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Looking for tutorial/description on audio amplifier input circuit operation

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dicky96:
Hi guys
As the title says, I'm trying to find a detailed description of a circuit I see quite often on the input of Class A/B or G amplifiers.  I can find lots of amplifier circuit descriptions on the net, but not this one made up of six transistors

See attached schematic.  It's the front end I want to understand thoroughly

I can kinda figure some of it out, but the details escape me - so what I can figure is:
A. The audio signal comes in to Q115 base and Q114 base

B: Q115 and Q117 form some kind of differential circuit, handling the negative audio half cycle

C: Q114 and Q116 form a similar differential circuit, handling the positive audio half cycle

D: The other inputs to these differential circuits (Q116 base + Q117 base) have some sort of varying DC voltage on them generated across C115 via R132/R131 which form some sort of voltage divider dependent on the average output voltage (to the speaker) and R132/R131 are critical (1% tolerance).

E: The collectors of Q114 and Q115 drive the next stage in the amplifier

F: This amplifier has +/- 90V supply rails (+B1 / -B1). OK it also has +/- 45V but they are only used for the Class G output stage

However I have various questions about the circuit and so far google did not find a description of the operation

1. I'm not sure why the input waveform is 'compared' with the output and how that affects the signal passed on to Q118 base / Q119 base (the next stage of the amplifier)?

2. Q112 somehow sets the emitter voltage for the differential pair Q115/Q117 and likewise Q113 sets the emitter voltage on Q114/Q116 but why do we need two more transistors (Q112/Q113) and such a sophisticated circuit to do this?  Is this some sort of constant current circuit and what should the DC emitter voltages of Q114/Q116 and Q115/Q117 pairs be with no audio input?

3. The voltage divider made up of D105+D106+R119+R120+D107+D108 in series sets the base voltages for Q112/Q113 but the pairs of diodes in series would suggest the base of Q112 is about 1.1V negative of the emitter and the base of Q113 is about 1.1V positive of the emitter - but obviously that can't be the case.  What are these four diodes doing?

4. What DC voltage should on base of Q118 and Q119 with no audio input?

Hopefully someone can help me understand the circuit better, or explain where I already have some errors in my understanding

TIA

dmills:
D is wrong, they get a divided down sample of the audio output (closing the negative feedback loop that sets the closed loop gain), C115 forces the gain to unity at DC to reduce DC offset at the speaker terminals.

1: Negative feedback is one of the foundational methods in modern electronics, by giving the circuit massive gain and then subtracting a fraction of the output from the input you both reduce distortion and stabilise the gain (It does not matter what the open loop gain is, so long as it is large). This is used in essentially EVERY modern audio amplifier that is not tweaky audiophile bullshit).

2: These transistors do not set emitter voltage, they set emitter CURRENT, consider Q113, the base is held two diode drops above the negative rail (D107,D108) so the emitter is one diode drop above the negative rail, so the emitter current is (0.7/150 (R122)) = ~5mA.

3: The diodes set the base voltages and hence the emitter voltages to cause a constant current to flow in the emitter resistor, see 2, this is a totally standard 'constant current' source, if not a particularly good one, you can do better with a transistor instead of the diodes.

4: With no signal (and the feedback in balance) the current in each of the long tailed pairs collectors will be approximately equal to half the emitter current (as defined by the current source), so 2.5mA, which across a 1k load resistor will develop 2.5V. Putting the emitter of Q118 1.8V below the rail, and thus causing approximately 8mA to flow in collector circuit Q118. By the same argument Q119 also passes about the same current.
Once there is a difference in the voltages at the base of the long tailed pairs, one of these transistors will be turned on harder and the other one turned off further, the current difference driving the following stage (Q123,Q124).

dicky96:

2. Yes I did consider in the OP that I thought Q112 / Q113 quote: 'Is this some sort of constant current circuit ' and thank you for explaining to me in more detail how this works. I had clearly missed how R121/R122 also are part of the base-emitter circuits and that is how we have the correct B-E voltage drop even with the two diodes.  So the pair of diodes in series are effectively acting as voltage references in this circuit, yes?

As you know this circuit well, what voltage should I expect to see on the emitters of the of the pairs Q114/Q115 and Q115/Q117 with no input?

best regards

TimFox:
A good monograph on solid-state linear audio amplifiers:
G R Slone, "High-power Audio Amplifier Construction Manual", McGraw Hill, 1999.  ISBN 0-07-134119-6
Chapter 4 discusses input stages.

--- Quote from: dicky96 on November 29, 2022, 07:01:49 pm ---B: Q115 and Q117 form some kind of differential circuit, handling the negative audio half cycle

C: Q114 and Q116 form a similar differential circuit, handling the positive audio half cycle

--- End quote ---
Nope to both of those. (The "half cycle" stuff).