What's the idea behind this, assuming it is ideal?

Question

The circuit is a biased push-pull. The diodes keep the input signal above and below the input signal so it is always conducting. The problem is that I don't see why one of the two transistors would not conduct in a semicycle.

If the input is \$5.3V\$ then above D1 the voltage is \$5.9V\$. The output after the drop is then \$5.3V\$. On the other side, below D2 the voltage is \$4.7V\$, which is below the output voltage, making the php transistor Q2 conduct. I know I'm missing something, but I don't know where. In the book (The Art of Electronics) it says that in the positive swing Q2 shouldn't conduct, but I can't figure out why.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Because it's an output stage, it won't make sense without a load. Try looking at it with a 100 ohm resistor from V_out to ground.

When Vin is 0, Vout will also be zero, but there will be a (hopefully small) amount of current flowing through Q1 and Q2.

Now let Vin jump to 1 volt. The anode of D1 will be at about 1.6 volts, and the output at about 1 volt. The cathode of D2 will be at about 0.4 volts, and the emitter-base of Q2 will be reverse biased, so no current will flow through it.

Likewise, if Vin equals -1.0, the output will be pulled down to -1, and Q1 will be cut off.

The advantage of putting the diodes in, rather than just tying the two bases together, is that the output will respond smoothly to Vin variations of less than +/- .6 volts. This is critically important in avoiding crossover distortion when the output moves through zero.

EDIT - Although I realize that this still doesn't explain why one transistor conducts but the other doesn't. Think of it this way. With an appreciable load, the emitter current of the transistor which supplies the load current at the specified polarity is greater than the draw with zero Vin. This means that the emitter/base voltage is greater, too. In turn, this reduces the base/emitter voltage at the other transistor. Since base currents are exponential wrt base voltage, this reduction in base voltage causes a major reduction in emitter current. Thus, only the transistor supplying load current conducts much, and in the process starves the one not conducting.

Answer 2

The bias network (resistors and diodes) makes a current flow through both transistors at 0V out. If the diodes and transistors are thermally closely coupled, the current is reasonably stable with temperature (the diodes change similarly to the Vbe of the transistors).

When the amplifier is producing a small to moderate output, the both transistors still conduct significant current at all times (one more than the other, but some bias current flows through both). If the input approaches the rails, then the respective transistor will stop conducting.

Here is what the collector current in the PNP transistor in the above schematic (diodes replaced by diode-connected transistors) looks like with +/-12V supplies and a 5Vpp 1kHz input and 500 ohm load (10K resistors used so bias current is about 1.5mA).

Here is what it looks like with a 10Vpp input:

And the output waveform with 10pp input:

Answer 3

Part of your confusion might come from the fact that it is not a very good circuit. The bias current in the transistors is not well-controlled. As drawn, shoot-through current from one transistor to the other is determined by the difference between the diode forward voltages and base-emitter forward voltages divided by the transistor intrinsic emitter resistances. These are not well-controlled.

Do a Google search of 'class a-b rubber diode' the show how it is done properly. Oh - I call them V_be multipliers, not rubber diodes.

Answer 4

It's an output stage. The diodes are for a bias voltage, so the transistors work in "class AB" mode.