[R-390] 6080 subs for 6082 in R390

[Roger L Ruszkowski]

Fellows,

I agree that,
Voltage is 25.2 Volts
6080 is 6.3V 2.5A
6082 is 26.5V .6A

I calculate
The resistance of a hot 6080 filament is 6.3 / 2.5 =3D 2.52 ohms.
Two of the filaments in series yield a resistance of 5.04 ohms.

I  can  not find any data that supports the power dissipation capabilit=
y of
the  6080  filaments. We expect them to be able to handle 6.3 V at 2.5 =
A or
15.75  W  for  their  rated  life. This does not make the filaments 15.=
75 W
resistors.  Clearly they need to handle with out loss of expected life =
span
more  power  during the turn on warm up period. I find no limit on how =
many
times  a  filament  can  be  switched on and off from a cold start. Cle=
arly
starting,   stopping,  reversing,  starting,  stopping  and  repeating =
 the
alternations  of  AC  current  across the hot filament is not the same =
as a
power on cold start of the filament.

I  expect  the resistance of the 6080 tube filament (or any filament) t=
o be
constant  regardless  of  the applied voltage, applied current or perce=
ived
calculated  power.  As  long as the resistor (load device, filament) is=
 not
actually  going  up in smoke under the applied force, the resistance of=
 the
load device is for the duration remaining constant.

And  we  want  to get a diode in the mix. Not just any where in the mix=
. It
must  be  in  series  with two 6080 tube filaments. These three devices=
 are
placed  as  a  specific  load  across  a  specific  power transformer. =
That
transformer  use  to give up 25.2 volts at 1.2 amps to a pair of 6082 t=
ubes
wired in parallel.

The  new  load on that transformer is now pulsing 5 Amps only one way (=
half
cycle).  Before  I  worry  about  the load resistors going up in smoke =
when
pulsed  with  25.2  V  at  5  Amps  every  1  /  2 cycle, I worry about=
 the
transformers  ability  to  source 25.2 V at 5 Amps every 1 / 2 cycle. W=
hile
the  transformer winding is rated at 5 A, there are other filament load=
s on
the transformer winding using some of that 5 amps..

Do we have the same dead horse here?

I understand the transformer provides a 25.2 volt potential to the circ=
uit.
I  understand  the evil diode limits the 25.2 volt potential applied to=
 the
load (filament resistance) to a 50 % duty cycle.
I  understand  the effective potential 50 % duty cycle limited and aver=
aged
over time can be expressed as a 12.6 volt average voltage potential.
I  understand  the  filaments  offer  a  5.04 ohm resistance to the flo=
w of
current when some voltage potential is applied to the filaments.
I  understand  that  when the peak voltage is applied to the filament (=
some
thing  that  now  only  happens  1  /  2  as  often  as  it  did before=
 the
introduction  of the evil diode) the 5.04 ohms of effective resistance =
will
limit the current to 25.2 volts / 5.04 ohms =3D 5 amps.
I  understand  the  instantaneous power calculated for this instant in =
time
(and repeated 60 times a second) is 25.2 volts * 5 amps =3D 126 watts
I understand real 6080 tubes have been placed this hellish circuit adja=
cent
to  an  evil  diode and have endured this powerful bombardment of elect=
rons
without  dieing  a  premature  death. A true testament to the 6080 tube=
 and
speaks  well  for  its inclusion into the scared hollows of R390 receiv=
ers.
While  two 6080 tubes do bring an evil diode and a wiring changes with =
them
into  the scared hollows of a R390 receiver The trio stand in for very =
rare
6082 tubes and keep a R390 alive and glowing. But I digress.

I  understand (because I read the mail) that some would grab on that nu=
mber
126  above,  point  to  the evil diode and curse this whole dialog with=
 bad
logic  stacked on false assumptions. Some assume the evil diode acts on=
 the
126  to  reduce  it  by half. And push on with logic to have me believe=
 the
filaments  of  the 6080 tubes get bombarded with twice as many electron=
s as
they  deserve  to  be  bombarded  with.  I do not buy into this new fan=
gled
theory  nor  do  my  ancient  (1958) texts support this theory that an =
evil
diode  acts  on  the hypothetical instantaneous value to double the flo=
w of
electrons in a circuit.

I  understand  the two 6080 tubes wired with filaments in series would =
glow
longest if a continuous potential of 12.6 volts at 2.5 amps were applie=
d to
the  filament  circuit.  There  are  words  in the text that an alterna=
ting
current  would  be  preferred  by  the  6080 tubes over a direct curren=
t. I
accept  this  as an article of faith that this is what tubes desire. I =
have
no  preference  for  either  alternating or direct current being applie=
d to
myself and seek insulation from the experience.

The R390 with its 25.2 volt potential does not readily offer the 6080 t=
ubes
this ideal environment. One alternative is to add a resistor in series =
with
the filaments and limit the bombardment on the filaments. I understand =
this
resistor  would  have  an  ideal  value 5.04 ohms. I understand the vol=
tage
applied  to  the  series  circuit  would  be  25.2  volts  the total se=
ries
resistance would be 10.08 ohms I understand the current that flows woul=
d be
2.5 amps and the power dissipated in the 5.04 ohm ideal resistor would =
be a
continuous  31.5  watts.  Also 2.5 amps flow in the filament resistance=
 and
each tube dissipates a continuous 15.75 watts.

I  read  from  web  page legend that once upon a time an engineer with =
true
understanding  of  the  electric mysteries correctly stated that it was=
 not
necessary  to  install  a  31.5  watt  electric power space heater into=
 the
scared hollows of a R390 receiver so as to accommodate the 6080 tubes b=
eing
ask  to  stand in for the more rare and expensive 6082 tubes of old. I =
read
that  the  engineer  proposed  the  installation  of an evil diode into=
 the
scared  hollows of a R390 receiver. The diode receives all the evil bec=
ause
tubes  are  scared and sand state stuff is evil. The engineer explained=
 the
original  circuit radiated 30.24 watts of power from the 6082 filaments=
. He
further  explained  the  changed  circuit would radiate 31.5 watts of p=
ower
from  the  6080 filaments. If the transformer were rated at 26.5 volts =
then
the  change would have been a zero sum change. Pear review of the engin=
eers
proposed  change  stood the test or reason and was advanced through the=
 web
pages as an acceptable thing to let happen to an R390.

I  have  read  this  pear  review  my  self  on  web  pages  and accept=
 the
assumptions  and  logic put forth to explain the electric mysteries of =
this
specific circuit.
I understand the transformer provides 25.2 volts of potential all the t=
ime.
With  out potential nothing happens. I understand the evil diode limits=
 the
potential  to  the tube filament to 50 % or the time. I understand the =
tube
filaments  offer  5.04  ohms  of resistance to the flow of current. I u=
nder
stand  that  when  the  evil  diode  allows  potential to be applied to=
 the
filaments  that  5  amps of current will flow. I under stand the evil d=
iode
will  limit the flow of current to 50 % of time. I understand that when=
 the
evil  diode  permits  (time  dependent)  potential  of 25.2 volts to dr=
aw a
pulsing  (time  dependent) current of 5 amps across the filament resist=
ance
of 5.04 ohms an instantaneous (time dependent) power is dissipated as h=
eat.
A bad math value that ignores time, power factor, duty cycle, diode vol=
tage
drop,  exact  resistance,  exact  voltage potential and exact current i=
s an
exact  power dissipation value of 126 watts. The math is perfect. The l=
ogic
is  impeccable.  The  assumption  is  false.  I  under stand the 25.2 v=
olts
applied  part  time  over  time  has and average over time of 12.6 volt=
s. I
understand  the  5  amps that flows part time over time has an average =
over
time of 2.5 amps. Consulting my 1956 text I find "P =3D EeffIeff I unde=
rstand
the  Eeff  =3D  12.6 and Ieff =3D 2.5 and that P =3D 31.5. I understand=
 that 31.5
watts  radiated  from 2 filaments averages 15.72 watts per filament. I =
find
noting  in  the  texts  that  leads me to believe that this amount of p=
ower
pulsed  from  the  filaments  at  a  50%  duty  rate  in exceeds the de=
sign
limitations  of  the  6080  tubes. While the filaments may prefer a the=
rmal
cycle  that  radiates power in 12.6 volt 2.5 amp jolts 60 times a secon=
d, I
can  not  find  any text to support an aversion to 25.2 volt 5 amp jolt=
s 30
times a second.

This  brings  me  to  a second legend I read from web page that once up=
on a
time  an  engineer with false understanding of the electric mysteries c=
ried
out  loudly, 12.6 * 2.5 *60 =3D 1890 while 25.2 * 5 * 30 =3D 3825 think=
ing this
was  a  meaning  full  statement.  And  this engineer picking up this f=
alse
assumption  and  stacking  much  logic on it never the less reached a f=
alse
premise.  The  false  assumption was to think that 60 and 30 can be equ=
ated
any old time. This engineer wrote in a text that was latter reproduced =
on a
web  page  the  following  statements.  25.2  volts  applied  to a 5.04=
 ohm
resistor 50 % of the time produces a 5 amps current to flow 50% or the =
time
which  causes  the  resistor  to dissipate 126 watts 50% of the time fo=
r an
average power of 63 watts. 25.2 volts times 5 amps equals 126 watts div=
ided
by  half  time  equals  63 watts all the time. Again the math works but=
 the
assumption going in is false. Any good math book says order of operatio=
n is
equal  for  multiplication or division. (25.2 /2 ) * (5/2) =3D (25.2 * =
5) / 2
=3D  63  any  way  you  display  it  on  a computer screen. However the=
 false
assumption  is  that  the  math  above  correctly  models the problem u=
nder
examination.

I do not understand how the square root of (63 watts * 5.04 ohms or 317=
.52)
=3D  17.819  volts and 17.819 volts / 5.04 ohms =3D 3.535 amps, can be =
balanced
with  the  square root of (63 watts / 5.04 ohms or 12.5) =3D 2.480 amps=
 =3D and
2.40  volts  *  5.04  ohms =3D 12.5 volts. 17.819 volts are not equal t=
o 12.5
volts and 3.535 amps are not equal to 2.480 amps.

I  do  understand how the square root of (31.5 watts * 5.04 ohms or 158=
.76)
=3D  12.6  volts  and 12.6 volts / 5.04 ohms =3D 2.5 amps. Can be balan=
ced with
the square root of (31.5 watts / 5.04 ohms or 6.25) =3D 2.5 amps and 2.=
5 amps
*  5.04  ohms =3D 12.6 volts. 12.6 volts are equal to 12.6 volts and 2.=
5 amps
are equal to 2.5 amps.

If  you incorrectly jump into this problem and fall onto a false assump=
tion
the  logic of the math will lead to a false conclusion. Clearly you can=
 not
look for math logic to carry you to the real true product. You can not =
wily
nily  disregard  time, average, order, and units and declare a results =
that
will not stand to counter checking.

Why  am  I  dragging  this  dead  horse  around the track for a second =
lap?
Because once an engineer got it right. Then a second engineer got it wr=
ong.
Then  it  went  into  the  archives  as  6080 can not be used with diod=
e to
replace  6082 as advertised. Bad engineering analysis follows to prove =
dead
horse can not run. All the counter points are in the archives as dead h=
orse
is  beaten  again  with at least three re-entries declaring original au=
thor
may  not  have high school equivalency or skills needed to activate gra=
mmar
checker on an E-mail app.

And  then  we  wonder why kids exposed to this trash can not wade throu=
gh a
rational  problem.  Go  back  and read the archives since the first of =
this
year. Where has our on line archive advanced our understanding with con=
cise
solid  resolve  that  establishes  a firm foundation to work forward on=
. We
have  a  lot of hype, hypothetical and half truth in the archive that d=
o to
its low humor value does not even rate as good entertainment.

In  this  post  there was one sentence of relevance. Response to any of=
 the
other  humor  wrapped  around that sentence will be of entertainment va=
lues
only and will be replied back to as such by the author. Any response to=
 the
editorial  comment  of  this  post will be ignored by the author. I did=
 not
belong  in  the  original  and  will  not  be knowingly propagated into=
 the
future.

Roger KC6TRU


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