Call Search
     

New to Ham Radio?
My Profile

Community
Articles
Forums
News
Reviews
Friends Remembered
Strays
Survey Question

Operating
Contesting
DX Cluster Spots
Propagation

Resources
Calendar
Classifieds
Ham Exams
Ham Links
List Archives
News Articles
Product Reviews
QSL Managers

Site Info
eHam Help (FAQ)
Support the site
The eHam Team
Advertising Info
Vision Statement
About eHam.net

   Home   Help Search  
Pages: [1] 2 Next   Go Down
  Print  
Author Topic: Swan 350 800ma Cathode Current? HELP  (Read 1471 times)
AC0FA
Member

Posts: 298




Ignore
« on: March 28, 2007, 10:15:39 AM »

I have a Swan 350 tube rig made around 1965.
I have a question about tuning and running in SSB mode.
 
In the tune position. With all the controls peaked on the external watt meter I get about 200w out with the AVERAGE button pushed in on the watt meter.
I show 795ma on the cathode current meter.

I am concerned that this might be too much CATHODE current for normal operation. Or something wrong with the meter.

In SSB mode with the PEP button Pushed on the external watt meter I constantly bump 290 watts PEP output.

The mike gain is set to 11:00 and the average  cathode current is 200ma in SSB.

The manual mentions not being discouraged if you can't load to 500ma.

My 350 is loading to 800ma. If I back off on the load control a couple notches to 500ma I get 100w average and 150W PEP.

Has my resistor gone really high in the meter or something?

Any thoughts from someone who has owned a swan 350, or something simillar,and knows how it is supposed to run would be really great.

Yours in the Boat Anchor brotherhood

AC0FA

     
     

 
Logged
AD4U
Member

Posts: 2153




Ignore
« Reply #1 on: March 28, 2007, 01:28:26 PM »

I do not own nor have I ever owned a SWAN.  I am interested in finding, restoring, and using older rigs.  I recently bought a rather "mint" Yaesu FTDX 570 transceiver.  This is a late 1960's / very early 1970's rig  that uses two 6KD6 TV sweep tubes in the final amp.  I think your SWAN uses the same tubes.  I can easily get 250 - 275 watts CW output on all bands including 10 meters.  The plate current at full bore runs around 800 mA.  Since the plate voltage is just above 700 VDC under load, I guess that is why the rig is called a 570 (800 mA X 700 VDC = 560 watts input).  Any way fully loaded the Yaesu draws around 800 mA plate current.  I think your meter is reading correctly, assuming the SWAN uses two 6KD6 or two 6LQ6 tubes.  For long tube life I would recommend running less than 800 mA unless you have a stash of tubes.
Logged
K3HVG
Member

Posts: 149




Ignore
« Reply #2 on: March 28, 2007, 02:54:39 PM »

I don't believe that a Swan 117XC power suppy will provide the current you're talking about without failing. I think you should measure the voltage and current via an external measuring method and then make your determinations.
Logged
KA5N
Member

Posts: 4380




Ignore
« Reply #3 on: March 28, 2007, 04:16:53 PM »

I think you have cooked one or more resistors in the cathode circuit metering circuit.  
73 Allen
Logged
AC0FA
Member

Posts: 298




Ignore
« Reply #4 on: March 28, 2007, 05:03:40 PM »

FB all and thanks Allen;
The first poster was thinking plate curent and we are talking CATHODE current.  

I will explore the resistors in the cathode meter circut when I have some time.

I wanted to double check and make sure I have all the figures right before zeroing in on a faulty meter circut.

I feel the need  to repair it soon. I just don't feel right operating a new (to me) rig with a bum meter. I like to keep a CLOSE eye on the grids.

Thank to all AC0FA        
Logged
KA5N
Member

Posts: 4380




Ignore
« Reply #5 on: March 29, 2007, 06:58:53 AM »

Erik,
While the cathode current and the plate current are different, the difference is slight when you are tuning the plate circuit.  The replies implied that if the current was 800 ma the plate would be red hot and the glass envelope would begin to melt.
73 Allen
Logged
KI4QWX
Member

Posts: 3




Ignore
« Reply #6 on: March 29, 2007, 09:49:29 AM »

I am not seen a schematic of a Swan radio.  However, many of the tube final rigs sense the current in the finals by measuring the voltage drop in the cathode resistor.  Note that these resistors are usually in parallel because the final tubes are running in parallel.  In units of that era, the cathode resistors were usually carbon composition that, when overheated, go UP in resistance.  THe net effect is that you will end up with reduced output, but the plate current meter (which actually measures cathode current) will read higher current than it actually is.  This is easy to test.  Measure the resistance of the cathode resistors and see if they are at spec.  If they are not, replace them.  Don't use a higher wattage rating if you do. This failure mode is a means of protecting your finals if they are mistuned for a significant period of time.
Logged
AC0FA
Member

Posts: 298




Ignore
« Reply #7 on: March 29, 2007, 02:02:17 PM »

Yes I can see the light.
You are correct about the cathode resistors. After checking them closely. The 4.7 0hm 1 watt ones that go from the paralell cathodes to ground have small cracks and are bulging. They are at 4.9 cold. I have no confidence in them warm. I will need to replace them and check the meter calibration.

Swan 350 plain has two resistors in parallel in the lead from the cathodes in the finals to the cathode meter.

I discovered 2K Ohm factory resistor and a 15k ohm
SELECTED resistor SELECTED BY SOME ONE ATTEMPTING TO CALIBRATE THE METER.

This results in a strange value of 1730 actual ohms.  

These are used to restrict the cathode voltage to less than 1ma so the meter can read it.  

I have good tubes, good supply, and good 120v
AC power at the mains. I am only 4 miles from the power plant.

I have seen 300w pep output on the meter.
I would expect to see 200ma on voice peaks.

Using full carrier in the tune position I expect to see a maximum of 500-600ma. Not 800ma

At 600ma I don't think 1730 ohms is right. Someone may have been trying to compensate for the cathode resistors going bad. As soon as I replace the 4.7 Ohm
cathode resistors. I know the selected resistor has to change But I am not sure how much.          

So for the calibration trial and error.

After setting the bias on the meter to read 200ma when I key the mike I should be able to read .1v
DC at the cathode resistors.

Does anyone have a guess what value of selected resistor I should start with.

I could be at this for hours. A little guidance from the elmers would be great.

Thanks AC0FA  
Logged
KI4QWX
Member

Posts: 3




Ignore
« Reply #8 on: March 29, 2007, 08:08:41 PM »

OK
First you have to determine what the meter SHOULD be reading.  Those resistors sound like they have been abused, so I would surely replace them.  After that, you can determine what current the tubes are actually drawing by measuring the voltage drop across the cathode resistors (you may have ti do the same for the screen grid resistors too depending on how much current the screens draw).  Using Ohm's law, we know that I=E/R.  You will know the value of the resistors and you can measure the voltage drop, hence you can then calculate the current.  The caveat I mentioned above is that you may need to calculate the screen current and subtract it from the total to get the plate current.
Logged
KI4QWX
Member

Posts: 3




Ignore
« Reply #9 on: March 30, 2007, 06:19:30 AM »

A further thought.  If you know the current draw of the meter movement (typically this is 50-100 microamps, you can calulate the necessary resistance values using ohms law.  Even better:  remove the old resistors for the meter and substitute a pot, say a 25K lnear - then adjust it so the meter reads the correct value.  Then measure the setting on the pot and replace it with a resistor of the appropriate value.  Be careful though.  Turn the pot too low and you could fry the meter movement.  Start at the highest setting. Then go down slowly.   The absolute best would be to use a ten turn pot, but a regular one will do.
Logged
AC0FA
Member

Posts: 298




Ignore
« Reply #10 on: March 30, 2007, 08:03:27 AM »

Cool beans;
Now were making progress. The meter is 0 to .1ma.
The rig uses the 6h5f finals.
It would seem there were 2 types of meters installed in the 350. The first type, installed in my rig pre 1965. Is the one that reads the SWR in the OPPOSITE  direction from the catode current. I believe this one uses the 2k resistor factory resistor in paralell with the 15k resistor for 1763 ohms.

The other type of meter used in the 350 in 1965 displays the SWR going in the SAME direction as the
S Meter. From what I have seen in schematics and photographs the factory resistor is 470 ohms and the selected is 1k to 1.2k  That gives me a range of 314 to 336 ohms so approxamately 50 ohms depending on the quality of the resistors.

If it is reading way high at 1763 I would need more resistance maybe 1800 ohms.

Assuming the 1500ohm diference is the resistance in the meters.

The pot is a great Idea TNX
ACOFA      

Logged
AC0FA
Member

Posts: 298




Ignore
« Reply #11 on: March 30, 2007, 11:15:51 AM »

oops:
I meant to say 1400 ohm difference. where's my coffee?
ACOFA  
Logged
KA5N
Member

Posts: 4380




Ignore
« Reply #12 on: March 30, 2007, 04:19:00 PM »

You put a variable resistor in series with the meter and a low voltage in series with both (1.5 volts is good) and then adjust the pot until the meter reads exactly half scale.  Then take the pot out of the circuit and measure the resistance.  This resistance is the internal resistance of the meter (which you need to know).  The full scale deflection is supposed to be 1000 ma (right?).  If you have two 4.7 Ohm resistors in parallel then 1000 ma across 2.35 Ohms is 2.35 volts.  Now the current through the series resistors (you have to calculate them) and the meter resistance must equal the meter full scale current 0.1 ma when 2.35 volts is applied.  The meter resistance is probably around 2000 Ohms.  So the series resistor should be around 20 K Ohms.  Of course if any of the existing value are incorrect, then all bets are off.  The schematic I have seen shows two 1 Ohm resistors in parallel which would make the selected resistor much smaller.  1 Ohm resistors would also disapate much less heat.
Chin up, it's only an Ohm's law problem.
73 Allen
Logged
AC0FA
Member

Posts: 298




Ignore
« Reply #13 on: March 30, 2007, 07:12:24 PM »

Well;
Got her done! It was no major deal. The meter was accurate in its previous configurtation Till the 4.7 ohm cathode resistors let go.

As soon as I replaced those the meter started to make sense. 475ma on the cathode meter in tune and 200ma peak in SSB and 300W pep output.
So all is well.
Thanks AC0FA  
Logged
WA7KKP
Member

Posts: 9




Ignore
« Reply #14 on: April 07, 2007, 12:42:35 PM »

I concur on the value of the cathode metering resistors having changed.  This is a common problem in many rigs.  I'd figure out how to add an externa. 500 ma or 1A panel meter (in a box) to read cathode current.

Also read carefully the manual.  Swan says that with proper operation, you should see no more than 100-125 ma on peaks on the meter in the rig.  

Back in those days, those @$%^ sweep tube finals were grossly overrated, and Swan was one of those.  Not having any sort of ALC, they could be "talked" upwards of 400-500 watts input but with a broad signal to boot.  Rinning them gingerly gave one a clean signal, albeit at a power level about the same as other rigs running a pair of 6146's . . . about 100 watts out in tune, and not much more than that in SSB peak power.

Gary WA7KKP
Logged
Pages: [1] 2 Next   Go Up
  Print  
 
Jump to:  

Powered by MySQL Powered by PHP Powered by SMF 1.1.11 | SMF © 2006-2009, Simple Machines LLC Valid XHTML 1.0! Valid CSS!