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This question refers to audio amplifier design, using BJTs only.

I am wondering how differential stage amplifier plays an important role in hi-fi audio amplifiers. According to this book (where picture was copied from) I have, a part of output signal is feeding the base of Q2. Q1 and Q2 somehow amplify input signal and at the same time cancel out some of distortion that is being brought back from the output to the differential input stage of amplifier.

  • I want to understand the process of this differential stage and how it manages to cancel out some of the output distortion.

enter image description here

I almost completely understand functions of other transistor stages so there is no need for explaining how the whole circuit works.

*Source: G. Randy Slone : High Power Audio Amplifier Construction Manual

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  • \$\begingroup\$ Possible duplicate of Illustrating op amp feedback without control theory. Your question is really about what negative feedback is and how it works. See my detailed answer to the above question. \$\endgroup\$
    – Olin Lathrop
    Commented Oct 23, 2017 at 13:08
  • \$\begingroup\$ @OlinLathrop Maybe just a part of my question refers to nfb. I was mostly interested of how Q1 and Q2 prevent distortions coming from the output. \$\endgroup\$
    – lucenzo97
    Commented Oct 23, 2017 at 13:43
  • \$\begingroup\$ The negative feedback reduces the distortion, not the differential pair by itself. The differential pair act as an error amplifier. He compares the input signal (not distorted one) with the output signal. So, the error signal is created to cancel the distortion. Exactly in the same way as a negative feedback "brings back" the output signal to the equilibrium state (V"+" = V"-"). The negative feedback does not distinguish between types of disturbances. \$\endgroup\$
    – G36
    Commented Oct 23, 2017 at 14:18
  • \$\begingroup\$ If for some reason the Vout is different than expected the negative feedback will correct it. If Vout is larger than expected (Vin) the Q2 will reduce his IC current. But in the same time, the Q1 Ic current will goes up. Q3 and Q4 current will also increase so the Vout will drop. \$\endgroup\$
    – G36
    Commented Oct 23, 2017 at 14:32
  • \$\begingroup\$ Note how the name "differential amplifier" relates to the fact that the amplifier (with a differential pair at its input) amplifies the difference (voltage) at its input. By amplifying the difference and feeding it back as negative feedback, the difference is minimized. And zero difference means no distortion. \$\endgroup\$
    – Bimpelrekkie
    Commented Oct 23, 2017 at 14:46

1 Answer 1

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The original question asked about "cancelling output distortion". This answer shows the diffpair ADDS its own distortion, uncancelable distortion.

with 200 milliwatt power dissipation in your diffpair, you will experience lots of thermal distortion. Put in a clean square wave (into the actual circuit) and examine the overshoots (or undershoots) on pulse leading edges.

You can model this in SPICE, using voltage-controlled-voltage-sources to modulate a DC voltage in series with base or with emitter, that source changing -2 milliVolts per degree Centigrade of collector heating.

Now if you just had a good thermal model of those transistors.

Assume each is 100 micron cube. 1u cube is 11.4 nanoseconds 10u cube is 1140 nanoseconds 100u cube is 114000 nanoseconds (114 microSeconds)

Given 4+ milliamp through R1, and 50 volts Vce across each of Q1 and Q2, there is at least 200milliWatts total in those 2 devices.

You care about the transient thermal resistance.

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    \$\begingroup\$ I didn't downvote but this needs a bit of a clean-up. Mind the units too: "V" or "volt" (units named after a person are capitalised only when abbreviated), "s" or "second" and "centigrade". \$\endgroup\$
    – Transistor
    Commented Oct 23, 2017 at 17:44
  • \$\begingroup\$ I don't know where you got 300mW from. None of this has anything to do with the question that was asked. \$\endgroup\$
    – user207421
    Commented Oct 23, 2017 at 20:00
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    \$\begingroup\$ This has nothing to do with my question.. \$\endgroup\$
    – lucenzo97
    Commented Oct 23, 2017 at 20:53

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