Gain in Common Emitter Amplifier

Question

I need to design an amplifier having a Vin=150mVp, 1khz; Vcc=3V that can run an 8ohms, 0.25 watts speaker, with a gain for about 7 without noticeable distortion.

Below is my designed voltage divider biased common-emitter amplifier with Vin=150mVp, 1khz; Vcc=3V.

But here is my concern: I need to have a gain of about 7, or having Vout for about 750mV

The first schematic I design having no load resistor of 8 ohms, when I simulated I measured a Vout of about 762mV, as you can see below:

But when I inserted a Load Resistance of 8ohms (assuming this is the speaker), I only got Vout of about 46mV which is very low, as you can see below:

****My question is what can I do to increase the Vout for about 750mv having the Load Resistance of 8ohms, at the same time without having noticeable distortion?**

Any help will be appreciated. Thank You.

Answer 1

The gain of a true common-emitter (emitter at AC common or "ground") is approximately:

$$A_v = -g_mR_C||R_L = -g_m 7.5 \Omega$$

for your resistor values. For a (magnitude) gain of 7, the required transconductance is:

$$g_m = \frac{7}{7.5} = \frac{I_C}{V_T} \Rightarrow I_C \approx 23mA$$

But, according to your schematic:

$$I_C \approx 8mA$$

Back to the drawing board!

Also, do realize that the above calculation is for small-signal gain. Since you have the emitter at AC ground, this amplifier is quite non-linear for large signals and the distortion is quite evidently large in both transient simulations you provide.

Answer 2

The speaker must have current both "sourced" and "sunk" from it, alternating with the positive and negative cycles of the audio signals.

Imagine 750mV appearing at an instant across the 8Ω speaker. That would require 0.09375A of current per Ohm's Law (V=IR).

If that 0.09375A were coming from the 3V Vcc, the collector resistor, Rc, would have to be 24Ω or less (in series with 8Ω speaker, ignoring current through transistor for this instant) at that instant. I cannot read your Rc value that well, but it appears to be over 100Ω, which is severely limiting your circuit to where it cannot reach your design goals.

If you choose to change the collector resistor to 24Ω or less, and scale the other resistors down as well to keep the circuit functioning, you will end up with an extremely low input impedance looking into the transistor base, which is probably not an OK solution, although input impedance was not specified in your design goals. (10k is considered a low input impedance for most pro and consumer audio applications.)

If your load impedance is 8Ω, then the voltage source impedance driving it must be near 0Ω to avoid significant voltage division, which is what you are getting with your current solution.

My answer to you is: Redesign this circuit for an output impedance at least 10x smaller than 8Ω. Your goal is an output impedance of 0.8Ω or less. The simplest solution is to replace the single transistor with an operational amplifier IC that can operate at 3V and supply enough current to the 8Ω load without failing. The TI LME49726 has a minimum voltage supply of 2.7V and an output current rating of ">300mA" (quote from data sheet).

EDIT: Per the comment (I tried the resistor...), if you want to see this work as a common emitter stage, use a 0.8Ω collector resistor, a 0.11Ω emitter resistor, a 6.5Ω base to +3V resistor, and a 3.5Ω base to ground resistor. Use 1000µF input and output coupling capacitors. I simulated this circuit and achieved voltage amplification from 150mV p-p to 885mV p-p with no distortion. I'm sure fine tuning the calculations could achieve a gain of 7, but the point is that this circuit is completely unpractical.