Lesson 5.3 Power Supplies and Batteries

5 Responses to Lesson 5.3 Power Supplies and Batteries

1. Richard A Purdy says:

   September 25, 2014 at 1:32 pm

   Dave,
   Tnx for your video lectures. They fill in the gaps in the ARRL License Manual.
   Re: Antenna length. Thx for the concept of ‘velocity factor’.
   For Example: Regarding the 2014 exam question T8A09 — The 2014 Tech Q&A answer book mercifully omits this complication by implying that there are 3.1 feet/m.
   — A useful shortcut for the majority of students.
   I appreciate hearing ‘the rest of the story’ about ‘velocity factor’ from your video.
   Thanx Again.
   Richard
2. Dave says:

   April 28, 2014 at 5:57 pm

   Hi Arthur, I guess the answer is “both.” My BYU undergraduate degree is in Math. My Cal State Northridge MS is in Engineering. After a stint in the Air Force, I’ve worked for either aerospace or IT companies. I’ve taught classes at work and at church and (for ham radio) as a hobby. I think I picked up my love of teaching from my father, a career teacher, who instilled in me the idea that my education is never finished and that helping others understand how man and nature work is one of the highest forms of service.
3. Arthur Charap MD Ph.D. Professor of ophthalmology UC Irvine [ret] says:

   April 28, 2014 at 5:45 pm

   your answer was perfect[Aristotelian!] . I now understand both the facts and the fudge factors. Were you a teacher or an engineer or both?
   Either way your explanation was a pleasure to read and internalize. Surprisingly ,I have not found your explanation in any of the online courses and if it is in the ARRL book I must be getting Alzheimer’s. I don’t mind memorizing rules etc. but the formula seemed to be pulled out of thin air[pun intended] . Once more ,thanks for your amazingly prompt and enlightening reply to my query
   Sincerest regards, Arthur
   P.s. I don’t know if you discuss the antenna theory you gave to me in your lectures but it certainly should be there for anybody else who failed to find 468 as intuitive as pi or relativity theory
4. Dave says:

   April 28, 2014 at 5:20 pm

   Arthur, the key is to understand the relationship between frequency and wavelength. It’s one thing to say that as frequency goes up the wavelength goes down, but why is that so? Look at it this way: light (or a radio wave) travels approximately 300,000,000 meters in one second in free space. If a radio wave vibrates at 7200 kHz, that means in one second the wave has cycled 7,200,000 times. So, in the one second that it travels 300 megameters, it has completed 7.2 million cycles. We have units of distance and time here. One wave length is how far the radio wave has traveled while it completes one cycle. That would be 300 megameters divided by 7.2 megacycles, or 300/7 = 41.7 meters. Now, a half-wave dipole, which is the basic antenna from which pretty much all other antennas are built, would be 41.7/2 = 20.83 meters. We can convert that to feet by multiplying meters by 3.28, or 20.83 meters times 3.28 feet per meter is 68.33 feet. But…and here’s where the black magic takes place…a radio wave does not travel through a metal wire as quickly as it travels through free space! The speed at which radio waves travel through transmission lines and wire antennas is the speed of light multiplied by the “velocity factor” of the particular medium. For some coax cables, the velocity factor can be as low as 0.75. For open wire, such as in an antenna, the velocity factor is about 0.95. That means our 68.33 feet is a little long–our signal can’t quite get to the ends of the wire in one cycle. So we need to shorten the wire by the velocity factor. Thus our wire is 68.33 time 0.95, or about 64.9 feet. Now, let’s work this backwards. We want a number such that when we divide it by the frequency in megahertz, we get the length of the wire in feet. Let’s call that number x. So x/f=length. or x/7.2 = 64.9. So x is 64.9 times 7.2, or 468. Presto! There’s the 468 so beloved of amateur dipole enthusiasts! To recap, radio waves don’t travel through wire at the same speed they do through free space; they travel slower. The time-honored (and reasonably accurate) formula of 468/f for the length of a half-wave dipole (in feet) works pretty well. In practice, I cut dipoles a couple feet long and then trim them down to resonance. So, there you have it: a derivation from first principles. Is that helpful?
5. Arthur Charap MD Ph.D. says:

   April 25, 2014 at 6:46 pm

   I am a retired surgeon, former dirt bike explorer and in 1955 was the youngest person to have novice license wvn2fco. Need my Technician license to use radio frequencies for FPV ,first person vewing and control of my radio controlled aircraft and drones. I still have mental block regarding antenna technology. I can memorize the questions but I haven’t a clue where 464/490 in the dipole equations come from. Any suggestions on resources that really get to the basics of antenna theory[either on line or in book form}. Sincerely Arthur. Enjoy your photos and teaching style very much

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