"Henry Kolesnik" wrote in message
...
Well I learnt how to do Bessel functions 40 years ago!
But today, who knows, I think Lipitor took some brain cells.
Go ahead with the complex variables..
I could be the lost son of the square root of minus one!
73
hank wd5jfr
Since I don't have the QST article with me to use as an example, lets just
start with a 10 Ohm resistor like in your OP. You really want 50 Ohms. So
what happens if I put a 1.137 nF capacitor (bear with me) in series with it?
The reactance of the capacitor is 20 Ohms at 7 MHz. The impedance now looks
like 10-j20, right?
20 Ohms
||
---||-------.
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.-.
| | 10 Ohms
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'-'
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------------'
Well, you don't want the -j20 but, if you add +j20 in series to get rid of
it, you're just back where you started with 10 Ohms resistive. So, what if
you put in a shunt inductor to compensate for the cap?
20 Ohms
||
o-----o----||-------.
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| |
C| .-.
C|25 Ohms | | 10 Ohms
C| | |
| '-'
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o-----o-------------'
created by Andy´s ASCII-Circuit v1.24.140803 Beta
www.tech-chat.de
It is easy to see how much inductance you need if you take the reciprocal of
the impedance. That is, find the admittance.
Y = 1/(10-j20) but here is where you do the complex conjugate bit.
Y = (1/10-j20)*(10+j20)/(10+j20) = 10+j20/(100+400) = 0.02+j0.04 and what
you really want is .02+j0. (The reciprocal of 50 Ohms).
So, if you add a shunt inductor which is 1/.04 or 25 Ohms at 7 MHz, then you
will have only the resistance left.
Once you do that, your equation will be
Z = (10-j20)*(0+j25)/(10-j20+0+j25) = 50+j0
I'll let you prove it to yourself.
If the load is a combination of resistance and reactance, similar procedures
can be applied but it sometimes takes some concentrated thought.
If your load resistance is less than the resistance you want, you can use
this to find the required series reactance:
X = sqrt(Rload*Rline-Rload*Rload) where the X is either capacitive or
inductive (whichever you choose, use the opposite for the shunt). This will
give the amount of series reactance needed so that, when you add the shunt
reactance as shown above, the resistance will be what you desired.
You can develop a similar approach for working with loads which are higher
than desired. Just start by assuming you need a shunt reactance at the load
and then a series compensating reactance.
I'm a lousey teacher, but I hope this has helped. If not, there are some
great guys here who will come to our aid.
73,
John
"John Smith" wrote in message
...
"Henry Kolesnik" wrote in message
...
In the Feb. 2004 QST there's an article by AD5X for a mobile antenna
using
a
Hamstick or Bugcatcher with shunt capacitor feed or an L match. Can
someone
please explain, including the math, on how a ~500pf capacitor
transforms
10
ohms to 50 ohms for 40 meters? I get no understand from reading either
my
ARRL Handbook or ARRL Antenna Handbook. There are no stupid
questions,
only
stupid people asking!
tnx
hank wd5jfr
Hi, Hank -
If you have a low value resistor and you want to make it appear to your
source to be a higher value resistor, you can add a series resistor or
reactance. If you add a resistor, it will use up some of the available
power
which you intended for the original resistor. So, you use a series
reactance
instead. The only problem is that you now have some (usually)
undesirable
reactance to contend with. But, you can get rid of the series reactance
by
adding a shunt reactance of the opposite type. That is, if you added a
series capacitor, you can put in a shunt inductor to compensate. Note
that
you could have added a series inductor and compensated with a shunt
capacitor. Which one you use is a matter of convenience sometimes.
Before I go into the math, I should ask if you are comfortable with
complex
variables. You know, complex conjugate, R+jX and all that stuff?
Besides,
the delay will give someone else an opportunity to put in something I've
missed or messed up. There are a lot of very knowledgeable guys on this
group, you know.
John