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FM transmitter oscillator question

Hi,

I was trying to design an FM transmitter and there is one thing that is puzzling me somewhat.

I know that I will have to build an oscillator running at a certain frequency. I want to make a BJT based hartley oscillator. I will have to give an audio signal as an input to the RF oscillator. Meaning that the oscillator will change its frequency. However, the tank circuit in the hartley would remain at more or less the same resonating frequency: 1/(2*pi*sqrt(LC)). So why:

  1. would the oscillating frequency of the oscillator change in the first place.

  1. if it would change then there ought to be a loss in efficiency for input frequencies that make the oscillator shift too far from its resonant frequency. Am I right in assuming that this loss would be negligible since we are RF frequency would be much larger than the audio frequency?.

Also, I have been trying to find some site which would explain hartley and colpitts oscillators in detail but I havent been able to...could somebody help in this regard.

thanks...

Reply to
shehry
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As it turns out, you're in luck. The base-collector capacitance of your typical transistor, it varies quite a bit with bias.

So you can simply feed some audio to the base and you'll get plenty of FM.

The typical circuit circa 1968 for this was a 2N706, LC circuit in the collector, a 5pf capacitor from collector to emitter, a 270ohm emitter resistor, and a 100K resistor from Base to +. This could be easily tucked into a plastic olive

Reply to
Ancient_Hacker

The site you seek is called a library. They are made up of "webpages" called books, which cover this territory well.

Michael

Reply to
Michael Black

FM transmitters work by vaying the capacitance, and therefore the resonant frequency, with the audio signal. You can do this using a varicap diode, or rely on the variation of transistor junction capacitance with bias point, as suggested by another poster.

No, as I've already stated the resonant frequency changes; but you should also be aware of the difference between deviation and modulating frequency in FM. Deviation of the carrier from its mean frequency depends on the amplitude of the modulating audio. The frequency of the audio determines how fast the carrier frequency changes NOT the amount of change.

Reply to
Andrew Holme

On a sunny day (19 Mar 2007 11:13:05 -0700) it happened "shehry" wrote in :

There are sevral ways. One is to use a varicap, a capacitance diode, that is a diode whos capaciance depends on the reverse voltage, all diodes have this more or less, as part of your tuned circuit. But with many BJTs modulating Uc wil lchange Ccb enough to FM modulate your transistor oscillator. It can even be as simple as putting a dynamic mike in series:

L Dynamic mike, about 200 Ohm. ___________ = | | )________ O_____ +9V ===C1 |C2 ) | === ) |--R1 ---|----- | | | c --------b BC107 e |------------------- 0V ///

C1 = 10pF C2 = tol trimmer L is 7 turns on 8mm diameter tube, tapped somewhere so your feedback is in the right phase. R1 is 470k or so.

Design from end of fifties.... Worked every time, connect short wire to collector as antenne.

Reply to
Jan Panteltje

I don't know. Lots of times people just know they have the solution and need a little help, when the reality is they don't have a clue and if they started from that premise they'd get better answers.

Just like that question in sci.electronics.basics the other day, when someone wanted to know about germanium transistors, without any real clue that their time is long in the past.

Yes, the poster is a beginner, because they'd not make the leap to knowing they need a hartley without knowing anything about that type of oscillator. But that doesn't make it a homework question.

Even if it is, you do have to give them credit, since most homework questions are immediately recognizable because someone just types in the actual question. This one is (if it's homework) masked by someone at least putting the question into their own words.

Michael

Reply to
Michael Black

the old Capacitance between the Base and emitter trick.

capacitance in a transistor plays a roll in the resonant circuit. By altering the current flow in the transistor (especially the Ibe) the capacitance changes.

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Reply to
Jamie

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