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I'm studying AOE but i can't figure out how this circuit even works(also i'm confused if Q1 and Q2 make a sziklai pair). any help with the circuit description will be great.

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    \$\begingroup\$ 30 years in electronics and never heard of a Sziklai pair so I looked it up and learned that it's a bit like a Darlington. Q1's collector current goes to Q2's base so yes looks like a Sziklai pair to me. \$\endgroup\$ – Bimpelrekkie Feb 28 '17 at 14:37
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    \$\begingroup\$ The circuit is an amplifier with feedback. The feedback works like this: assume Vin increases then Vbe of Q1 increases and Q1's Ic increases, this increases V(R3) and thus Vbe of Q2 making the Ic of Q2 increase lifting the output voltage. This makes I(R5) increase and this current flows into R4 making V(R4) increase. That increase in V(R4) decreases Vbe of Q1 making the negative feedback as we saw that the input voltage tries to increase Vbe of Q1. The ratio R5/R4 roughly sets the gain. \$\endgroup\$ – Bimpelrekkie Feb 28 '17 at 14:49
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I'm studying AOE but i can't figure out how this circuit even works

If the 10 volt output rises a little bit then Q1's emitter is also raised a little bit and this has the effect of turning Q1 off a little bit. As Q1 turns off a little bit, its collector voltage rises. This has the effect of slightly shutting-down Q2 and therefore the slightly risen 10 volt output falls back to 10 volts.

It's negative feedback.

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  • \$\begingroup\$ " Q1's emitter is also raised a little bit and this has the effect of turning Q1 off a little bit." I have to ask for some details here , Does Q1 turns off a little bit because its Vbe is reduced(ebers moll)? \$\endgroup\$ – iMohaned Feb 28 '17 at 15:04
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    \$\begingroup\$ Mohannad Maklad, yes - of course, that is how negative feedback works. An increase of Ic1 (Q1) - either wanted (because of rising Vin1) or unwanted (Ic1 increase due to temperature effects) - increases also the voltage at the emitter of Q1, thereby reducing again Vbe1 (which, in turn, reduces Ic1 again). For negative feedback, the loop gain must be negative (phase inversion). That is the case in your circuit because the loop contains a noninverting common base stage (in the loop, input to Q1 is at the emitter) and an inverting common emitter stage (Q2) \$\endgroup\$ – LvW Feb 28 '17 at 15:36
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    \$\begingroup\$ @MohannadMaklad yes it does - BE voltage is reduced by increasing the emitter voltage. \$\endgroup\$ – Andy aka Feb 28 '17 at 16:11
  • \$\begingroup\$ Perhaps one should add that R4 alone provides already negative feedback. This effect is enhanced (supported) by the second stage which feeds its emitter current Ie2 also into R4 (same polarity as Ie1) \$\endgroup\$ – LvW Feb 28 '17 at 17:56
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    \$\begingroup\$ @MohannadMaklad, the ratio R5/R4 is a rough approximation (simplification) to the correct gain expression which includes all the transistor parameters. Hence, you must not arbritrarily set R5=0 and draw the (false) conclusion that that the gain would be be zero. \$\endgroup\$ – LvW Mar 1 '17 at 11:07
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Overall feedback is between R5 (10K) and R4 (1k), which yields an overall voltage gain near ten. And because feedback is to the emitter of Q1, input impedance becomes quite high. Like FakeMoustache, I've never heard of "Sziklai pair", and have "invented" this circuit for my own use. Kudo's to Sziklai for publishing first.

Q1 converts its base input voltage to current. It has less-than-one voltage gain, but a portion of its collector current change feeds into the base of Q2. In this case, the base of Q2 presents a dynamic load resistance to Q1 of about 26 ohms times its Beta of about 100 (because its emitter current is about one milliamp). So dynamic input resistance of Q2 is near 2600 ohms. That's in parallel with R3.
Open-loop gain is quite high, compared to closed-loop gain of ten. So Andy's approach is a good one. Here's a slightly different approach..consider what happens if you raise the input source a small amount (say 0.1v) from 2.6v to 2.7v. If Q1's emitter stayed at 2.0 v, Q1 would conduct very much more current, pulling its collector voltage down. But it must also pull down Q2's base voltage too. That will increase Q2's base current a great deal, which multiplies by perhaps one-hundred at Q2's collector. That increase in Q2's collector current must flow through R4 and R5, pulling up output voltage.
How high will output voltage rise? It will rise until the voltage drop across R4 nearly counteracts the initial disturbance of 0.1v. Because R5 is ten times larger (and carries the same current change as R4), output voltage must rise nearly one volt. Voltage gain is nearly ten, since a 0.1v input change causes a one volt output change (slightly less than one volt). I'm assuming that output is taken from Q2's collector.

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  • \$\begingroup\$ Stupid question alert: is it allowed to say " collector current change feeds into the base of Q2" because -you know- the current doesn't go this direction, or is it allowed to say this in terms of absolute thinking of input impedence and things like that? i hope i cleared what i mean, great explanation though. \$\endgroup\$ – iMohaned Mar 1 '17 at 7:23
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    \$\begingroup\$ @MohannadMaklad Don't forget that for changing currents & voltages, Vcc and ground act identically. Q1's increasing collector current causes Q2's base current to increase. Q1's increasing current divides between R3 (620 ohm) and Q2's approximate 2600 ohm. 81% goes to R3, 19% goes into base of Q2. All this current is absorbed by Vcc supply (which never changes). Hope this helps. \$\endgroup\$ – glen_geek Mar 1 '17 at 16:34
  • \$\begingroup\$ Today i tried to prove that mathematically and i got open loop gain of about 198 (by grounding R5), then i tried to calculate the colsed loop gain of this circuit using the general formula A/1+AB , but the output is referenced to ground so we shouldn't ignore R4 (Vout = Vr4+Vr5) so i calculated B= R4/R4+R5 (divider) and ignoring the intrinsic emitter resistance i got a closed loop gain of about 10.5 , is that right? \$\endgroup\$ – iMohaned Mar 2 '17 at 16:06
  • \$\begingroup\$ Excellent. I had used the gain approximation of A=R5/R4, which is ten. Should be eleven since it is non-inverting amplifier feedback. Since open-loop gain is less than infinite, closed loop gain will be less than eleven. A SPICE simulation shows closed-loop gain very close to 10.5 for generic transistors. \$\endgroup\$ – glen_geek Mar 2 '17 at 17:43
  • \$\begingroup\$ great :D now i can rest in peace \$\endgroup\$ – iMohaned Mar 2 '17 at 18:23

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