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I have been recently studying about the transistor RC phase shift Oscillator. I calculated the attenuation provided by the RC network to be equal to 29. But when we use a transistor to make the circuit the attenuation will not be equal to 29 because the input impedance of transistor is not very high. Correct me if i am wrong. I have been following this website. It mentions that the the gain of the transistor to be around 56 which is way off from 29 but it doesn't tell how to derive the formula. I have searched various but i couldn't find any source which explains it in detail.

  • Since i am very new to electronics this is very confusing to me. Can you please tell me how to derive the formula.
shahrOZe
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  • The number 29 comes from \$\mid H \left(j\:\omega_{_0}\right)\mid\approx\frac1{29}\$ where \$\omega_{_0}\$ is the solution to the denominator when the imaginary part is set to zero. You can do all this work on your own if you know how to set up the \$\tau=RC\$ low-pass (\$\omega_{_0}=\frac{\sqrt{6}}{\tau}\$) or RC high-pass (\$\omega_{_0}=\frac{1}{\tau\:\sqrt{6}}\$) passive filter transfer function for three stages. All this assumes that while successive passive stages load earlier stages, the amplifying stage does not add further loading. Opamp qualifies. BJT does not. So more is needed then. – jonk Mar 24 '21 at 04:24
  • See [here](https://electronics.stackexchange.com/a/371292/38098) for details on the low-pass form and using an opamp. That will tell you about your factor of 29, in more detail. It will also discuss computing \$\omega_{_0}\$. (I did not do a similar page for high pass, yet.) You can work out your own gain requirements if you include the BJT load into your transfer function. The value will then be different. For a BJT version that will run with reasonable values, see the very last schematic [here](https://electronics.stackexchange.com/a/337900/38098). (Your web page may be wrong about it.) – jonk Mar 24 '21 at 04:34
  • @jonk can you please give me some details how to calculate the BJT load. Since there the RC network connects back to the transistor itself (loop), I don't know how to approach the problem. – shahrOZe Mar 24 '21 at 04:43
  • Almost any BJT CE stage discussion will show you how. But [here](https://electronics.stackexchange.com/a/536779/38098) is an example by Andy. There's more to it if you insist on getting the same results that a simulator would provide. But for practical needs that's probably fine. However, you have to know how to properly apply it to the transfer function, too. Andy doesn't discuss that detail. – jonk Mar 24 '21 at 04:52
  • Finally, this is ***not*** a subject for people who are *very new to electronics*, unless you happen to be one of the very rare super-genius types. If so, I'm glad to meet you! Otherwise, plan some significant study time and make sure that you are relatively comfortable with complex numbers, Euler's, multiplication on the complex plane, etc. – jonk Mar 24 '21 at 04:56
  • @jonk i am actually familiar with complex numbers and all that stuff and had even completed a course in complex analysis. Its just that sometimes i get too confused how to approach such problems and with what methods. Thanks for you guidance – shahrOZe Mar 24 '21 at 05:01
  • Well, that's very good to hear. So you aren't new to the topics at hand. Just electronics. You probably should have said as much when writing out the question. I was totally blind as to what you might need or be able to absorb, so had no real idea what to say at first. From what you say, then I think you have enough to get moving. So long as you are aware that parts are NOT mathematically exact devices. Resistors are usually 2% or 1%, plus or minus, and capacitors are usually lots worse. Everything varies with temperature and drifts over time, too. And BJTs are yet another story. – jonk Mar 24 '21 at 05:05
  • Don't expect mathematical rigor from a realizable circuit. Most of what an electronics engineer does is about placing all of the vagaries of part variations, temperature variations, initial accuracy guarantees that drift more or less rapidly over time, vibration, interference, and general manufacturing issues under some kind of "management" (where it matters, anyway.) And a lot less about computing exact numbers from closed equations (which are essentially impossible to achieve.) It's numerical analysis to a larger degree, though the basic physics and math is also important as a guide. – jonk Mar 24 '21 at 05:09
  • @jonk I should have said "new" rather tha "very new". Anyways, I went through the answer of input impedance but notice that site which i have posted in my question has the rc network connected to the collector. Does this add to the input impedance? Also the site mention that hfe should depend on Rc. I don't know how these terms will end up in the calculation of the input impedance. Also it says that the gain should be around 56. Even a rough calculation about it will be very helpful – shahrOZe Mar 24 '21 at 05:12
  • The voltage or current gain depends on lots of things. You need to be able, on your own, to work out these values given any particular circuit configuration. You have to be able to navigate them. There is no such thing as a "stock formula" that applies everywhere. No universal answer. That's why you need to use your brain (if all you have is your finger and some sand to draw in) or a simulator (if you want to be mentally lazy.) When you looked at the last BJT circuit in the link I mentioned, how would you compute the gain? (It's not a match for most web site answers.) Why does it work at all? – jonk Mar 24 '21 at 05:16
  • The collector resistor in that example represents the *output impedance* of that circuit. And you'll have to incorporate that, as well. (I forgot to mention it.) The *input impedance* has to do with the biasing resistors, the emitter resistor and expected \$h_\text{FE}\$ at the operating point, temperature, and details of the particular BJT. (And capacitances, where explicit, plus trace parasitic capacitances, as well.) For example, your web site case includes an emitter capacitor that directly bypasses the emitter resistor. This is for high gain. But it is huge and must be accounted. – jonk Mar 24 '21 at 05:20
  • Unfortunately, I want to avoid writing a long discourse tonight on this topic. You have been given a lot to advance where you were, earlier. I don't think I'll have time in the next few days for more than I've given tonight. If you had written a lot more in your question, this would have allowed me to write a lot less. But your question is far far too short and leaves open far far too much ground to cover. Perhaps you could consider investing a lot more time in writing and refining the details of where you are actually stuck, with work product in view. That may help motivate me more. – jonk Mar 24 '21 at 05:26
  • @jonk got it. I will edit the question with my approach and effort to the problem later. Your comments have given me a head start and basic ideas. I just need more time to work on the problem. Thanks for guiding me. – shahrOZe Mar 24 '21 at 05:29
  • It was my pleasure. I'm glad (and lucky) to meet anyone who shows such a strong interest in mastering details. You have my better wishes! – jonk Mar 24 '21 at 05:32

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Shahroze Shabab, here is my short advice:

I assume in the feedback path you are using the C-R highpass chain (3 elements), right? You know that the required gain is -29 (in reality you need something more - perhaps 30..31).

Your problem is the influence of the finite input impedance r_in at the base. Hence, the "last" resistor in the chain is R||(r_in). Why not simply remove the last ohmic resistor R ? In this case, the input resistance r_in alone could "do the job". It should not be a big problem to make a good guess for h11=hie (input resistance at the base node) and you can use the voltage divider resistors to adjust the parallel combination to get the desired value of r_in.

Of course, a redesign of the whole C-R feedback chain would be necessary because ALL resistors should have the value of r_in. I already have built such a circuit with good success.

LvW
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  • but the sight says that the gain should be around 56? Can you explain this part. – shahrOZe Mar 24 '21 at 09:01
  • Do not mix current gain (hfe) with voltage gain. They speak about the required current gain hfe=56. In general: Do not blindly trust internet contributions. In the refrenced article both opamp based examples are wrong They do not work. – LvW Mar 24 '21 at 09:46
  • Ok now i got it. Thanks – shahrOZe Mar 24 '21 at 10:05