Based on the amplifier given in Fig. 2, draw the small signal equivalent and  derive the two time constants associated with capacitances C1 and C2. Assume  early voltage is infinity. Finally, indicate which formula out of the two should be  used for the lower 3 dB frequency calculation? Formulate the maximum gain.  I need help solving this problem. I asked a similar question

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Based on the amplifier given in Fig. 2, draw the small signal equivalent and 
derive the two time constants associated with capacitances C1 and C2. Assume 
early voltage is infinity. Finally, indicate which formula out of the two should be 
used for the lower 3 dB frequency calculation? Formulate the maximum gain. 

I need help solving this problem. I asked a similar question before but after trying to get some clarification on my values and my instructor said that it was incorrect. The values that I have attached as screenshots. Is there something wrong with the math. The main issue that came up is the value for the time constant for C1.

 

V2 R6
10002
1.4mVpk
1kHz
0°
C1
HH
1µF
R1
10k
R2
1kQ
R3
1.5kQ
Q2
2N3904
R4
470
wi
C2
HH
10μF
vout
R5
*100ΚΩ
_V1
-9 V
Transcribed Image Text:V2 R6 10002 1.4mVpk 1kHz 0° C1 HH 1µF R1 10k R2 1kQ R3 1.5kQ Q2 2N3904 R4 470 wi C2 HH 10μF vout R5 *100ΚΩ _V1 -9 V
-Vth + RthlB + VBE + R4le = 0
-Vth + RthlB + VBE + R₁(1 + B)IB = 0
Vth - VBE
Rth + (1+B)R4
IB =
0.82V - 0.7V
.91k + (1 + 100)47
Ic = BIB Ic= 0.0212m x 100 → Ic= 2.121mA
IB =
VT
1π = →=
Ic
IB 0.0212mA
0.026V
2.121mA
T= 12.260
T₁ = C₁Req1
T₁ = C₁ x ([Rth ||{ßr + (1 +ß)R₁}] + 100)
T₁ = 1μ × ([0.91k||{100(12.26) + (1 + 100) × 47}] + 100)
T₁ = 1μ x {0.79k + 0.1k} → T₁ = 8.9 x 10-4sec
T₂ = C₂Req2
T2
10μ x {R3 + R5}
T₂ = 10μ x {1.5k + 100k
T₂
1.015sec
0.79k
0.79k + 0.1k
Vin = 1.126Vp → Vin = 1.126 x 5.973kib → Vin = 6.7kib
V₂ =
X Vin
Vout = -(1.5k||100k)ẞib → Vout= -1.48kẞi,
-1.48ßib
6.7kip
→ Ay = -22.1
Vout
Ay = → Ay =
Vin
Transcribed Image Text:-Vth + RthlB + VBE + R4le = 0 -Vth + RthlB + VBE + R₁(1 + B)IB = 0 Vth - VBE Rth + (1+B)R4 IB = 0.82V - 0.7V .91k + (1 + 100)47 Ic = BIB Ic= 0.0212m x 100 → Ic= 2.121mA IB = VT 1π = →= Ic IB 0.0212mA 0.026V 2.121mA T= 12.260 T₁ = C₁Req1 T₁ = C₁ x ([Rth ||{ßr + (1 +ß)R₁}] + 100) T₁ = 1μ × ([0.91k||{100(12.26) + (1 + 100) × 47}] + 100) T₁ = 1μ x {0.79k + 0.1k} → T₁ = 8.9 x 10-4sec T₂ = C₂Req2 T2 10μ x {R3 + R5} T₂ = 10μ x {1.5k + 100k T₂ 1.015sec 0.79k 0.79k + 0.1k Vin = 1.126Vp → Vin = 1.126 x 5.973kib → Vin = 6.7kib V₂ = X Vin Vout = -(1.5k||100k)ẞib → Vout= -1.48kẞi, -1.48ßib 6.7kip → Ay = -22.1 Vout Ay = → Ay = Vin
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