Ham Radio Blog
I’m not particularly interested in WAJA (Worked All Japan prefectures Award), but this is to show how my 40m band dipole is working. The RF power is mostly between 20 W and 30 W.
My secret wish is to get WAS (Worked All States), but there is a long way to go. Currently, 0 out of 50. hi
Note: the map generated by http://aoki2.si.gunma-u.ac.jp/map/map.html
I am not a QSL bureau member and do not send QSLs direct. Therefore my QSOs can be confirmed only by electronically. Actually, I do not have any printed QSL cards neither my own nor received ones.
There are various ways in enjoying ham radio. I am not a card collector or award chaser, but the number of entities could be some indication of my activities.
Obviously, there is a lot of room to keep me interested in this wonderful hobby for a very long time.
The signal was 599+ even with my dipole. I was running at 13W, so it was not a QRP QSO. I should have started calling him from my usual 2W, but the temptation was too strong. Was I cheating?
We have two blue dots, which one should I use?
If you want to make the cable length to be minimum, you should definitely choose the closer one to the red point.
You mean closer when we go clockwise starting from the red point.
Correct.
What happens if I do not care for the cable length?
In that case, you can choose whichever you like. If you choose the blue dot on the upper half plane, the impedance Z there will be 50+jX [ohm], where X is some positive number.
Or if you choose the blue dot on the lower half plane, the impedance Z will be 50-jX [ohm], where X is the same as before.
Which means?
Which means, you put some capacitance in series in the former case, and some inductance in the latter case.
So far so good, but what happens if I want to jump off at the second blue dot?
Since your second blue dot is at five o’clock, you must rotate 150 degrees from the red dot, which means the cable length now should be half the wavelength times (150/360).
It is 5.5m for 40m band and for a coax cable with the velocity factor 0.66.
We jump off from the carousel at the right point, but what cable length does it correspond to?
Well, in this case your carousel rotates clockwise, or “towards generator”, technically speaking. You start from the red dot, which is approximately at twelve o’clock, and let’s say your first blue dot is at one o’clock, which means you must rotate 30 degrees on a carousel.
So?
Since the transmission line with the length of half the wavelength gives you a 360 degrees rotation of the carousel, you length should be half the wavelength times (30/360), which is the wavelength divided by 24.
If we are talking about an antenna for 40m band, it is about 1.66m, right?
Right. But, when you are using a coax cable, do not forget the velocity factor, which is around 0.66 for a cable with polyethylene dielectric.
So, it will be about 1.1m.
O.K. Now we are on a carousel, but how we go round?
By changing the length of a transmission line.
Oh?
Each time you go round, you cross with the green circle twice, shown as two blue dots. That is the point you jump off from the carousel.
(picture from [1], but the color added by the author.)
One way to solve the problem of moving from an arbitrary point to somewhere on the green circle is to use a transmission line, a coax cable, for example.
How?
There is another family of circles called SWR circles, which means the SWR is constant anywhere on a particular circle. The SWR circles are concentric circles centered on the prime center, z=1+j0. One example is the brown circle on the chart. On the brown circle, the SWR is 4.265 everywhere.
How do you know that?
Well, the brown circle is drawn so that the red point showing your antenna impedance of Z=25+j50 [ohm], or equivalently, z=0.5+j1, is on the circle.
OK. I have an antenna, measure its impedance, put a point on a Smith Chart, and then draw a circle.
Now, watch what I do. First, the reflection coefficient, gamma:
gamma=(z-1)/(z+1)=0.07692+j0.61538.
then the absolute value of the reflection coefficient, rho:
rho=abs(gamma)=0.62017.
then the SWR, or the VSWR to be exact:
SWR=(1+rho)/(1-rho)=4.265.
Of course, you are not going to add any resistors, I understand. For example, if I add a 25 [ohm] resistor in series to the antenna, moving from the red point to the pink point, the impedance becomes Z=50+j50 [ohm], or z=1+j1, and we are on a green circle, but that is not what you recommend to do.
No, there are two ways, but both does not use a resistor.
(picture from [1], but the color added by the author.)
You said “once you are on a green circle”, but what happens if I’m not.
O.K. Let’s assume you have an antenna with its impedance Z=25+j50 [ohm], which is z=0.5+j1 on the Smith Chart (the red dot).
So we start from the point in red, and somehow must go to somewhere on the green circle. How can we do that?
Well, there are two ways…