transistors - For the following amplifier, why is the output voltage 0 when input voltage is 0?

The amplifier is the following class AB amplifier:

For an input voltage of 0V, the output voltage is 0V. However, I don't understand why. My textbook does not delve into why this is the case.

(Also VBB/2 is large enough to turn on both transistors with no input voltage.)

Answer

Whenever you have this type of question, usually the best approach is to compute the base-emitter voltage of the transistors. This is what you tried to do, judging from your comments.

The base-emitter voltage of the NPN transistor is \$V_\mathrm{BB}/2 - v_\mathrm{O}\$. The base-emitter voltage of the PNP transistor is \$-V_\mathrm{BB}/2 - v_\mathrm{O}\$. Applying Kirchhoff’s current law at the output junction and substituting the diode equation yields

$$i_\mathrm{N} - i_\mathrm{P} = I_s e^{V_\mathrm{BB}/(2V_\mathrm{T})} (e^{-v_\mathrm{O}/V_\mathrm{T}} - e^{v_\mathrm{O}/V_\mathrm{T}}) = v_\mathrm{O} / R_\mathrm{L}.$$

The trivial solution of this equation is \$v_\mathrm{O} = 0\$. It is straight forward to show that this is the only solution. Note that the gradient for the LHS is decreasing while the gradient of the RHS is increasing.

Although it is difficult to see, the LHS (blue) has the opposite sign of the RHS (red).