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I stumbled across this circuit to generate sin wave using the RC Phase shift in this texas instruments paper https://www.ti.com/lit/an/snoa665c/snoa665c.pdf?ts=1729278724739&ref_url=https%253A%252F%252Fwww.google.com%252F

I Understand the concept of the phase shift to generate sin wave and why we need three High Pass Filters so the phase shift is equal to zero and the concept of unity gain my problem with this circuit is that I see 1 Low Pass Filter and and 2 High Pass filters or you can correct me if I am wrong and the capacitor at the top where's its resistance?

Also what are the operation modes of these transistors they all seem to be in saturation So What's the Point of having them?

enter image description here

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  • \$\begingroup\$ Have you tried simulating the circuit to gain insight. That would be top of my list. \$\endgroup\$
    – Andy aka
    Commented Oct 19 at 8:50
  • \$\begingroup\$ Hassan, start all analysis by setting down and redrawing the schematic. (It looks old.) For example, what is the 2k resistor doing with the lm313? What does this mean for q3's emitter? What does that imply for the feedback to its base and by implication to its collector current being applied against the 27k resistor? That should tell something about what must then happen to q1's emitter. Start with basic identifiable bits and hold them firmly in mind as you move around. \$\endgroup\$
    – periblepsis
    Commented Oct 19 at 10:32
  • \$\begingroup\$ hey easy guys, i did all that \$\endgroup\$
    – Hassan Mehriz
    Commented Oct 19 at 14:49
  • \$\begingroup\$ @HassanMehriz I don't see a redrawn schematic. That schematic is very poorly drawn. \$\endgroup\$
    – periblepsis
    Commented Oct 19 at 19:03

1 Answer 1

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Here's the schematic, redrawn in one fashion:

schematic

simulate this circuit – Schematic created using CircuitLab

What makes this more useful is that it is now much clearer how the RC phase shift passive staging has been designed. You can see that \$C_3\$, \$C_4\$, and \$C_5\$, plus \$R_6\$ and \$R_7\$ are obvious parts. The only remaining non-obvious part is \$R_4\$. Given that \$\beta\approx 200\$ for \$Q_2\$ then the input impedance for \$Q_2\$ will also be approximately \$\approx 200\cdot 10\:\Omega=2\:\text{k}\Omega\$. So that is your "third resistor" in the passive part of the 3-stage RC phase shifter.

It's also obvious now that \$Q_3\$'s emitter is being held tight at about \$1.2\:\text{V}\$ by the temperature compensated zener (now very obsolete.) So \$Q_3\$ is being used as a comparator, with its base being fed by the RC filtered peak positive-going pulses at the output of the LM386. (\$D_2\$ only feeds over the positive half-cycle into that filtering.)

The \$Q_3\$ behavior is highly non-linear (exponential), as a \$60\:\text{mV}\$ change will cause a \$10\times\$ change in its collector current. That highly non-linear collector current will work against \$R_2\$ to develop a rapidly altering voltage at the base of \$Q_1\$, whose emitter will follow as a low-output-impedance to form the positive side voltage for \$R_3\$.

All that this amounts to is that there is a very narrow range of allowable voltage at the \$R_9\$/\$C_7\$ filter over which this circuit operates. Given the idea that everything interesting happens within a factor of 10 and given that \$Q_2\$'s collector current varies by \$10\times\$ with only \$60\:\text{mV}\$ change at \$Q_3\$'s base, this means that for all intents and purposes the operating point of this system will be held within about \$\pm 30\:\text{mV}\$ around whatever operating point this circuit ends up at. (Likely with the base of \$Q_3\$ about \$1\:\text{V}\$ above the LM313 reference voltage, at a guess.)

It is for this reason that the comments, in what used to be Nat-Semi's AN-263, talk about how well this operates with changes in the supply rail voltage. In essence, the feedback will automatically find a valid operating point. It's actually pretty nice. And all this using only NPN transistors and the same part number, too.

Once you re-draw a schematic, a lot of things become much clearer.

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    \$\begingroup\$ Yeah it’s much clearer now, as i said i’m new at all this and i couldn’t redraw it like this thanks a lot \$\endgroup\$
    – Hassan Mehriz
    Commented Oct 21 at 2:12
  • \$\begingroup\$ Can I ask you one last question, I don’t understand the rationship between β and Re why you multiplied them. Know that Ie= β Ib \$\endgroup\$
    – Hassan Mehriz
    Commented Oct 21 at 5:29
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    \$\begingroup\$ @HassanMehriz Imagine a very tiny signal change at the base; call it \$\Delta v\$. Assuming it is very small then \$V_{_\text{BE}}\$ doesn't change. Which means the same \$\Delta v\$ change also appears at the emitter. This invokes a change in current in the emitter resistor of \$\Delta i_e=\frac{\Delta v}{R_{_\text{E}}}\$. But the base only requires a current that is \$\Delta i_b=\frac{\Delta i_e}{\beta+1}=\frac{\Delta v}{\left(\beta+1\right)R_{_\text{E}}}\$. So the signal source at the base sees an impedance of \$\frac{\Delta v}{\Delta i_b}=\left(\beta+1\right)R_{_\text{E}}\$. \$\endgroup\$
    – periblepsis
    Commented Oct 21 at 5:48
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    \$\begingroup\$ @HassanMehriz And that's why I multipled the emitter resistor by \$\beta\$. (Ignoring the +1 for now.) There is also another dynamic impedance that is added to this called \$r_\pi\$. And I glossed over that, despite it being significant here, for simplicity's sake. \$\endgroup\$
    – periblepsis
    Commented Oct 21 at 5:49

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