Use DC equivalent circuit for each BJT. (Ignore capacitors in this moment).
Later use KVL\KCL in circuit to find all curents and voltages needed.
Ic= beta Ib
Ie= (Beta +1)Ib
Vbe=0.7V
Beta = 100
Example for Q1
Ir3=Ir2+ Ib1, Ur2=Ube=0.7V > Ir2=0.7/47k= 0.01489 mA [similar for Ur6=Ube=0.7V Ir6 = 0.7V/33k= 0.02121mA]
I= Ic+Ir3 = beta*Ib1 + Ib1+Ir2 = Ib1*101 + 0.01489mA
Vcc=5V= I (R4+R5 ) + Ir3 R3 + 0.7 = (Ib1*101 + 0.01489mA)( 1k+0.47k) + (Ib1+ 0.01489) 150k + 0.7
Ib1( 101*1,47k+ 150k) = 5- 0.7 – 0.01489*(1,47k +150k)
Ib1= 2.0456/298.47=0.00685mA=6.85uA
Ic=0.685mA
I= 0.7067mA
Uce(Q1)= Vcc- I(R4+R5)= 5-0.7067mA(1.47k)= 3.96V
.............
IcQ3=0 and IcQ4 =0
Use KVL for D1, D2, Ube3, R9, R10, Ube4.
Ud2 + Ud1= UbeQ3 + IeQ3*R9 + IeQ4*R10 + UbeQ4
we have
Ud2=Ud1=Ube=UbeQ3=UbeQ4=Ube= 0.7V
and
IeQ3=IeQ4
So
IeQ3(R9+R10)=0 => IeQ3=0, IeQ4=0
And in this way we can ignore IbQ3=beta*IcQ3 =0 and also IbQ4 = beta* IcQ4= 0
…………….
Apply KVL to find IcQ2 in a similar way as was solved for Q1.
Read here: **broken link removed**
for AC analysis
1) Kill all DC sources
2) Coupling and Bypass capacitors are short. The effect of there capacitors is to set a lower cut-off frequency.
3) Replace BJTs with hybrid small signal model (You can use either model to analyze a small signal BJT circuit, but the hybrid-pi model is most commonly used)
If we ignore early effect, you may use the simplified form, ro= resistance is infinite
And need some time to solve for voltage and current transfer function, i/o and o/p impedances... :roll: