On Mon, 03 Nov 2008 03:08:55 GMT, Owen Duffy wrote:
Jeff Liebermann wrote in
:
...
However, if you wanna run 50 ohm coax, the mismatch loss at the 75 ohm
antenna is about:
reflection_coef = (75-50)/(50+75)= 0.20
voltage = 1 - (0.2^2) = 0.96
20 * log(0.96) = 0.35 dB mismatch loss.
The analysis you give assumes that the notional 'reflected power' is lost
from the system as heat. The old 'reflected power is dissipated as
increased heat in the PA' line.
In the real world, the power that a transmitter delivers to a non ideal
load is not so simply predicted, and it is entirely possible that it
delivers more power to the mismatched load.
Owen
I beg to differ somewhat. In order for the reflected power to
contribute to the incident power, the reflected power would first be
attenuated by the coax loss. It would then require a substantial
mismatch at the transmitter, which is unlikely. However, assuming
there is a mismatch at the source, some of the reflected power will be
sent back to the load (antenna), after getting attenuated by the coax
for a 2nd time. There may be some contribution, but it will very very
very very small.
Let's try some more or less real numbers. I kinda prefer doing
everything in dBm but hams have this thing about using watts...
Start with a 50 watt xmitter and 20 meters of LMR-240 coax at
0.09dB/meter for an attenuation of 1.8dB.
The power delivered to the antenna is:
50 watts / ((1.8/10)^10) = 50 / 1.5 = 33 watts
The 1.5:1 VSWR reflects 4% of 33 watts for 1.3 watts reflected.
The 1.3 watts is again attenuated by the 1.8dB coax loss resulting in:
1.3 watts / (1.8/10)^10) = 1.3 / 1.5 = 0.87 watts
Now, lets assume that the xmitter has a broadband output stage,
optimized for 50 ohms and lacks the ability to properly match 75 ohms.
Once again, 4% of the power if reflected, resulting in:
0.87 watts * 4% = 0.035 watts reflected
Once again, the power reflected from the source end (xmitter end) is
attenuated by the 1.8dB coax loss for:
0.035 watts / ((1.8/10^10) = 0.035 / 1.5 = 0.023 watts.
Therefore, you're correct. It's possible that some of the reflected
power adds to the incident power. However, it's a really small
amount. In this case, it's only 23 milliwatts added to 33 watts
delivered to the antenna.
--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060
http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558