Electronics > Beginners

why is class A audio amp saturating?

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fourfathom:

--- Quote from: cvanc on August 14, 2019, 08:29:00 pm ---Shouldn't the speaker be capacitively coupled to the amp output?

--- End quote ---

In the real world yes, but then you would be using a push-pull output stage.  The OP's circuit is an attempt to avoid that.  A practical class-A would probably use an output transformer driving the speaker.

magic:
Transformer won't help, the primary would be a short at DC.

In a more proper single ended class A stage, the speaker would get a series capacitor and additional power resistor or active current sink would go from the emitter to ground to provide negative drive.

fourfathom:

--- Quote from: magic on August 15, 2019, 05:22:19 am ---Transformer won't help, the primary would be a short at DC.
--- End quote ---
Well you wouldn't use the same driver design.  There would be feedback to set the transformer current, and probably voltage feedback to control the AC gain.  This is how the old-fashioned class-A amps were designed.  Some still are, but they usually use vacuum tubes.


--- Quote from: magic on August 15, 2019, 05:22:19 am ---In a more proper single ended class A stage, the speaker would get a series capacitor and additional power resistor or active current sink would go from the emitter to ground to provide negative drive.
--- End quote ---
That's the resistor (class A) or push-pull (class AB) design I mentioned.  And yes, in a Class-A design a resistor could be replaced with an active current sink, but at that point you already have the makings for a well-performing class-AB stage.  I suppose we could argue about distortion...

Audioguru again:
I think the class-A amplifier produced a low loudness in the speaker because the very high DC current in the speaker caused its coil and cone to be pushed hard against one side of its magnet structure.
Then the circuit does not amplify anything, it simply heats everything.

Zero999:
I think I should highlight the point made about biasing, as I don't think everyone has picked up on it. In this case, biasing the DC point at half the supply voltage is undesirable. The output should be biased at half the maximum voltage swing of the amplifier output. If the maximum positive output swing of the amplifier is 6.6V (1.5V loss for the op-amp and 0.9V for the transistor, as emitter followers drop a lot more than 0.7V, at high currents) and the negative is 0V, then bias point needs to be half way between these voltages, which is 3.3V.

Using a pull down resistor will reduce the power dissipated by the speaker, but also efficiency of the circuit. If the resistor is the same value as the speaker's impedance, then the maximum power delivered to the speaker will be reduced by a factor of four, because it will drop half the voltage as P = V2/R. Using a lower value resistor will drastically increase the static power dissipation and using a higher value will reduce the maximum power output even more. Replacing the resistor with a constant current sink, but it would require an additional transistor and it will be cheaper to just use a class AB push-pull topology. Exercise for the original poster: work out the optimal current for the sink, given a voltage swing of 6V peak-to-peak (the current sink will prevent the output voltage reaching 0V) and an 8Ω speaker.

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