The AM Forum
December 12, 2024, 01:09:23 PM *
Welcome, Guest. Please login or register.

Login with username, password and session length
 
   Home   Help Calendar Links Staff List Gallery Login Register  
Pages: [1]   Go Down
  Print  
Author Topic: Purely Resistive versus Highly Reactive Loads: a mechanical analogy.  (Read 16175 times)
0 Members and 1 Guest are viewing this topic.
k4kyv
Contributing Member
Don
Member

Offline Offline

Posts: 10037



« on: March 16, 2012, 04:54:16 AM »

This occurred to me the other day, while I was doing some emergency repairs after our recent windstorm, attempting to nail back a piece of wood siding that had blown loose.

Whenever I attempted to drive a nail through the plywood sheathing directly over a stud, the nail went firmly and soundly into the wood, with little effort.  But between the studs, the sheathing is springy and has "give" to it, and most of the energy from the hammer blows tends to momentarily displace the wood instead of driving the nail.  The wood reflects the energy back to the hammer and it bounces.  Often one ends up bending the nail and/or splitting the wood and sometimes mashing one's finger.

It occurred to me that nailing directly over solid studs is analogous to working a feedline into a purely resistive load. The energy exerted by the hammer simply overcomes the resistance of the wood to the nail, and the nail is forced into the wood, with little reaction or bounce. But over the springy part between studs, the sheathing "reacts" to the blows of the hammer, and "reflects" most of the momentum back to the carpenter. With each blow the nail may go slightly deeper into the wood, but most of the energy is reflected back to the sender.  If enough energy is reflected back unexpectedly, it can cause injury.

Hammering over a stud is analogous to what happens when a flat feedline is terminated into a purely resistive load: no energy is reflected back; it all goes to heat up the resistance or radiate into space.  Hammering between studs is analogous to terminating the feed line into a highly reactive load. Most of the energy is reflected back and only a portion goes into the load.  Undesirable results may occur if the system is not properly tuned to cancel out the reactance as seen by the transmitter: the final tank may be thrown off resonance causing the tube to overheat, air dielectric capacitors may flash over, and coils and mica capacitors may heat up due to high circulating currents.
Logged

Don, K4KYV                                       AMI#5
Licensed since 1959 and not happy to be back on AM...    Never got off AM in the first place.

- - -
This message was typed using the DVORAK keyboard layout.
http://www.mwbrooks.com/dvorak
k4kyv
Contributing Member
Don
Member

Offline Offline

Posts: 10037



« Reply #1 on: March 16, 2012, 12:12:22 PM »

Undesirable results may occur if the system is not properly tuned to cancel out the reactance as seen by the transmitter: the final tank may be thrown off resonance causing the tube to overheat, air dielectric capacitors may flash over, and coils and mica capacitors may heat up due to high circulating currents.

This has lead to a long-time but frequent misunderstanding by amateurs: a mistaken belief that the final tubes may turn red with high SWR on the antenna feedline because the reflected power feeds back into the final stage and overheats the tube(s).

The tubes may overheat because the final is working into a reactive load, and this reactance seen by the transmitter has detuned the final tank from resonance - exactly the same thing as if one had rotated the final tank capacitor away from the dip. The operator needs only to re-resonate the final with the "plate tuning" control to correct the problem. The reflected power does not feed back into the final stage; the tank circuit or tuning network re-reflects it back down the feed line towards the load where more of it is radiated and the process is repeated until all the energy is expended. This oscillation back and forth, to and fro, between the transmitter and load is what forms the standing wave. Most of the energy eventually gets radiated, and the rest is wasted in feed line loss, accounting for any measured loss in signal strength. The reflected power does not come back into the final amplifier stage and cause the tube to overheat.

An open-wire tuned feed line may operate with a SWR of 10:1 or more, and yet be more efficient than a well matched coax line of the same length. The final runs perfectly cool. Obviously, that reflected power is not coming back to the final tube and heating it up. Even when using coax, a SWR of up to 3:1 or so will cause negligible loss if the coax is good and everything is properly tuned to resonance.
Logged

Don, K4KYV                                       AMI#5
Licensed since 1959 and not happy to be back on AM...    Never got off AM in the first place.

- - -
This message was typed using the DVORAK keyboard layout.
http://www.mwbrooks.com/dvorak
Pages: [1]   Go Up
  Print  
 
Jump to:  

AMfone - Dedicated to Amplitude Modulation on the Amateur Radio Bands
 AMfone © 2001-2015
Powered by SMF 1.1.21 | SMF © 2015, Simple Machines
Page created in 0.045 seconds with 19 queries.