Using one antenna for RF transmitter and receiver
SZ_78
Posts: 1
I just wanted to know is it possible to connect only one antenna to both RF transmitter and RF receiver. I'm using a TWS434A (6 pin package) and an RWS 434A with a Whip antenna.
Thanks
Thanks
Comments
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Manetheren
If the transmitter and receiver are in close together the signal will still induct into the receiver so if that's a problem you will have to use 2 different frequencies.
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 4/25/2009 3:54:00 AM GMT
If on the other hand the frequencies are close together, you'll need a different sort of duplexer, large and expensive and requiring adjustment with exquisite precision.· It won't fit in a shoe box, but (depending on frequency) might fit in the shipping carton for, say,·a microwave oven.
Jmalaysia suggests that you could use diode switches, but this is true only at very low power levels; and if you use it with a transmitter it carries the risk of generating spurious signals that would interfere with other radio services.
Me, I'd use a simple relay.
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· -- Carl, nn5i@arrl.net
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· -- Carl, nn5i@arrl.net
The trick is to assure that the DC current you impose (to turn the switch on) is much larger than the instantaneous peak current of the radio-frequency signal.· "Much larger than" means, say, twice as large anyway.
So you must·be able to calculate that instantaneous·peak RF current.··This is not too tough if you know the carrier power and the kind of modulation.·
Here's an example:
Assume an amplitude-modulated signal, at 1 watt carrier power, in a 50Ω system.
Because it's AM, the rms power at a modulatiion peak envelope will be 4x the carrier power, or 4 watts.
The signal voltage (rms) at a modulation peak·we calculate from p=e2/Z so with 4·watts in 50Ω· we get 14 volts rms.
The peak signal voltage·is 'prox 1.414 times this, or about 20 volts.
The peak·signal current, then,· is 20v/50Ω or 0.4 amperes.· To switch the diode on you need at least twice this, or 0.8 amperes.· Not too tough.
For other modulation types·the calculation of peak RF power would be different.· For FM, as an example, the peak and average RF power are identical, so you wouldn't multiply by 4.· For radar signals, usually the peak RF power is what's specified anyway.· And like that.·
All of that is if my arithmetic is correct.· No guarantees.
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 4/27/2009 3:27:27 PM GMT
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Manetheren