another FM transmitter project

[audioguru]

and.....what happens at output when I increase the supply voltage at oscillator?

Then the frequency changes because the capacitance of the transistor changes (which is why my oscillator is powered from a voltage regulator) and the output level and output power increases.

 

[Mikebits]

audioguru,
-Can I use smaller inductors than your Oscillator's with same inductance? Like:- small inductor but lots of turns that provides same inductance (0.1 nH). Or smaller/tiny inductors used ferrite and same inductance 0.1 nH. (Smaller inductor but same inductance) What happens if I used it instead of your big inductor? Does it critical in output power?
- And what happens if I used such smaller inductors in your amplifier but same inductance 0.1nH?

By changing the size but keeping same inductance will affect the Q factor of the tuned circuit.
https://en.wikipedia.org/wiki/Q_factor

 

[audioguru]

The AC gain of a transistor has nothing to do with the base biasing resistors.

With a 5.0V supply, a 2.2M resistor in series with a 270k resistor produces an unloaded voltage of only 0.55V which is too low to turn on a BC547 that needs a loaded base voltage of 0.6V.
Your transistor probably does not have an important emitter resistor to ground then it will not work properly.

Assuming that the collector is 3.0V then the collector current is 3V/2.7k= 1.1mA. The 270k resistor has about 0.6V but it changes with temperature changes. Then the current in the 270k resistor is 2.22uA. The 2.2M resistor has a voltage of 4.4V across it then its current is 2uA. So the base of the transistor has a bias current of only 0.22uA.

The DC current gain of a BC547 is from about 110 to 800.
If the current gain is only 110 then for a collector current of 1.1mA the base current must be 1.1mA/110= 10uA but your circuit does not have enough current so the transistor will be cutoff producing severe distortion most of the time.

You should learn how to bias a transistor. It is simple.

 

[Willen]

By changing the size but keeping same inductance will affect the Q factor of the tuned circuit.
https://en.wikipedia.org/wiki/Q_factor

Generally can I say:- Large diameter inductor has high Q factor?

 

[Willen]

Your transistor probably does not have an important emitter resistor to ground then it will not work properly.

This pre-amplifier is doing its job with very high gain in this biasing:-
- 6 voltage supply
- BC547 or 9014 transistor
- +ve to mic = 15K resistor
- Parallel on mic = 103 capacitor
- 1.5K resistor and DC blocking capacitor 103 in series to the base from mic
- +ve to base= 2.2M resistor
- Ground to base= 270K resistor
- Collector resistor 2.7K
- Emitter has directly connected to ground (Zero)

You should learn how to bias a transistor. It is simple.

OK I am trying my best, and searching a perfect books suitable for me with detailed and clear discription but having difficult to find such books... Still searching.

 

[audioguru]

You used a BC547 transistor that has a very wide range of DC current gain from 110 to 800. My simulation shows a BC547B that has a range of DC current gain from 200 to 450.
Your transistor is biased so it is almost cutoff because the base current is too low. It will be worse if the transistor has a DC current gain of only 110.
Your transistor has an AC voltage gain of about only 10 or less.
Your transistor will be cutoff if the transistor base-emitter voltage is a little high and/or if the transistor is cold.

My transistor is biased with enough base current and enough current in the voltage divider so that its collector voltage averages half the supply voltage so it can swing equally up and down.
My emitter resistor cancels most changes in base-emitter voltage and differing current gain.
My collector resistor is 10k (instead of 2.7k like yours) to stop wasting current.
My transistor has a voltage gain of 80.

An electret mic needs 0.5mA of current at about 3V so the resistor feeding it from 6V should be 3V/0.5mA= 6k (use 6.2k) not 15k.

The 10nF capacitor parallel to the mic cuts audio frequencies above 6khz. FM radios are supposed to play frequencies up to 15kHz.

The 103 (10nF) coupling capacitor value is too small and will cut low audio frequencies. I use 330nF.
The 1.5k resistor is not needed.


The values for your base bias resistors are much too high for any transistor unless the collector current is very low.

 

[Willen]

*Can I make larger amplitude (little high gain) of audio by decreasing the value of collector resistor from 10K to 5K?
* For why you used base biasing capacitor (C5, 471) at oscillator? In some oscillators I didn't find.

 

[audioguru]

*Can I make larger amplitude (little high gain) of audio by decreasing the value of collector resistor from 10K to 5K?

That is backwards. The voltage gain of a transistor is Rc/Re. Rc is the collector resistor in parallel with its load and Re is the unbypassed emitter resistor in series with the transistor's internal emitter resistance which is in a fortmula involving collector to emitter current.
So a high voltage gain is when the collector resistor value is high, the supply voltage is high (for a higher current) and the transistor is biased correctly.

But a transistor produces severe distortion when its voltage gain is high and its output level is fairly high.
Here is a simulation of the distortion which is heard and shows on an oscilloscope:

* For why you used base biasing capacitor (C5, 471) at oscillator? In some oscillators I didn't find.

The oscillator is a Colpitts type. The transistor is a common-base amplifier (The emitter is the input and the collector is the output).
At 100MHz the 470pF capacitor has a reactance of only 3.4 ohms so it shorts the base to RF ground.

 

[Willen]

An electret mic needs 0.5mA of
current at about 3V so the resistor
feeding it from 6V should be
3V/0.5mA= 6k (use 6.2k) not 15k.

but supply voltage is 6, why you use 3 devided by 0.5? Give me another example:- if suppy has +15 voltage then how to calculate resistor of mic? Please

 

[audioguru]

but supply voltage is 6, why you use 3 divided by 0.5? Give me another example:- if suppy has +15 voltage then how to calculate resistor of mic? Please

The mic needs 3V at 0.5mA. The supply is 6V so the resistor has 6V - 3V= 3V at 0.5mA. Then the resistor is 3V/0.5mA= 6k ohms.

If the supply is 15V then the resistor has 15V - 3V= 12V across it and a current of 0.5mA. Then the resistor is 12V/0.5mA= 24k ohms.

 

[Willen]

Then the pre-emphasis will be removed and it will sound awful like your stereo with its treble tone control turned all the way down or like a telephone or AM radio.
Instead, use the entire circuit but add two or three resistors to mix stereo into mono and attenuate the input signal.


ALL semiconductor manufacturers make many 5V low-dropout voltage regulator ICs.

talking about Audio input option of your Mod-4:-
there is a attenuator (I think) by 3 resistor. Can I use small audio source (cell phone, PC) and loud audio source (some general amplifiers) without modifying this attenuator? I think this current attenuator is for Cell phone/PC audio input, and if want use loud audio source then I should decrease 1K to 500 ohm (?) OR just only to increase the value of 27K to almost 60/70K?
Demo please

 

[audioguru]

An audio amplifier output probably has much more signal voltage than a cell-phone or pc. Then modify the attenuator by reducing the value of the resistor to ground. Use Ohm's Law to calculate the amount of attenuation needed.

If the resistor to ground value is reduced to half then the attenuation will be half which is a small volume change. Your hearing's sensitivity to sound is logarithmic so 1/10th the level sounds half as loud. Therefore volume controls have a logarithmic (audio) taper.

If the transmitter's audio input level is too high then a radio will produce severe distortion. Make the volume sound the same as a radio station.

 

[Willen]

Then it is better to use potentiometer ot 1k. Thanks

 

[audioguru]

The input level for my preamp at 400Hz is about 10mV for a very loud transmission.

The two 27k resistors to a 1k resistor forms an attenuator that cuts the 280mV from an MP3 player down to 10mV.
But the level from a 20W into 8 ohms amplifier is 12.7V RMS. So the 1k resistor must be reduced to 21 ohms which will be impossible with a linear 1k ohms potentiometer unless it is a logarithmic volume control.

 

[Willen]

But the level from a 20W into 8 ohms amplifier is 12.7V RMS. So the 1k resistor must be reduced to 21 ohms which will be impossible with a linear 1k ohms potentiometer unless it is a logarithmic volume control.

Please design one another attenuator suitable for 20 Watt or 40 watt! I have a amplifier as a source for Mod 4

 

[audioguru]

Please design one another attenuator suitable for 20 Watt or 40 watt! I have a amplifier as a source for Mod 4

The problem with using the output of an amplifier is that its volume control changes the amount of signal.
You should use an output that does not change its level like an output that feeds a recorder.

 

[Willen]

The problem with using the output of an amplifier is that its volume control changes the amount of signal.
You should use an output that does not change its level like an output that feeds a recorder.

Can I use 1/2 or 1/4 watt small resistor in your 20 or 40 watt attnuator? (specially in 22 ohm and 15 ohm)

 

[audioguru]

Can I use 1/2 or 1/4 watt small resistor in your 20 or 40 watt attnuator? (specially in 22 ohm and 15 ohm)

Do the calculation for the tiny amount of power in each resistor:
0.01V squared/22 ohms= 0.0000045W.
0.01V squared/15 ohms= 0.0000067W.
A 1/4W resistor will not get warm.

 

[Willen]

I never found low drop-out 5V regulator but found only big 7805 and used it in your Tx. It won't regulate at 7V or lower. So I have planned to add a 5V zenior diode Base to Ground at oscillator to solve the problem. Can I do this?

 

[audioguru]

My oscillator is powered from regulated 5V. The base of the oscillator is about +3.33V, not +5V.
A 5.1V zener diode can be used (with a series resistor from +9V) to power the oscillator and preamp. But a zener diode wastes battery power.

All Western semiconductor manufacturers make low-dropout 5V regulators.