Good day, we were tasked to design a transistor based RF oscillator to be used as a VCO for our design experiment. In the picture shown below I have seen many tutorials on how to set the L1, C2, C3 values. I also already know how to set R1 and R2 as they are only used to bias the transistor. I would just like to ask

What are the C1 and C5 capacitors, and the L2 inductor for and how do I set them to a certain value?

enter image description here


2 Answers 2


C5 is relatively large, it decouples the emitter to keep the AC gain high.

L2,C2,C3 form a parallel resonant circuit. L1 is simply an RF choke (high impedance at oscillator frequency).

C1 can be relatively small as it provides positive feedback to the transistor's base which is a relatively high impedance node - its actual value will depend on Cbe and (Ccb * voltage gain) - the latter being the Miller capacitance. These (Cbe and Miller capacitance) are effectively in parallel, and will attenuate the base signal.

C1 will also depend on the ratio of C2 to C3 which provide the ground tap on the resonant circuit: if C2 > C3 then X(C2) < X(C3) and the tap is closer to L1, reducing the positive feedback voltage via C1.

  • \$\begingroup\$ How can C1 provide "positive feedback to the transistor`s base"? \$\endgroup\$
    – LvW
    Feb 18, 2016 at 9:50
  • \$\begingroup\$ @LvW : Because the C2/C3 tap is grounded, the voltage above C2 is out of phase with the collector thus in phase with the base voltage. \$\endgroup\$
    – user16324
    Feb 18, 2016 at 9:53
  • \$\begingroup\$ ....in phase with the base voltage? But the base voltage is inverse (180deg) with respect to the collector! With other words: What is wrong (in your view) with my explanation? \$\endgroup\$
    – LvW
    Feb 18, 2016 at 11:45
  • \$\begingroup\$ Exactly. The base voltage is 180 degrees out of phase with the collector. Now look at the tuned circuit - C2,C3 and L2 (not L1). Imagine there was a tap somewhere on L2, and that tap was grounded. What would that imply about the phase of the voltage at the top of L2? It is 180 degrees out of phase with the collector and in phase with the base. (The grounded tap is actually between C2 and C3). \$\endgroup\$
    – user16324
    Feb 18, 2016 at 12:56
  • \$\begingroup\$ Brian D, perhaps I misunderstood your description. But that`s not the point. The point is the following: The L2-C2-C3 combination is - without any doubt - a third order lowpass. Driven with a non-ideal current source (BJT with internal resistance ro) we get the classical 3rd-oder lowpass function between the collector node and the common node of C2 and C1. And this lowpass has a phase shift of -180 deg at wo=1/SQRT(LC) with C=C2||C3. I admit, this formula LOOKS like a bandpass resonant frequency. But it is not! It is a 3rd-order lowpass. \$\endgroup\$
    – LvW
    Feb 18, 2016 at 15:24

My interpretation/explanation is a bit different from Brian Drummond`s description. For my opinion, we have no "resonant circuit" (L1, C2, C3) because this would not explain why we have positive feedback. The common node between C2 and C3 is grounded - and therefore, the working principle is as follows:

  • The feedback path resembles a third-order lowpass which can produce a phase shift of -180deg at the common node between L2 and C2 for the desired oscillation frequency. Together with the inverting characteristic of a common-emitter stage (between B and C) we have a total phase shift of -360 deg (positive feedback) .
  • Lowpass: It is a simple task to see that the lowpass function is realized as a classical 4-element ladder circuit: ro-C3 (grounded), L2-C2 (grounded). The resistance ro is the dynamic output resistance of the transistor.
  • The task of C1 is to couple the feedback signal to the base node and L1 decouples the feedback node from Vcc - otherwise we would have no feedback at all. The capacitor C5 cancels the negative feedback effect of R3 for the oscillation frequency (and, thus, allows a larger gain).

EDIT1: Keeping in mind that neither L1 nor C1 (should) contribute to the feedback path the lowpass function between output (collector) and the upper end of C2 can be derived. For this purpose, the transistor is assumed to be a current source I with an internal dynamic resistance ro:

H(s)=Vout/I=ro/[1 + s*(C2+C3)*ro + s²*L2*C2 + s³*C2*C3*L2*ro)

It can be shown that this lowpass has an amplitude peak with a phase shift of exactly -180deg for w=1/SQRT(L2*Cp) with Cp=C2C3/(C2+C3).

EDIT2: With respect to Brian Drummond`s last comment - I agree that two alternative methods exist to explain the working principle of the circuit - however, finally both approaches join in a common loop gain function. Let me explain:

In my detailed answer I have, from the beginning and based on a visual inspection, the feedback path treated as a 3rd-order lowpass. This can explain why such a network produces a 180deg phase shift at one single frequency.

However, another view is possible - as proposed by Brian Drummond: The collector path contains a parallel LC tank circuit. It is well-known that at the resonant frequency wo=1/SQRT(LC) the voltage across the tank circuit assumes a maximum and the CURRENT through each of the parallel branches also is at its maximum value. If we treat the transistor as anon-ideal current source (internal resistance ro) the current through the capacitive branch is

Ic=I*[sro*C2/D(s)] with denominator D(s) as given in the function H(s) above (EDIT1). This current resembles a bandpass function.

The signal which is fed back to the base node is the voltage V2 across C2. This voltage is nothing else than V2=Ic*(1/sC2) . Therefore, we arrive again at the lowpass function H(s).

Summary: The tank circuit contains a current function which has a bandpass character. Using the voltage across a capacitor as a feedback signal we divide the current function by sC2 and arrive at a lowpass function with a peak and a phase shift of -180deg at w=wo. This is the way to explain the relation between bandpass and lowpass functions (and the role of the resonant frequency wo) in the shown circuit.


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.