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Common emitter BJT Amplifier - maximum input voltage

  • Thread starter etf
  • Start date

etf

179
2
Here is my circuit:
β=100, VCEsaturation=0.2V, VBE=0.7V, λ->0, VT=26mV

kolo.jpg

After completing DC analysis I got these results :
IB=18.05μA, IC=1.805mA, IE=1.823mA, VCE=4.658V>VCEsaturation=0.2V, gm=0.06943, rCE=∞ Ω, rpi=1.44kΩ.
After completing AC analysis I got Av=-115.7, Ai=72.23, Rin=1.040kΩ, Rout=1.67kΩ.
How to calculate maximum input voltage for proper amplification (without distortion I mean) ?
 
Last edited:

berkeman

Mentor
55,128
5,356
Here is my circuit:
β=100, VCEsaturation=0.2V, VBE=0.7V, λ->0, VT=26mV

View attachment 68124

After completing DC analysis I got these results :
IB=18.05μA, IC=1.805mA, IE=1.823mA, VCE=4.658V>VCEsaturation=0.2V, gm=0.06943, rCE=∞ Ω, rpi=1.44kΩ.
After completing AC analysis I got Av=-115.7, Ai=72.23, Rin=1.040kΩ, Rout=1.67kΩ.
How to calculate maximum input voltage for proper amplification (without distortion I mean) ?
What are your thoughts on this? You have calculated the gain, and you have an idea of where saturation is -- how can you use this? Is the output centered naturally between saturation and cutoff? If not, that will decrease the range of signals that can be passed without distortion.
 

etf

179
2
I know that Q point of BJT should lie in middle of load line for maximum amplification and if input voltage v(t) is too high then BJT will operate in saturation and cutoff and therefore there will be distortion. I'm still not sure how to use this fact for my calculation :(
 
Last edited:

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