Transcript Document

NAVMARCORMARS COAXIAL CABLE TRAP DIPOLE FOR 3 & 5 MHZ
THE CONCEPT – HOW IT WORKS
• TRAPS ARE PARALLEL RESONANT CIRCUITS:
 PARALLEL CIRCUITS EXHIBIT VERY HIGH IMPEDANCE AT THE RESONANT FREQ.
THE 5 MHZ DIPOLE DOESN’T “SEE” WIRE BEYOND THE POINT OF THE TRAP.
 THE 3 MHZ DIPOLE “SEES” THE ENTIRE LENGTH OF WIRE.
• BOTH THE COIL AND CAPACITOR CAN BE MADE FROM THE SAME LENGTH OF COAXIAL
CABLE WOUND ON A LENGTH OF PVC PIPE :
 THE CAPACITOR IS FORMED BY THE CAPACITANCE BETWEEN INNER AND OUTER
CONDUCTORS OF THE COAX CABLE.
 THE WINDING OF THE CABLE ON THE FORM ALSO CREATES AN INDUCTANCE.
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NAVMARCORMARS COAXIAL CABLE TRAP DIPOLE FOR 3 & 5 MHZ
COAX CONNECTION DETAIL
• THE COAX (RG-58/U IS WOUND ON A 1/8-IN PVC PIPE FORM 1-1/2 IN IN DIAMETER.
• THE TRAP WILL RESONATE AT THE LOW END OF 5 MHZ WITH 13 TURNS OF
COAX CLOSE-WOUND ON THE FORM.
• USE BRAIDED CENTER CONDUCTOR COAX VS SOLID –IT’S EASIER TO WORK WITH !
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NAVMARCORMARS COAXIAL CABLE TRAP DIPOLE FOR 3 & 5 MHZ
THE LENGTH OF A HALF-WAVE DIPOLE IS CALCULATED BY THE FORMUA:
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LENGTH IN FT = -----------------F (MHZ)
• CALCULATE THE LENGTH OF THE 5 MHZ DIPOLE FIRST:
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L = ------------------- = ------------------ = 86.988 OR 87 FT
F (MHZ
5.380 MHZ
• CONNECT 43.5 FEET OF WIRE TO EACH SIDE OF THE CENTER INSULATOR AND
THEN CONNECT THE TRAPS.
• CALCULATE THE LENGTH OF THE 3 MHZ DIPOLE NEXT:
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L = ----------------- = -------------------- = 143.1 OR 143 FT
F (MHZ)
3.270 MHZ
• SUBTRACT THE LENGTH OF THE 5 MHZ DIPOLE FROM THE 3 MHZ DIPOLE:
143 FT - 87FT = 56 FT
• ADD 28 FEET OF WIRE FROM EACH OF THE TRAPS TO THE END INSULATORS.
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NAVMARCORMARS COAXIAL CABLE TRAP DIPOLE FOR 3 & 5 MHZ
• TRIM THE DIPOLES TO THE DESIRED CENTER FREQUENCY AND LOWEST SWR.
• START WITH THE 5MHZ DIPOLE THEN TRIM THE 3 MHZ DIPOLE.
• TIP: PRUNE ONLY 6 INCHES OF WIRE AT A TIME.
• EXACT LENGTH DEPENDS ON MANY FACTORS:
 THE QUALITY OF THE COAX USED
 THE HEIGHT OF THE DIPOLE ABOVE GROUND
 NEARBY OBJECTS: HOUSE, TOWER, TREES, SWING SETS, ETC.
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