One of the concepts which I’ve been exploring on my blog, concerns tuned circuits, and another concerns Voltage-Controlled Oscillators (VCOs). As one type of voltage-controlled oscillator, I have considered an Astable Multivibrator, which has as advantage a wide frequency-range, but which will eventually have as disadvantage, a limited maximum frequency, when the supply voltage is only 3V. There could be other more-complex types of VCOs that apply, when, say, 200MHz is needed, but one basic type of oscillator which will continue to work under such conditions, which has been known for a century, and which will require an actual Inductor – a discrete coil – is called the Colpitts Oscillator. Here is its basic design:

In this schematic I’ve left out actual component values because those will depend on the actual frequency, the available supply voltage, on whether a discrete transistor is to be used or an Integrated Circuit, on whether a bipolar transistor is to be used or a MOSFET… But there are nevertheless certain constraints on the component-values which apply. It’s assumed that C1 and C2 form part of the resonant “Tank Circuit” with L1, that *in series, they define the frequency*, and that *they are to be made equal*. C3 is not a capacitor with a critical value, instead to be chosen large enough, just to act as a coupling-capacitor at the chosen frequency (:2) . R2 is to be made consistent with the amount of bias current to flow through Q1, and R1 is chosen so that, as labelled, the correct bias voltage can be applied, in this case, to a MOSFET, without interfering with the signal-frequency, supplied through C3.

I’m also making the assumption that everything to the right of the dotted line would be put on a chip, while everything to the left of the dotted line would be supplied as external, discrete components. This is also why C3, a coupling capacitor, becomes possible.

The basic premise of this oscillator is that C1 and C2 do not only act as a voltage-divider, but that, *when the circuit* that forms between L1, C1 and C2 *is resonant* with a considerable Q-factor (>= 5), *C1 and C2 actually act as though they were a centre-tapped auto-transformer*. If this circuit was not resonating, the behaviour of C1 and C2 would not be so. But as long as it is, it’s possible for *a driving voltage*, together with a driving current, to be *supplied to the connection between C1 and C2*, in this case by the Source of Q1, and that the voltage which will form where C1 connects with both L1 and the Gate of Q1 (that last part, through C3), will essentially be *the former, driving voltage doubled*. Therefore, all that needs to happen on the part of the active component, is to form a voltage-follower, between its Gate and Source, so that the voltage-deviations at the Source, follow from those at the Gate, with a gain greater than (0.5). If that can be achieved, the open-loop gain of this circuit will exceed (1.0), and it will resonate.

It goes without say that C1 and C2 will also isolate whatever DC voltage may exist at the Source of Q1, from the DC voltage of L1.

There is a refinement to be incorporated, specifically to achieve a VCO. Some type of varactor needs to be connected in parallel with L1, so that low-frequency voltage-changes on the varactor will change the frequency at which this circuit oscillates, because by definition, a varactor adds variable capacitance.

What some sources will suggest is that, the best way to add a varactor to this circuit will be, to put yet-another coupling capacitor, and a resistor, the latter of which supplies the low-frequency voltage to the varactor. But I would urge my reader to be more-creative, in how a varactor could be added. One way I could think of might be, to get rid of R1 and C3, and instead of terminating L1 together with C2 to ground, to terminate them to the supply voltage, thus ensuring that Q1 is biased ‘On’, even though the coupling capacitor C3 would have been removed in that scenario. What would be the advantage in this case? The fact that The varactor could be implemented on-chip, and not supplied as yet-another, external, discrete component, many of which would eat up progressively more space on a circuit-board, as a complex circuit is being created.

I should also add that some problems will result, if the capacitance to be connected in parallel with L1 becomes as large, *as either C1 or C2*. An eventual situation will result, in which C1 and C2 stop acting, as though they formed a (voltage-boosting) auto-transformer. An additional voltage-divider would form, between C1 in this case, and the added, parallel capacitance. And this gives more food for thought. (:1)

(Possible Usage Scenario : )

(Updated 7/29/2019, 14h45 … )

(As of 7/27/2019 : )

This circuit might be useful, as long as the range of frequencies required is narrow. For example, an FM radio receiver could be tuned in to the band from 88MHz to 108MHz, and, because it’s a superheterodyne receiver, it could require that a local oscillator run over a range of frequencies from 98Mhz to 118Mhz. What this would suggest is that the range of frequencies is ‘only’ ~20% of the centre-frequency, which means that *the possibility might just exist*, for the required varactor to be implemented.

If the assumption was that a 4:1 range of frequencies was needed, then this requirement could never be fulfilled.

**1:)**

I have gathered some experimental results, using the circuit-simulator ‘NG-SPICE’, which suggest that the available range for the added capacitor, which is not shown above, is slightly more generous than what I just wrote. In certain situations where the resonance of the tank circuit follows the driving voltage with a 90⁰ phase-shift, the amplitude at which the circuit resonates, actually forms as the product of the Q-factor, with the voltage-division. Thus, if the Q-factor was in fact (10), but if the added capacitance formed (4/5) the total capacitance connected to L1, meaning that it was 2 times C1, then the overall amplitude at which the circuit resonates *could* end up being (10 * 0.2 * 2 = 4).

*The addition of any capacitance in parallel with ‘C1′ and ‘C2′ above, will cause a phase-delay to appear, in the resonating voltage, with respect to the driving voltage*.

This could ultimately prove useful because on-chip varactors tend to have a limit in capacitance-ratio. The mere possibility that the variable part of the capacitance connected to L1, was only allowed to amount to half the total capacitance L1 ‘sees’, then this could further narrow the possible frequency-range, to a point where 20% could no longer be achieved.

However, if there was in fact a 90⁰ phase-shift between the driving voltage and the resonating voltage, then neither the Source nor the Drain of Q1 above, nor the Emitter nor the Collector of a bipolar transistor, would be appreciably in-phase, in order to provide the positive feedback that an oscillator requires. A more-complex oscillator would need to follow, that also applies a phase-shift to the feedback-voltage, so that at least some component of that would amplify and reinforce oscillation. And at that point, the oscillator would no longer be a Colpitts Oscillator. The circuit shown above will not generate any phase-shifts.

**2:)**

A variant that does qualify as a Colpitts Oscillator is one, in which the combination of C3 and R1 form a high-pass filter, the corner-frequency of which is approximately the expected resonant frequency, so that at that frequency a phase-advance of 45⁰ is also introduced into the loop. At that point, the value of C3 needs to become precise. But the advantage of doing that would in fact be, that some phase-delay in the resonating voltage-curve is still supported, such as for example because a discrete, external varactor diode has been connected in parallel with L1. There would be two facts to observe about that circuit:

- C3 cannot be omitted, and therefore, parallel capacitance added by the chip will have diminished or no effect on the frequency,
- The simple introduction of a 45⁰ phase-shift will also imply that only sqrt(1/2) times the amplitude can really contribute to positive feedback. Additionally, the presence of the high-pass filter itself reduces that to (1/2).

I suppose that, if the Q-factor was really (10), its effect will be to assure that the oscillator will work anyway.

(Update 7/29/2019, 9h35 : )

Given that I had a special scenario in mind, which called for the elimination of the capacitor ‘C3′ above, then the question can also be asked, as to why the following circuit is unsuitable:

The answer is really straightforward. If the circuit was running at 100MHz, and if the value of ‘C3′ was in the vicinity of 5pF – which are both valid assumptions – then, the associated value for (R1 + R2) in this case, that would give a high-pass filter with the correct corner-frequency, would be ~320Ω. This is an amount of resistance too low, to be inserted as a Source-Follower, or as an Emitter-Follower. It can be used where bias is supplied either to the Gate of a MOSFET, or to the Base of a bipolar transistor, but would set incorrect bias-levels if used, as shown in the second diagram above.

And in fact, this latter observation also explains, why a bipolar transistor, that already has low impedance at its Base, is used more often, than a MOSFET, the latter of which really assumes high-impedance conditions.

With or without a non-trivial value for ‘C3′, when operating at 100MHz, this oscillator connects a virtual impedance of 320Ω in various places, and whichever type of transistor is being used must be prepared for that…

(Update 7/29/2019, 14h45 : )

This should work:

Dirk