How transmitters work.

[Jason]

Please forgive the pure ignorance of this question. I understand that
this is very basic stuff that I should probably know, but I have never
been to class to learn about this stuff, nor have I found any tutorials
that explain this stuff in simple everyday terms. I passed by
technician and general class amateur license exam by memorizing the
right buzzwords for this stuff, but not actually understanding the
concepts. I admit, I am ashamed of this fact. Here's what I have
gathered so far.

1. DC current flows through a crystal.
2. The resonance of the crystal causes the current to form waves around
the frequency of crystal
3. This new "pulsating" current is then amplified and then sent through
the crystal again. (This is called positive feedback)
4. The pulsating current is sent through a low-pass filter to "filter
out" unwanted frequencies. (That's why I said "around the frequency"
of the crystal)
5. From the filter, the current flows to an antenna which converts the
current to RF radiation.

Comments and help would be greatly appreciated.

 

[Andrew Holme]

Jason wrote:

1. DC current flows through a crystal.

Crystals, like capacitors, don't pass DC.

2. The resonance of the crystal causes the current to form waves
around
the frequency of crystal
3. This new "pulsating" current is then amplified and then sent
through
the crystal again. (This is called positive feedback)

The oscillator is an amplifier with a freqyuency-selective positive
feedback network around it. Oscillation starts by amplifying noise.

4. The pulsating current is sent through a low-pass filter to "filter
out" unwanted frequencies. (That's why I said "around the frequency"
of the crystal)

The crystal is a band-pass filter not a low pass filter. A low-pass
filter is also required on the output to remove harmonics.

 

[Jason]

Thanks for all of the help. I know that I do need to find a good book
and just start from there. The problem is finding one that doesn't
require huge leaps in understanding. For example, my ARRL handbook
begins slowly enough, but then almost immediately jumps into formulas
that describe what it going on. I recently found
http://www.ibiblio.org/obp/electricCircuits/ and I hope that helps out.
I tried electronics-tutorials.com but didn't find it too useful. It
wasn't straightforward like a textbook.

Next time, I will try to make sure that my questions are a little more
intelligent.

 

[BobG]

While we're talking about transmitters, let me ask this question: Lets
say I have a video opamp that will put out +-10V at 1 MHz into 377
ohms... can I hook this to an antenna? P=E^2/R=100/377= 260mw In
other words, the 'traditional' approach of making everything 50 ohms
and using tuned class c amps is just how it was done with tubes in the
40s right? If the impedance of the antenna is 377 ohms, and I can drive
it direct, thats all it takes, right?

 

[Andrew Holme]

BobG wrote:

While we're talking about transmitters, let me ask this question:
Lets
say I have a video opamp that will put out +-10V at 1 MHz into 377
ohms... can I hook this to an antenna? P=E^2/R=100/377= 260mw In
other words, the 'traditional' approach of making everything 50 ohms
and using tuned class c amps is just how it was done with tubes in
the
40s right?

No.

If the impedance of the antenna is 377 ohms, and I can drive
it direct, thats all it takes, right?

If there is such a thing as a single-ended (unbalanced) antenna with an
impedance of 377 ohms, you could drive it direct. If not, you can
still drive it - through a matching network. Typically, though, your
transmitter is not physically located at the antenna feedpoint - it is
located at the end of a transmission line.

The standard coax feedline impedance is 50 ohms, but this is not
necessarily the impedance "seen" by tube or transistor output stage.
The transmitter output contains an impedance matching network which
transforms the 50 ohms up or down to the right impedance to suck the
right amount of power from the transmitter.

For class C, P = V^2/2R

Where P = power "sucked" from PA stage
V = DC power supply voltage
R = Transformed load impedance as seen by PA transistor/tube

This formula assumes the peak-to-peak AC voltage developed at the
collector / drain is not far shy of 2*V.