Electronic – When multiplexing audio do I need a pre-amplifier for each channel

No. In fact, the mixer usually functions as the central grounding tie point for the entire system.

For the inputs in particular, which are mostly connected to microphones that have no other ground reference, this is particularly important.

Only in special cases, such as taking an input from a separately-powered instrument amplifier, would you consider using a transformer or other form of isolation (i.e., a "direct box") to separate the grounds.

Make the 3 capacitors at least 100X bigger.

And remove that 8 ohm load.

Then resim.

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

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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)

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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

or

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.