Hello , the circuit in the photo below taken from the manual depicts an amplifier circuit.
regarding the subcircuit depicted in red:
1.What is the logic of these Bjt structure?
2.How can I know in what region these bjt's are operating in?
Thanks.

 

the amp works as usual. but since it does not have enough power on its own, external transistors are added. so when amp draws more current, voltage drop across R3 is sensed and transistors pass through current (amplified signal). since signal is AC, two such branches exist - one por positive half period (top R3, transistors 5,6) and same thing for the negative half period (bottom R3, and transistors 8,9).since transistors are non linear and their "support" is intermittent (only for large peaks), output would be distorted. but thanks to negative feedback (R2) that is negligible, opamp compensates for this.
one problem with such high power output is that if the speaker is shorted, transistors would blow up. to prevent that, current limited is added. if current through Rx is large enough (0.6V) transistors 2 and 9 kick in and short the input of the transistor pairs (5/6 by 2, and 8/7 by 9).

 

Hello Panic mode , I am trying to analize this situation in a more basic way.
Suppose the output is reduced current is starts flowing towards -Vs(R10) which increases the voltage drop across Q8 and RE,11 so Ic8 is increased..
How can I deside regarding the Vbe of Q7 from this?
Thanks.

 

Hello Panic mode , I am trying to analize this situation in a more basic way.
Suppose the output is reduced current is starts flowing towards -Vs(R10) which increases the voltage drop across Q8 and RE,11 so Ic8 is increased..
How can I deside regarding the Vbe of Q7 from this?
Thanks.
View attachment 365070

@panic mode explained it pretty well. Just walk through what is happening as you go from small-output to large-output, to overloaded-output.

At small outputs, the voltage drop across the R3 resistors is too small to turn on the middle transistors (Q5, Q8), which in turn keep the output transistors (Q6, Q7) off. Hence, all of the output to the load comes from the opamp, which means that it must come through one of the R3 resistors (the top one for positive output and the bottom one for negative output). This results in increased voltage drop across the corresponding R3, but in this regime, not enough to turn on the middle transistors.

As the signal increases, eventually the drop across one of the R3 resistors gets large enough to start turning on the associated middle transistor, which turns on the corresponding output transistor, which then starts supplying the bulk of the current to the load. This results in a voltage drop across the associated Rx transistor, but in this regime that drop is not enough to start turning on the clamp transistors (Q2, Q9).

Eventually, the voltage drop across Rx does become sufficient to turn on the associated clamp transistor, which shunts current around its R3 resistor, which reduces the voltage drop across it. In this mode, the system will limit the output to right about the boundary of the large signal and overload regimes.

As for the Vbe of Q7, in the small signal regime, it will be less than the Vbe needed to turn it on. Depending on the transistor, this will be less that about 0.5 V to 0.7 V. In either of the other two regimes, this transistor will be on and so its Vbe will be very close to its turn-on voltage.

The boundary between the small-output and large-output regimes will be, roughly, when the voltage across R3 reaches the turn-on voltage of the middle transistors. Since these are probably not power transistors, that will probably be somewhere in the 650 mV to 700 mV range. The current limiting will occur when the voltage drop across Rx reaches the turn-on voltage of Q2, which again will not be a power transistor and so, again, will be about 650 mV to 700 mV (this is assuming we are talking about silicon BJT transistors, which is probably a good assumption).

 

Hello WBahn, I am trying to understand your words with the theory. few questions:

"At small outputs, the voltage drop across the R3 resistors is too small to turn on the middle transistors (Q5, Q8)"
- for Q5 and Q6 to be open their Vbe needs to be 0.65V how can you see that by a small output the Vbe for Q5 and Q6 will not be 0.65?

"all of the output to the load comes from the opamp, which means that it must come through one of the R3 resistors (the top one for positive output and the bottom one for negative output)"
- what is "it" what comes threw R3 ?

" Eventually, the voltage drop across Rx does become sufficient to turn on the associated clamp transistor, which shunts current around its R3 resistor, which reduces the voltage drop across it. In this mode, the system will limit the output to right about the boundary of the large signal and overload regimes."

-there is current is parralel and current in series, how does the current will flow?

 

don't skip small things. try to understand how things work and build on previous knowledge.
for example are you sure you understand how basic amp works?
are you sure you understand correlation between Vi and current draw of the amp?
the larger Vi amplitude, the more current is drawn but the Amp.



to measure current draw, you can insert ampmeter. can you guess where the good place would be?

yes you can put one in series with V+, or V- or both...

do you understand what is the equivalent circuit of the ampmeter? it is a low value resistor. and voltage drop across it is proportional to current.

now add transistor to respond to that voltage drop. for values smaller than some 600-650mV, transistor response is almost non-existent. but if current exceeds that threshold, suddenly a much larger current can be observed. so transistor does not amplify everything, only peaks that are above certain level.
the output signal is jig-saw style composite of signal amplified by amp, top transistor circuit, and bottom transistor circuit. but negative feedback compensates for that. for small signal that does not trigger transistors, speaker get signal only from Amp itself, through unmarked resistor.
btw, if you want help with circuit, please make sure everything in circuit is marked correctly (and no duplicate labels).

 

Hello Panic mode, I did what you asked I know that when output goes up its because the current is flowing from postitive rail to the output.
When output voltage decreaing or going negative its beucase we have current flow and discharge form the output to the negative power rail.
Also I simulated what you asked and saw the current consumption rises as the power rail needs to work harder.
If I have done it correctly, could you give the next step I need to understand of seeing the dynamics?
Thanks.

 

Ups and downs - how the positive part of the input audio signal becomes a positive voltage across the load.

You have one unmarked part, the resistor between the opamp output and the load. Call this Rout.

Input signal goes up.
Opamp output goes up.
Current through Rout increases.

That current comes from the opamp positive power rail pin. The current into that pin comes from two places: R3(1) and Q5's base. This base current causes Q5 to conduct, which pulls up the Q6 base. Q6 is an emitter-follower, so its emitter goes up and the voltage across the load goes up.

What about Q2? Q2 is there to protect the output transistor Q6 from a short-circuit across the load. Rx4 is a current-sensing shunt resistor, probably a very low value. As the voltage across the load increases, so does the current through R4X, increasing the voltage drop across it. That voltage is across the Q2 base-emitter junction. When this voltage reaches 0.6 V, Q2 begins to conduct - Its collector begins to pull up. This pulls up the Q5 base, decreasing that transistor's conduction. This decreases the current through the Q5 base, beginning to turn it off. This decreases the drive to Q6, reducing its collector-emitter current. This circuit loop causes the output stage to go into constant-current limiting when the output current tries to exceed a certain value.

All of this is true for the negative part of the audio signal, but the currents and voltages are reversed.

ak

 

you are simulating using just an OpAmp (very low current), which is also without load connected to its output. so current you are sensing is very small (mA). and since your resistors are only 1 Ohm, voltage drop across them will be in mV. you may want to bump those R3,R4 values to perhaps 100 Ohm. then you will get something usable and matching the original post.

then it should quickly become apparent that adding Q1 does mean business - BJT output current is significantly larger than what OpAmp could do on its own.
if that output current was fed to a speaker (Rload) you would see massive power gain... at least on positive half wave (for now).



suppose we add the other side:


notice that there is a discontinuity at zero. that is expected because both transistors only turn on when OpAmp current exceeds some threshold. for an audio amp that would be unacceptable distortion...

so we combine the three currents (R3,R5,R6) an we see something like this. note that now current through R5 becomes distorted too. and it has to be because the load current is matching input signal (thanks to feedback).


just to be sure, compare input and output. they are out of phase because you opted to connect signal to inverting input

 

you will notice that reducing load impedance (for example 4 Ohm) will result in clipping. not surprising since there are pretty large values (R3.R4,R6,R7). when load is low impedance, driver need to be lower impedance too. one obvious option is to add another pair of even bigger transistors (push more current through the load).

 

you are simulating using just an OpAmp (very low current), which is also without load connected to its output. so current you are sensing is very small (mA). and since your resistors are only 1 Ohm, voltage drop across them will be in mV. you may want to bump those R3,R4 values to perhaps 100 Ohm. then you will get something usable and matching the original post.

then it should quickly become apparent that adding Q1 does mean business - BJT output current is significantly larger than what OpAmp could do on its own.
if that output current was fed to a speaker (Rload) you would see massive power gain... at least on positive half wave (for now).

View attachment 365102

suppose we add the other side:
View attachment 365105

notice that there is a discontinuity at zero. that is expected because both transistors only turn on when OpAmp current exceeds some threshold. for an audio amp that would be unacceptable distortion...

so we combine the three currents (R3,R5,R6) an we see something like this. note that now current through R5 becomes distorted too. and it has to be because the load current is matching input signal (thanks to feedback).
View attachment 365106

just to be sure, compare input and output. they are out of phase because you opted to connect signal to inverting input
View attachment 365107

This is a masterful progression of ideas- really helps me understand this circuit - one I have never seen before.
Thank you!

 

this was very common in audio amplifiers using ICs. hanging couple of transistors onto existing amp allowed for more power than IC itself could handle. the idea is simple - the more volume , the higher the current draw, once you reach some reasonable limit (say 50% of power that IC could handle on its own), let the transistors kick in. they do so by measuring current draw of the IC. they are only boost for high power (high volume), at low volume all output is done by IC alone. today audio amps are using class D and can can reach higher output power level than analog ICs could.

note, this is not an OpAmp. this is an actual audio amplifier IC, capable of tens of Watts of output. as a result current draw is much larger so current sensing "shunt resistors" are much lower value. pair of 1.7 ohms in parallel is 0.7 Ohm. so at 700mV for Vbe current draw of the amp is about 1A. when the Amp draws 1A (or more) transistors kick in.
in my previous post OpAmps was used and current was in the mA range. hence those resistors (R1,R2) were chosen to be higher value so example would work. R5 was just a copy or R1, only used to easily identify current from the opamp output. otherwise, it is not really needed.


if one only compares currents R5 and Rload it is easy to see why R5 current is distorted.
portion between blue lines is entirely due to Amp... but this works only in narrow range, amp alone would never reach peaks shown in red (Iload).
but once the transistor do their part, OpAmp current is only a small portion of the output.. so the peaks of current through R5 are much lower.
if the mid section was removed and


if that center band is removed, red curve looks a lot more like transistor current (not exactly the same since the green curve is not subtracted).
it does show the discontinuities near zero.