From what I am able to understand, a voltage controlled oscillator is "controlled by a voltage" when this voltage is applied to vary the value of the tank circuit: e.g. this could happen by applying this voltage on a varicap diode, which change its capacitance according to the voltage.

However I also noticed another thing: I have assembled a very simple oscillator based on a transistor and a handful of components: it is very basic and weak (I found the schematic in a book for beginners,) and I have noticed that if I vary the value of the power supply, the oscillator will vary/drift the working frequency (I have verified this, since a friend of mine has good equipment at his lab.)

So, can this oscillator be called VCO, or is just a "feature" due the weakness/instability of the circuit? Also, why will the frequency vary according to the value of the power supply? I was also thinking, if I apply a sawtooth waveform as power supply, will the oscillator automatically drift on its frequency range? I ask this also because I'm studying this kind of modulation.

Please consider the fact that I am a beginner, so I will appreciate an exhaustive answer also with practical examples/schematics.

Unfortunately, since the schematic of my oscillator is only in the book, and I don't have a scanner in my printer, I can't post the exact schematic.

However my circuit looks very similar to this one, except for the fact that I don't have a mic, and in parallel with C1 there is another resistor.

Oscillator schematic

  • \$\begingroup\$ Of course your circuit is a VCO, but it is probably not a very good (wide range, linear, stable, reproducible, etc.) one. There are various other ways to make a VCO, often based on charging a capcitor with a fixed current up to the controlling voltage, or to a fixed voltage by a controlled current. \$\endgroup\$ Commented May 28, 2015 at 15:49
  • \$\begingroup\$ @WoutervanOoijen I know that there are various other ways to make a VCO, as I've said in my question; I'd just like to know why my oscillator acts in this way, if this were a kind of VCO, and what happen if i attempt to modulate it using a sawtooth signal as power supply. Thanks. \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 15:52
  • \$\begingroup\$ if the frequency changes with power supply voltage that's probably because of the transistor working in a different bias point. \$\endgroup\$ Commented May 28, 2015 at 15:54
  • 1
    \$\begingroup\$ You can create a copy of the schematic with the schematic editor up the top of the post edit panel. \$\endgroup\$ Commented May 28, 2015 at 16:14
  • \$\begingroup\$ Which book (title/author) is the schematic in, just out of interest? Can't you take a photo of it and post it? \$\endgroup\$ Commented May 5, 2019 at 12:25

3 Answers 3


The problem with varying power supply is that you'll vary amplitude as well, which will create a big deal of intermodulation and parasitic frequency modulation; an undesirable heterodyne effect. So, therefore a good VCO should have the amplitude independent from the controlling voltage.

  • \$\begingroup\$ What would be a viable/good way way to achieve an amplitude independent from the controlling voltage? Can you provide me some practical example? Thank you. \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 16:31
  • \$\begingroup\$ @MisterD One may use varicaps, digital synthesis etc. It is much, much easier though, to build a VCO for rectangular pulses than for the sine wave. \$\endgroup\$
    – ilkhd
    Commented May 28, 2015 at 16:47

The general answer is "yes", you can call it a VCO. It may well not be a very useful one, but that's another question. Almost any circuit will change its operating parameters slightly with operating voltage. In op amps, you'll find the number PSRR - Power Supply Rejection Ratio specified. All sorts of oscillator configurations will produce a frequency which varies with supply voltage. Whether this is a bug or a feature depends on what you want the oscillator to do. For a crystal oscillator which provides a time base, any change in frequency is a problem.

However, some variations are more useful than others. In your case, I suspect that the variations you see are not well-controlled. For instance, your oscillator probably also drifts with temperature, and will even vary if you put your finger close to the circuit. So it's not very useful for producing a controlled frequency deviation.

In order to be marketed as a VCO, an oscillator must be well-behaved, which means that as far as possible the frequency ONLY varies with the control input. An obvious problem with your oscillator is that, as you change the power supply voltage, not only does the frequency change but so does the output amplitude. Sometimes this is a problem, and sometimes not - it depends on what you are using the circuit for.

  • \$\begingroup\$ Thank you. However, since I've said that I've found the schematic on a book for beginners, the oscillator itself is obviously useless; I've assembled it just for didactical reasons and I don't care if is unstable :) However, what about the modulation using a sawtooth as power supply? This will cause a wide oscillation? \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 16:05
  • \$\begingroup\$ There are two problems. First, if the sawtooth voltage gets too low, the oscillator will just quit, so a simple 0 to x volt sawtooth will not produce a simple deviation curve. Second, any VCO will have a linearity spec, and for some circuits the deviation is wildly non-linear. Worse, high-quality time base oscillators typically take a while to start oscillating, and have limited frequency response to a control input, so for some VCOs the modulation frequency is quite limited. \$\endgroup\$ Commented May 28, 2015 at 16:10
  • \$\begingroup\$ Ok, I suspected these troubles (if the sawtooth voltage gets too low). In my case I noticed that if the range of the power supply will go from 4 volt to 9 volt, I will obtain some different frequencies; so, assuming this, if i provide a sawtooth which goes from 4v to 9v, this will be the same as if I change rapidly the power supply voltage? \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 16:14
  • \$\begingroup\$ Sorry, I don't understand the question. Please try again. \$\endgroup\$ Commented May 28, 2015 at 16:17
  • \$\begingroup\$ Don't you worry: I am sorry, since English is not my main language. I mean: If i vary the power supply from 4 volt to 9 volt I notice different frequencies. If iI will provide a sawtooth power supply which goes from 4 volt to 9 volt, I will obtain automatically this range of frequencies? \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 16:21

The original FM transmitter circuit in the question relies on the audio modulating signal (from the microphone) altering dynamically the voltage between base and collector - this in turn modulates the so-called "miller capacitance" and, in turn modulates the oscillation frequency. It's not a great FM modulator but will do the job as a demo transmitter to an FM broadcast band receiver.

why the frequency will vary according to the value of the power supply?

If you replace the microphone with a resistor, there will still be modulation as you raise and lower the power rail. The average collector voltage equals the positive battery voltage (due to the inductor). The voltage on the base is some fraction of the supply voltage (say) 20%.

If the battery voltage is 4V then Vcb(average) is 4-0.8 volts = 3.2 volts

If the battery voltage is 9V then Vcb(average) is 9-1.8 volts = 7.2 volts

This difference voltage modulates the frequency of oscillation. It will alter the amplitude too but I have previously considered using this technique to make a VCO and may yet do that. The job I may put it on doesn't care much about side effects of amplitude modulation of course.

  • \$\begingroup\$ Many thanks to you too, for the comprehensive reply. In anyway I was reading, by a search on internet, that would be also possible apply the sawtooth signal on the base of the transistor to get another modulation; I think that this way would be different than apply the sawtooth signal as power supply, right? What will happen in that case? \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 17:16
  • \$\begingroup\$ Either way will work fairly effectively and both modulate the base-collector capacitance aka "miller capacitance". \$\endgroup\$
    – Andy aka
    Commented May 28, 2015 at 17:24
  • \$\begingroup\$ Eg: I feed the oscillator with a "regular" voltage, from a battery or an external adapter. In the meantime i will apply a sawtooth signal (from 4 volt to 9 volt) to the base of the transistor. But in this case two different voltages (the sawtooth signal and the one from battery/ext adapter) will be present, at the same time, in the circuit. This sum of two voltages may cause the risk to overload/damage the transistor? \$\endgroup\$
    – Mister D
    Commented May 28, 2015 at 17:32
  • \$\begingroup\$ Not 4V to 9V - that will be too high - I estimate 0.8V to 1.8 volts but that depends on the resistor you replaced the microphone with. It might be 1V to 2V or 1V to 3V maybe but not 4V to 9V on the base. \$\endgroup\$
    – Andy aka
    Commented May 28, 2015 at 17:48

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