-
C=1/(2πFXc) is the common formula we need to find the Xc for each each capacitor
Xc1-The reactance of the capacitor connected to the Ac source
Xc2-The reactance of the capacitor connected to the load
Xce-The reactance of the capacitor connected to the emitter
Xc1≤0.01*(βRe||R1||R2)→Xc1≤23Ω
Xc1=20Ω
Xc2≤0.01*Rl→Xc2≤100Ω
Xc2=90Ω
Xce≤0.01*Rl→Xce≤42Ω
Xce=32Ω
Take the least frequency for calculations:-
C1=1/(2*π*f*Xc1) =79μF≈80μF
C2=1/(2*π*f*Xc2)=1.76μF≈20μF
Ce=1/(2*π*f*Xce) =49.7μF≈500μF
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That's all the calculations.
Check the rms of input and output voltage
Av=393mV/7.07mV=55.58, but we assumed Av=60, This is due to some minor value errors, the capacitance needed to be slightly higher and in most of the cases Av(we get) ≥Av(given or ideal) /2
If the calculations are mostly correct.
I think i covered everything, you can ask any questions, I will be happy if you see my other circuitries also. I will try to do every circuitry with its calculations.
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