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Author Topic: Antenna Matching Using A Coaxial Stub  (Read 2356 times)
TANAKASAN
Member

Posts: 933




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« on: November 16, 2012, 09:18:29 AM »

This matter is currently the source of considerable argument in our radio club so I'm looking for some opinions. If nothing else I need someone to check my math:

Suppose we have an electrically short antenna on 160m and using the VNA we measure an impedance of 350 ohms. After adding various lengths of coax at the rig end as a stub we find that an 18 foot length brings the impedance down to 50 ohms. So, using the formula for resistors (or impedances) in parallel we have

350 ohms antenna impedance
58.33 ohms stub impedance

50 ohms impedance at the rig

However, what happens if we feed 500W of RF into the network? The HF P.A. puts out 500 watts into 50 ohms, this is 158 volts RMS at 3.16 amps

The antenna is being fed with 158 volts and therefore 0.45 amps flows
through the impedance of 350 ohms. 71.42 watts are transmitted.

The 58.33 ohm stub is also being fed with 158 volts but its lower
impedance means that 2.71 amps flow through the cable. 428.6 watts are
dissipated across the length of the cable which will be enough to warm
the room.

Is this correct and if not where am I going wrong?

Tanakasan
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AA4PB
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Posts: 12836




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« Reply #1 on: November 16, 2012, 09:23:33 AM »

The stub impedance consists of reactance and resistance. Only the resistance part disipates power as heat.
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W5DXP
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Posts: 3581


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« Reply #2 on: November 16, 2012, 10:09:59 AM »

350 ohms antenna impedance, 58.33 ohms stub impedance

350 ohms at what phase angle? What is the resistance? What is the reactance?

58.33 ohms at what phase angle? What is the resistance? What is the reactance?

You have an antenna feedpoint complex impedance in parallel with a stub complex impedance. DC electrical math will not work.
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
TANAKASAN
Member

Posts: 933




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« Reply #3 on: November 16, 2012, 12:13:47 PM »

See, I knew I was missing something Roll Eyes

Tanakasan
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WB6BYU
Member

Posts: 13242




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« Reply #4 on: November 16, 2012, 12:57:07 PM »

As others explained, you can't use DC math to solve the problem:  you have to consider
the impedances as vector quantities containing both reactance and resistance.

But the length of the stub (assuming it gives an exact match) is sufficient to solve
the problem. You can follow along and change any of the parameters as needed
if I have made a wrong assumption.

Let's assume that you are using RG-213 coax for the stub. Using VK1OD's Transmission
Line Loss Calculator
for 18' of RG-213 terminated by 0.01 ohms (close to a short circuit)
at 1.8 MHz, the impedance looking into the stub is 0.55+j16.21 ohms, or .002098-j0.0616 S
in parallel form.  You should get basically the same result using a coil with a reactance of 16.21
ohms (about 1.3uH) in place of the stub.  (Any analysis that doesn't also work using such a
coil in place of the coax stub is suspect.  The residual 0.55 ohms of resistance is due to the
imperfect short circuit and the cable losses.)

Now if this gives you a perfect match to 0.02 S (50 ohms) it must be that the impedance
at the end of the coax is (0.02 +j0 S) - (0.002098 - j0.0616 S) =  0.017992 + j0.0616 S,
or 4.37 - j14.96 ohms, so the SWR without the stub should be about 12.5 : 1.  (If you don't
have a calculator that goes between parallel and series notation, you can plug one or the
other into VK1OD's calculator with a very short bit of transmission line.)

In fact, knowing the impedance at the transmitter, VK1OD's calculator will give you the
actual impedance at the antenna if you know the type and length of feedline.


Why did the analyzer give an impedance of 350 ohms for this load?  I have no idea.  Perhaps
it is out of the range over which the analyzer is accurate, or the stub didn't really give a
perfect match to 50 ohms.  (There is no guarantee that it will for random impedances.)


So all that we did was to put 0.55+j16.21 ohms in parallel with 4.37 - j14.96 ohms to get
50 ohms.  We know the voltage across the stub from the output power (since it is at
a 50 ohm point in the system).  If we used a coil instead, the losses would be based on
the Q of the coil.  The losses will be somewhat higher using a coax stub due to the
losses in the coax.  

But power is ONLY dissipated in the resistive portion of impedance, not in the reactive
part.  That's why you can't use the total impedance to calculate power lost.


(And say "Hi" to your club members from me.)


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