How to distinguish between FM and AM transmitter circuits

[walid1]

hi
please look at the following circuit, the author claim it is AM transmitter:

it is from: http://www.zen22142.zen.co.uk/Circuits/rf/amtx.htm

it is look exactly FM Transmitter, the only difference is in the values of capacitors and coils!!!
Is there another difference I did not watch?
Are both circuits were the same?
thanks alot.

 

[Hero999]

That's because cheap FM transmitters built with a single transistor and no varactor diode produce both AM and FM. AM is produced because the transistor's gain alters as the base current is varied and FM because the capacitance between the PN junctions also alters as the voltage varies.

The circuit you've posted will also create some FM but because the value of C1 is so high compared the the transistor's capacitance, it will be negligible.

To build an FM transmitter which doesn't produce much AM, you need to use a varactor diode as part of the tuned circuit, rather than altering the bias point of the base.

 

[walid1]

Thank you Hero999, you helped me

To build an FM transmitter which doesn't produce much AM, you need to use a varactor diode as part of the tuned circuit, rather than altering the bias point of the base.

Like this circuit?

Another question:
Why the transmitter frequency change when the battery goes down, despite that the calculation of LC tank frequency does not doesn't take the value of the battery into account?
thank you alot

 

[Hero999]

Think about it. Re-read my previous post paying close attention to what I said about the FM transmitter.

 

[walid1]

Ok I get it,

"AM is produced because the transistor's gain alters as the base current is varied and FM because the capacitance between the PN junctions also alters as the voltage varies.
"

BUT: FM is alters as the voltage varies, you mean the voltage of the voice signal, not the voltage of the battery?

in the circuit above, if we omit the voice section with its preamplifier, will remain only the oscillator section
will it change its frequency with change in battery voltage?
thank you

 

[Hero999]

Yes, the frequency will go down as the battery discharges.

You need to use a voltage regulator to stop the frequency being dependant on the battery voltage.

 

[walid1]

Yes, the frequency will go down as the battery discharges

Why scientists not put the battery voltage in the famous expression used to calculate any oscillator frequency?

 

[Hero999]

There are three reasons I can think of:

F= 1/(2π√(LC)) is only a an approximation. The resistance of the coil and load will also make a difference, a higher resistance or more loading will reduce the frequency.

There will be some parasitic capacitance and inductance in the circuit so the resonant frequency will be slightly lower than expected.

The capacitance of the transistor adds to C because it is in parallel with it. The capacitance will vary from transistor to transistor and is not always listed on the datasheet. Calculating the capacitance of the transistor will be similar to the formula for a varactor diode.
http://www.tpub.com/content/neets/14179/css/14179_131.htm
http://www.qsl.net/in3otd/electronics/varactor_model/varactor_fitting.html

 

[walid1]

Good
I've implemented Hartley oscillator using a simulation program multisim 10
the first figure was with 6v batt

the freq = 1.15MHz as shown
then i used 9 v batt:

i had the same freq.
please explain that
thank you

 

[audioguru2]

The sim program did not read the details of the datasheet of the transistor so it doesn't know that the capacitance of the transistor changes when its voltage changes.

But your frequency is so low that the very small change of the capacitance of the transistor causes a very small change in the frequency. Now try a 100MHz or 200MHz oscillator to see big changes in frequency.

 

[walid1]

The Circuit does not respond when I reduce the values of capacitors and inductor for higher frequency, but I am confident that you are true.
Thank you and to  Mr. Hero999

 

[Hero999]

That's not a Hartley oscillator, it's a Colpitts.

I calculated the theoretical frequency to be 1.39MHz so it must be taking the transistor capacitance into account but whether it knows that the capacitance varies at different voltages is another matter.

Yes, you need to increase the frequency to see a change. Maybe you're pushing the transistor too far?

Try using 47pF and 100nH which should give a theoretical resonant frequency of 104MHz, although if it works, I'd expect it to be nearer 86MHz.

I wouldn't expect to see much of a change in frequency vs the supply voltage in a simulation, unless you're sure the transistor model accounts for the change in capacitance at different voltages.

 

[walid1]

Yes, it is not Hartley it is colpitts
I have not built this circuit, but I've found it within the ready-made examples in the program
I'm not confident in the results because the frequency stops always at 50 MHz and does not exceed that value no matter how reduced the values of capacitors
Here are pictures of results

thank you Hero999

 

[Hero999]

Reducing the values of R1 and R2 should increase the maximum frequency slightly, try changing them to 4k7 and 470R respectively.

 

[walid1]

Hi Hero999
I try 4.7k and 470 ohm but the freq remain at 50 MHz
I then changed the transistor to 2N3904 but it still 50MHz
thank you