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I'm rather curious as to how oscillating conditions are achieved in this oscillator. My attempt at an analysis follows.

The BJT is operating in common base mode. As the base resistors are equal, the base is biased to half of supply voltage - \$3V\$. Let's say \$\beta=100\$: then current into the base is \$I_{b}=\frac{3V-0.7V}{\beta * 2.2k\Omega} \approx 10\mu A\$ and \$I_{C} = 100 * I_{b} = 1mA\$. At room temperature \$V_{T} = 25mV\$, so \$r_{e} = \frac{25mV}{1mA} = 25\Omega\$. This is negligible compared to \$R_1\$, so we can say that \$V_{EQ} \approx 2.2V\$.

The capacitors \$C_1\$ and \$C_2\$ are in series and their equivalent capacitance is \$50pF\$. The resonant frequency \$f=\frac{1}{2\pi \sqrt{L_1C}} \approx 58MHz\$. At this frequency the impedance of the inductor is \$Z_{L_1} \approx 55\Omega\$. The open gain of common base amp is \$A=\frac{Z_{L_1}}{r_e} \approx 2.2\$.

The input impedance of the amp is just \$r_e\$. At \$58MHz\$ a \$100pF\$ cap has an impedance of about \$27\Omega\$. Therefore there is \$27\Omega\$ between the output and input of the amp (\$C_1\$). The input impedance of the amp in parallel with \$C_2\$ to ground gives \$r_e||27\Omega \approx 13\Omega\$. So if we image a voltage divider from the BJT collector (amp output) to BJT emitter (amp input) to ground, we get circuit attenuation of \$\frac{13\Omega}{13\Omega+27\Omega} = 0.325\$. This attenuated voltage is amplified by \$2.2\$ to get \$2.2*0.325 = 0.715.\$

But to oscillate, the loop gain should be slightly over one. So by my analysis, the circuit should not oscillate. However, I have built this and it does oscillate. So what is the flaw in my analysis?

  • \$\begingroup\$ You first stated (I believe correctly) that the BJT is operating in common base mode. Then you go on to calculate the gain based on it being a common emitter amp. I think that discrepancy alone would cause your gain to be much lower than it actually is. A ~= gm*R_c en.wikipedia.org/wiki/Common_base \$\endgroup\$
    – horta
    Aug 19, 2019 at 22:18
  • \$\begingroup\$ @horta Oops, I meant to say common base. The formula is the same for both, as r_e is the reciprocal of gm. \$\endgroup\$
    – S. Rotos
    Aug 19, 2019 at 22:36
  • \$\begingroup\$ Go look up colpitts oscillator theory. \$\endgroup\$
    – Andy aka
    Aug 20, 2019 at 7:15
  • \$\begingroup\$ The power supply is part of the circulating (resonant) current path. \$\endgroup\$ Aug 21, 2019 at 6:27

1 Answer 1


You are failing to take into account the fact that the capacitors are reactive.

Ignoring the coil, the load seen by the collector is this \$-j 27\Omega\$ in series with the parallel combination of \$-j 27\Omega\$ and \$25 \Omega\$.

That appears to the collector as a \$129 \Omega\$ resistor in parallel with a cap having a reactance of \$-j 44 \Omega\$. Cancel that reactance with the right inductive reactance, and the collector sees just the \$129 \Omega\$, for a gain of around 5 -- which is (A), not unreasonable for an oscillator (you want lots of excess gain on startup), and (B) probably a significant overestimate, because it's not taking the transistor's losses into account.

  • \$\begingroup\$ S.Rotos, try reading zen22142.zen.co.uk/Analysis/analysis.pdf for its design example using this capacitive-tap transformer. \$\endgroup\$
    – glen_geek
    Aug 20, 2019 at 12:16
  • \$\begingroup\$ Why does the \$-j44\$ cap cancel with the inductor? Aren't they in parallel since the power supply is short for the AC signal? \$\endgroup\$
    – S. Rotos
    Aug 23, 2019 at 14:10
  • \$\begingroup\$ Because inductive and capacitive reactance cancel. That cap has an admittance of \$j 23\mathrm{m}\mho\$ (note the change in sign); an inductor that resonates at that frequency would have an admittance of \$-j 23\mathrm{m}\mho\$ (note the minus sign). That adds to zero. \$\endgroup\$
    – TimWescott
    Aug 23, 2019 at 14:55

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