FM transmitter

[Robert C Monsen]

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402130622.3b67aaad@posting.google.com...

http://www.geocities.com/x_file_space/fmt-1.gif

This circuit looks too simple to be any good, but just
out of interest I was wondering how the fm modulation
is occuring. Obviously the tuned circuit will determine
the main frequency, but how varying the base of Q2
varies the frequecy this I don't see, any ideas?

Here is my long, boring guess about how this oscillator works:

http://www.meridianelectronics.ca/circ/fmt1.htm

I don't think its a colpitts oscillator, at least as I understand them.
A
colpitts oscillator will use two series caps in the tank as a phase
shifting
device, to allow the phase of the excitation waveform to vary by 180
degrees
from the feedback waveform. This is done so it can be fed back into an
inverting transistor amplifier, such as a common emitter amplifier.

I agree, also in Colpitts, the center tap of the two capacitors
is connected to ground whereas here it's connected to the
emitter (which is not bypassed to ground).

Kevin pointed out that colpitts is really just a 'mode of oscillation', as
opposed to any particular configuration that produces it. I'll buy that. So
Kevin's assertion that this is a 'colpitts oscillator' is probably true,
although its unlike any of the colpitts oscillators I've seen described.

For this one, however, I think that the way to analyze this is to see
that
Q2 is configured as a common base amplifier, which is NOT an inverting
amplifier. Also, the capacitors that influence the frequency are all in
parallel, not in series.

I agree, this is a common base. I don't think all the capacitors
are in parallel though, C4 and C5 are in parallel? unless you are
ignoring R7, are you ?

Actually, I believe that the capacitance is Ct = C4 + C5||Cbe + Cbc. The
capacitance out of the junction is the important thing, I think, at least in
this application. Whether it goes to the transistor's base, or to Vcc (both
of which's voltages are stationary in relation to the other due to C3)
doesn't really matter all that much, at least as far as how the junction
sees the capacitors. Both have a nearly constant voltage in this system in
relation to ground.

I want to see Kevin's response to my question, since he clearly knows alot
more about the internal capacitances of a transistor than I do, due to his
work on Spice. With that additional help, we may be able to predict where
the modulation will go based on changes to the base voltage, which would be
useful. Without that, one can't tell how wide the band of the transmitter
is, given an input signal specification. However, since the oscillation is
based on fairly unpredictable properties of the transistor, it may not be
possible to predict this in any useful way. I hope Kevin has the answer to
this.

One useful thing I've discovered with this analysis is that if you stiffen
up the voltage at Q2's base by adding a 0.04uF cap to ground, you can get
alot more power into the antenna. The new cap value is designed so that the
reactance is 1/4 the reactance of C3. Biasing the transistor properly helps
as well; changing R6 to 47k, and adding a 10k resistor to ground helps.

I also think that moving the antenna to the emitter of Q2 will allow it to
produce quite a bit more power, because a common base amplifier won't
produce much power at the collector. I don't know how to quantify this
effect, though.

I've learned alot about common base amplifiers with this exercise that I
didn't know before. One interesting this is that I was able to simulate Paul
Burridge's 40.5Mhz to 41Mhz VCO with this circuit pretty easily. By moving
around the base voltage, one gets to mess with the output frequency. Its
not, unfortunately, a linear effect (which also probably means that the
output of the transmitter will be distorted unless this is compensated.
Another reason for knowing the exact relation between Cbe and Vb.)

Regards,
Bob Monsen

 

[Kevin Aylward]

Robert C Monsen wrote:

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in
message news:Sp0Xb.50$1e2.15@newsfep3-gui.server.ntli.net...
Robert C Monsen wrote:
"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402091712.579aaa9d@posting.google.com...
I have found an fm radio transmitter circuit diagram,
can't remember where it was so I put it here:

http://www.geocities.com/x_file_space/fmt-1.gif

This circuit looks too simple to be any good, but just
out of interest I was wondering how the fm modulation
is occuring. Obviously the tuned circuit will determine
the main frequency, but how varying the base of Q2
varies the frequecy this I don't see, any ideas?

Here is my long, boring guess about how this oscillator works:

http://www.meridianelectronics.ca/circ/fmt1.htm

I don't think its a colpitts oscillator, at least as I understand
them.

You misunderstand colpitts oscillators. It is indeed a colpits
oscillator. It is not debatable.

Draw a floating transistor. Put 3 impedances across it, Zce, Zbe,
Zbc. Now do a nodal analysis, e.g. equate real and imaginary terms
for there to be a sustained current. You will find that in this
completely general circuit there are only *two* possible ways for
such a transistor circuit to oscillate. One where Zce and Zbe are
capacitors with Zbc an inductor, and one where Zce and Zbe are
inductors with Zbc a capacitor. The first is a Colpits, the second
is the Hartley. It is trivial that Q2 *topologically* satisfies the
Colpits connection. Don't be confused as to where the ground is
connected. It makes no difference.

Thanks for attempting to educate me. I see that your view of
colpitt/hartley is far more general than mine.

Yep

A colpitts oscillator will use two series caps in the tank as a
phase shifting device, to allow the phase of the excitation waveform
to vary by 180 degrees from the feedback waveform. This is done so
it can be fed back into an inverting transistor amplifier, such as a
common emitter amplifier.

For this one, however, I think that the way to analyze this is to
see that Q2 is configured as a common base amplifier, which is NOT
an inverting amplifier. Also, the capacitors that influence the
frequency are all in parallel, not in series.


Ho hummm....


Sorry, my misunderstanding of your definition of colpitts as one of
the two modes of oscillation of a transistor. I'm unfortunately
thinking from a few examples.

The definition is topological, the details can vary a it.

The gain of the amplifier increases with the impedance of the tank,
and the tank impedance increases as the frequency aproaches the
resonant frequency.

Completely irrelevant.

1
f0 = ----------------
2.pi.sqrt(L1.Ct)

where Ct is the total capacitance in the tank... Thus, the system
will oscillate at frequencies near the point where there is maximum
impedance, ie, the tank resonance frequency.

No. Go and do the actually sums as I suggest above.


I'd like to see your closed form analysis of the circuit that
predicts the frequency, given in terms of the capacitance and
inductance values, Vcc, and the base voltage. That might help me
understand what you are trying to say.

For me, given my simpleminded analysis above, it seems fairly easy to
predict the oscillation frequency using the formula

f0 = 1/(2.pi.sqrt(L1.(C4 + C5||Cbe + Ccb)))

This matches pretty well with what spice predicts for a similarly
configured circuit, when using Cbe=CJE and Ccb=CJC. I don't know what
the relation betwen Cbe and Vb is, though, so I can't really predict
the frequency given the base voltage. Thus, I'm not able to predict
the modulation.

Let Vo be the ac voltage across the collector and emitter. The voltage
at the base is then, by potential divider action:

Vb = Vo.Zbe/(Zcb + Zbe).

The load on the transistor is the parallel combination of the series
impedances, that is:

ZL = (Zce *(Zcb + Zbe))/(Zce + Zcb + Zbe)

Now:

Vo = gm.ZL.Vb


Therefore:

Vo = gm.Vo.Zbe/(Zcb + Zbe).(Zce *(Zcb + Zbe))/(Zce + Zcb + Zbe)

or

1 = gm.Zbe .(Zce *(Zcb + Zbe))/(Zce + Zcb + Zbe)

(Zcb + Zbe).(Zce + Zcb + Zbe) = gm.Zbe .(Zce *(Zcb + Zbe))


with gm = 40.Ic

You should be able to do the rest, after you plug in for the Z's. You
will get the conditions for vo to exist. It is an exact relation.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Kevin Aylward]

Robert C Monsen wrote:

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402130622.3b67aaad@posting.google.com...
http://www.geocities.com/x_file_space/fmt-1.gif



I don't think its a colpitts oscillator, at least as I understand
them. A colpitts oscillator will use two series caps in the tank as
a phase shifting device, to allow the phase of the excitation
waveform to vary by 180 degrees from the feedback waveform. This is
done so it can be fed back into an inverting transistor amplifier,
such as a common emitter amplifier.

I agree, also in Colpitts, the center tap of the two capacitors
is connected to ground whereas here it's connected to the
emitter (which is not bypassed to ground).

Kevin pointed out that colpitts is really just a 'mode of
oscillation', as opposed to any particular configuration that
produces it. I'll buy that. So Kevin's assertion that this is a
'colpitts oscillator' is probably true, although its unlike any of
the colpitts oscillators I've seen described.

The reason is geared around what type of reactance that is necessary at
each node at the oscillation frequency. Its a +jX1, +jX2 and -JX3 or
a -jX1, -jX2 and +JX3.

For this one, however, I think that the way to analyze this is to
see that Q2 is configured as a common base amplifier, which is NOT
an inverting amplifier. Also, the capacitors that influence the
frequency are all in parallel, not in series.

I agree, this is a common base. I don't think all the capacitors
are in parallel though, C4 and C5 are in parallel? unless you are
ignoring R7, are you ?


Actually, I believe that the capacitance is Ct = C4 + C5||Cbe + Cbc.
The capacitance out of the junction is the important thing, I think,
at least in this application. Whether it goes to the transistor's
base, or to Vcc (both of which's voltages are stationary in relation
to the other due to C3) doesn't really matter all that much, at least
as far as how the junction sees the capacitors. Both have a nearly
constant voltage in this system in relation to ground.

I want to see Kevin's response to my question, since he clearly knows
alot more about the internal capacitances of a transistor than I do,
due to his work on Spice.

Actually, I knew the theory on this way before I knew anything about
spice. Its pretty much standard.

With that additional help, we may be able
to predict where the modulation will go based on changes to the base
voltage, which would be useful. Without that, one can't tell how wide
the band of the transmitter is, given an input signal specification.
However, since the oscillation is based on fairly unpredictable
properties of the transistor, it may not be possible to predict this
in any useful way. I hope Kevin has the answer to this.

I did state what the equation was for the diffusion capacitance.

Cbe' = gm/2.pi.ft. With gm=40.IC. IC will be modulated by Vin/RE, which
modulates the capacitance. There is also a static Cbeo which effects the
results a bit.

Ft changes quite a bit from transistor to transistor. The tuning of the
tank will compensate a for this, hopefully.

I gave an outline of the full equations in another post.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[fred]

Actually, I believe that the capacitance is Ct = C4 + C5||Cbe + Cbc. The
capacitance out of the junction is the important thing, I think, at least in
this application. Whether it goes to the transistor's base, or to Vcc (both
of which's voltages are stationary in relation to the other due to C3)
doesn't really matter all that much, at least as far as how the junction
sees the capacitors. Both have a nearly constant voltage in this system in
relation to ground.

I still can't see why C5 is parallel with Cbe. C5 is between c & e
while Cbe is from b to e. b for ac is also the same point as top
of tank circuit, so Cbe is from the top of tank circuit to *emitter*
and not to bottom of tank circuit, so surely it's not a simple as
adding it to C4 ?

 

[fred]

The capacitance of the tank is due to the parallel combination (ie, the sum)
of C4, C5, and the capacitance between the collector and emitter of the
transistor. Thats C4+C5+Cce. The last term, Cce, is a series term of Cbe,
which is proportional to Ic, as Kevin Aylward mentioned in his message. Ic
is modified by the base voltage. Consequently, the total capacitance of the
tank, Ct, is influenced by the bias voltage of the transistor, which is how
the output signal is modulated.

o.k. I starting to understand now, but still they are not simply
all in parallel. I can't see why Ct=C4+C5+Cce should be, but rather

Ct= C4 || ( C5||Cce + Cbe )

This implies that Cbe needs to be significantly smaller
than C5||Cce, otherwise changes in Cbe will be "hidden"
due to the other term. Is that the right way ?

 

[Robert C Monsen]

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402140440.2e12528@posting.google.com...

The capacitance of the tank is due to the parallel combination (ie, the
sum)
of C4, C5, and the capacitance between the collector and emitter of the
transistor. Thats C4+C5+Cce. The last term, Cce, is a series term of
Cbe,
which is proportional to Ic, as Kevin Aylward mentioned in his message.
Ic
is modified by the base voltage. Consequently, the total capacitance of
the
tank, Ct, is influenced by the bias voltage of the transistor, which is
how
the output signal is modulated.

o.k. I starting to understand now, but still they are not simply
all in parallel. I can't see why Ct=C4+C5+Cce should be, but rather

Ct= C4 || ( C5||Cce + Cbe )

This implies that Cbe needs to be significantly smaller
than C5||Cce, otherwise changes in Cbe will be "hidden"
due to the other term. Is that the right way ?

Right, I got my notation screwed up. The parallel capacitance we are
computing is that which goes from the collector node to a node that doesn't
change is relation to Vcc (which is true of the base of the transistor,
because of C3.)

Ct is the result of C4 parallel (C5 series Cbe) parallel Cbc

I was using the || to mean series, because mathematically its the same
function with caps as parallel is with resistors (but, of course, it
confuses the issue completely, as I wasn't seeing.)

So, mathematically, we have

Ct = C4 + C5*Cbe/(C5+Cbe) + Ccb

Kevin mentioned that Cbe is 40.Ic/2.pi.Ft (where I'm guessing Ft is the
current gain bandwidth product, like 300Mhz for a 2N3904).

In the given configuration, Ib = (Vcc - Vb)/10k, so if you put it all
together, in terms of Vb, you get

Cbe = 40*Beta*(Vcc - Vb)/(10k*2*pi*Ft)

as the capacitance between base and emitter.

From that, and assumptions about beta, Ft, Cbc, etc, you can calculate the
min and max values of resonant frequency, based on the min and max values of
Vb. The difference between those two will give you the bandwidth of the
transmitter.

Regards,
Bob Monsen

 

[fred]

Right, I got my notation screwed up. The parallel capacitance we are
computing is that which goes from the collector node to a node that doesn't
change is relation to Vcc (which is true of the base of the transistor,
because of C3.)

You don't need to say this. Just say "for ac the positive rail
and the transistor base are the same point because C3 is a short
circuit"

Ct is the result of C4 parallel (C5 series Cbe) parallel Cbc

I was using the || to mean series, because mathematically its the same
function with caps as parallel is with resistors (but, of course, it
confuses the issue completely, as I wasn't seeing.)

So, mathematically, we have

Ct = C4 + C5*Cbe/(C5+Cbe) + Ccb

Kevin mentioned that Cbe is 40.Ic/2.pi.Ft (where I'm guessing Ft is the
current gain bandwidth product, like 300Mhz for a 2N3904).

Can you clarify this, even if it's only a guess.

In the given configuration, Ib = (Vcc - Vb)/10k, so if you put it all
together, in terms of Vb, you get

Cbe = 40*Beta*(Vcc - Vb)/(10k*2*pi*Ft)

o.k. this is much clearer now. It is a Colpittz. What is
a Colppitz, a Colppits has an L in parallel with two C's
which are center tapped. So here our center tap is the
emitter.

But is it also a common base ? Can we have both ?
Common base, because for ac the base is bypassed to Vcc ?

b.t.w. this may seem stupid, but is the audio signal
considered as dc or ac? If ac, doesn't
that mean that Vb in the above formula is zero ?

as the capacitance between base and emitter.

From that, and assumptions about beta, Ft, Cbc, etc, you can calculate the
min and max values of resonant frequency, based on the min and max values of
Vb. The difference between those two will give you the bandwidth of the
transmitter.

Regards,
Bob Monsen

 

[Robert C Monsen]

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402151642.667b3307@posting.google.com...

Right, I got my notation screwed up. The parallel capacitance we are
computing is that which goes from the collector node to a node that
doesn't
change is relation to Vcc (which is true of the base of the transistor,
because of C3.)

You don't need to say this. Just say "for ac the positive rail
and the transistor base are the same point because C3 is a short
circuit"

Yes, thats a better way to say it. For AC, base, VCC, and GND are equal.

Ct is the result of C4 parallel (C5 series Cbe) parallel Cbc

I was using the || to mean series, because mathematically its the same
function with caps as parallel is with resistors (but, of course, it
confuses the issue completely, as I wasn't seeing.)

So, mathematically, we have

Ct = C4 + C5*Cbe/(C5+Cbe) + Ccb

Kevin mentioned that Cbe is 40.Ic/2.pi.Ft (where I'm guessing Ft is the
current gain bandwidth product, like 300Mhz for a 2N3904).

Can you clarify this, even if it's only a guess.

Ft is a value on the datasheets. I have no idea of the derivation of the
formula Kevin quoted. It probably has something to do with the fact that for
a particular current, the size of the depletion region changes, but I can't
say for sure. Its unexpected that Ft shows up in the formula, though.

In the given configuration, Ib = (Vcc - Vb)/10k, so if you put it all
together, in terms of Vb, you get

Cbe = 40*Beta*(Vcc - Vb)/(10k*2*pi*Ft)

o.k. this is much clearer now. It is a Colpittz. What is
a Colppitz, a Colppits has an L in parallel with two C's
which are center tapped. So here our center tap is the
emitter.

But is it also a common base ? Can we have both ?
Common base, because for ac the base is bypassed to Vcc ?

Well, the amplifier formed by Q2 is definitely configured as a common base.
The definition of colpitts oscillation seems to be 'an oscillation mode in
which there is a single inductor', as opposed to a hartley oscillator, which
is defined as 'a mode of oscillation where there are two inductors'.

b.t.w. this may seem stupid, but is the audio signal
considered as dc or ac? If ac, doesn't
that mean that Vb in the above formula is zero ?

No, Vb is the DC bias of Q2. Since the carrier is so much faster than the
audio, from Q2's point of view, the audio is just moving the bias point,
which modulates the carrier.

as the capacitance between base and emitter.

From that, and assumptions about beta, Ft, Cbc, etc, you can calculate
the
min and max values of resonant frequency, based on the min and max
values of
Vb. The difference between those two will give you the bandwidth of the
transmitter.

Regards,
Bob Monsen

 

[Watson A.Name - \"Watt Su]

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402091712.579aaa9d@posting.google.com...

I have found an fm radio transmitter circuit diagram,
can't remember where it was so I put it here:

http://www.geocities.com/x_file_space/fmt-1.gif

This circuit looks too simple to be any good, but just
out of interest I was wondering how the fm modulation
is occuring. Obviously the tuned circuit will determine
the main frequency, but how varying the base of Q2
varies the frequecy this I don't see, any ideas?

Voltage changes the capacitance of the transistor and hence the
frequency.

One problem with this particular circuit is the bias arrangement of the
mic preamp. The bias is alright at 9V, but the battery voltage drops
down to 7 or 8V and the bias changes and the audio is not good, and
maybe no audio at all.

The single 10k resistor to the base of the RF oscillator is too low, it
should be 5 to 10 times higher, 100k is common. But this stage should
use the kind of bias that's on the preamp, with two resistors. The
common values are 8.2k and 4.7k. The 1k R7 could be reduced to get more
output at the expense of shorter battery life.

 

[fred]

Ct is the result of C4 parallel (C5 series Cbe) parallel Cbc
But is it also a common base ? Can we have both ?
Common base, because for ac the base is bypassed to Vcc ?

Well, the amplifier formed by Q2 is definitely configured as a common base.
The definition of colpitts oscillation seems to be 'an oscillation mode in
which there is a single inductor', as opposed to a hartley oscillator, which
is defined as 'a mode of oscillation where there are two inductors'.

I have never seen that, but I shall have a look at some books in the
library and post if I find this definition. The thing is, if you are
defining Colppits as one L and Hartley as 2 L's, then basically you
are saying the any LC tank circuit is Colppitz. I believe the centre
tap thingy is necessary.

b.t.w. this may seem stupid, but is the audio signal
considered as dc or ac? If ac, doesn't
that mean that Vb in the above formula is zero ?

No, Vb is the DC bias of Q2. Since the carrier is so much faster than the
audio, from Q2's point of view, the audio is just moving the bias point,
which modulates the carrier.

"sigh of releif"

 

[Kevin Aylward]

fred wrote:

Ct is the result of C4 parallel (C5 series Cbe) parallel Cbc
But is it also a common base ? Can we have both ?
Common base, because for ac the base is bypassed to Vcc ?

Well, the amplifier formed by Q2 is definitely configured as a
common base. The definition of colpitts oscillation seems to be 'an
oscillation mode in which there is a single inductor', as opposed to
a hartley oscillator, which is defined as 'a mode of oscillation
where there are two inductors'.

I have never seen that, but I shall have a look at some books in the
library and post if I find this definition. The thing is, if you are
defining Colppits as one L and Hartley as 2 L's, then basically you
are saying the any LC tank circuit is Colppitz. I believe the centre
tap thingy is necessary.

The defining feature is the necessity to have either two +jx impedances
with one -jx impedance, or two -jx impedances with one +jx impedance, at
the relevant nodes, in order to achieve a net positive feedback
condition. The individual impedances can be pretty much anything
whatsoever, so long as at the oscillating frequency, they have the
desired sign of reactance.

The use of a tuned circuit simply gives a faster rate of change of phase
with frequency, which means that the circuit usually achieves better
frequency stability.

I personally, don't consider the topology associated with any "tap" as
that relevent. For me, its simply 3 impedances across the nodes of the
transistor.


Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[fred]

I have never seen that, but I shall have a look at some books in the
library and post if I find this definition. The thing is, if you are
defining Colppits as one L and Hartley as 2 L's, then basically you
are saying the any LC tank circuit is Colppitz. I believe the centre
tap thingy is necessary.

The defining feature is the necessity to have either two +jx impedances
with one -jx impedance, or two -jx impedances with one +jx impedance, at
the relevant nodes, in order to achieve a net positive feedback
condition. The individual impedances can be pretty much anything
whatsoever, so long as at the oscillating frequency, they have the
desired sign of reactance.

The use of a tuned circuit simply gives a faster rate of change of phase
with frequency, which means that the circuit usually achieves better
frequency stability.

I don't wish to be "difficult" , but you said it was a Colppitz, now
you are saying it's a tuned circuit again, or did you mean the same
thing?

I personally, don't consider the topology associated with any "tap" as
that relevent. For me, its simply 3 impedances across the nodes of the
transistor.

Again, it is difficult to guess what you mean by "For me, its simply 3
impedances across the nodes of the transistor."

 

[Watson A.Name - \"Watt Su]

"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:dOUWb.17563$uV3.36193@attbi_s51...

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402091712.579aaa9d@posting.google.com...
I have found an fm radio transmitter circuit diagram,
can't remember where it was so I put it here:

http://www.geocities.com/x_file_space/fmt-1.gif

This circuit looks too simple to be any good, but just
out of interest I was wondering how the fm modulation
is occuring. Obviously the tuned circuit will determine
the main frequency, but how varying the base of Q2
varies the frequecy this I don't see, any ideas?

Here is my long, boring guess about how this oscillator works:

http://www.meridianelectronics.ca/circ/fmt1.htm

I don't think its a colpitts oscillator, at least as I understand
them. A
colpitts oscillator will use two series caps in the tank as a phase
shifting
device, to allow the phase of the excitation waveform to vary by 180
degrees
from the feedback waveform. This is done so it can be fed back into an
inverting transistor amplifier, such as a common emitter amplifier.

For this one, however, I think that the way to analyze this is to see
that
Q2 is configured as a common base amplifier, which is NOT an inverting
amplifier. Also, the capacitors that influence the frequency are all
in
parallel, not in series.

For a common base amplifier, the signal is applied to the emitter, and
the
output is at the collector, in phase (mostly) with the input. (There
is a
slight phase shift through the transistor, which I believe will limit
the
max output frequency.)

The gain of the amplifier increases with the impedance of the tank,
and the
tank impedance increases as the frequency aproaches the resonant
frequency.

The capacitor C5, in parallel with the capacitance between the
collector and
the emitter in the transistor, form the feedback of the system. They
modify
the voltage at the emitter, which is then amplified back to the
collector,
forming a feedback loop.

The gain of the amplifier is maximum when the impedance of the tank is
maximum. That will happen when the tank is in resonance, ie, when the
frequency is equal to

1
f0 = ----------------
2.pi.sqrt(L1.Ct)

where Ct is the total capacitance in the tank... Thus, the system will
oscillate at frequencies near the point where there is maximum
impedance,
ie, the tank resonance frequency.

Ok, that allows it to oscillate, and sets the carrier frequency. Now
lets
talk about modulation.

Since the inductance of the tank is known, varying the capacitance of
the
tank will vary the carrier frequency, thus modulating the output.

The capacitance of the tank is due to the parallel combination (ie,
the sum)
of C4, C5, and the capacitance between the collector and emitter of
the
transistor. Thats C4+C5+Cce. The last term, Cce, is a series term of
Cbe,
which is proportional to Ic, as Kevin Aylward mentioned in his
message. Ic
is modified by the base voltage. Consequently, the total capacitance
of the
tank, Ct, is influenced by the bias voltage of the transistor, which
is how
the output signal is modulated.

Unfortunately, the capacitance of the transistor is hard to predict.
Consequently, the variable cap is needed to tune the system to a
particular
frequency. I believe Cce will be influenced by temperature as well, so
the
system will require tuning at different temperatures, and probably as
it
warms up as well.

Additionally, I believe the antenna will also contribute slightly to
the
capacitance of the tank in this transmitter. However, I don't have any
way
to predict how much of an effect that will have.

Comments requested...

Regards,
Bob Monsen

Right on, Bob. This circuit has been around for an awful long time,
since the early days of silicon transistors. I built one back when I
was in high school, and I put it up on a ham friend's tower a block or
so away from my house. I used a couple pieces of 14 gauge wire for an
antenna and tried to pick it up at my house about 650 feet (200m) away.
I could just barely pick it up with my sooper dooper Heathkit FM tuner,
which was really sensitive. I used a 2N706 which was the bestest
silicon transistor I could beg, then I got a 2N834, which helped a bit.
I didn't know much about it, but it was cool to make something that I
could hear a whole city block away.

Then I got this ingenious idea that I could do the same thing over a
light beam so I built a light beam modulator and mounted the light in a
parabola made from plaster of paris on the back of an auto sealed beam
headlight, lined with aluminum foil. On the other end I used a solar
cell for a detector, and a microphone preamp. It worked somewhat. But
the ham friend had an old tube rig with 11 meters on it, so we resorted
to talking to the CBers around town. Ran 65 watts into a long wire,
every time he keyed the mic we would get RF burns from anything metal we
touched in the shack!

Anyway, that was in the '60s, so this circuit has been around at least
40 years. It took a while for the average hobbyist to get a silicon
transistor for a reasonable price that could do 100 MHz or so. Five
bucks was a lotta money back then; you could get a hamburger and fries
for under a buck. Nowadays, you can get a hundred 2N3904s for the price
of a 'meal'.

Thanks for the explanation.

 

[Watson A.Name - \"Watt Su]

"fred" <fred222@btinternet.com> wrote in message
news:2685a85f.0402121706.5d02674c@posting.google.com...

I don't actually see, but not to worry i think this
circuit is too complicated for a on-the-hoof analysis
in a newsgroup, probably the complete solution would
be spread over a whole chapter in text books. I doubt
if this is even tought in text books because this
circuit is a "trick circuit" because it has hidden
capacitors, and so is a bad example.

Well, then all RF transistor circuits are trick circuits and bad
examples, then, because they all have hidden capacitance that has to be
taken into account.

 

[Kevin Aylward]

fred wrote:

I have never seen that, but I shall have a look at some books in the
library and post if I find this definition. The thing is, if you are
defining Colppits as one L and Hartley as 2 L's, then basically you
are saying the any LC tank circuit is Colppitz. I believe the centre
tap thingy is necessary.

The defining feature is the necessity to have either two +jx
impedances with one -jx impedance, or two -jx impedances with one
+jx impedance, at the relevant nodes, in order to achieve a net
positive feedback condition. The individual impedances can be pretty
much anything whatsoever, so long as at the oscillating frequency,
they have the desired sign of reactance.

The use of a tuned circuit simply gives a faster rate of change of
phase with frequency, which means that the circuit usually achieves
better frequency stability.

I don't wish to be "difficult" , but you said it was a Colppitz, now
you are saying it's a tuned circuit again, or did you mean the same
thing?

I said it was a Colpits, which it is. It is also a tuned circuit.
However, the tuning is *not* at *resonance*. Just off resonance there is
a fast rate of change of phase with respect to frequency, which results
in better frequency stability. The tuned circuit is tuned so that it has
a net +jx impedance. It is not a pure real impedance.

I personally, don't consider the topology associated with any "tap"
as that relevent. For me, its simply 3 impedances across the nodes
of the transistor.

Again, it is difficult to guess what you mean by "For me, its simply 3
impedances across the nodes of the transistor."

Is the glass half empty or half full?

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Watson A.Name - \"Watt Su]

"Robert C Monsen" <rcsurname@comcast.net> wrote in message
news:2w1Xb.24122$jk2.65355@attbi_s53...

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in
message

[snip]

I'd like to see your closed form analysis of the circuit that predicts
the
frequency, given in terms of the capacitance and inductance values,
Vcc, and
the base voltage. That might help me understand what you are trying to
say.

For me, given my simpleminded analysis above, it seems fairly easy to
predict the oscillation frequency using the formula

f0 = 1/(2.pi.sqrt(L1.(C4 + C5||Cbe + Ccb)))

This matches pretty well with what spice predicts for a similarly
configured
circuit, when using Cbe=CJE and Ccb=CJC. I don't know what the
relation
betwen Cbe and Vb is, though, so I can't really predict the frequency
given
the base voltage. Thus, I'm not able to predict the modulation.

Regards,
Bob Monsen

FWIW, a few URLs:
http://www.electronics-tutorials.com/oscillators/colpitts-oscillators.ht
m

http://www.mitedu.freeserve.co.uk/Analysis/colfr.htm
"Colpitts", but this uses two tank capacitors of equal value, like usual
colpitts.

http://members.tripod.com/rclindia/trans.html Weird four stage design-
waste!

http://home.pacbell.net/lengal/ip/fmtheory.pdf

http://www.electronic-projects.net/Schematics/Bug/Bug.htm

http://www.electronic-projects.net/Schematics/Bug/Bug.htm

http://sound.westhost.com/project54.htm

http://www.geocities.com/tomzi.geo/1-transistor/1-transistor.htm

 

[fred]

I said it was a Colpits, which it is. It is also a tuned circuit.
However, the tuning is *not* at *resonance*. Just off resonance there is
a fast rate of change of phase with respect to frequency, which results
in better frequency stability. The tuned circuit is tuned so that it has
a net +jx impedance. It is not a pure real impedance.

Are you saying it is a tuned circuit because C4 is a variable
capacitor? I think it would be simpler to say it is a tunable
colppits because one of the C's is variable, otherwise people
(me) might think "tuned circuit" as in "pure LC circuit". This
circuit is so bad in the way it is drawn, the LC circuit gives
the impression it's a pure tuned circuit. The circuit is
difficult enoght without this added confusion. The person who
drew this circuit probably didn't quite understand it.

 

[Kevin Aylward]

fred wrote:

I said it was a Colpits, which it is. It is also a tuned circuit.
However, the tuning is *not* at *resonance*. Just off resonance
there is a fast rate of change of phase with respect to frequency,
which results in better frequency stability. The tuned circuit is
tuned so that it has a net +jx impedance. It is not a pure real
impedance.

Are you saying it is a tuned circuit because C4 is a variable
capacitor? I think it would be simpler to say it is a tunable
colppits because one of the C's is variable, otherwise people
(me) might think "tuned circuit" as in "pure LC circuit". This
circuit is so bad in the way it is drawn, the LC circuit gives
the impression it's a pure tuned circuit. The circuit is
difficult enoght without this added confusion. The person who
drew this circuit probably didn't quite understand it.

We have two concepts here. One is tunability in the sense of varying
components, one is whether or not a circuit is at resonance in some
manner. One point on this circuit that I was addressing was that,
although it looks like there is an LC circuit, that would otherwise
might be expected to be tuned to its resonance, it isn't. The connection
from the collector back to the base via the + power supply line, has to
be as a +jx impedance, (i.e net inductive), not a resistive impedance
as would be the case if the tank was "tuneded" to resonance. There is
clearly some sort of tuning going on for the circuit to oscillate at one
frequency, but its given by a more complicated function of all the c's
and l's in the circuit.


Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

"That which is mostly observed, is that which replicates the most"
http://www.anasoft.co.uk/replicators/index.html

"quotes with no meaning, are meaningless" - Kevin Aylward.

 

[Dbowey]

Fred posted:
<< Are you saying it is a tuned circuit because C4 is a variable
capacitor? I think it would be simpler to say it is a tunable
colppits because one of the C's is variable, otherwise people
(me) might think "tuned circuit" as in "pure LC circuit". This
circuit is so bad in the way it is drawn, the LC circuit gives
the impression it's a pure tuned circuit. The circuit is
difficult enoght without this added confusion. The person who
drew this circuit probably didn't quite understand it.
I think the circuit is drawn quite well, and that you need to study a bit more
before you critique other's work.

The L and C form a "tank circuit" which is tunable by varying the value of C4.
I could also tune it by varying the inductance by compressing or stretching
the coil. I can also tune it by varying the applied voltage to the circuit.
Whether it is "pure" or all screwed up, is all in your mind.

Whether the type of circuit is a Colpits or something else, is not particularly
relevant to how the circuit tunes.

However, regardless of all the previous posts, this is not a Colpits. It looks
like a form of Pierce oscillator.

Don

 

[Dbowey]

kevin posted:

Oh never mind, it's trivial. No wonder the OP is confused.