# How transmitters work.

Discussion in 'Electronic Basics' started by Jason, Apr 20, 2005.

1. ### JasonGuest

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
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

Comments and help would be greatly appreciated.

2. ### Andrew HolmeGuest

Crystals, like capacitors, don't pass DC.
The oscillator is an amplifier with a freqyuency-selective positive
feedback network around it. Oscillation starts by amplifying noise.
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.

3. ### Rich GriseGuest

More or Less.
Not Exactly.
In a way, yes.
http://www.electronics-tutorials.com/basics/basic-electronics.htm

Have Fun!
Rich

4. ### Michael BlackGuest

No. A crystal is made of quartz, which isn't a conductor. It will
never pass DC. Electrically, a crystal is the equivalent of an inductor
and capacitor in series, hence it's ability to control frequency.

is a big difference between asking questions based on points you
don't understand, and asking us to write that material down here.

You don't have the basics to understand how it works, and you

Here's one place to start. Take an amplifier of some sort (even
a computer with speakers connected to a soundcard), and plug
in a microphone. Hold the microphone near the speakers, and
start turning up the volume. You'll start getting a squeal,
that's the positive feedback, ie oscillation.

The signal goes through the amplifier and then is fed back to
the input, not just once but an infinite amount. It's no longer
amplifying an external signal, it is oscillating.

The frequency of the howl is determined by the inherent characteristics
of the setup, things like the response of the microphone and the speaker,
and the response of the amplifier, and even the microphone's distance
from the speaker. You don't have much control over the frequency, but
it is a good demonstration of oscillation.

Usually, one wants something more definite. So instead of the
speaker/microphone combination, the output of the amplifier feeds
the input directly. And between the output and the input is
some selective element so the feedback is at that frequency. Put
a bandpass filter in there that has a center frequency of1KHz,
and the thing will oscillate at 1KHz.

The crystal acts as the selective element, to ensure that the
oscillator runs at that frequency.

A crystal is one of the better filtering devices available, it
doesn't generate unwanted signals.

But the active elements are not so perfect, and they will often
add unwanted signals. Indeed, the design of some transmitters
will count on generating multiple signals, with the wanted
one filtered out.

Michael

5. ### JamieGuest

lets break it down in a different form.
forget the crystal idea for now, that is only a source to generate
a stable frequency reference!. there are other ways.
any ways.
This generator will produce AC (alternating current/Voltage) at a rate
per second that is equal to the frequency of what you want to transmit on.
for example:
1.6 Mhz, = 1600000 times per second will an alteration of a signal
going from 0 Volts, to Full Volts+, back to 0 volts, then Full Volts-,
and then back to 0 volts completes a full cycle.
this signal is most likely then past to an amplifier to increase the
output power abilities.
it then makes it to the antenna as a sine wave (AC), at that point
the AC will generate an electro magnetic field via the antenna..
and since its now magnetic, the principles of Unlike and like
polarities apply..
minus and minus for example repel the magnetic energy that
was there before gets pushed away.. this is how the energy actually
travels in space.

Known as a simple carrier, it really does nothing for you other than
allows another
CW (Morse code)is a good example
systems like AM (amplitude modulation) will modify the level of this
sine wave power output to reflect the patterns of the audio frequencies.
freq, many cycles of your carrier freq will be sent out in space
changing power levels in very gradual steps! but the receiver at the
other end is converting this carrier into a DC (direct Current signal)
where is the level of voltage that is changing depending on the Level of
the signal which is influenced by the audio//
etc.//

if you need a deeper detailed than that i think i can also supply that
also. this looked as simple as i could explain it.

6. ### JasonGuest

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.

7. ### BobGGuest

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?

8. ### John SmithGuest

Impedance mismatch will mean that the antenna will not radiate 260mW, a
significant amount will be reflected back to the op-amp. That is why the

9. ### Andrew HolmeGuest

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.