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The Development of a Transmitter


  • pg 1
									 The Evolution
of a Transmitter

   By Wayne McFee NB6M

      Salmoncon 2010
          Designing a Simple
           QRP Transmitter

• Let’s say we have a circuit we like for a 40
  Meter QRP Transmitter, but would really
  like to put it on a different frequency band.

• What do we do?
          The Answer……….

• Modify the existing circuit so that it will
  perform well on the band of choice.
            Resource Material
• The transmitter circuit we will use as an example
  is the “First Transmitter” circuit from Experimental
  Methods in RF Design, which was explored in
  yesterday’ Ham Radio Coloring Book lesson.

• This circuit, from Wes Hayward, W7ZOI, is
  discussed in detail in the first chapter of EMRFD,
  pages 1.17 through 1.22.

• EMRFD, published by the ARRL, is truly a
  “Must Buy” for anyone wanting to learn about
  radio and electronic Circuitry.
     What is the objective
       of this lesson?

• To make the necessary changes to this
  40 Meter Transmitter to make it work well
  on the band of choice.

• We will first look at putting it on 20 Meters.
How do we go about this?
• Every transmitter has certain parts with
  values selected to optimize performance
  on the desired frequency band.

• The first thing we need to do is to identify
  those parts.

• The second part of the process is
  replacing those parts with part values that
  will optimize performance on the band of
       “First Transmitter” Stages
• Although relatively simple, this transmitter is
  thoughtfully designed for good performance.

• It contains a Variable Crystal Controlled Oscillator
  (VXO), Buffer Amplifier with Power Output control,
  a Driver, and a Power Amplifier.

• It also contains a Keying Circuit and a manual T/R

• Let’s take a look at a block diagram of the circuit.
• Now let’s take a look at the circuit.

• You will note that there is a mistake in the
  drawings you have. Q5, 2N5321, is an
  NPN transistor, not a PNP as is shown.

• The schematics shown on the screen are
Now let’s take a look at the
circuit with the individual
stages outlined.

Using your colored pencils
from yesterday, outline the
stages on your own diagram.
Let’s identify the parts that are
           specific to
       40 Meter operation.
    Frequency Sensitive Parts
• Crystal
• Feedback Capacitors in the Oscillator
• Coupling Capacitor to Buffer Amp
• Impedance transforming “L” network
  between Driver and PA
• Impedance transforming “PI” network
  between PA and 7 Element Filter
• 7 Element Output Filter
• The Crystal is really easy, we just remove it and
  replace with a 20 Meter Crystal.

• The rest of the parts we are concerned with are
  either capacitors or inductors.

• What do we need to know in order to make the
  appropriate changes to these parts?

• In order to answer that question, we need first to
  understand why these capacitor and inductor
  values were chosen.
 The answer is Reactance

• Capacitors and Inductors have
  Reactance to radio frequencies.

• These component values were chosen
  because they exhibit the needed
  amount of reactance at 7 MHz.
Now let’s look at the easy part!

• Do we have to know advanced
  electronics theory, or be able to
  work with advanced mathematics
  to be able to put this transmitter
  on the 20 Meter Band?
               NOT AT ALL
• Since all of that advanced work has already
  been done by Wes, W7ZOI, we need only the

• The ability to multiply and divide simple numbers

• The ability to determine the number of turns
  needed on toroids to arrive at a desired amount
  of inductance.
                 Coils - 2010
• I know that we can all multiply and divide simple
• And we can either work out the necessary
  number of turns on toroids ourselves, using
  published formula, or use a computer program
  such as “Coils – 2010”, by Wes Hayward,

• It is contained in a zip file, coils– and is
  available here:
    Stage by Stage changes

• Let’s take this stage by stage, starting with
  the oscillator
    VXO Circuit Changes
• Change the crystal to a 14 MHz unit.

• Change the value of the feedback caps.

• Change the value of the VXO coupling
• Everyone knows this means:

 “Keep It Simple, Stupid”.

• How do we apply that theory here?
   Use the Available Info
• Wes has already worked out the part
  values required to provide the necessary
  amount of capacitive reactance at 7 MHz.

• Do we need to do any of that work?

• Do we even need to know what the
  reactance of a particular part is?
• All we really need to know is the simple
  numeric relationship between 7 and 14.
Firmly Grasp the Obvious

• OK, so 14 is two times 7, or twice the
  original number.

• How do we use that fact?
Capacitive Reactance is Inversely
   Proportional to Frequency

        Xc = ____________

What is Xc of 390pF at 7 MHz?

6.28 X 7 X .00039uF (390pF) = .0171444

   _________       =   58.32 Ohms

What Capacitance Reactance does
     390pF have at 14 MHz?

  6.28 X 14 X .00039 = .0342888

_________     =    29 Ohms

   What does this mean to us?
• This means simply that if we want a
  capacitor value that will provide the same
  amount of Xc at 14 Mhz that a specific
  capacitor value did at 7 MHz, all we have
  to do is divide the capacitor value by 2.

• 390pF divided by 2 = 195pF
  Xc of 195pF at 14 MHz

 6.28 X 14 X .000195 =      .0171444
                         (look familiar?)

_____________   =   58.32 Ohms

• To optimize the oscillator performance at
  14 MHz, we change the feedback cap
  values to 195pF, or the closest standard
  value that provides quick and reliable
  starting and running of the oscillator.

 This could be either a 180pF or 220pF cap.
• The 7 MHz coupling capacitor between the
  oscillator and buffer amp is now a 100pF
  value, which works out to have an Xc at 7
  MHz of 227.47 Ohms.

• What capacitance value will provide that
  amount of Xc at 14 MHz?
• Yes! All we do is halve the
  current 7 MHz value to arrive at
  the 14 MHz value.

• ½ of 100pF is 50pF, the closest
  standard value is 47pF.
         Moving On

• Let’s apply our KISS method
  to the next set of components
  that need to be changed.
This L Network transforms 200
 Ohms to 50 Ohms at 7 MHz

• What new Capacitance value is
  needed for 14 MHz operation if
  the 7 MHz value is 200 pF?
   Piece of cake, right?

• 200pF divided by 2 is 100pF

• What new inductance value is
  needed if we use 2.028 uH at 7
Inductive Reactance

     Xl = 2πFL
6.28 X 7 X 2.028   =   89.15 Ohms

6.28 X 14 X 2.028 =    178.30 Ohms

6.28 X 14 X 1.014 =    89.15 Ohms

 Amazing! Once again, all we need to
 do is divide the current inductance
 value by two.
       What’s Next, this is FUN!

• Let’s look at the PA Circuit, specifically at
  the impedance transforming PI network
  and 7 Element Output Filter.
           Test Time
• Here is the circuit for the PA,
  Impedance Matching PI Network, and
  7 Element Output Filter.

• Your test is to correctly identify the
  parts which need to be changed, and
  the new capacitor and inductance
  values for 20 Meters.
           Final Challenge
• If you have a 1000pF cap in a circuit
  designed for 7 MHz, and you want to
  change it to a value suitable for 10.1 MHz,
  what would the new value be?

• Hint - use the KISS method…….
        OK, a little help here
• What percentage of 10.1 is 7?

• Let’s see, 7 divided by 10.1 = 69.3

• So what’s 70% of 1000?

• New value 700pF - Closest standard value
  is 680pF
    Turn the Question Around
• What capacitance value would be needed
  for 3.5 MHz if the 7 MHz value is 1000 pF?

• Think about it - the original frequency is
  double the desired frequency.
               You Got it
• That’s right, the value for 3.5 MHz would
  be 2000 pF.

• We could use a pair of 1000pF in parallel
  if we needed that exact value.
See How Simple This Is?

• To determine the new value, use the
  percentage relationship between the
  old and new frequencies.
            Rule of Thumb
• For New L or C value, ALWAYS divide the
  original frequency by the new frequency.

7 ÷ 3.5 = 2        multiply old value by 2
7 ÷ 10.1 = ± .70   multiply old value by .7
7 ÷ 14 = .5        multiply old value by .5
7 ÷ 21 = .33       multiply old value by .33
7 ÷ 18 = .39       multiply old value by .39
Questions or comments?
Thank you all for your kind attention!

          Happy Building!

           Wayne NB6M

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