# Transistor as an amplifier

#### GastonJam

Joined Jun 25, 2020
11
I'm currently working on this circuit and I run upon this question which I'm not sure I understood it correctly :

For starters I have these values below - UCE=5V

• R1=6.1kΩ
• R2=1.35kΩ
• RE=1kΩ
• R3=1.5kΩ
• V1=15V
• IC=2mA
• UBE=0.4V
• B=180
Given formula rBE≈40mV / IB First I calculated CE with the use of this formula τE≈(rBE⋅CE)/β=1/(2πf)

These are my calculations:
IC=B.IB

rBE=40mV/IB=40mV.B/IC=40.180/2mA≈4kΩ
Let's say the frequency of this amplifier is 1kHz

=>τE=1/(2πf)=1/(2π1000)=1.519.10^(−4)s

Which makes

=>CE=B.τE/rBE=1.519.10−4.180/3600=7.6uF
And also
τE=(rE/B).C1
rE=(rBE+BRe)||R1||R2=(4kΩ+180∗1kΩ)||6.1kΩ||1.35kΩ=1100Ω=1.1kΩ

C1=τE.B/rE=>C1=(1.519.10−4s.180)/(1.1∗103)=0.26uF
Then with this question I could calculate C_2.
What is the minimum size of capacitor C2 if the lower limit frequency should be approximately 1000 Hz when the output is loaded with 100 kΩ?

τE=R∗C2=>C2=τE/R=1.519∗10−4/100kΩ=1.519nF

(I don't think that these calculations are related to my my question but I still wanted to write them down to check whether it's correct or not)

And now this question : What values do you expect for the AC voltage gain with and without CE? (Note: rCE >> RC; rBE ≈ 40 mV / IB)
Apparently there's no need for calculations here : Expected values will be probably in mV but I don't understand the relation of rBE ≈ 40 mV / I
and rCE >> RC to this question

Last edited:

#### Papabravo

Joined Feb 24, 2006
20,384
At low frequencies the impedance to GND in the emitter circuit is 1KΩ || CE. CE represents an open circuit at DC so the impedance will be effectively 1K. As the frequency increases the impedance of CE gets smaller and smaller. At some frequency the impedance will be equal to 1KΩ, and at higher frequencies it will get smaller yet. This means the AC gain of the amplifier will change as well.