# FM transmitter

Discussion in 'Electronic Basics' started by fred, Feb 10, 2004.

1. ### fredGuest

I have found an fm radio transmitter circuit diagram,
can't remember where it was so I put it here:

http://www.geocities.com/x_file_space/fmt-1.gif

This circuit looks too simple to be any good, but just
out of interest I was wondering how the fm modulation
is occuring. Obviously the tuned circuit will determine
the main frequency, but how varying the base of Q2
varies the frequecy this I don't see, any ideas?

3. ### Kevin AylwardGuest

Not really. That is, it does effect the frequency of oscillation, but it
is not part of the LC tank.
The circuit is a colpits oscillator. There is a capacitor from collector
to emitter. There is a capacitor from base to emitter (the transistor
diffusion capacitance). At the oscillation frequency the tank is
inductive (it is not at resonance) and is connected from the collector
to the base via C3.

So...changing the bias, changes the bias...very informative

Ice changes thus effecting not only the Z on the C5
The base emitter diffusion capacitance, Cbe is a direct function of the
emitter current, i.e. Cbe=40Ic/2.pi.ft, hence the oscillation frequency
will change with input signal.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

4. ### JamieGuest

C5 is the Cap that is making the majority of the freq shift.
as C5 is part of the tuned circuit. with out it you can
no oscallation.
since the circuit biasing the emitter with the Tuned Tank
giving you the phase shift needed you get your oscallation state.
if you start varying the bias on the transistor it will cause this
emiter bias to change thus causing the rate of cycle time to change
due to current Ice changes thus effecting not only the Z on the C5 but
changing its return path Z..

5. ### Sir Charles W. Shults IIIGuest

Shouldn't you capitalize "i"?
What is a "lamen"? Some sort of mammal from South America? Or maybe
you mean "layman".
Still short of capital letters, I see. I will send you a shift lock
code.
"Ediot"?? Man, you should have finished third grade.
I looked all over my jar of peanut butter but could not see your origin.
I'm not certain if you meant "Jif" or "gif". I suppose I should give you
the benefit of the doubt and assume that English is not your first language.

Putting on the asbestos tee-shirt...

Chip Shults

6. ### Kevin AylwardGuest

Jamie wrote:

Don't top post.

Obviously, not. What you said was incorrect.
I did.
It isn't part of the oscillator frequency setting components. As I
already stated in my first post, it connects the LC tank circuit from
the collector to the base. The tank operates off resonance so that the
collector is connected to the base via an effective inductive reactance.
I never said it was. What part of "..is connected from the collector
to the base via C3..." did you not understand?
You manage that quite well on you own, with no help from me.
I'm sure they will.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

7. ### JamieGuest

I may not be an expert on the subject at hand but
i do know enough that about what i said while i was
trying to put it lamens terms.
http://www.geocities.com/x_file_space/fmt-1.gif
tell me that C3 is part of the osc and not a
simple RF by pass.
as you notice the value is 0.01 ! kind of a long
shot for being used as part of the resonant circuit.
so since you like chopping others messages and trying to make
i hope those that know better after looking at the
Gif can see where i am coming from.

8. ### JamieGuest

i rest my case.

9. ### fredGuest

Good discussion, one last point:

since the LC circuit is not a tuned tank circuit,
why do we need C4 anyway ?

10. ### DboweyGuest

Fred.... posted:
<< Good discussion, one last point:

since the LC circuit is not a tuned tank circuit,
why do we need C4 anyway ? >>

Wow! The discussion must not have been that good. C4 IS used to tune the
circuit.

Don

11. ### fredGuest

Wow! The discussion must not have been that good. C4 IS used to tune the
It was established that the LC tank circuit does not
before you comment.

12. ### Kevin AylwardGuest

Although the tank is not in resonance, the capacitor is still used to
"tune" the oscillator frequency. The tank needs to be +jX as far as its
reactance goes, and this reactance is a function of both L and C.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

13. ### DboweyGuest

Fred posted:
<<
It was established that the LC tank circuit does not
before you comment.--------
Y'all may "establish" anything you like, but it does not make it correct. The
inductor and C4, being in a state of Xc = Xl, is what permits the circuit to
oscillate at a particular frequency.

Don

14. ### Robert C MonsenGuest

Here is my long, boring guess about how this oscillator works:

http://www.meridianelectronics.ca/circ/fmt1.htm

I don't think its a colpitts oscillator, at least as I understand them. A
colpitts oscillator will use two series caps in the tank as a phase shifting
device, to allow the phase of the excitation waveform to vary by 180 degrees
from the feedback waveform. This is done so it can be fed back into an
inverting transistor amplifier, such as a common emitter amplifier.

For this one, however, I think that the way to analyze this is to see that
Q2 is configured as a common base amplifier, which is NOT an inverting
amplifier. Also, the capacitors that influence the frequency are all in
parallel, not in series.

For a common base amplifier, the signal is applied to the emitter, and the
output is at the collector, in phase (mostly) with the input. (There is a
slight phase shift through the transistor, which I believe will limit the
max output frequency.)

The gain of the amplifier increases with the impedance of the tank, and the
tank impedance increases as the frequency aproaches the resonant frequency.

The capacitor C5, in parallel with the capacitance between the collector and
the emitter in the transistor, form the feedback of the system. They modify
the voltage at the emitter, which is then amplified back to the collector,
forming a feedback loop.

The gain of the amplifier is maximum when the impedance of the tank is
maximum. That will happen when the tank is in resonance, ie, when the
frequency is equal to

1
f0 = ----------------
2.pi.sqrt(L1.Ct)

where Ct is the total capacitance in the tank... Thus, the system will
oscillate at frequencies near the point where there is maximum impedance,
ie, the tank resonance frequency.

Ok, that allows it to oscillate, and sets the carrier frequency. Now lets

Since the inductance of the tank is known, varying the capacitance of the
tank will vary the carrier frequency, thus modulating the output.

The capacitance of the tank is due to the parallel combination (ie, the sum)
of C4, C5, and the capacitance between the collector and emitter of the
transistor. Thats C4+C5+Cce. The last term, Cce, is a series term of Cbe,
which is proportional to Ic, as Kevin Aylward mentioned in his message. Ic
is modified by the base voltage. Consequently, the total capacitance of the
tank, Ct, is influenced by the bias voltage of the transistor, which is how
the output signal is modulated.

Unfortunately, the capacitance of the transistor is hard to predict.
Consequently, the variable cap is needed to tune the system to a particular
frequency. I believe Cce will be influenced by temperature as well, so the
system will require tuning at different temperatures, and probably as it
warms up as well.

Additionally, I believe the antenna will also contribute slightly to the
capacitance of the tank in this transmitter. However, I don't have any way
to predict how much of an effect that will have.

Regards,
Bob Monsen

15. ### fredGuest

I don't actually see, but not to worry i think this
circuit is too complicated for a on-the-hoof analysis
in a newsgroup, probably the complete solution would
be spread over a whole chapter in text books. I doubt
if this is even tought in text books because this
circuit is a "trick circuit" because it has hidden
capacitors, and so is a bad example.

16. ### Kevin AylwardGuest

You misunderstand colpitts oscillators. It is indeed a colpits
oscillator. It is not debatable.

Draw a floating transistor. Put 3 impedances across it, Zce, Zbe, Zbc.
Now do a nodal analysis, e.g. equate real and imaginary terms for there
to be a sustained current. You will find that in this completely general
circuit there are only *two* possible ways for such a transistor circuit
to oscillate. One where Zce and Zbe are capacitors with Zbc an inductor,
and one where Zce and Zbe are inductors with Zbc a capacitor. The first
is a Colpits, the second is the Hartley. It is trivial that Q2
*topologically* satisfies the Colpits connection. Don't be confused as
to where the ground is connected. It makes no difference.
Ho hummm....

Completely irrelevant.
No. Go and do the actually sums as I suggest above.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

17. ### Kevin AylwardGuest

Not at all. Its a very simply circuit.
I agree that it is a somewhat subtle circuit, but once one is told how
it works, it is trivially obvious after the fact. The relation between
Cbe and Ic is a very basic one. Colpits oscillators with tuned circuits
are also very well covered.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

18. ### Robert C MonsenGuest

Thanks for attempting to educate me. I see that your view of colpitt/hartley
is far more general than mine.
Sorry, my misunderstanding of your definition of colpitts as one of the two
modes of oscillation of a transistor. I'm unfortunately thinking from a few
examples.
I'd like to see your closed form analysis of the circuit that predicts the
frequency, given in terms of the capacitance and inductance values, Vcc, and
the base voltage. That might help me understand what you are trying to say.

For me, given my simpleminded analysis above, it seems fairly easy to
predict the oscillation frequency using the formula

f0 = 1/(2.pi.sqrt(L1.(C4 + C5||Cbe + Ccb)))

This matches pretty well with what spice predicts for a similarly configured
circuit, when using Cbe=CJE and Ccb=CJC. I don't know what the relation
betwen Cbe and Vb is, though, so I can't really predict the frequency given
the base voltage. Thus, I'm not able to predict the modulation.

Regards,
Bob Monsen

19. ### fredGuest

http://www.geocities.com/x_file_space/fmt-1.gif
I agree, also in Colpitts, the center tap of the two capacitors
is connected to ground whereas here it's connected to the
emitter (which is not bypassed to ground).
I agree, this is a common base. I don't think all the capacitors
are in parallel though, C4 and C5 are in parallel? unless you are
ignoring R7, are you ?

Again, as I mentioned above this is not clear to me unless R7 is ignored.

series? series with what you forgot to mention ?

20. ### Kevin AylwardGuest

See my other post on this.

You need to see the *big* picture.
It is also common emitter and emitter follower. The oscillator is all
configurations at once!
You need to actually do the full, general sums of the small signal
equivalent circuit. I have given a bigger outline in the other post.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.