CALCULATING COILS FOR THE HF BANDS

                    

Several people have asked me about determining the number of turns of wire
for a particular frequency when building QRP radio projects.   I have many
times given the ROUGH "starting" point:-

            Capacitance(pf) = Wavelength in meters
            Coil (turns)    = Wavelength in meters

Naturally the size of the coil former you use will affect the turns needed
for a particular frequency, so this "starting"  point is very rough and is
only intended to get you in the right vicinity.   A GDO is one of the most
tools I posess, and I will describe one of these in another posting in the
near future. To answer the first question  for those who are interested in
calculating coil turns, read on.

There are several formulas for calculating the number of turns required in
any given application, but the results seem to differ quite a lot from one
formula to another. I use two variants of Wheelers formula, one for single
layer coils and another for multi-layer (pile wound) coils.   I shall give
the formulas twice, in case some of these  "special"  ASCII characters are
printed as garbage on the text editor you use (I use XTGOLD).

The formulas assume that there is NO ferrite, or brass core to the coil. A
ferrite core will reduce the number of turns by 10% to 25%, depending upon
the size of the ferrite lump, and how far it is screwed into the former. A
long ferrite rod can reduce the number of turns by as much as  50% to 60%. 
A brass tuning slug will INCREASE the turns required by typically 5% - 10%

Just in case you have forgotten,   you can calculate the value of required
inductance, capacitance and impedance with these formulas,  but I will not
insult you intelegence explaining them.


                               1
Capacitive reactance (Xc) =  ------
                             2 f C

Incuctive reactance  (Xl) =  2 f L

                                1
Tuned circuit frequency   =  ------
                             2 LC

At resonance Xc = Xl

Tap inductors;   Impedance ratio = turns ratio squared
--------------------------------------------------------------------------
SINGLE LAYER COILS

              L (9r + 10l)                              L (9r + 10l)
    N   =    ------------                  N x N   =   ------------
                   r                                       r x r

  where  L = inductance in micro-henries
         l = length of the winding in inches
         N = the number of turns
         r = outside radius of the coil in inches

EXAMPLE. I want a coil to have 20uH inductance in an HF PA valve anode for
         30 MHz.  I shall use a toilet roll tube for a former (2 inches in
         diameter).   I will space the coil out so that the winding length
         is one inch.

              20 x (9x1 + 10x1)           20 x 19              380
     N   =   -----------------     =     -------      =       ---
                    1 x 1                    1                  1

     N =  380   therefore    N  =  19.4935 turns       (20 turns)

Even 4mm Dia. Cu wire will allow 20 turns per inch, take your pick.  If it
burns when you apply power then use a thicker wire.
--------------------------------------------------------------------------
MULTI-LAYER COILS

             L (3a + 9b + 10c)                          L (3a + 9b + 10c)
    N   =   -----------------              N x N   =   -----------------
                  0.2a                                   0.2a  x  0.2a

  where  L = inductance in micro-henries
         a = winding thickness + former diameter in inches
         b = length of winding in inches
         c = winding thickness in inches
         N = number of turns

EXAMPLE. I want a coil for a 1MHz medium wave QRP pirate-radio transmitter
         amplifier stage tuned circuit.    The inductance I need is 200uH.
         I will use a plastic bottle-top former, 0.5" Dia.   I can allow a
         a total of 1.0" for the coil diameter, and, it must be 0.5" long.
                 (L = 200,   a = 0.75",   b = 1",   c = 0.25")

             200 (3x0.75 + 9x0.5 + 10x0.25)         200 (2.25 + 4.5 + 2.5)
    N   =   ------------------------------    =    ----------------------
                    (0.2 x 0.75)                          0.0225

             200 x 9.25        1850
    N   =   ----------   =   ------   =   82222.222r
               0.0225         0.0225

    N   =   82222.222r   therefore   N = 82222.2222   =   286.744 turns.

This winding is .25" thick by .5" long.  Total cross-sectional area of the
coil is 0.125".  I therefore need a wire thickness that will fit into this
area = 2293.953 turns per square inch or thinner.  29 SWG (0.35mm Dia.) is
the max thickness of wire I can use, according to the following table.
--------------------------------------------------------------------------
AWG   SWG   (mm)    t/i     t/i     /100m         COPPER WIRE TABLE
---   ---   ----   -----   ------   -------     -------------------------
 --    47   0.05   444     197000   894         AWG    = American Wire
 38    42   0.10   151      22900   224                  Guage.
 35    38   0.15   109      12000    99
 32    36   0.20    85.7     7300    55.8       SWG    = Standard Wire
 30    33   0.25    74.6     5550    35.7                Guage.
 29    31   0.30    67.6     4550    24.8
 27    29   0.35    56.5     3190    18.2       (mm)   = Diameter of wire
 26    27   0.40    51.5     2650    14.0                in millimeters.
 25    26   0.45    46.5     2070    11.2
 24    25   0.50    42.4     1789     8.9       t/i    = Number of turns
 23    23   0.60    38.3     1513     6.21               per inch.
 21    22   0.70    29.2      852     4.54
 20    21   0.80    26.0      676     3.49      t/i   = Number of turns
 19    20   0.90    23.5      550     2.76               per square inch.
 18    19   1.00    19.7      388     2.26
 12    14   2.00     9.3       86     0.567     /100m = Ohms per 100meter
  9    11   3.00     7.2       51     0.252              length.
  6     8   4.00     6.1       38     0.142
--------------------------------------------------------------------------
0.05 mm wire will handle     3 mA
0.10 mm wire will handle    12 mA
0.50 mm wire will handle   300 mA
1.00 mm wire will handle  1.25 amperes
4.00 mm wire will handle 20.10 amperes

Further reading:

1. "Radio and Telecommunication Engineers Design Manual" by R. E. Blakley
   (Pitman).                              [Quite detailed]
2. "Coil Design and Construction Manual" by B. B. Babani
   (Bernard Babani Publishing)            [Simple presentation]
3. "The ARRL Handbook For Radio Amateurs" by the ARRL
   (ARRL)                                 [Information a little limited].

Have fun, de HARRY, Upplands Vasby, Sweden.


I am very grateful to Harry, SM0VPO for giving me permission to use his circuits and ideas. He can be reached by Email at harold.lythall@era.ericsson.se

Frank, G3YCC

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