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Johnson Messenger 223 crystal question

May 25, 2011
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Does anyone know which leg on the crystals is the hot side and which side is the ground on a Johnson Messenger 223. Also does the crystal osillator circuit use a floating ground? I am trying to install a Pal VFO. Thanks
 

CK is correct, again.
The crystal works in series with the circuit and is not polarized.
Love those old Johnsons, I had 2 loaf of bread radios back in the day.
Johnson Messenger 2`s , had 8 plug in ( 16 slots, 8 for TX and 8 for RX) crystal channels AND a you could switch in the 23 channel RX separately.
Fun old school radio.

73
Jeff
 
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I have never known any crystal to be polarized. They work either way.
I was confused because the VFO installation instructions say to put the center conductor to the hot side of the crystal socket and the shield into the ground side unless it has a floating ground in the crystal oscillator circuit. Then they say to connect the shield to the chassis
 
That is because of the circuit design. One side of the crystal is grounded and the other side is connected to the circuit itself. The crystal itself has no polarity to worry about. Plug it in either way. Some crystals have a third wire connected to the metal case which is grounded to shield the crystal.
 
I guess what I'm asking is how can I tell which side is the circuit and which side is the ground. Maybe it will be self evident when I get into it?
 
Do you have a schematic and if so do you know how to read it? That is the easy way to find out what you want to know. If not then trace the circuit. One side of the crystal should go to ground either directly or via a capacitor. The "hot" side of the crystal should go to the base of a transistor, again either directly or likely via a small capacitor.
 
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Do you have a schematic and if so do you know how to read it? That is the easy way to find out what you want to know. If not then trace the circuit. One side of the crystal should go to ground either directly or via a capacitor. The "hot" side of the crystal should go to the base of a transistor, again either directly or likely via a small capacitor.
Thanks CK, that answers my question!
 

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