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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.
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  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Mar 24, 2021 at 4:24
  • \$\begingroup\$ See here 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. (Your web page may be wrong about it.) \$\endgroup\$
    – jonk
    Mar 24, 2021 at 4:34
  • \$\begingroup\$ @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. \$\endgroup\$
    – shahrOZe
    Mar 24, 2021 at 4:43
  • \$\begingroup\$ Almost any BJT CE stage discussion will show you how. But here 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. \$\endgroup\$
    – jonk
    Mar 24, 2021 at 4:52
  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Mar 24, 2021 at 4:56

1 Answer 1

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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.

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  • \$\begingroup\$ but the sight says that the gain should be around 56? Can you explain this part. \$\endgroup\$
    – shahrOZe
    Mar 24, 2021 at 9:01
  • \$\begingroup\$ 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. \$\endgroup\$
    – LvW
    Mar 24, 2021 at 9:46
  • \$\begingroup\$ Ok now i got it. Thanks \$\endgroup\$
    – shahrOZe
    Mar 24, 2021 at 10:05

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