Make the 3 capacitors at least 100X bigger.
And remove that 8 ohm load.
You have about 0.3 volts across the Remitter. Thus about 1mA.
That makes the 'reac', which is 1/gm, be 26 ohms.
Now divide that into the collector resistor.
- Gain = 680 / 26 = 6 * 4 = 24X (about 22 dB).
Once you are seeing about 24X as the out/in gain, then evaluate the distortion with
1millvolt PP in
10millivolt PP in
100 millivolt PP in
Additional content to the answer, as requested.
How to get 0.3 volts across the Remitter (470 ohms)?
the VDD is 12 colts, the base bias chain is 47K/(479K + 47K) = 1/11.
Without the transistor (no base current), we expect 12 * 1/11 ~~ 1.3 volts where the input comes thru that input capacitor.
We do have a transistor, that pulls base current. Thus Vbase will be lower than 1.3 volts.
The bias chain is about 2 uAmps per volt (1v/500K is exactly 2uA/volt).
We have 12 volts, so 24uA flows thru the bias (divider chain).
As current is also pulled into the base (to cause emitter charges to chase the base charges in an attempt to annihilate by combination, but most of the emitter charge MISS and move across the very thin base region because of the relatively high electric field that accelerates those charges, to be
COLLECTED), the Vbase drops and drops. To find a good approximation will require an iterative solution.
And I think my math was wrong. 0.3v/470 ohms is 600 uA (0.6mA)
How to compute the gain?
For small signals on the base (100uV or 1 millivolt qualify as small), we can use the calculus derivative of the diode equation, which gives us the very useful
transconductance, or amps out per volts in
(more accurately, this is delta_I_out per delta_V_in)
from vacuum tube days, this was conductance_mutual, or gm
The 'gm' of a bipolar is very temperature sensitive, but is VERY PREDICTABLY TEMPERATURE SENSITIVE, and is
gm = [Iemitter (or Icollector) / 0.026] at room temperature
Thus at 0.026 amps thru the bipolar transistor, the gm = 0.026 / 0.026
or 1.0 amps out per volt incoming (on the base).
I simply remember the very useful value at 1 milliamp
gm[1milliAMP] = 0.001/0.026 = 0.039 ampsvolt = 1 / 26 ohms.
Using the collector resistor to convert the delta_collector current back into a voltage, we find
Voltage_gain = gm * Rcollector = (Iemitter/0.026) * Rcollector
and you will notice this Voltage_gain, at 1mA, is [Rcollector/0.026] * 0.001
Rcollector / 26 ohms
Now at 0.6 milliamp (600uA), the division becomes
Rcollector / ( 26 ohms * 1/0.6) or about Rcollector/40.
Thank you for asking me to explain this. Its good practice to run the mind over this, several times a year. Hopefully I slowly become better at explaining.