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I have recently started to study about class AB amplifiers online, as a newbie, i have found every schematic of the class AB amplifier is in common collector circuit just like this

Schematic i find everywhere in internet...

schematic

simulate this circuit – Schematic created using CircuitLab

But i eager whether the circuit can be used in common emitter mode, so i have drawn my own circuit below , in which, everything is same but the connection of the transistors. I dont know whether my made circuit will work or not, but i don't want any harms to happen while i am testing the circuit so , i considered that it would be better if i ask this topic on stack exchange before implementing.

My drawn circuit below...

schematic

simulate this circuit

So here are my questions ...

  1. Why is class AB amplifier used in common collector method?

  2. Can class AB amplifier be implemented according to my drawn schematic in which transistors are in common emitter ? If my drawn circuit cant be applied, then please explain why .

  3. Why is common collector chosen best for class AB amplifiers?

Thank you so much !

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    \$\begingroup\$ It can be arranged either way. But the 2nd case won't work as shown. You'll need some different way of driving the power output pair of BJTs. A Sziklai pair comes immediately to mind. But that's not the only approach. \$\endgroup\$
    – jonk
    Apr 23, 2021 at 8:10
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    \$\begingroup\$ Not an expert, but it has to do with cross conduction issue. The sequence NPNPNP is more beneficial than PNPNPN.. \$\endgroup\$ Apr 23, 2021 at 8:12
  • \$\begingroup\$ Your common collector circuit has very low base bias because the 11k resistor values are much too high resulting in low maximum output power. Your common emitters circuit will draw a massive DC current all the time and not amplify anything (Positive supply, PN, diode, diode, PN, Negative supply). An audio amplifier has an extremely low output impedance so it can damp speaker resonance. \$\endgroup\$
    – Audioguru
    Jun 6, 2022 at 19:42
  • \$\begingroup\$ The swap of D1 with R2, and D2 with R1, will keep the second schematic from burning up; those diodes are base-emitter connected for thermal control, connecting them from base to collector has no purpose. \$\endgroup\$
    – Whit3rd
    Aug 3, 2023 at 18:25

7 Answers 7

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  1. Another name for that config is emitter follower. It provides current drive ability with voltage gain of (almost) 1. It basically just buffers the signal.

  2. Nope. The semiconductors would blow up. Assume that each semiconductor junction (diodes and transistor B-E) has a voltage drop of 0.7V when forward biased. Try and simulate it.

  3. It may not be the best, but most simplest and practical two transistor push pull configuration that is good enough for whatever the puropose is - such as learning the basics from textbooks. If you need better performance for another purpose then you can easily find better but more complex configurations that will fullfill the requirements of those other purposes.

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  • \$\begingroup\$ I still have not understood "why the common collector is chosen best for class AB amplifiers" (3)... \$\endgroup\$ Aug 4, 2023 at 9:28
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The premise is false.

You can make class B or Class AB amplifiers without emitter followers; indeed that was the norm for at least the first thirty to forty years.

Here's a counter-example.

schematic

simulate this circuit – Schematic created using CircuitLab

This is only a representative, with bias details and overall negative feedback omitted, but it is a classic example of a Class AB or Class B amplifier, depending on the voltage applied at Vbias.

Older engineers may laugh to see it drawn with transistors, but (a) Circuitlab, inexcusably, doesn't offer valves, and (b) this is actually the way early transistor amplifiers were built (usually with PNP transistors, since NPN Germanium transistors were rare beasts) when transistors were new and this topology was familiar. So you will find it in 1960s transistor radios, identifiable by the output transformer.

Q1 and Q2 are the output devices, both in common emitter configuration. (It would be common practice to add small emitter resistors for better current control, bypassed with large capacitors. This matches "automatic bias" in valve circuits, where Vbias would then be 0V). With Vbias up to 0.6V this is Class B, with some element of crossover distortion : above 0.7V it's into AB, and above 0.7V plus (Vin peak * Q4 gain) pure class A.

Q3 is a phase splitter : Q1 is driven by its emitter follower output (gain = 1) while Q2 is driven by its common emitter output with gain = -R1/R = -1. (Approximately; I did say I was omitting details). Another common phase splitter used a long tailed pair input stage as seen in most opamps, providing both true and inverted outputs.

Q4 is the main voltage gain stage though the output pair also provide voltage gain.

The reason this configuration fell into disuse is obvious : the output transformer. But feel free to invent transformerless configurations (perhaps based around full bridges) if you like; just don't blame us fi it turns out to be more difficult than at first glance.

The basic circuit configuration lives on where transformers are unavoidable; typically in cheap 12V to mains voltage inverters, with MOSFETs, and either square wave drive, or PWM drive to produce a "pure" sinewave.

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  • \$\begingroup\$ This is not a push-pull stage, but simply two separate common-emitter stages driving two separate loads (the transformer halves). The OP's question is about a push-pull stage where the two transistors are connected via their collectors to a single load... and that's something entirely different... \$\endgroup\$ Aug 4, 2023 at 9:24
  • \$\begingroup\$ @Circuitfantasist This configuration has been known as a push pull amplifier for about a century now. en.wikipedia.org/wiki/Push%E2%80%93pull_output#/media/… \$\endgroup\$
    – user16324
    Aug 4, 2023 at 12:50
  • \$\begingroup\$ Thanks for the response. I do not mean the name, but the essence of the circuit solution. This is more "pull-pull" than "push-pull" because both transistors (tubes) are "pulling" only. In addition, as I already wrote, they pull different loads and not a common one. In a true push-pull stage, the output parts of the transistors are connected in series and the load is connected to the midpoint. \$\endgroup\$ Aug 4, 2023 at 13:16
  • \$\begingroup\$ No they do not pull different loads, because the two halves of the primary are tightly coupled. A "pull" on one leg (Q1 collector) produces a "push" on Q2 collector despite (in class B operation) Q2 being off. \$\endgroup\$
    – user16324
    Aug 4, 2023 at 13:19
  • \$\begingroup\$ "Push" and "pull" are in relation to the load. The transformer configuration is more like a bridge amplifier because the load is "floating" and driven simultaneously on both sides (differentially). In transformerless push-pull stages, the load is controlled from one end only, and the other is fixed. Obviously, in the transformer stage, the inversion of one "pull" with respect to the other takes place through the inductive connection. \$\endgroup\$ Aug 4, 2023 at 13:36
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The popularity of common-collector output stages comes from the ease of use: particularly biasing, and insensitivity to load condition.

Also to fill out the list: common-base requires full drive current (input current is slightly higher than output) -- something of a non-starter. So the only viable alternative is common-emitter.

CE has substantial voltage gain, and a high output impedance. Meaning, the gain is high, but it's also proportional to load impedance (which is much lower than the amplifier's intrinsic source impedance, thus dominating, and the equivalent circuit looks like a Norton source, a current into a load impedance).

On top of this, we need considerable gain in the input stage, to get sufficient overall gain to keep distortion low and bandwidth high. (The fact that amplifiers regularly have < 0.01% THD is a direct consequence of having loop gain of several thousands -- millions, even.)

A consequence of high gain, is low bandwidth at that gain. A CE or CB stage's bandwidth is limited by capacitance at its output: at DC, Early effect and load resistance dominate, but at AC, node capacitances quickly take over. And with that capacitance, comes a 90° phase shift. (A resistive-loaded 2N3904 in CE, might have a gain of 10 up to a couple MHz, then fall off above there; a CCS-loaded one might have a gain of 2000, but rolling off at say 10kHz -- it has the same GBW (gain-bandwidth product), just different gain at DC. An op-amp might have a gain of 10^5 at DC, dropping off at single digit Hz -- but still a GBW of some MHz!)

Too much phase shift stacked up, causes oscillation. More than 180° means the feedback signal comes in positive rather than negative, thus increasing rather than decreasing the signal. We must cut the loop bandwidth below the point where this happens. If we have two stages cascaded with say gain = 100 and GBW = 1MHz, then the total phase shift is close to 180° above 50kHz or so. (Note they roll off at GBW/gain = 10kHz, at the same time transitioning from 0° phase shift at DC-1kHz say, to 45° at 10kHz, to ~90° at >>10kHz. See standard Bode plots for a visualization.)

If we have two or three stages cascaded, with high gain and thus significant phase shift at signal frequencies, we simply get an oscillator. We need to reduce the phase shift (which is to say, increase GBW) of at least one such stage, within the bandwidth of the other, until stable loop gain and phase shift is met.

Conventional amplifier designs use a three-stage cascade: the input stage (typically a differential pair) has high gain and bandwidth, so that it contributes little phase shift within the design bandwidth; the middle ("VAS" volt-amp stage) has high voltage gain (typically near rail-to-rail), adequate drive current for the output, and modest bandwidth (it sets the "dominant pole" limiting loop gain and roll-off); and the output is a simple emitter follower, taking the voltage from the middle stage (which is easy enough to cover) and amplifying its current capability (enough that the output voltage gain is fairly independent of load). Thus, phase shift is modest within the required bandwidth, and compensation can be tested and set fairly easily, without much trouble from pathological load impedances.

Finally, a CE output stage is difficult to bias. You can set quiescent current alright, but the transistors are pulling from opposite rails, no communication between each other, just a solid tug-of-war between their collectors. It's very easy to design a circuit which wastes a lot of current on shoot-through (especially dynamically, as it typically takes more time for one side to turn off (storage time) than the other to turn on). And, as they heat up, you need to somehow sense that current and throttle them down, or compensate beforehand (by setting bias voltage and thus current, lower as temperature rises).

Whereas, these are all very easy to do in the complementary emitter follower design: a fixed voltage between bases, will set the bias voltage, and by making it proportional to a diode drop (stack of diodes, or a "Vbe multiplier") and thermally coupling those junction(s) to the output heatsink, it tracks output transistor Vbe just fine. A little emitter resistance sets the current, and you're done.


The flip side of this is shown in strong relief, when it's unavoidable: low-dropout (LDO) regulators are of such a design, that is, an error amp followed by a common-emitter/source stage. They are notoriously sensitive to load impedances, typically demanding output bypass capacitance, with ESR, in a certain range. (Only in recent decade(s) have designs improved to the point where they are stable with very low ESRs (ceramic capacitors), as well as eliminating other pathological behaviors like ground-pin current draw in dropout -- a classic issue of PNP types, and solved with more careful design, or with CMOS based designs.) Early rail-to-rail op-amps suffered similar problems; some types were completely unstable with any kind of capacitive load. Robust types (from both product families) are widely available today, so it's less of an issue now, but always something you need to be wary of when selecting a new part.

Note that regulators are just half an amplifier: typically they only source current, not sink it, so they actually have the added complication that bias current is externally controlled (load current), and thus also falling slew rate depends.

How did they solve the problem? One way is to put all the gain at the output. Construct the output stage, not as a single CE transistor, but a compound "super transistor" with a wide(r) bandwidth gain/driver stage in front, while still looking like a transconductance amplifier overall (that is, some input signal voltage is transformed to some proportional output current). This pushes the gain node (and thus the dominant pole) straight to the output pin itself: while this makes the amplifier indeed maximally sensitive to load variation, compensation is still alright, as long as the phase shifts in the rest of the circuit are low enough. Basically, the input and driver stages need to perform that much better. So, it's not ideal, kind of suboptimal. (They have many hacks, to save on current consumption, to short-circuit the phase shift through certain paths (feed-forward), nonlinear adjustments (slew rate boosting) to fake having higher speed, etc. Lots of stuff I don't know about, too, tucked away inside patents or trade secrets.)

In short, these are techniques that aren't very useful for audio amplification purposes -- it takes tons more transistors to implement (which cost approximately zero on chip!), while you have more than enough voltage available (30V+) that losing the 1-2V drop in a conventional design is inconsequential. So these rail-to-rail designs aren't worth the trouble.

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  • \$\begingroup\$ Remarkable overeview! \$\endgroup\$ Aug 4, 2023 at 9:44
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But i eager whether the circuit can be used in common emitter mode

Yes. You don't see it as often as the emitter follower topology, but it is common. Here is an example, where it is used to boost the output of an opamp:

enter image description here

http://www.seekic.com/circuit_diagram/Basic_Circuit/POWER_BOOSTER_1.html

Here it is at a much higher power level:

enter image description here

http://easycircuit012.blogspot.com/2012/10/1000watt-audio-power-amplifier-blazer.html

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  • \$\begingroup\$ Your opamp circuit is comparatively obscene since it uses the supply current of the opamp as input to the power transistors. That requires rather particular relations of quiescent current and load currents, meaning that replacing the 741 with "equivalent" or "better" opamps may blow up around your ears. \$\endgroup\$
    – user107063
    Aug 3, 2023 at 17:34
  • \$\begingroup\$ First, that's not "my" amplifier; it is an innergoogle grab to illustrate the point raised by the OP. Second, the circuit is much more tolerant of component variations than you think due to negative feedback. Outside of that range, R1 and R2 have to be adjusted. \$\endgroup\$
    – AnalogKid
    Aug 3, 2023 at 21:34
  • \$\begingroup\$ Third, note that the output transistors are not on all of the time. For zero and low signal amplitudes, both output transistors are completely off and the opamp drives the load directly. It is the signal current into the 5.1K resistor and the load that modulate the output transistors. This is a variation of a Sziklai pair, with the opamp acting as the emitter follower (with gain). \$\endgroup\$
    – AnalogKid
    Aug 3, 2023 at 21:34
  • \$\begingroup\$ You got the idea very well. I saw for the first time this exotic circuit solution in the late 80s, in the NSC Linear Applications Handbook (AN 29, written by Widlar). Then I was deeply impressed by it and even formulated a circuit principle. The trick here is that a power transistor common-emitter stage monitors the load current and connects in parallel to the weak op-amp to "help" it... just like in life... \$\endgroup\$ Aug 4, 2023 at 17:06
  • \$\begingroup\$ I have two copies (you always need a spare) plus a PDF. Note that while the circuit greatly eases the output stage static current problem, it adds stability problems caused by the fact that the output transistors now have voltage gain - and, by definition, phase shift. \$\endgroup\$
    – AnalogKid
    Aug 4, 2023 at 17:59
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The upper circuit will have a very small output power because the base resistors have much too high resistances. The bottom circuit will draw massive DC currents.AB

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  • \$\begingroup\$ A string of four forward-biased diodes in series stretched between +12 V and -12 V supply rails?!? \$\endgroup\$ Aug 4, 2023 at 12:06
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I have recently started to study about class AB amplifiers online, as a newbie, i have found every schematic of the class AB amplifier is in common collector circuit

If you have studied class AB amplifiers then you should already have studied class A and class B amplifiers. So you could already have the answers.

A Class A amplifier can provide high voltage gain along with low distortion, but it has very high output impedance. And thus it's not suitable to drive a low-impedance load such as a speaker. Still a class A amplifier can drive a low-impedance load but requires a transformer for impedance conversion (Google "transformer coupled class A amplifier").

A class B amplifier can provide high current gain along with unity voltage gain and very low output impedance thanks to the emitter-follower configuration, but it has poor distortion performance because of the unbiased transistors. It's ideal to drive low-impedance loads but still requires the input voltage's peak to be greater than VBE. That's why class B amplifiers have poor distortion performance.

The idea behind a class AB amplifier is to provide a class A amplifier's low distortion and a class B amplifier's low output impedance (i.e. high drive capability). That's why a class AB amplifier has a common collector (a.k.a. emitter follower) output configuration (answer 1 and answer 3). However, it's possible to build a common-emitter push-pull stage but we may not name it a class AB amplifier. Google "common emitter push-pull amplifier". Your configuration will not work, or even the transistors can break down (answer 2).

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  • \$\begingroup\$ A Class A amplifier...And thus it's not suitable to drive a low-impedance load such as a speaker. I'm not sure I agree with that, I can take a class AB with common collectors (1st schematic in question) and then change it such that the output transistors never have Ic = 0. In essence, I increase their bias current by a lot. Then I can call this amplifier "class A" since the output transistors never switch off (Ic = 0). \$\endgroup\$ Apr 23, 2021 at 8:48
  • \$\begingroup\$ @Bimpelrekkie Then I can call this amplifier "class A" since the output transistors never switch off (Ic = 0). True. The "Class A amplifier" mentioned in my answer is the basic one with only one switch that the output is taken from its collector/drain/anode, of course. Even with that one, it's possible to drive a low-impedance load, but it'll be way more impractical and inefficient. And we're talking about the practical/applicable models. \$\endgroup\$ Apr 23, 2021 at 9:34
  • \$\begingroup\$ Not sure what the classes have to do with the load driving ability or output impedance. \$\endgroup\$
    – tobalt
    Jun 7, 2022 at 6:07
  • \$\begingroup\$ @tobalt Typical discrete class A preamp circuits (out of vogue for several decades) have a single BJT acting against a resistor. That resistor of course limits the open loop output impedance you can achieve and delivers asymmetric slew rates on capacitive loads. Power stages, however, will usually have some push/pull pair, and then the load driving ability is merely limited by having to account for the wasted power of class A. \$\endgroup\$
    – user107063
    Aug 5, 2023 at 11:08
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It is absolutely feasible to construct a class AB or B output stage with common emitter topology. This is usually called Rail-to-rail output in the context of opamps, and is frequently done. In the context of CMOS, most gates have a common source type output arrangement, too.

enter image description here

Image taken from bobflux's post here

The reason why you don't see this more often in discrete designs, is that it is more difficult to drive correctly, especially with Class AB. If you watch the image above carefully, you notice that the drive input is tied together for the common-collector type (A), but two different drive inputs are necessary for the common-emitter arrangement (B). If you tie them together such as in your suggestion, you short the supplies through the diodes.

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  • \$\begingroup\$ It's easy to drive by just adding a complementary transistor in front, forming a Sziklai pair that has its joint "collector" at the point of the emitter of the power transistor. When power transistors were most powerful as NPN (the heydays of the 2N3055), it was pretty customary to use two NPN power transistors in complementary output stages, one in a Darlington pair with an NPN driver BJT, one in Sziklai pair with a PNP driver BJT. \$\endgroup\$
    – user107063
    Aug 3, 2023 at 17:42
  • \$\begingroup\$ @user107063 While you can put a complementary transistor in front of the power transistors above in image "B)", to form Sziklai pairs, that does make it easy to drive, but it also removes one of the main benefits of the simple common emitter output, namely that you can drive it close to the rails. That's not possible with Sziklais, which saturate at ~1 diode drop. Using Sziklai for more voltage gain is usually overkill. \$\endgroup\$
    – tobalt
    Aug 3, 2023 at 18:29

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