[Asterisk-Users] [OF] Cable Pinouts

[Jon Pounder]

if you are adventurous, find the line termination resistors on the board. 
The transmit one will be coming from the low impedance transmit pin on the 
driver through likely a 100ohm resistor, possibly fuses or thermistors to 
the output pin in the rj45. The receive one will be similarly in series 
with the high protectors and in parallel with the high impedance pin of the 
receiver.

Due to the high/low impedance of the driver pins, you can normally simply 
replace the line termination resistors with the correct values (anything 
less than 1k is not going to change the rest of the circuit parameters 
enough to matter in well designed hardware)

I can't say on the digium stuff but in some cases a jumper will be provided 
to series or parallel up more resistance to change between common 
impedances easily. eg: components with an in and out jack often make the 
termination resistor selectable in the circuit and only use it when they 
are at the end of the line.


At 10:59 PM 5/27/2003 +0800, you wrote:
>Eduardo Goncalves wrote:
>
>>Hi,
>>
>>Digium's E400P has RJ45 conector and my E1 link has BNC concetor. Could 
>>someone tell me the cable pinouts to make this conection?
>Your BNC is a 75ohm unbalanced line. The E400P expects a 110ohm balanced 
>line. If you are lucky you can just fudge the connections. When the line 
>is longer you normally need a little balun gadget to interface them. These 
>contain almost nothing, but can cost quite a lot. A cheaper solution is to 
>go to a place that sells CCTV for security. They usually have cheap BNC to 
>twisted pair adaptors. These are not quite the right impedance match for 
>the E1 application, but usually work OK.
>
>The pinout for the RJ48C (it looks and feels like an RJ45, but it is 
>really an RJ48C) on the E400P is:
>
>pin 1 receive ring
>pin 2 receive tip
>pin 3 N/C
>pin 4 transmit ring
>pin 5 transmit tip
>pin 6 N/C
>pin 7 frame ground
>pin 8 frame ground
>
>Regards,
>Steve
>
>
>
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