Looking for tutorial/description on audio amplifier input circuit operation - Page 1
Forum section: electronics / projects

author: dicky96
author_rank: Regular Contributor
timestamp: November 29, 2022, 07:01:49 pm
content: Hi guysAs 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 transistorsSee attached schematic. It's the front end I want to understand thoroughlyI 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 baseB: Q115 and Q117 form some kind of differential circuit, handling the negative audio half cycleC: Q114 and Q116 form a similar differential circuit, handling the positive audio half cycleD: 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 amplifierF: This amplifier has +/- 90V supply rails (+B1 / -B1). OK it also has +/- 45V but they are only used for the Class G output stageHowever I have various questions about the circuit and so far google did not find a description of the operation1. 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 understandingTIA

author: dmills
author_rank: Super Contributor
timestamp: November 29, 2022, 08:27:42 pm
content: 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).

author: dicky96
author_rank: Regular Contributor
timestamp: November 29, 2022, 10:09:39 pm
content: Thanks dmills for your advice2. 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

author: TimFox
author_rank: Super Contributor
timestamp: November 29, 2022, 10:59:04 pm
content: A good monograph on solid-state linear audio amplifiers:G R Slone, "High-power Audio Amplifier Construction Manual", McGraw Hill, 1999. ISBN 0-07-134119-6Chapter 4 discusses input stages.

author: wasedadoc
author_rank: Super Contributor
timestamp: November 30, 2022, 12:17:28 am
content: Quote from: dicky96 on November 29, 2022, 07:01:49 pmNope to both of those. (The "half cycle" stuff).

author: dmills
author_rank: Super Contributor
timestamp: November 30, 2022, 12:25:30 am
content: With the bases at ground (no input) the emitters will clearly be ~0.7V below (NPN or above (PNP) ground, because BJT in the active region.And I agree with dicky96, if either input pair cuts off it all goes sideways, you generally design everything up to the output stage to run in class A (And the output stage to run in class AB with as much bias as your heatsinking can stand.I like either Bob Cordell or Doug Selfs book on power amplifiers, they come from slightly different places, but both are good to the point of being canonical references on this subject, I have not tried Slones book.

author: magic
author_rank: Super Contributor
timestamp: November 30, 2022, 09:36:58 am
content: Quote from: TimFox on November 29, 2022, 10:59:04 pmI recall skimming that book and leaving with impression that it's basically a rip-off of Self, but with somewhat less details and explanation.It even starts with a rant about "subjectivists"...

author: Terry Bites
author_rank: Super Contributor
timestamp: November 30, 2022, 12:07:42 pm
content: Q112 and Q113 set the current in the "tails". You have two differential (long tailed) pairs, you only need to analyse one. The other one will work the same way.Q118 etc increases the total gain seen at the phase splitter. From there on its unity gain so you can substitute an Av=1 amp fo rthe purpose of analysis.The overall voltage gain is set by R132/R131, this is 1+(11000/620) =19

author: TimFox
author_rank: Super Contributor
timestamp: November 30, 2022, 02:43:47 pm
content: Quote from: magic on November 30, 2022, 09:36:58 amSlone's background is in sound reinforcement (e.g., musicians on stage).His rant about "subjectivists" is very like the attitudes common on this forum about "audiophools".He has a good discussion about the characteristics and pros/cons about BJT and MOSFET power stages, and he has a useful discussion of construction techniques without PCBs.

author: David Hess
author_rank: Super Contributor
timestamp: November 30, 2022, 05:33:14 pm
content: Quote from: dicky96 on November 29, 2022, 07:01:49 pmIt is not just "some kind of differential circuit". Q114/Q116 are one differential pair and Q115/Q117 are another differential pair.In this configuration, sometimes called complementary push-pull or double input stage push-pull, both sides are active with positive and negative signals which doubles the transconductance and helps suppress second harmonic distortion. In practice the advantages are negligible.QuoteThat is the feedback signal from the output going to the inverting input. This amplifier is configured as a non-inverting amplifier with gain.QuoteThe collectors drive the voltage amplification stages made up of Q118 and Q119 which are Miller integrators with their outputs tied in parallel.QuoteThat is how negative feedback works.QuoteQ112 and Q113 set the tail current of each input differential pair. The voltages at the collectors of Q112 and Q113 follow the non-inverting inputs at the bases of Q114 and Q115, with a DC offset from the constant current through the resistors. R195 through R198 are not strictly needed but do lower the power dissipation in Q112 and Q113.QuoteThe diodes set the voltages at the bases of Q112 and Q113, which sets the emitter voltages of Q112 and Q113, which places a fixed voltage across resistors R121 and R122, which sets the emitter currents, which results in the collector currents. These are just current sources.QuoteThe voltage is set by the current through R123 and R126. The tail current of the differential pairs is about 4.7 milliamps, so the current through Q114 and Q115 is half that or 2.3 milliamps, so the voltage across R123 and R126 is about 2.3 volts, which incidentally sets the current at idle through Q118 and Q119 to about 7.4 milliamps.Douglas Self discusses several variations and the history of this input topology in his book Audio Power Amplifier Design Handbook.

author: Cliff Matthews
author_rank: Unknown
timestamp: December 01, 2022, 06:13:44 pm
content: Very nice to watch this guy at work and a lot of folks learn great stuff on the channel

author: coromonadalix
author_rank: Super Contributor
timestamp: December 01, 2022, 08:01:59 pm
content: Im personally intrigued by q106 q109Since they power up the output transistors, not sure if they act as anti-pop or act as thermal protection or do both, since you see thermistors in their circuit drivers

author: dmills
author_rank: Super Contributor
timestamp: December 01, 2022, 10:16:56 pm
content: Q106,109 are the rail switching transistors that switch the supply to the output stage to a higher voltage when the output approaches clipping on the lower voltage rail.For typical audio, most of the time the system runs quite happily off the lower rail, with only relatively occasional need for more voltage, and using a rail switching design like this makes for a substantial reduction in heat produced and a massive improvement in safe operating area.Rail switching is tricky and a better approach is usually to have a linear arrangement that boosts the rail once the output stage gets within a few volts of saturation,much better behaved that way, and spreads the heat between the output stage and the passbank, but a rail switcher is cheap.This is class G (Or H, the terminology gets rather confused).