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There is the power amplifier circuit here:

enter image description here

  1. How 18k/1k/.0012uF filter works in the input circuit?

  2. It's clear for me that Q1/Q2 is a transconductance stage: voltage in - current out. But where is that "current" (that has been amplified and moving through Q9-Q8-Q7) are transformed back to voltage (to form a transimpedance stage)? How that transimpedance stage can be identified in that amplifier where there is no resistor in collector of voltage amplifier stage because of current source is used instead of that resistor?

  3. Why 100 pF capacitor used at cascode? Is it not enough to use cascode to compensate Miller effect?

  4. What is the purpose of 100 Ohm resistors at Q11/Q12 bases?

  5. Why two different voltage sources are used for pre- and out- stages? And what is disatvantage if I will use one?

MaxMil
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3 Answers3

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1 - 18kΩ resistor together with 2.2µF capacitor form a high pass filter. The Fc frequency is equal \$F_C\approx\frac{0.16}{RC} = 4\textrm{Hz}\$

Additional 1kΩ resistor form a low pass filter together with 1.2nF

\$F_C\approx\frac{0.16}{RC} = 134\textrm{kHz}\$. And this filter stops any RF signal, so the RF signal is not being amplified by the amp.

2 - In the VSA stage the current is transformed back to voltage. Because due to large Miller (100 pF) capacitance. The VAS stage work as a transimpedance amplifier (current-to-voltage converter).

3 - Because we want a dominant pole compensation (big Miller capacitance).
And in cascode stage Q7 does not provide any voltage gain (no voltage gain no Miller effect). This is why we put a 100 pF capacitor at the Q8 collector.

4 - These resistors separate the output stage (emitter follower) from the VAS collector. Because sometimes the output stage can cause HF oscillation. And we can stop this oscillation from happening by adding a base stopper resistor.

So the VAS will always see a "real" resistance (emitter follower can have a negative input resistance).

http://www.ti.com/lit/an/snoa737/snoa737.pdf

5 - The voltage headroom is the answer.
The voltage swing at the VAS stage (Q7, Q9) is much lower than this +/-55V. So to be able fully "diver" the output stage. The input stage + the VAS stage is supplied from the higher supply voltage.

G36
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  • Is it correct to think that in VSA stage the current is transfomed back to voltage during drop of that current on all resistors from VSA stage output to the amplifer output load which the current see on his path? – MaxMil Nov 21 '17 at 17:54
  • Where I can see that "emitter follower can have a negative input resistance" (book or link if you know)? – MaxMil Nov 21 '17 at 17:57
  • @MaxMil I think he's talking about the effect of the output stage's capacitive load. Note that the output of the VAS *(not VSA, it's **V**oltage **A**mplification **S**tage and, as stated in the awesome answer, it transforms current to voltage)* is directed to the input of the output stage. As you might know, speaker is a reactive load. So the capacitive part of the output stage's load (i.e. speaker) can cause the input **conductance** of the stage to be **negative**. But this might occur if the capacitance is high enough (say, in nano-Farads range). – Rohat Kılıç Nov 21 '17 at 18:08
  • @MaxMil This is not the easiest subject to understand. But you can try to read about this here: http://audioworkshop.org/downloads/AMPLIFIERS_OSCILLATION_BJT_CIRCUITS.pdf Or you could accept this simplified explanation. At higher frequencies the BJT's beta start to drops. So, the emitter follower output resistance increases his value ( ro = rs/(beta +1) ). The rising output resistance in some frequency region acts like an inductor and can form a resonant circuit with load capacitance. – G36 Nov 21 '17 at 21:53
  • As for the VAS. At first you can try to treat the VAS stage as CE amplifier with active load (Q9). CE stage is a inverting amplifier. So we can treat it as a inverting op-amp. Hence by adding big external Miller capacitance we create Miller integrator. Some kind of a voltage to current converter. http://uploads.tapatalk-cdn.com/20170108/a45ae07b23c9b7d319b76d29449a5ab3.png and this http://www.ecircuitcenter.com/Circuits_Audio_Amp/Miller_Integrator/Miller_Integrator.htm – G36 Nov 21 '17 at 22:04
  • @G36 Ok about VAS. When I speak about active load I suppose that the output of the stage not a voltage - it's a current! But that current must be converted back to voltage if it's a voltage amplifier stage. Where that conversion is take a place? Or it's not necessary to convert VAS input current back to voltage immediately on output and it may be done on the next stage. Or if speak in terms of transconductance it's no difference where this convestion actually take a place I'll always think that voltage in - current out. – MaxMil Nov 22 '17 at 07:09
  • @MaxMil The BJT alone is a voltage (Vbe) controlled current source (Ic) and the gain is Iout/Vin = gm = transconductance. And this output current (Ic) is converted back to the voltage in the collector resistance Rc. In this audio amp CE stage is loaded by Q9 ro and the load resistance (beta*bete*RL). And the input stage output current is converter into the voltage via Miller capacitor. – G36 Nov 22 '17 at 14:54
  • And for low frequency Q7 and his input resistance (47 + beta*(re7)) is the load for the input diff stage. – G36 Nov 22 '17 at 16:13
  • @G36 If I add two additional BJT's to Q1/Q2 (equal to Q1/Q2) to form the darlington input stage to increase the input resistance of that amplifier is it necessary to change surrounding elements values or it willn't have infuence to circuit behaviour excluding increasing the input resistance? – MaxMil Nov 23 '17 at 09:45
  • @MaxMil First you need to ask yourself a question why you want to increase the input stage resistance. In the amplifier, the input resistance is set 18k ohm's resistor Rin = 18k. Because the Q1 and Q2 input resistance is incense by the negative feedback loop gain (1 + Aol*1.2/(1.2+18)). Also if you add a Darlington pair at the input the gain of an input stage decreases by the factor of 2. So you still want to add a Darlington pair? – G36 Nov 23 '17 at 15:21
  • @G36 Not fully understand. I found from your comment that input resistance will be determined by 18k Ohm resistor which top junction is connected to 2.2uF input capacitor. And the input resistance of that amplifier is determined by feedback loop resistance. But can't understand why with using Darlington the input resistance will be decreased by factor of 2. I think that double BE-resistance with using Darlington will form more resistance from emitter to base and will shunt less the surrrounding input circuit. – MaxMil Nov 24 '17 at 07:43
  • @MaxMil The input stage gain will drop by a factor of 2. Not the input resistance. In your audio amplifier, we have a negative feedback loop (via 18k and 1.2k and Q2 base) and this type of a feedback is we call Voltage Series Feedback ( series- series). And ins short this type of a feedback inverses the input resistance seen from the Q1 base into the base terminal. This is why the input resistance in the audio frequency band is set mainly by external resistor resistance (18k in this case) because Q1 input resistance is much higher this 18k. – G36 Nov 24 '17 at 16:14
  • @G36 Ok, I wrote with a mistake, of course the "gain", not the resistance. But why it will drop? It seems the input stage gain in very easy approximation equal to Rc/re and if the "re" will be doubles then the gain must drop by factor of 2. Is it so about what you speak? And give me please the link to any material where I can found the derivation of input/output resistance equations for non-inverting operation amplifier. For example Sedra Smith doesn't show that equations. – MaxMil Nov 24 '17 at 17:10
  • As for the Darlington gain. We simply have two Vbe, so the input voltage will split between this two BJT's try it in simulation and you will see. As for the feedback https://electronics.stackexchange.com/questions/177007/input-impedance-of-a-non-inverting-op-amp?rq=1 and this https://electronics.stackexchange.com/questions/297409/confusion-with-output-impedance-of-non-inverting-and-inverting-opamp-circuit/298428#298428 and this https://masteringelectronicsdesign.com/the-non-inverting-amplifier-output-resistance/ – G36 Nov 24 '17 at 18:28
  • @G36 It seems that the resistor at the base of Q8 equal to 8.2k is used to limit the Q8 base current. What value it must have, how much the Q8 base current must be? I know that as a part of cascode the Q8 allow to increase output cascode resistance to not allow next stage to shunt the Q5/Q6 current mirror and decrease the input stage gain significantly. – MaxMil Nov 25 '17 at 17:33
  • @MaxMil The Q8 collector current is "set" by Q9 current source circuit. Ic9 = 0.7V/47Ω = 15mA. So the Q8 base current is around Ic8/β = 15mA/50 = 0.3mA. But the main job for this 8.2k resistor is to provide a current for the Zener diode. And typical value for Zener diode current is Izk = 5mA hence R < (55V-3.3V)/(5mA + 0.3mA) = 10k. As you can see we do not care much about Q8 base current. – G36 Nov 25 '17 at 18:02
  • @G36 Is it necessary to fix the Q8 base voltage with 3.3V zener or it can be another voltage drop like 5V? – MaxMil Nov 26 '17 at 08:17
  • @MaxMil The higher the Q8 base voltage the lees the headroom you have for a voltage swing at the collector. But for sure you can use the 5V Zener diode. You set the voltage at Q8 base so that Q7 is not in saturation. – G36 Nov 26 '17 at 08:24
  • @G36 In this situation the Q7 emitter will be at -55+(15ma*47Ohm) = -54.295V and collector at -55+3.3-0.7 = -52.4V, so the swing will be less than 2V, is it enough to drive next stages? – MaxMil Nov 26 '17 at 15:03
  • @G36 Why it's necessary to use Zener here to fix the Q8 base voltage? Is there any BJT schematic may be used here? – MaxMil Nov 26 '17 at 15:10
  • @MaxMil Here you can read about cascode https://www.passlabs.com/sites/default/files/cascode.pdf As for the Zener at the Q8 base. It is desirable to have constant voltage at the Q8 base. And Zener diode does this just fine. Also, Zener diode voltage does not change much with supply voltage which is very good. Voltage divider does not provide this kind of a "stiffness". – G36 Nov 26 '17 at 15:58
  • @G36 Why the schematic input is grounded through 10 Ohm resistor (is the voltage drop on it have any meaning?), why not directly? – MaxMil Feb 26 '18 at 17:37
  • This resistor separates the "signal GND" (low current) form the "power GND" (high current). – G36 Feb 26 '18 at 19:36
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I'll just write out the first few answers I was working on, before seeing G36's response (which you should definitely read.)

  1. The leftmost two form a high pass filter with about \$4\:\textrm{Hz}\$ as the corner frequency. The other two form a low pass filter with more than \$100\:\textrm{kHz}\$ as the corner frequency. Together, I'd have to play with how the low pass filter loads the high pass filter to work out the actual bandpass (and I'm not going to bother separating the damping part (\$\sigma\$) from the oscillatory part (\$\omega_d\$) for now.) Just call it a "band pass filter" that performs DC blocking and elimination of higher frequencies (such as RF, but also lower in frequency due to unwanted oscillations in whatever is driving the amplifier) that might otherwise just waste power without contributing to the desired output.
  2. \$Q_1\$ and \$Q_2\$ are a "long-tailed pair BJT differential amplifier." It needs a current source at the shared emitters and the output is taken from one or both of the collectors. In this case there is a current mirror, \$Q_5\$ and \$Q_6\$, in the collector pair, which if a perfect mirror would have infinite impedance in both legs of the diff-amp and, without some added path, would fight any diff-amp pair attempts to direct current from the source between its legs. Luckily, there is a connection to \$Q_7\$, so the base of \$Q_7\$ receives the differential current, directly. The mirror will mirror the smaller of the two diff-amp leg currents with the remainder going to \$Q_7\$ (which is part of the VAS.)
  3. The pole compensation capacitor (the Miller capacitor of (\$100\:\textrm{pF}\$)) reduces the VAS output impedance, so that the non-linear load caused by the output stage generates less distortion.
  4. Damps oscillation. BJTs are known for oscillation when their bases are driven directly by a good voltage source. Placing some small resistance in the base helps damp out this problem. You will see this technique used often. You could remove them and see if it isn't a problem, but including them is usually a good idea.

Some things I also wanted to add are:

\$Q_3\$ and \$Q_4\$ and surrounding parts help create the current source needed by the diff-amp pair. In addition, there is a fairly reliable base voltage of \$Q_3\$, which when coupled with \$Q_9\$ as an emitter follower can create a semi-reliable current source needed for the \$V_{BE}\$ multiplier (\$Q_{10}\$ and surrounding parts) and the VAS section, in order to control the output stage.

There is very little signal voltage at \$Q_7\$, due to its current input, and therefore very little on the first stage output. (This minimizes Miller phase shift and possible Early effect.) \$Q_8\$ is a cascoded BJT, whose base voltage is held steady by the zener/resistor pair and \$Q_7\$ is the VAS BJT.

jonk
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  • I can't fully understand "would fight any diff-amp pair attempts to direct current from the source between its legs" in "2" – MaxMil Nov 21 '17 at 18:01
  • Is it wd is equal to filter "resonant frequency"? – MaxMil Nov 21 '17 at 18:36
  • @MaxMil Not understanding everything is fine. It's a process. One thing at a time. But the mirror -- mirrors. It's not going to have one current in one leg that is different than in the other. Meanwhile, the diff-amp *demands* that the leg currents are different, if the inputs are different. So the upshot is that you need a path where the difference current can "squirt." That's all I meant. – jonk Nov 21 '17 at 21:03
  • May you clarify what about 4x1.5 Ohm resistors purpose. It's clear that total resistance will be 1.5/4Ohm, but what about power of each resistor. If I'll use one resistor instead 4 it must be 4W power? – MaxMil Nov 23 '17 at 07:08
  • @MaxMil Those are used because different BJTs in each pairing will have different behaviors, even if drawn from the exact same lot number. They provide some negative feedback to help distribute the current loading equally, despite individual and different behaviors. It's easy to work out the appropriate values. But now you are asking yet another question and you have already selected an answer here. – jonk Nov 23 '17 at 07:27
  • Strongly agree with you. Thanks for the answers. They as always are very useful! – MaxMil Nov 23 '17 at 09:47
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And a free suggestion about this LD low-distortion amplifier. The collector voltage of Q1 varies dramatically. The thermal dissipation of Q1 will vary also dramatically.

Assume the die size of this discrete transistor is 1mm by 1mm area, with thickness of 0.3 mm (the default wafer thickness in some foundaries. As the heat varies, the thermal timeconstant leads to thermal distortion; the tau of 1mm is 11.4 milliSeconds (the tau of 1meter is 11,400 seconds) and thermal tau varies as the square of the size change.

This thermal distortion modulates ALL the other tones.

Your power dissipation in the diffpair is about 200 milliwatts, or 4mA * 50volts. Or 100mW per transistor. Expect serious overshoot or undershoot on leading edges of squarewaves into this amplifier. Model it in SPICE, using voltage-controlled-voltage-sources to feedback the collecto dissipation into the base voltage.

Cure? Use PNP cascodes on the diffpair, with bases approx. -5 volts.

Notice Q7 has a cascade, to minimize the change in Collector voltage along with the unavoidable large changes in collector current. This is to minimize distortion from the self-heating of Q7 EB junction.

The Q1 has the same issue, because during times when the feedback loop is not controlling the output, the linearity of Q1 is very important.

analogsystemsrf
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  • Why there are too many schematics that prefer PNP input stages instead of NPN? There are too many texts where it can be seen that PNP BJT's have a worse frequency characteristics in comparing with NPN BJT's. – MaxMil Nov 22 '17 at 07:12
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    @MaxMil In the old days the PNP transistor was more popular than the NPN. Also only good "high voltage" BJT' for the VAS stage exist in NPN type. So to eliminate the voltage stacking (NPN stage + NPN stage) the input stage needs to be a PNP type. And the PNP BJT's are usually have lower base bulk resistance (lower noise). But in full complementary input stage you will need both transistor polarities. – G36 Nov 22 '17 at 15:04
  • @analogsystemsrf May you clarify what is "voltage stacking (NPN stage + NPN stage)"? – MaxMil Nov 22 '17 at 15:48
  • @MaxMil As for the "voltage staking" DC voltage shifting. Think of two DC coupled NPN transistor stages. The collector node of a first transistor is directly coupled to the base of a second transistor. This will result in voltage swing reduction due to DC "voltage staking". But if we replace the second stage with PNP transistor we can have the collector of a PNP transistor at a lower voltage. Do you understand this? Or you need a further explanation? – G36 Nov 23 '17 at 15:44
  • @G36 Yes, Thanks! I was found a good illustration of that concept and your explanation is a good addition for that. – MaxMil Nov 24 '17 at 07:27