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I'm trying to understand how the following FM radio schematic works.

enter image description here

Specifically, I want to know how the carrier wave is generated. I understand the concept of an LC tank and I think I see it there in the upper right, but what I don't understand is how the oscillation / resonance gets started. All of the examples that I'm seeing online show the use of a frequency generator to make a LC tank "go". Obviously there is no frequency generator attached to this small (simple) circuit.

I asked a friend and he told me he suspected that the transistor(s) were involved, which makes sense, but I'm hoping someone can either explain that to me in more detail or if it's too involved to answer here, point me at some resources (books, web sites, videos, etc.) to get me moving in the right direction.

Thanks!

Update
Many thanks for all of the great information. After learning that this is a Colpitts Oscillator I was able to find the following resources that give even more details. I'm posting here for my future reference and for those that might find this question useful:
Wikipedia
Learn About Electronics
YouTube Video
A breadboard based example
Falstad Circuit Simulator
Learn about electronics

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  • \$\begingroup\$ For such simple circuits it often makes sense to simulate it to understand them. \$\endgroup\$ – PlasmaHH Sep 5 '15 at 13:10
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    \$\begingroup\$ Are you suggesting to simulate it with software? If so, what package do you suggest? Spice? \$\endgroup\$ – Matt Ruwe Sep 5 '15 at 13:13
  • \$\begingroup\$ I had somewhat the same question. electronics.stackexchange.com/q/86100/22607 \$\endgroup\$ – Parth Parikh Sep 5 '15 at 15:21
  • \$\begingroup\$ @ParthParikh Your question is similar but focused on frequency modulation whereas my question is about the carrier wave generation. \$\endgroup\$ – Matt Ruwe Sep 5 '15 at 16:27
  • \$\begingroup\$ @MattRuwe: I am not sure what else than software one could use. And use whatever is reasonable realistic, most spice packages will work, I personally often use ltspice. \$\endgroup\$ – PlasmaHH Sep 5 '15 at 20:32
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Q2 and the circuit around it form a Colpitts oscillator. This makes use of the fact that a transistor in the common base configuration can have voltage gain from emitter to collector. Consider this simple circuit:

When IN is biased so that OUT is near the middle of its range, then small voltage changes in IN cause large voltage changes in OUT. The gain is in part proportional to R1. The higher R1, the larger the resulting voltage change from a small current change. Note also that the polarity is preserved. When IN goes down a little, OUT goes down a lot.

A Colpitts oscillator exploits this greater than unity gain of a common base amplifier. Instead of the load being R1, a parallel resonant tank circuit is used. A parallel resonant tank has low impedance except at the resonant point, at which it has infinite impedance in theory. Since the amplifier gain is dependent on the impedance tied to the collector, it will have a lot of gain at the resonant frequency, but that gain will quickly fall below 1 outside of a narrow band around that frequency.

So far, that explains Q2, C4, and L1. C5 feeds a little of the output voltage of the common base amplifier from OUT to IN. Since the gain at the resonant point is greater than one, this causes the system to oscillate. Some of the change in OUT appears at IN, which is then amplified to make a larger change in OUT, which is fed back to IN, etc.

Now I can hear you thinking, but the base of Q2 isn't tied to a fixed voltage as in the example above. What I showed above works at DC, and I used DC to explain it because that's easier to understand. In your circuit, you have to think about what happens at AC, particularly at the oscillating frequency. At that frequency, C3 is a short. Since it is tied to a fixed voltage, the base of Q2 is essentially held at a fixed voltage from the point of view of the oscillating frequency. Note that at 100 MHz (in the middle of the commercial FM band), the impedance of C2 is only 160 mΩ, which is the impedance the base of Q2 is being held constant with.

R6 and R7 for a crude DC bias network to keep Q2 close enough to the middle of it's operating range for all the above to be valid. It's not particularly clever or robust, but will probably work with the right choice of Q2. Note that the impedances of R6 and R7 are orders of magnitude higher than the impedance of C3 at the oscillating frequency. They don't matter to the oscillations at all.

The rest of the circuit is just a ordinary and not particularly clever or robust amplifier for the microphone signal. R1 biases the (presumably) electret microphone. C1 couples the microphone signal into the Q1 amplifier while blocking DC. That allows the DC bias points of the microphone and Q1 to be independent and not interfere with each other. Since even HiFi audio only goes down to 20 Hz, we get to do what we want with the DC point. R2, R3, and R5 form a crude bias network, working against the load of R4. The result is that the microphone signal is amplified, with the result appearing on the collector of Q1.

C2 then couples this audio signal into the oscillator. Since the audio frequencies are much lower than the oscillating frequency, the audio signal passing thru C2 effectively perturbs the bias point of Q2 a little. This changes the driving impedance seen by the tank slightly, which slightly changes the resonant frequency the oscillator runs at.

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  • \$\begingroup\$ I think I'm going to have to read this a few times to fully understand, but this looks like the answer I wanted. The other answers are also good, but alas, I can only accept one. \$\endgroup\$ – Matt Ruwe Sep 7 '15 at 11:57
  • \$\begingroup\$ @Matt: If you explain exactly what you don't understand, I can maybe elaborate on that point. \$\endgroup\$ – Olin Lathrop Sep 7 '15 at 13:47
  • \$\begingroup\$ It all makes sense, I just need to do some experimentation to apply everything you mentioned. I'll let you know if I still have questions after that. \$\endgroup\$ – Matt Ruwe Sep 9 '15 at 10:20
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In that schema Q1 is a class A audio amplifier with a gain about 50-100. It is used to drive the oscillator stage — I've never been very good at recognizing oscillator types [turns out Q2 is a Colpitts oscillator] with C4/L1 @ ~110 MHz. If my memory serves me right C5 increases the miller effect to bring Q2 in an unstable, self-oscillating state.

EDIT: See Kevin White's response on how modulation works in this circuit.

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  • \$\begingroup\$ Isn't it a collpits oscilator? hartley is 2L 1C. collpits is 2C and 1L. \$\endgroup\$ – Bruce Sep 6 '15 at 7:48
  • \$\begingroup\$ Hence why I'm not good at recognizing oscillator types :-D . \$\endgroup\$ – user59864 Sep 6 '15 at 10:18
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    \$\begingroup\$ trick to remember them: colpits starts with a C(apacity) so 2*C and Hartley starts With H(enry) so 2*L \$\endgroup\$ – Bruce Sep 6 '15 at 10:40
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Q2 is configured as what is know as a Colpitts oscillator. C5 feeds the signal from the collector to the emitter. One important component in a Colpitt's oscillator is a second capacitor that doesn't exist as a physical component and is the emitter to base capacitance of Q2.

As you mention the LC tank forms a resonant circuit at the frequency of transmission.

To make an oscillator needs more than just a resonant circuit though, it needs an amplifier to make up for losses due to the resistance of the inductor and the fact that some of the power is radiated away.

The transistor Q2 forms an amplifier by taking some of the signal through C5 to the emitter an amplified version of the signal then appears at the collector back into the LC tank. This signal is then fed back to the emitter to be amplified further and so on.

This is called positive feedback and the signal will continue increasing until it is limited by something such as reaching the amplitude of the power rail or non-linearity in Q2 that limits the amplitude. It only needs an infinitesimal signal to start things going and the oscillations will quickly build up.

How do things start? As Martin states it can begin from the disturbance caused when power is turned on but that is not necessary. Any practical electronic circuit generates what is referred to as noise (the hiss in the background of audio for example). Even if this is only a few millionths of a volt it will built up as I described in the previous paragraph.

What does Q1 do?

Q1 amplifies the signal from the microphone to a level of 10's or 100's of millivolts that is fed to the oscillator Q2. Although I stated the frequency of oscillation is determined by the LC tank it is also affected by the characteristics of the transistor Q2. As the input voltage from Q1 is fed to Q2 it changes its characteristics slightly and will vary the frequency of oscillation causing FM.

It will also vary the amplitude of oscillation as well causing amplitude modulation (AM) but an FM receiver will ignore that.

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  • \$\begingroup\$ Are you certain the small modulation caused by varying Q2 characteristics is the one tuners will catch? From what I recall of my signal theory courses (and it struck me at that time) the teacher told us since AM and narrowband FM spectra are practically identical (I remember studying the equations though I no longer remember them, I only remember they were identical) you can actually demodulate an amplitude-modulated signal on an FM tuner switched to narrowband. \$\endgroup\$ – user59864 Sep 5 '15 at 18:37
  • \$\begingroup\$ The spectra of AM and narrowband FM are very similar but not identical; the two sidebands are opposite in phase with FM. I suspect that you could receive the AM signal on the FM receiver either because it didn't have very good AM rejection or you had to mistune it slightly. \$\endgroup\$ – Kevin White Sep 5 '15 at 19:02
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Regarding the start up of the oscillator circuit, I suspect C3 to be the important part. In the first moment while power is applied, C3 is basically a short circuit and turns on Q2. This provides power for the initial oscillation. C5 then provides positive feedback to sustain oscillation.

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