The simplest FM transmitter and receiver module

Probably it was more of a case of deciding that 100nf was a nice capacitor value, then selecting the resistor to give 50us timeconstant.

If the gain is to be ~20 times, that sets the ratio Rc/Re.
If the collector is to be at 2.5V that sets the operating current and hence the bias point.

Sometimes you have to run around this loop a few times trying different values and trading things off against each other, but eventually you will come to a solution you like.

Regards, Dan.

Thank you for your help gentlemen. I am very grateful to both of especially.

Now, kindly tell me how to find and decide the values of the C1, C2, C3 and C4.
Using both procedure, quantitatively and qualitatively.

Thank you all.

Hi Princess,
I explained that if the collector/emitter current is higher at 2mA then the pre-emphasis capacitor value would be high requiring an electrolytic type of capacitor which has a poor tolerance of 50%. If the current is lower at only 20uA then the output impedance of the preamp would be too high and will have its level reduced when the lower impedance of the oscillator input loads it down. Therefore 200uA collector/emitter current was selected.

Pre-emphasis: America invented and began FM radio broadcasting a long time ago before other countries. In those days there were few high frequencies reproduced in music and speech. They wanted hiss from the radio system reduced a lot so they selected a lot of pre-emphasis, 75 micro-seconds.
Years later the rest of the world began trying FM but found that audio had improved and had much more high frequencies that caused the transmitters to produce distortion so they reduced the pre-emphasis to 50 micro-seconds. America already had many radios made with 75 micro-seconds de-emphasis so they did not change it and now high tech transmitter equalization makes it sound fine.
That is why my circuit has two values for C4.

C4 * R5 is the timeconstant of the preemphasis network (50us for europe, 75us for america), so that is easy.

C1 and C3 are audio coupling caps and form high pass filters with their load impedances, each having a 3dB point of 1 / (2 * pi * R * C), with both load impedances being of the order of at least a few K ohms, 330n is more then large enough, giving into 10K ohm (Lower then any of the loads really are) a 3dB point of 50Hz.

C2 we discused further up the thread.

I worry that a third year EE cannot figure this on their own.

Regards, Dan.

Teachers and professors do not teach anything anymore. Instead they let the students copy the work of others or find somebody in a forum to teach them.

Hi Princess,
I explained that if the collector/emitter current is higher at 2mA then the pre-emphasis capacitor value would be high requiring an electrolytic type of capacitor which has a poor tolerance of 50%. If the current is lower at only 20uA then the output impedance of the preamp would be too high and will have its level reduced when the lower impedance of the oscillator input loads it down. Therefore 200uA collector/emitter current was selected.

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Hello sir.
Yes, I remembered that you taught me this. I understand this.

Pre-emphasis: America invented and began FM radio broadcasting a long time ago before other countries. In those days there were few high frequencies reproduced in music and speech. They wanted hiss from the radio system reduced a lot so they selected a lot of pre-emphasis, 75 micro-seconds.
Years later the rest of the world began trying FM but found that audio had improved and had much more high frequencies that caused the transmitters to produce distortion so they reduced the pre-emphasis to 50 micro-seconds. America already had many radios made with 75 micro-seconds de-emphasis so they did not change it and now high tech transmitter equalization makes it sound fine.
That is why my circuit has two values for C4.

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Thank you for the history. I was not known to this. I was thinking why did you choose for values for C4.

C4 * R5 is the timeconstant of the preemphasis network (50us for europe, 75us for america), so that is easy.

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I have calculated the value for Europe. Thank you again.

C1 and C3 are audio coupling caps and form high pass filters with their load impedances, each having a 3dB point of 1 / (2 * pi * R * C), with both load impedances being of the order of at least a few K ohms, 330n is more then large enough, giving into 10K ohm (Lower then any of the loads really are) a 3dB point of 50Hz.

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Did not get it. Sorry teacher :-(

C2 we discused further up the thread.

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where did we discussed it sir? :???:

I worry that a third year EE cannot figure this on their own.

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These all work only can be seen by an experienced eye. when I will become an expert then I will tell others and other will ask me how do you know that we should use this value here blah blah.. then I will tell them it is my experience but I will try to teach them just like a child step by step.

Teachers and professors do not teach anything anymore. Instead they let the students copy the work of others or find somebody in a forum to teach them.

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I agree. And during exam time they just ask MCQs while in my view they must give designing questions in the exam and teach design techniques in the class. But they don't.

Thank you both of you.

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I see. We discussed C2 but that was used with 100MHz of frequency. Is it corner frequency? If so then why did you say to use 50Hz?

- - - Updated - - -

I assume all capacitors C1, C2 and C3 are calculated on the same frequency. Isn't it teacher?

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I calculated value of coupling capacitor below,
C = 1/(2 * pi * f * Xc)
C = 1/(2 * 3.142 * 50 * 10K)
C = 318.309nF
or
C = 330nF
Is it correct sir?

But still not getting why 3dB point lies at 50Hz of frequency. And why choose 10K or impedance. Why not above or below the 10K?

Thank you again

The electret mic and the 10k resistor that powers it have a total impedance of about 2.5k ohms. R2 and R3 are effectively in parallel so their total is 25.3k. The transistor has an input impedance of (hfe x 470 ohms=) 94k at low frequencies. The 25.3k and the 94k are in parallel with a total of 19.9k ohms and the 2.5k is in series with it for a total of 22.4k ohms.
C1 is the audio input coupling capacitor so it has an audio cutoff frequency of 1 divided by (2 x pi x 330nF x 22.4k ohms=) 21.6Hz.

C2 is a short at the RF frequency of 100MHz to stop the preamp from being overloaded by RF and causing severe audio distortion. C2 has no effect at audio frequencies.

C3 is an audio coupling capacitor from the preamp to the oscillator. It has the 10k output resistance of the preamp in series with the input resistance of the oscillator (hfe x 220 ohms=) 44k for a total of 54k ohms. The audio cutoff frequency is 1 divided by (2 x pi x 330nF x 54k ohms=) 9Hz. A 150nF capacitor could have been used instead of 330nF for a cutoff frequency of 19.8Hz.

C1 and C3 both reduce low audio frequencies so each one was calculated to have a cutoff frequency lower than 50Hz which is the lower limit for FM radio station's audio.

Instead of 1 divided by ( 2 x pi x R x C) I use 0.16 divided by (R x C). I corrected it.

Sorry for late reply sir.
I am busy in university study. I will post here but not daily because I have regular classes now. But I want to learn this completely so kindly attend my post and leave reply.

The electret mic and the 10k resistor that powers it have a total impedance of about 2.5k ohms.

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How 10K resistor + electret mic resistance = 2.5k?

C1 is the audio input coupling capacitor so it has an audio cutoff frequency of 1 divided by (2 x pi x 330nF x 22.4k ohms=) 21.6Hz.

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I got calculation. But I am confused.
The formula you used here is f=1/(2*pi*Xc*C). This is the formula actually for reactance of the capacitor.
But you used the value of the total resistors i.e. 22.4k and it is not Xc. Is this formula can be used with this total resistor value? Isn't for Xc?

It has the 10k output resistance of the preamp in series with the input resistance of the oscillator (hfe x 220 ohms=) 44k for a total of 54k ohms

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Sorry sir, did not get how 54k come. I mean is this the approximate value you chose nearer to the 44k?

A 150nF capacitor could have been used instead of 330nF for a cutoff frequency of 19.8Hz.

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Why? Any specific reason?

C1 and C3 both reduce low audio frequencies so each one was calculated to have a cutoff frequency lower than 50Hz which is the lower limit for FM radio station's audio.

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50Hz? But I have studied the range of FM Band is 87MHz to 108MHz. You are saying 50Hz which is quite low. Or may be I am mistaking in understanding your quote.

Thank you very much for your great help sir.

Eshal said:

How 10K resistor + electret mic resistance = 2.5k?

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The output of an electret mic is the drain of a Mosfet that has an output resistance of about 3.3k ohms which I measured on some electret mics. This 3.3k resistance is parallel to the 10k resistor that powers it and the total is 2.5k ohms.

I got calculation (of C1 the input coupling capacitor). But I am confused.
The formula you used here is f=1/(2*pi*Xc*C). This is the formula actually for reactance of the capacitor.
But you used the value of the total resistors i.e. 22.4k and it is not Xc. Is this formula can be used with this total resistor value? Isn't for Xc?

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I wanted a cutoff frequency at about 25Hz. Calculating for the value of the coupling capacitor results in 286nF which is not available so I used 330nF which is common and results in a cutoff frequency of 21.6Hz which is fine.

Sorry sir, did not get how 54k come (the resistance fed by C3). I mean is this the approximate value you chose nearer to the 44k?

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The output resistance of the preamp transistor is its collector load resistor which is 10k ohms. The input of the oscillator is 44k ohms. Their total is 54k ohms.

Why could 150nF be used instead of 330nF? Any specific reason?

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C1 and C3 reduce low frequencies. We want 50Hz or less to be -3dB the cutoff frequency of the entire circuit so the cutoff frequency of each capacitor should be 25Hz or less.
150nF will have a cutoff frequency of 19.8Hz which is fine.

50Hz? But I have studied the range of FM Band is 87MHz to 108MHz. You are saying 50Hz which is quite low. Or may be I am mistaking in understanding your quote.

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87MHz to 108MHz is the RADIO frequency. It is 100 MILLION Hz in the middle. 50Hz is the low AUDIO cutoff frequency produced by an old FM radio station. I think modern FM radio stations produce lower audio frequencies today so that a sub-woofer can shake your bones. The high AUDIO cutoff frequency of an FM radio station is 15kHz. Some people can hear as high as 20kHz. When I was young I could hear ultrasonic signals maybe as high as 26kHz.

Thank you very much sir.
Very nice comments and I appreciate your help to me.

Now we should go further. The next block is RF OSCILLATOR
I want to learn its designing. Kindly, start it that how could you chose the specific values of the components in it. And what precautions you keep in mind during RF oscillator designing.

Thank you once again.

My RF oscillator is a Colpitts type that is used in hundreds of simple FM transmitter circuits shown in Google. I didn't design it, instead I simply copied it.
The base of the transistor has a 470pF capacitor C5 to ground to allow the transistor to be a common-base amplifier and it oscillates with positive feedback through the capacitor from its collector to its emitter. The 470pF capacitor has low reactance at the 100MHz radio frequency but has a high reactance at the 15kHz highest audio frequency.

The audio modulation changes the conduction of the oscillator transistor that changes its capacitance that changes its frequency creating FM (and some AM).
The RF parts are close together so that the inductance of the wiring is low and the capacitance between parts and wiring is low. It probably will not work if it is built on a solderless breadboard.

Hello sir.

I have question here. May I use crystal oscillator here? If so then which frequency?

Regards,
Princess

A crystal oscillator is a good idea to have a stable RF frequency.

I think the highest frequency crystal is about 40MHz. You will need to select a frequency that is not used by an FM radio station and build an oscillator that works with a 3rd, 5th, 7th or 9th overtone crystal to reach that frequency.
Modulate the original crystal frequency to obtain a wide enough FM frequency deviation at the final frequency.

I think the highest frequency crystal is about 40MHz. You will need to select a frequency that is not used by an FM radio station and build an oscillator that works with a 3rd, 5th, 7th or 9th overtone crystal to reach that frequency.

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Sir, did not understand this. What is meant by 3rd, 5th, 7th or 9th overtone crystal?

I also want to know two things here. What is the amplitude of the input signal to the RF oscillator in your circuit which is posted in the post#3? And what is the frequency output of your RF oscillator?

Thanks a lot sir.

Eshal said:

Sir, did not understand this. What is meant by 3rd, 5th, 7th or 9th overtone crystal?

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The maximum fundamental frequency of a crystal is about 40MHz I think. To make a higher frequency then a lower frequency crystal is used and the output of the oscillator is tuned to one of its harmonics which are also called "overtones". When an amplifier symmetrically clips a signal then odd harmonics are produced at 3, 5, 7, 9 etc. times the fundamental frequency. Overtone crystals are made to produce these harmonics and a tuned LC at the output of the oscillator selects one harmonic. Look at Crystal Oscillator in Google for more information.

I also want to know two things here. What is the amplitude of the input signal to the RF oscillator in your circuit which is posted in the post#3? And what is the frequency output of your RF oscillator?

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The mic produces about 10mV p-p and the mic preamp amplifies it 17.9 times so the input to the RF oscillator is about 0.179V p-p for the loudest sounds of speech about 10cm from the mic.

C6 is a trimmer capacitor and it adjusts the RF frequency of the oscillator from about 85MHz to higher frequencies. My FM dial is full of near and far FM radio stations mainly at the high frequency end of the dial so I did not try my FM transmitter at the high end. Then maybe it does not go as high as 108MHz. Stray capacitance of the wiring of the circuit limits the highest frequency.

Very nice explanation sir.

Here are few questions.
1) How to know what value or range of trimmer (C6) should choose? Means on what it should be dependent, frequency?
2) How to choose L1's value? And explain its working briefly. I know its working but I want to listen it from an engineer like you.
3) What is the output voltage's amplitude of the RF oscillator?

Thank you sir.

Eshal said:

1) How to know what value or range of trimmer (C6) should choose? Means on what it should be dependent, frequency?

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C6, C7, stray capacitance and the capacitance of the oscillator transistor are the "C" of a tuned LC that resonates with the inductance of L1. Many years ago I bought those very small trimmer capacitors so I used them in my FM transmitter.

2) How to choose L1's value? And explain its working briefly. I know its working but I want to listen it from an engineer like you.

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I simply copied L1 from another FM transmitter project I found. It resonates as above.

3) What is the output voltage's amplitude of the RF oscillator?

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I cannot measure a frequency that high. In a simulation it is about 1.4V p-p at the emitter of Q2 and is very distorted.
The output of the project into a 75 ohm 80cm long antenna is 6.8V p-p which is 2.4V RMS which is a power output of 77mW.

Hello sir,
I calculated the value of Colpitts oscillator for C6 maximum at 35pF, below image.


I also calculated the value of Colpitts oscillator for C6 minimum at 5pF, below image.


So what does it mean?
Does it mean when C6 is tuned to 35pF then the transmission will occur at 247.247MHz? Means, when I will tuned my receiver to this frequency then I will get my audio signal on the radio in mobile or etc?

Similarly, I also calculated frequency to 5pF too. Does it mean when C6 is tuned to 5pF then the transmission will occur at 323.349MHz? Means, when I will tuned my receiver to this frequency then I will get my audio signal on the radio in mobile or etc?

In short, if my C6 is set at 5pF then which frequency at FM radio I choose to get my voice over it?

Thank you sir.

By the way, is R8 the output resistance of the RF oscillator?

I used a different LC calculator website:
88MHz, 0.1uH. Ct= 32.7pf.
100MHz, 0.1uH. Ct= 25.3pf.
108MHz, 0.1uH. Ct= 21.7pF.

A simulation gives completely different results to yours or the above. Maybe my inductance is not 0.1uH.

You forgot that C7, stray capacitance of the wiring and the capacitance of the transistor reduces the required capacitance from the trimmer capacitor.

The RF oscillator has the output from its emitter which is a low impedance (emitter-follower). When I tried the output from its collector then it stopped oscillating.
R8 biases the RF amplifier transistor with base current (177uA) so it operates in class-A.