L = (a^{2}N^{2}) ÷ (9a + 10b)

Where N = Total Turns

a = radius length in inches

b = winding length in inches

Turns may be space wound or close wound.

The INDUCTANCE of coils with air cores is:

- Proportional to the square of the number of turns if the length and diameter of winding are kept constant as the number of turns are altered.
- Proportional to the size (length or shape) of coils having same shape (b/a) and same number of turns.

Thus two coils; one twice as big as the other with same number of turns has twice the inductance of the smaller.

Q = 2πfL ÷ R

Where f is frequency in Hertz and R is the Resistance of the coil.

The Q of a coil is dependent on a number of relative factors such as physical size, wire diameter, form factor, etc.

To obtain highest Q:

- Choose a size of coil where the length is moderately greater than the diameter;
- then choose wire size so that wire diameter occupies 0.5 to 0.75 of spacing between the turns.
- The LARGER the physical size of the coil, the HIGHER the Q will be- the fewer the number of turns required to obtain a given inductance- the larger will be the optimum wire diameter.

The coefficient of coupling (K) between two coils:

K = M ÷ √L_{1}, L_{2}

Where M is mutual inductance L_{1} is inductance of first coil, L_{2} is inductance of second coil.

M, L_{1}, and L_{2} can be in any units as long as they are the same in henries, micro-henries, etc.

Under conditions of resonance, the load, antenna, dummy, etc. couple resistance into the tank circuit in the amount:

(2πfM)^{2} ÷ R

where f is frequency in Hertz and R is the Resistance of the *LOAD*. Hence the LOADED Q of your tank circuit can be determined by knowing M if you know K.

- K varies from 0.35 to 0.37 if the coupling link is in the center of the tank but of the same diameter.
- When the link is over one end, the K is approximately 0.3.
- When the link is of the same diameter as coil but spaced slightly off the end, the K is approximately 0.2
- There is little reduction in K if the link in the center is wound over the tank coil on a large diameter form.
- When the link is of smaller diameter, K decreases in proportion to the reduction in diameter.

If the tank coil is overheating it may *not* need larger diameter wire- excessive circulating current in the tank may be the cause. If this is the case your final is POORLY designed and only a SMALL PORTION of your power is ending up at the antenna. Now, we could suggest using a very large conductor, but this would not SOLVE the problem. This large circulating current is passing through your switches and capacitor which is *not* good and eventual failure is probable.

Here are some points to keep in mind:

- Tank (pi network) circuit inductance is inversely proportional to EFFECTIVE Q of circuit, other things (load V, power output and frequency being equal.

Q_{EFFECTIVE}= ωL ÷ (R_{COIL}+ R_{coupled from load}).

The EFFECTIVE Q is the ratio of the coil reactance (ωL) to the SUM of the resistance inherent in the coil and resistance coupled into the tank circuit by the load.

- A high effective Q corresponds to a low L/C ratio with a resultant large circulating current, which, flowing through the resitance inherent in the coil causes heating. Conversely a lower effective Q corresponds to a high L/C ratio and a lower circulating current.

Capacitor spacing based on representative voltages of input capacitor of a pi network or tank capacitor in a parallel tuned tank.

BAND MHz | POWER INPUT | WIRE SIZE | CAPACITOR SPACING |
---|---|---|---|

3.5 | 1000 W | 10 | 0.25" |

7.0 | 1000 W | 8 | 0.25" |

14.0 | 1000 W | 8 or 1/4" tube | 0.25" |

28.0 | 1000 W | 8 or 1/4" tube | 0.25" |

3.5 | 500 W | 14 | 0.08" |

7.0 | 500 W | 12 | 0.08" |

14.0 | 500 W | 12 | 0.08" |

28.0 | 500 W | 8 or 1/4" tube | 0.05" |

3.5 | 150 W | 18 | 0.03" |

7.0 | 150 W | 14 | 0.03" |

14.0 | 150 W | 14 | 0.03" |

28.0 | 150 W | 12 | 0.03" |

3.5 | 75 W | 18 | 0.02" |

7.0 | 75 W | 18 | 0.02" |

14.0 | 75 W | 18 | 0.02" |

28.0 | 75 W | 14 | 0.02" |

The output capacitor of pi network need not be more than 0.015" for a 50 ohms load at 1000 W input. For lower powers, receiving capacitors may be used in most cases.