I have designed an RC phase shift oscillator and it works in LT spice but i want to know if the analysis i have did on this circuit is correct or not. I want to calculate the "gain" as well as the "input impedance". Please guide me where i am wrong because i am a beginner in electronics. It oscillates at around 500Hz. enter image description here

  • AC analysis

\$|Z_{in}| = |(X_{C1}||R3)| \approx 315\Omega \$

\$|Z_{in_{base}}| = |(R1||R2)||\beta*Z_{in}| \approx 13K\Omega\$

\$|Z_{out}|= |Z_{RCnetwork}||R4| \approx 8K \Omega\$

\$\text{Voltage Gain} = Z_{out}/Z_{in} \approx 25.4\$

  • DC analysis

\$ V_{b} = 1.62V \$ \$ I_e \approx 0.4mA \$ \$ V_c \approx 5V \$

My questions are -

  • Why is there sustained oscillations even when the voltage gain is around 25. Shouldn't the voltage gain be around 29. Where am i wrong in my analysis.Is my AC analysis correct?
  • Is my Dc analysis of the circuit correct since i want the signal to swing around the midpoint i.e 4.5V
  • \$\begingroup\$ You haven't stated what frequency it oscillates at. \$\endgroup\$
    – Andy aka
    Commented Mar 25, 2021 at 9:13
  • \$\begingroup\$ It oscillates at around 500Hz. Sorry i forgot to write it. \$\endgroup\$
    – shahrOZe
    Commented Mar 25, 2021 at 9:46
  • \$\begingroup\$ @shahroze shahab, your gain formula is wrong. The gain can be approximated using Zout/RE with emitter resistor R3. However, in your case R3 is bypasssed with a C and there is no signal feedback. Therefore, you must use the correct formula gain Av=-Zout*gm (gm=transconductance=Ic/26mV) \$\endgroup\$
    – LvW
    Commented Mar 25, 2021 at 10:22
  • \$\begingroup\$ @Lvw But the oscillations are low at around 500 hz. So the reactance of the capacitor C1 is finite. Shouldn't we take the combined impedance with R3 into account?? \$\endgroup\$
    – shahrOZe
    Commented Mar 25, 2021 at 10:26
  • \$\begingroup\$ Oh yes, you are right - for 500 Hz the capacitor C1 has a finite impedance of app. 320 ohms. So your calaculation is OK - I was confused because you wrote "Zin". Therefore, it was not clear to me that you spoke about the emitter leg. Sorry. \$\endgroup\$
    – LvW
    Commented Mar 25, 2021 at 10:47

1 Answer 1


Please guide me where I am wrong because I am a beginner in electronics.

You have miscalculated the parallel impedance of the RC network and the 10 kΩ collector resistor (R4). It's more like 9540 Ω than 8 kΩ. The mistake you have made is that you have assumed the impedances are just paralleled like parallel resistors but this isn't the case when the first 10 nF capacitor is dominant.

So, if you assumed that the 10 nF was directly loading the 10 k&ohm R4 resistor, the parallel impedance would be this produced by this calculator: -

enter image description here

So now, the gain will be dominated by that impedance and the 1 uF capacitor in the emitter. That yields a gain value of 30 i.e. suitable for oscillation.

But, of course, the 10 kΩ resistor isn't directly loaded by a 10 nF capacitor (worst case) and so, the net collector impedance will be a little higher and there will be more gain.

  • \$\begingroup\$ Thanks but i have one doubt, What do you mean by the 10kOhm resistor is directly loaded by the 10nf capacitor? What does the word "loaded" means in this context?. Also is my calculation of input impedance correct. Once again thanks for the answer. \$\endgroup\$
    – shahrOZe
    Commented Mar 25, 2021 at 10:12
  • \$\begingroup\$ I have assumed that the worst case impedance scenario for the collector is 10 kohm in parallel with 10 nF. Input impedance calculation is correct. \$\endgroup\$
    – Andy aka
    Commented Mar 25, 2021 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.