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RF Transmitter

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dark_soul

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Hi everyone, I'm very interested to understand how make design about the RF Transmitters but the search has been difficult across the web, since alone I find pure circuit diagrams. I will try to understand one of them that I put in this message.

My doubts are the following:

-Which is the function of C3 and C5, and how it is calculated.
- If I want more power (watts) i need only to replace a Transistor or needs more considerations.

For all thank you.
 

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The purpose of C3 is a stabilizing capacitor that helps to absorb the ripple caused by the audio signal. C5 is a bypass gain capacitor that slightly increases the power output of the transmitter. If you want more power, then you can try to boost the voltage supply to the circuit but I doubt that it will make much of a difference. The two 2N3904 transistors that they used in the circuit are capable of handling up to 40V without any problems. This is a crappy transmitter and will probably get you busted for broadcasting on a restricted frequency without a licenses. Just try to find a better transmitter that will have less drift. Audioguru's transmitter is a good start if you are looking for something simple. If you want a high quality transmitter, Google BA1404, or BH1417. Both are good quality transmitters and both have their advantages.

The BA1404 has a crystal that it uses to flip between the left and right input to send them over a single antenna. The advantage of this circuit is that the circuit is still tunable via a variable capacitor. The bad news is that the circuit is needs a very low 3V max in order to operate, and it is likely to drift frequencies as the battery dies.

The BH1417 has an advantage in the sense that it still uses a crystal to switch channels, but unlike the BA1404, it doesn't drift, it has a higher supply voltage (5V), but the drawback is that it isn't tunable. It is set to a single frequency and has to be left there.

Hope this helped :)

Vince
 
Hi Vince, thak you for help me, only have a question, the value of C3 and C5 are a random value or there are a method to calculate it?.

Thank you.
 
[FONT=&quot]I have fully explained how RF transmitters work on my website. I have used very simple language that will not confuse the beginner.

In exactly the same way I will describe how to work out the value of C3.
C3 is present to hold the base of the RF transistor steady (fixed) so that the base does not move up and down. This allows the transistor to operate as a common-base stage.
This means the collector of the transistor must be able to move up and down and also the emitter.
In the collector circuit is a coil and capacitor. These two components form a circuit called a resonant circuit and when we talk about transmitters, we call the two components a TANK CIRCUIT.
When we bias the transistor, by putting a voltage on the base and connecting the emitter to the 0v rail via a resistor, the transistor is turned on and it produces a small amount of noise on the collector. This is called junction noise and the resonant circuit in the collector receives this noise and the two components create something that is amazing.
Even though these two components are classified as passive (non amplifying) items, they work together to create a waveform that is a sinewave and can have an amplitude that is higher than the noise or voltage that is being delivered to it.
This is exactly what happens and this waveform is picked up by the 5p6 capacitor connected to the collector and passed to the emitter. If this 5p6 is a larger value, it will “suck away” nearly all the waveform and the circuit will not gradually increase in amplitude to a point where it will start to “really work.”
The small waveform that passes through the 5p6 moves the emitter up and down and this is detected by the transistor.
Now we come to another point.
There are two ways of getting a transistor to amplify – to reproduce a waveform. One is to hold the emitter steady (fixed) and deliver the waveform to the base. The other is to hold the base steady and move the emitter up and down.
That’s what we have done.
The 5p6 is just like a shock-absorber.
Imagine I am holding one end and you are holding the other.
If I push towards you, you will move backwards a small amount and the shock absorber will reduce in length, when I pull, the shock absorber will extend and you will move slightly forward.
This is what the 5p6 does.
When a rising voltage is present on the collector of the transistor, this rising voltage is passed through the 5p6 and raises the emitter.
When the emitter voltage rises, the difference between the emitter voltage and the base voltage is LESS and the transistor turns off slightly.
This makes the collector voltage rise and the action increases.
The transistor could actually do this very quickly but the capacitor in the collector-circuit has a period of time in which is will charge and discharge and this determines the timing of the whole circuit. The energy from the capacitor is passed to the coil and the coil has a certain period of time for it to absorb the energy and turn it into flux.
When the energy passes from the capacitor to the coil, the coil produces magnetic flux and when the capacitor cannot deliver any more energy, the flux collapses and cuts the turns of the coil to produce a voltage in the coil of OPPOSITE POLARITY.
This opposite voltage is detected by the 5p6 to turn the transistor off slightly and this is how the stage operates (oscillates).
This means the TANK CIRCUIT is producing its opposite polarity waveform when the transistor is completely tuned off and that why it can produce a waveform that is higher than the supply voltage.
The value of the capacitor on the base can be almost any value as long as it holds the base steady at 100MHz.
If you want to inject the stage will audio, the capacitor must allow the audio waveform to rise and fall.
But audio is say 5kHz and the oscillator 100MHz so the reactance (resistance) of the capacitor at these two frequencies is 20:1 and it will have a very big effect at 100MHz and very little effect at 5Hz.


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Hi Colin55, Thank you for his help, I believe now I understand this design already I gave a step in my learning, graces again.
 
The purpose of C3 is a stabilizing capacitor that helps to absorb the ripple caused by the audio signal.
Wrong.
The value of C3 is far too low to affect audio frequencies. It shorts the base of the Colpitts oscillator transistor to ground so it is a common-base amplifier at 100MHz FM broadcast band frequencies.

C5 is a bypass gain capacitor that slightly increases the power output of the transmitter.
Wrong.
C5 applies positive feedback from the collector to the emitter of the oscillator transistor so it oscillates at about 100MHz.

This is a crappy transmitter and will probably get you busted for broadcasting on a restricted frequency without a licenses. Just try to find a better transmitter that will have less drift.
Correct. Its radio frequency drifts all over the place and it sounds bad on an FM radio.

If you want a high quality transmitter, Google BA1404, or BH1417. Both are good quality transmitters and both have their advantages.
No.
The BA1404 was another crappy transmitter with many problems including radio frequency drift. It has not been made for about 8 years when it was replaced by the much better BH141x line of FM stereo transmitter ICs.

The BH1417 has an advantage in the sense that it still uses a crystal to switch channels, but unlike the BA1404, it doesn't drift, it has a higher supply voltage (5V), but the drawback is that it isn't tunable. It is set to a single frequency and has to be left there.
No.
The BH1417 uses a DIP switch to select one of seven low radio frequencies or one of seven high radio frequencies in the FM broadcast band. Usually there are few radio stations at each end of the band.
 
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