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

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

• What V_ee value? During the positive phase, I am pretty sure that Q2 isn't in the saturated zone therefore it cannot conduct in theory.I would need to re-check my book to be sure. The idea behind this set-up from a Box POV, is that it give you a gain in currrent. Commented Aug 4, 2015 at 14:59
• @MathieuL I believe that the idea is to make a voltage buffer with (very) low power consumption, compared to a class A configuration. Commented Aug 4, 2015 at 19:03

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.

• Thanks. That might be my problem. On the second case, Q1 should shut off. But the collector is at V_cc, the base is at -0.4 and the emitter at -1. Doesn't that make the base-emitter voltage positive, and thus Q1 conducts? Commented Aug 4, 2015 at 5:19
• @OFRBG - see edit. Commented Aug 4, 2015 at 5:43
• Great. So what explains that one transistor conducts and the other one almost doesn't is the exponential relation between V_BE and the collector current, as long as the load draws as much current as possible? Commented Aug 4, 2015 at 5:49
• Maybe as a strange extra doubt, would you be able to tell why they determined so nonchalantly that the transistors were cutoff without using Ebers Moll? Just for curiosity. Commented Aug 4, 2015 at 5:52

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:

• So in the end both transistors are always sourcing/sinking current but just switch intensity depending on the polarity established with the load? (Assuming the load is not necessarily grounded.) Commented Aug 4, 2015 at 5:41
• The transistor that's not driving load current will conduct the bias current or less (down to about zero for large output voltages near the supply rails- as above in the middle graph). The other one will conduct that current plus the load current. If the load is not grounded then this bias scheme might not be ideal in general, but it's fine for grounded load or a bridging amplifier (where outputs are driven antiphase to double the output voltage across the load). Commented Aug 4, 2015 at 6:06

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.

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