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Author Topic: Incorrect capacitor for matching. problem?  (Read 295 times)
VK1KHV
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Posts: 73




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« on: September 08, 2009, 12:12:22 AM »

I have a 500PF doorknob i sourced today for free. and im wondering. i have a 3/8 wave vertical on 80 its 50 j60.

claculations ask for a 680PF capacitor on 3.8mhz

if i use a 500PF what will the 50 j60 turn into 50 j15?

73
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VK1KHV
Member

Posts: 73




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« Reply #1 on: September 08, 2009, 01:44:58 AM »

is there a way to calculate this "whatif"? scenario where your using what you have in the junkbox so to speak.
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N3OX
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Posts: 8853


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« Reply #2 on: September 08, 2009, 05:39:02 AM »

Yes.

The reactance of a capacitor in ohms is

-jXc = -j/(2*pi*f*C)

Where f is the frequency in Hz (so put 3,500,000 Hz) and C is the capacitance in Farads (500 x 10^-12 F).  If you use Google calculator or other "auto-converting" system you can enter the cap in pF and the frequency in MHz, but if you're doing it by hand you have to use base units of Hz and Farads or you'll be many orders of magnitude off ;-)

Then, since the capacitor is in *series* the reactance adds in series.  Only the imaginary part -jX is affected by the series addition of a pure reactance.

So for 500pF, -jXc = -j/(2*pi*f*C) = -j/(2*pi*3.5MHz * 500pF) = -j90.9 ohms

50 + j60 ohms - j90.9 ohms = 50-j30.9 ohms

I could write this as:

Zant + Zc = Zmatched

or

(Rant+jXant) + (Rc+jXc)

(50+j60 ohms) + (0-j90.9 ohms) = (50-j30.9 ohms)

You can do this with any sort of series/parallel RF circuit, although the parallel combination

1/Z = 1/(1/Z1 + 1/Z2)

has much more complicated algebra where the reactance can affect the parallel resistance and vice versa.

In the end, though, it's just a generalization of Ohm's law to include complex impedances... so series combinations add and parallel combinations add in inverse.  Just the inverse of a complex number is a bit weird.

73
Dan
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73,
Dan
http://www.n3ox.net

Monkey/silicon cyborg, beeping at rocks since 1995.
W5DXP
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Posts: 3546


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« Reply #3 on: September 08, 2009, 06:07:33 AM »

Monopoles shorter than 1/4WL exhibit a capacitive reactance. For that 3/8WL monopole, that impedance is probably 50-j60. A series or parallel capacitor will make that impedance worse. You need a series inductance to neutralize the series capacitance.

You may be using an MFJ-259B to read the reactance and it doesn't display the sign of the reactance.
--
73, Cecil, w5dxp.com
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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.
K1DA
Member

Posts: 473




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« Reply #4 on: September 08, 2009, 09:04:12 AM »

Is not a 3/8 LONGER than a 1/4 wave ..--..
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