Electronic – How do diodes and capacitors reduce Crossover distortion

amplifierdistortion

I found this diagram about Class AB Amplifiers and the reduction of crossover distortion: enter image description here

http://www.electronics-tutorials.ws/amplifier/amp_7.html

This pre-biasing voltage either for a transformer or transformerless amplifier circuit, has the effect of moving the amplifiers Q-point past the original cut-off point thus allowing each transistor to operate within its active region for slightly more than half or 180° of each half cycle. In other words 180° + Bias. The amount of diode biasing voltage present at the base terminal of the transistor can be increased in multiples by adding additional diodes in series. This then produces an amplifier circuit commonly called a Class AB Amplifier and its biasing arrangement is given below.

I don't understand the explanation of how the diodes and the capacitors reduce the Crossover distortion. Each transistor (npn and pnp) should cover 180 degrees a sine, why doesn't 180 + bias remove the complete distortion, what have the capacitors and diodes to do with this?
I read about the diodes compensating for the transistor voltage drop of twice 2× 0.6V How does this exactly work? How does the capacitor smooth the signal?

Best Answer

Cross over distortion of a class B amplifier: -

enter image description here

The top half of the waveform comes from TR1 conducting and the bottom half from TR2 conducting. At some point a class B amplifier changes from using the top transistor to the bottom transistor. When this happens there is insufficient voltage across base/emitter to activate either transistor hence there is a dead zone: -

enter image description here

The diodes turn a class B design into a class AB. Now, neither transistor is fully off therefore the dead zone is no more.

The capacitors are incidental - they allow the input signal to couple to both bases without the new biasing arrangement being affected.