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Help on how to understand this circuit (BJT amplifier, feedback)

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patroclus

New Member
Hi,
Up to date, I'm familiarized with simple transistor circuits, and I know how to work with small-signal equivalent circuit, to work out input/output impedance, current/voltage gain, .... but I get intro trouble when trying to fully understand how complex circuit work.

As an exercise, I set myself the goal with this circuit (a small amplifier for faint sounds). If any of you think it is definitely not a good example to learn from, please tell me. What I really want is to learn how to face the analysis of this kind of circuits.

I attached a LtSpice schematic I just done to aid me (but simulation doesn't really match with the circuit explanation given by its author). I got the idea from this web page:
Amplified Ear - RED - Page38

As far as I know, when there're several stages, you should work output impedance from right to left, and input impedance and gain from left to right. First you get the bias, or DC op point (this is not a problem for me) and then work out the AC equivalent circuit, and replace all transistors with a equivalent (I like and always use an hybrid-pi model, **broken link removed**).

But after hours, I just get stuck with this circuit.

The first thing I don't know how to handle is that negative feedback from Q3 to the input. I do understand, from a qualitative point of view, that it turns Q1 down when the signal goes up (negative feedback), but I don't see the full image clear.
Also, When biasing a common-emitter with a base-collector feedback resistor (like Q2 and Q4), I don't know how to work out the impedances (in the hybrid-pi model, that resistor appears between base and collector, that is, from input to output, and it confuses me).
Sure I'm not focusing the problem the right way, and that is what I would like to.
Any advice/help is appreciated.
I'm not looking for the solution, but for learning how to do myself, and I promise to work hard on it :)

Thank you!
 

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audioguru

Well-Known Member
Most Helpful Member
Q3 does not provide negative feedback to Q1. Q3 is just a simple rectifier. It conducts more when the signal level is higher.
See C4? It filters the rectified audio into DC. The DC is less when there is a higher signal level and since Q1 is biased from that DC then with less DC then Q1 has less gain.

I don't think the circuit simulates properly.
 

colin55

Well-Known Member
Q3 DOES provide negative feedback to Q1 but it is not negative-feedback of the normal type.
It is not fast-acting and when we talk about negative-feedback, we consider feedback that will operate as fast as the signal is flowing through a circuit and only be about half a cycle behind the progress of the signal.
In this case the feedback produces CONSTANT VOLUME and this is ideal in a hearing aid where the gain of the circuit is a maximum when the background noise is low and it gets turned down when a loud noise is detected. This makes it comfortable for the deaf person.
 

colin55

Well-Known Member
One of the biggest problems you will have with the circuit is determining the gain of the stages when the supply rail is so low.
Transistors operating on a 1v5 supply have a much lower gain than than even a 3v, 6v, or 12v supply.
 

patroclus

New Member
Q3 purpose is clear now. Still, I'd like to know how to work out the exact values, or at least, how to start to.
There's still one more thing I don't understand very well, and it's about R8 and C5. Is it some sort of "make smooth power level little changes from battery"? When analyzing each stage, should I consider R8 in series with Rc (collector resistor from each stage)? In other words, is it like if R8 was contributing to each Rc? I'm not sure about this, because current through R8 is higher than current through each Rc.
 

colin55

Well-Known Member
When a signal is picked up by V2, it gets amplified by Q1 Q2 and Q4.
Suppose a 1mV signal is amplified 500 times. This signal will cause a current to flow through the 64R load and cause a larger current to flow when the signal is a maximum.
As you know, when a battery is required to deliver a larger current, the terminal voltage of the battery reduces slightly.
If R8 and C5 were not included, the reduced battery voltage would appear on the positive rail and pass straight to V2 via R1 and instead of the 1mV swing produced by V2 being passed to the first amplifying transistor via C1, the waveform for the power rail will be detected and amplified by the circuit.
This will cause an even-larger signal to be amplified by the circuit. This is called positive feedback and is called distortion - sometimes called "motor-boating."
To prevent the signal on the output from passing back to the input we place a resistor and capacitor across the power rails.
These two components effectively create a separate power supply for the front end and separate the battery from a new battery across the front end and prevents instability.
The two components work like this:
The 470u takes a long time to charge and discharge. It charges-up as a miniature battery via the 100R.
When the battery voltage drops say 20mV, the energy in the electrolytic is able to maintain the same voltage at the front-end and thus either none or very little of the dip in voltage appears on the power rail supplying the front end.
 

patroclus

New Member
Much clearer now, thanks.
So, if I'm right now, all collectors "see" their own collector resistor only, and that's all.

So, aprox output impedance of Q1 is 4.7k in parallel with Q2 input impedance, Q2 output impedance is 3.9k in parallel with input impedance of Q3 and Q4, and so on,...

Q1 input impedance is aprox 10k
Q3 input impedance is aprox r-pi (Vt/IB)

Am I getting this right?
For Q2, what is the effec of feedback resistor R4 on input impedance?
 
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