fm transmitter ?

[aaa]

Hello to all,
Does anyone know if pantec hobby kits are still an active company ?
I have their fm transmitter kit (no. 11).

First of all, I was wondering if anyone can offer an explanation of
how the circuit works, in particular the oscillator part.

So here is the circuit:
http://www.geocities.com/x_file_space/fmcct.jpg

Tia
Addie

 

[Dbowey]

Addie posted:

Hello to all,
Does anyone know if pantec hobby kits are still an active company ?
I have their fm transmitter kit (no. 11).

First of all, I was wondering if anyone can offer an explanation of
how the circuit works, in particular the oscillator part.

So here is the circuit:
http://www.geocities.com/x_file_space/fmcct.jpg

Tia
Addie

I have no idea about Pantec.

The "oscillator" is just a multivibrator. The tapped coil, TR1 (a trimmer
cap), and (D and C7) form the tuned circuit and the point of entry for the
modulation voltage. D is a varacter diode and it changes capacitance by
changing the voltage to ground at the junction of D and C7. In the case of
this circuit the voltage is an amplified AC signal.

Since the combination of (d and C7) are in parallel with TR1 and the coil, the
frequency will change with the applied AC signal at the "input."

My guess is that the coil connected to TR2 is one or two turns of wire around
the center of the "tuning" coil.

Since the little hog draws 500 mA of current, it should put out a fairly strong
signal.

This info may be too much or too little. If you look at most any multivibrator
you will see the similarity right away.

Don

 

[aaa]

"Dbowey" <dbowey@aol.com> wrote in message
news:20040108001331.23043.00001602@mb-m27.aol.com...

Addie posted:
Hello to all,
Does anyone know if pantec hobby kits are still an active company ?
I have their fm transmitter kit (no. 11).

First of all, I was wondering if anyone can offer an explanation of
how the circuit works, in particular the oscillator part.

So here is the circuit:
http://www.geocities.com/x_file_space/fmcct.jpg

Tia
Addie

I have no idea about Pantec.

The "oscillator" is just a multivibrator. The tapped coil, TR1 (a trimmer
cap), and (D and C7) form the tuned circuit and the point of entry for the
modulation voltage. D is a varacter diode and it changes capacitance by
changing the voltage to ground at the junction of D and C7. In the case
of
this circuit the voltage is an amplified AC signal.

Since the combination of (d and C7) are in parallel with TR1 and the coil,
the
frequency will change with the applied AC signal at the "input."

My guess is that the coil connected to TR2 is one or two turns of wire
around
the center of the "tuning" coil.

The coil is in fact not a component at all, but simply a a few turns on the
printed
circuit board. However, I have not got the circuit board any more and I need
an alternative "real" component which I can solder onto a breadboard, any
ideas as to what I may need ? Also, can you explain how the circuit actually
oscillates. Assuming we have no input signal we will still have the center
fm
frequency. I understand the bit with the capaciative diode, but how does the
tuned circuit determine the frequency anyway? and what is the role of c9 &
c6 ?

Since the little hog draws 500 mA of current, it should put out a fairly
strong
signal.

This info may be too much or too little. If you look at most any
multivibrator
you will see the similarity right away.

Don

 

[Robert C Monsen]

"aaa" <aaa@aaa.aaa> wrote in message
news:btjnd6$nhp$1@newsg4.svr.pol.co.uk...

The coil is in fact not a component at all, but simply a a few turns on
the
printed
circuit board. However, I have not got the circuit board any more and I
need
an alternative "real" component which I can solder onto a breadboard, any
ideas as to what I may need ? Also, can you explain how the circuit
actually
oscillates. Assuming we have no input signal we will still have the center
fm
frequency. I understand the bit with the capaciative diode, but how does
the
tuned circuit determine the frequency anyway? and what is the role of c9
&
c6 ?

The two transistors form a 'multivibrator', which is a fancy way of saying
they oscillate. While one is conducting, the other one is not, or more
likely, their peaks of conducting alternate, one conducting more while the
other is conducting less.

They are connected by the two caps C6 and C9, which don't really affect the
frequency; they simply pass the waveform from one transistor's collector
through to the base of the other.

The frequency is affected by C7, D, TR1, and the inductors. If the
capacitance of D is Cd, then the total capacitance is

Ct = TR1 + Cd.C7/(Cd + d7)

The inductance is whatever the lower coils sum up to. Lets call it L. Then
the frequency of oscillation is something like

f = 1/(2.pi.sqrt(Ct.L))

As you can see, varying Cd by modifying the bias voltage on D will cause f
to change slightly (it will 'modulate' it), which is how FM works.

The reason it oscillates at that frequency is that the inductors and
capacitance in parallel acts like a bandstop filter near the resonant
frequency f. What this means is that other frequencies outside of the
stopband get passed through while the 'other' transistor is conducting, and
thus any energy at frequencies outside the stopband gets dissipated.

There are undoubtedly other effects at work here that will perturb the
frequency from the value specified above.

Regards,
Bob Monsen

 

[Dion]

Some questions:

1. How are C6 and C9 determined
2. Is it a parallel or a series tuned circuit. You said it's a band stop,
this implies a series circuit. At first glance it looks like a parallel
configuration, but is it? The coil is centre tapped so looking from
one transistors point of view, it may be in series. What are the
network terminals on the tuned circuit, is it where the two transistors
are connected, or is it the centre tap and each transistor alternately ?

 

[Robert C Monsen]

"Dion" <Dion@nomail.commm> wrote in message
news:btood2$a9v$1@newsg1.svr.pol.co.uk...

Some questions:

1. How are C6 and C9 determined

The value of C6 and C9 have to be big enough so as not to attenuate the
signal. That means the impedance should be 1/10 of the parallel impedance of
the base biasing resistors. (ROT)

2. Is it a parallel or a series tuned circuit. You said it's a band stop,
this implies a series circuit.

Its tuned by the 'tank', which is the parallel combination of the inductors
and capacitors. A 'tank' is a form of bandstop filter.

At first glance it looks like a parallel

configuration, but is it? The coil is centre tapped so looking from
one transistors point of view, it may be in series.

No, the reason the coil is center tapped is that doing that makes the
oscillations opposite on different sides of the inductor. Thats required,
because a common emitter amplifier (which is what the transistors each form)
is an inverting configuration. An oscillator needs feedback that is in phase
with its output, so that feedback is taken from the 'opposite' transistor.

What are the

network terminals on the tuned circuit, is it where the two transistors
are connected, or is it the centre tap and each transistor alternately ?

What? I'm not familiar with the terminology. Perhaps somebody else can help
here.

Regards,
Bob Monsen

 

[Dion]

Can you elaborate a little. Try to talk about the current. For example:
current goes down from +15V through the center tap, goes left and
down T3 and into TR1, if you can continue in this manner until you
fully describe a full cycle. This way I will know whats happening in
the electrons level, rather than talk about phase shift, inverting
configurations, and other terms which don't mean much to me.

2. Is it a parallel or a series tuned circuit. You said it's a band
stop,
this implies a series circuit.

Its tuned by the 'tank', which is the parallel combination of the
inductors
and capacitors. A 'tank' is a form of bandstop filter.

At first glance it looks like a parallel
configuration, but is it? The coil is centre tapped so looking from
one transistors point of view, it may be in series.

No, the reason the coil is center tapped is that doing that makes the
oscillations opposite on different sides of the inductor. Thats required,
because a common emitter amplifier (which is what the transistors each
form)
is an inverting configuration. An oscillator needs feedback that is in
phase
with its output, so that feedback is taken from the 'opposite' transistor.

What are the
network terminals on the tuned circuit, is it where the two transistors
are connected, or is it the centre tap and each transistor alternately ?

I was just referring to the tank circuit. If you consider it as a two port

network, where are the ports, at both sides, or at alternate sides and
center tap. If you still don't know what I mean, don't worry, as long
as you can explain about the currents as mentioned above I will be
happy to understand the circuit in that way. Just assume you are explaining
to someone who doesn't like maths.

What? I'm not familiar with the terminology. Perhaps somebody else can
help
here.

Regards,
Bob Monsen

 

[Robert C Monsen]

"Dion" <Dion@nomail.commm> wrote in message
news:btqbov$fnv$1@newsg1.svr.pol.co.uk...

Can you elaborate a little. Try to talk about the current. For example:
current goes down from +15V through the center tap, goes left and
down T3 and into TR1, if you can continue in this manner until you
fully describe a full cycle. This way I will know whats happening in
the electrons level, rather than talk about phase shift, inverting
configurations, and other terms which don't mean much to me.

Here is a shot. However, one of the more knowlegable folks may wish to put
forth a better description.

Capacitors are plates that are some given distance apart. They don't allow
electrons to flow across them, but act like they do; turns out that how fast
the electrons accumulate is linearly related to how quickly the voltage
changes across the plates. The constant that relates the rate of change of
voltage to the current is called the capacitance C of the capacitor.

Inductors are devices that want to maintain the flow of electrons through
them at a constant rate. If the rate of flow of electrons changes, the
voltage across the inductor will change linearly with the change. The
constant for a particular device is called the inductance L.

If an inductor and capacitor are in parallel, and a DC voltage is applied
across the two, then current (electrons) will flow through the inductor
faster and faster until the resistance of the wires limits them.

However, if an AC voltage is put across them, then an odd thing happens. One
side is increasing in voltage while the other is decreasing in voltage. If
the speed of this change (the frequency) is just right, no AC current will
flow through the inductor/capacitor combination. However, current will flow
back and forth between the capacitor and the inductor. This situation has
been compared to a spring and weight in dynamics. The AC voltage excites the
system, and the current flows back and forth like a spring moving back and
forth.

Oddly enough, however, if the AC voltage fluctuations are lower or higher
than that 'perfect' value, AC current will flow through the system. You've
probably heard of the concept of resonance. With a spring and weight, you
can 'hit' it, and it will usually oscillate at a frequency that is related
to the force in the spring and the mass of the weight. For AC, if the ac is
lower frequency, then the inductor doesn't oppose the current as much, and
the capacitor opposes it more, but the sum doesn't equal 0, so current flows
through rather than circulating. Same if its higher, although in that case
the inductor opposes the current more, and the capacitor less. Same thing,
though, the total will be less than the maximum at that 'perfect' frequency.

Set that aside for a moment.

Now, the gain of a common emitter amplifier (which is how the transistors
are hooked up in your circuit) is related to the resistance at the collector
(among other things). Larger resistance means larger gain, up to a certain
point. Resistance means the ability to block electron flow; its the same as
the AC resistance described above. Given random fluctuations, there are lots
of different frequencies in both the transistors initially. However, because
of the fact that fluctuations at the particular 'perfect' frequency are
blocked, this increases the gain of the amplifier for fluctuations at that
frequency, and decreases the gain at all other frequencies. Thus, the
natural fluctuations at that frequency are amplified into oscillations at
that frequency.

I don't usually think about these things in this way, so my descriptions are
based on half remembered physics lectures in the early 80s.

Regards,
Bob Monsen

 

[Dion]

Here are some points for you:

You did not explain about the centre tap. If this is indeed an parallel
LC tank circuit, then it is not a classic one, you must explain how
the centre tap works.

Will this circuit oscillate without the tuned circuit as well?

Here is my analysis, which you can confirm/dismiss/or add to :

Current goes from +15V into centre tap, turns left goes down and through

C9 and T4 base, to ground. This allows a current which is about 100 times
greater to go into centre tap, trun right and go down via T3 to ground.
This continues untill C9 cannot supply any more current to base of T3.
Then, T4 stops, and a much reduced current goes instead through C6
and into base of T3. This results in large current going through the left
side of the centre tap, through T3, and into ground. Again this continues
untill C6 is charged. During this time C9 has discharged through R8, and
then the cycle repeats itself.

You will notice that I have left out the effects of the tuned circuit,
hopefully you can complete the explanation for me.