ken scharf wrote:
Tim Wescott wrote:
On Thu, 04 Dec 2008 06:23:48 -0800, JIMMIE wrote:

After years of avoiding morse code I am finally getting into it. the
ideal of operating a very simple transmitter appeals to my junkbox/
trashcan construction mentality.
I am looking for a substitue for a 6L6 that has the plate brought out
the top. I was trying to build a little Glowbug transmitter but was
having all kinds of problems neutralizing the the thing. I think I would
have better luck if I can separate the grid and plate circuits from each
other. I am considering an 807 as a replacement Suggestions would be
appreciated.

Jimmie

Dunno if it's been mentioned yet -- 2E26. The 807 is _not_ a 6L6 in a
different envelope -- it's quit arguably the 6L6's big brother, but
it's got different ratings; it could probably be shoe-horned into a
circuit designed for the 6L6, but you'd be missing out on about 6dB of
final output power.

The 807 IS a 6L6G with a 5 pin base and the plate connected to the top.
The published ratings of the 6L6G look different than the 807 because of

-- snip --

The type 1614 is a metal tube based on the 6L6. It is a transmitting
version, and probably has additional shielding. Otherwise its internal
structure is the same as the 6L6 metal type.

How is the 6L6 metal type different internally from the 6L6G?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Dec 4, 9:26*pm, "NoSPAM" wrote:
"Stev eH"
StevehkhhSDJvhvbjjxbvvbhnvbhzjdnxzvzhzdshbvnjzvnb vnvjbvbcjbvvvvnmxvzjhjzsdgfgsfghgjsghgsljhglhdjfgh fufgfhzysgfhczgugfvzlvufzvllgfzlyfyvlgbylvdfghvbly
wrote in ...

My first amp had 3 PL519s with a lethal power supply using a voltage
tripp{l}er direct off the 240V mains, heaters were direct off mains via a
1n5406 diode. It's still up in the parents loft somewhere, must dig it
out over the holidays and give it a try....

Steve H

Do you realize that the RMS value of half-wave rectified AC is not one half
of the applied RMS voltage but rather 0.707 of the applied RMS voltage. *So
your three 40 volt filament PL519s actually saw about 56 volts RMS across
each filament.

Hey OM
I beg to differ. As any casual observer can see that peek is 330V and
the average would be 0.363 of the peak so that's 40 volts in my
calculator all day long, for a 3 tube string.

Lets take a square wave, the average would be 0.5 of peek, far lower
than 0.707 you are used to using. Even if it was full wave 120 cycle
dc, the average would only be 0.636 still lower than 0.707.

I tell you what. Take a diode and rectify the mains with it and
measure the DC voltage with a DVM, VTVM or a VOM and 0.363 of peak is
what you will get for a DC voltage on that meter. Or do the real
integral of the dv/dt and see for yourself.

The TV repair business I was in manufacturers put 50volt peak on a CRT
6.3volt filament, the average voltage was 6.3 volts. Lets see you RMS
that one.

73 OM

n8zu

"raypsi" wrote in message ...
On Dec 4, 9:26 pm, "NoSPAM" wrote:
"Stev eH"
StevehkhhSDJvhvbjjxbvvbhnvbhzjdnxzvzhzdshbvnjzvnb vnvjbvbcjbvvvvnmxvzjhjzsdgfgsfghgjsghgsljhglhdjfgh fufgfhzysgfhczgugfvzlvufzvllgfzlyfyvlgbylvdfghvbly
wrote in ...

My first amp had 3 PL519s with a lethal power supply using a voltage
tripp{l}er direct off the 240V mains, heaters were direct off mains via a
1n5406 diode. It's still up in the parents loft somewhere, must dig it
out over the holidays and give it a try....

Steve H

Do you realize that the RMS value of half-wave rectified AC is not one half
of the applied RMS voltage but rather 0.707 of the applied RMS voltage. So
your three 40 volt filament PL519s actually saw about 56 volts RMS across
each filament.

Hey OM
I beg to differ. As any casual observer can see that peek is 330V and
the average would be 0.363 of the peak so that's 40 volts in my
calculator all day long, for a 3 tube string.

Lets take a square wave, the average would be 0.5 of peek, far lower
than 0.707 you are used to using. Even if it was full wave 120 cycle
dc, the average would only be 0.636 still lower than 0.707.

I tell you what. Take a diode and rectify the mains with it and
measure the DC voltage with a DVM, VTVM or a VOM and 0.363 of peak is
what you will get for a DC voltage on that meter. Or do the real
integral of the dv/dt and see for yourself.

The TV repair business I was in manufacturers put 50volt peak on a CRT
6.3volt filament, the average voltage was 6.3 volts. Lets see you RMS
that one.

73 OM

n8zu
Ray,

You obviously do not understand the meaning of RMS as you are confusing the average voltage with the RMS voltage. In a resistive circuit, the RMS voltage of any waveform will produce identical _heating_ as a DC voltage of the same value. This is an important point which is often overlooked, and it is why measuring the RMS value of a waveform is difficult. Let's start with some numbers but avoiding the use of integration...

Assume we have an AC voltage described by:

Vac = 339.41 sin[(2*Pi*f)t] where (2*Pi*f) is 376.99 for 60 Hz or 314.159 for 50 Hz power
Vac = 339.41 sin(376.99*t)

Since this is a symmetrical waveform, the _average_ value is 0 while the _peak_ positive voltage is 339.41 volts.

Now let us apply this voltage to a 1000 ohm resistor. Using Ohm's
Law, we can calculate the current, I=V/R.

Iac = 339.41 sin(376.99*t) volts/1000 ohms
Iac = 0.33941 sin(376.99*t) amps

Again this a symmetrical waveform with an average of 0, and a peak positive current of 0.33941 amps.

The instantaneous power dissipated by this resistor can be calculated by multiplying the voltage by the current.

Pac = 339.41 sin(376.99*t) * 0.33941 sin(376.99*t) watts
Pac = 115.20 [sin(376.99*t) * sin(376.99*t)] watts

Using a trigonometric identity (http://en.wikipedia.org/wiki/List_of...ric_identities) this can be rearranged to:

Pac = 115.20 * [1 - cos(2*376.99*t)]/2 watts

It is important to understand this function. The value of this function goes from zero to a peak of 115.20 watts at twice the frequency of the original waveform. The average value is no longer zero but is now 57.6 watts.

Now let us go back to using RMS values. Since the waveform is sinusoidal, we can use some common formulas.

Vrms = 0.707107 * Vpeak = 0.707107 * 339.41 = 240 volts
Irms = 240/1000 = 0.240 amps
Prms = 240 * 0.24 = 57.6 watts

This is exactly the same as the average of the power calculated above. If we had applied 240 volts DC to the 1000 ohm resistor, the power would again be 57.6 watts.

Now let us half-wave rectify the original AC waveform. The peak voltage will be the same as before, 339.41 volts. But the waveform is no longer symmetrical about zero. The average value is:

Vavg = Vpeak/Pi = 339.41/3.14159 = 108.038 volts

The RMS value is:

Vrms = Vpeak/2 = 169.706 volts

Applying this to the same 1000 ohm resistor gives a current of

Irms = Vrms/1000 = 0.169706 amps

The RMS power dissipated by the resistor is

Prms = Vrms * Irms = 28.8 watts

Note that this exactly _twice_ the power if the waveform had not been rectified.
Now let us do the same calculations for a symmetrical square wave.

Vpeak = 339.41 volts

Vavg = 0 volts

Vrms = 339.41 volts

Ipeak = 0.33941 amps

Iavg = 0 amps

Irms = 0.33941 amps

Prms = 115.2 watts

If the square wave is half wave rectified,

Vpeak = 339.41 volts

Vavg = 169.71 volts

Vrms = 169.71 volts

Ipeak = 0.33941 amps

Iavg = 0.16971 amps

Irms = 0.16971 amps

Prms = 28.80 watts

Note that this is exactly the same as with the sinusoidal waveform. This is exactly the point, half wave rectification produces _one_half_of_the_power_ that would be produced by the original sinusoidal AC waveform. It does not produce half of the original peak voltage. It produces SQRT(2)/2 times the original RMS voltage.

As for your numbers, Ray, you need a refresher course in mathematics. As for DVM, VTVM, or VOM readings, you will have to specify the filtering of the waveform before I can tell you the value. In the case of the CRT filament
(and I really think you might mean the HV rectifier filament) power is supplied by a waveform having a high peak value but an RMS value that is low. Again you would need to specify the waveform exactly.

There is a very simple way to prove that my calculations are right. Get two incandescent light bulbs of the same type and wattage. Wire a diode of adequate voltage and current rating in series with one bulb and connect to the wall outlet. Use a variac to feed the second bulb. With the bulbs side by side, adjust the variac until both bulbs are the same brightness. Measure the variac output voltage. You will find that it is not 50% of the line voltage but instead very close to 70.7% of the line voltage.

In fact, the most accurate RMS voltmeters operate in a similar fashion. They let the unknown voltage heat a resistor. Then they use an adjustable DC voltage to heat an identical resistor. When both resistors are at the same temperature, the DC voltage is the RMS value of the unknown voltage. Consult the Linear Technology LT1088
datasheet for a description of one method of doing this. http://www.datasheetcatalog.org/data...ogy/lt1088.pdf

The same company also offers several other RMS-to-DC converters that operate on a delta-sigma computation technique.
http://www.linear.com/pc/downloadDoc...54,C1086,D4486

A general discussion of RMS measurements may be found at:
http://www.edn.com/contents/images/46857.pdf.

This last reference shows typical errors for different types of RMS measuring meters. I strongly suggest to anyone still confused to read it. And I really hope Phil tries the experiment I suggested.

73, Barry L. Ornitz, PhD WA4VZQ

"raypsi" wrote in message ...
On Dec 4, 9:26 pm, "NoSPAM" wrote:
"Stev eH"
StevehkhhSDJvhvbjjxbvvbhnvbhzjdnxzvzhzdshbvnjzvnb vnvjbvbcjbvvvvnmxvzjhjzsdgfgsfghgjsghgsljhglhdjfgh fufgfhzysgfhczgugfvzlvufzvllgfzlyfyvlgbylvdfghvbly
wrote in ...

My first amp had 3 PL519s with a lethal power supply using a voltage
tripp{l}er direct off the 240V mains, heaters were direct off mains via a
1n5406 diode. It's still up in the parents loft somewhere, must dig it
out over the holidays and give it a try....

Steve H

Do you realize that the RMS value of half-wave rectified AC is not one half
of the applied RMS voltage but rather 0.707 of the applied RMS voltage. So
your three 40 volt filament PL519s actually saw about 56 volts RMS across
each filament.

Hey OM
I beg to differ. As any casual observer can see that peek is 330V and
the average would be 0.363 of the peak so that's 40 volts in my
calculator all day long, for a 3 tube string.

Lets take a square wave, the average would be 0.5 of peek, far lower
than 0.707 you are used to using. Even if it was full wave 120 cycle
dc, the average would only be 0.636 still lower than 0.707.

I tell you what. Take a diode and rectify the mains with it and
measure the DC voltage with a DVM, VTVM or a VOM and 0.363 of peak is
what you will get for a DC voltage on that meter. Or do the real
integral of the dv/dt and see for yourself.

The TV repair business I was in manufacturers put 50volt peak on a CRT
6.3volt filament, the average voltage was 6.3 volts. Lets see you RMS
that one.

73 OM

n8zu


Ray,



You obviously do not understand the meaning of RMS as you are confusing the average voltage with the RMS voltage. In a resistive circuit, the RMS voltage of any waveform will produce identical _heating_ as a DC voltage of the same value. This is an important point which is often overlooked, and it is why measuring the RMS value of a waveform is difficult. Let's start with some numbers but avoiding the use of integration...



Assume we have an AC voltage described by:



Vac = 339.41 sin[(2*Pi*f)t] where (2*Pi*f) is 376.99 for 60 Hz or 314.159 for 50 Hz power

Vac = 339.41 sin(376.99*t)



Since this is a symmetrical waveform, the _average_ value is 0 while the _peak_ positive voltage is 339.41 volts.



Now let us apply this voltage to a 1000 ohm resistor. Using Ohm's

Law, we can calculate the current, I=V/R.



Iac = 339.41 sin(376.99*t) volts/1000 ohms

Iac = 0.33941 sin(376.99*t) amps



Again this a symmetrical waveform with an average of 0, and a peak positive current of 0.33941 amps.



The instantaneous power dissipated by this resistor can be calculated by multiplying the voltage by the current.



Pac = 339.41 sin(376.99*t) * 0.33941 sin(376.99*t) watts

Pac = 115.20 [sin(376.99*t) * sin(376.99*t)] watts



Using a trigonometric identity (http://en.wikipedia.org/wiki/List_of...ric_identities) this can be rearranged to:



Pac = 115.20 * [1 - cos(2*376.99*t)]/2 watts



It is important to understand this function. The value of this function goes from zero to a peak of 115.20 watts at twice the frequency of the original waveform. The average value is no longer zero but is now 57.6 watts.



Now let us go back to using RMS values. Since the waveform is sinusoidal, we can use some common formulas.



Vrms = 0.707107 * Vpeak = 0.707107 * 339.41 = 240 volts

Irms = 240/1000 = 0.240 amps

Prms = 240 * 0.24 = 57.6 watts



This is exactly the same as the average of the power calculated above. If we had applied 240 volts DC to the 1000 ohm resistor, the power would again be 57.6 watts.



Now let us half-wave rectify the original AC waveform. The peak voltage will be the same as before, 339.41 volts. But the waveform is no longer symmetrical about zero. The average value is:



Vavg = Vpeak/Pi = 339.41/3.14159 = 108.038 volts



The RMS value is:



Vrms = Vpeak/2 = 169.706 volts



Applying this to the same 1000 ohm resistor gives a current of



Irms = Vrms/1000 = 0.169706 amps



The RMS power dissipated by the resistor is



Prms = Vrms * Irms = 28.8 watts



Note that this exactly _twice_ the power if the waveform had not been rectified.

Now let us do the same calculations for a symmetrical square wave.



Vpeak = 339.41 volts



Vavg = 0 volts



Vrms = 339.41 volts



Ipeak = 0.33941 amps



Iavg = 0 amps



Irms = 0.33941 amps



Prms = 115.2 watts



If the square wave is half wave rectified,



Vpeak = 339.41 volts



Vavg = 169.71 volts



Vrms = 169.71 volts



Ipeak = 0.33941 amps



Iavg = 0.16971 amps



Irms = 0.16971 amps



Prms = 28.80 watts



Note that this is exactly the same as with the sinusoidal waveform. This is exactly the point, half wave rectification produces _one_half_of_the_power_ that would be produced by the original sinusoidal AC waveform. It does not produce half of the original peak voltage. It produces SQRT(2)/2 times the original RMS voltage.



As for your numbers, Ray, you need a refresher course in mathematics. As for DVM, VTVM, or VOM readings, you will have to specify the filtering of the waveform before I can tell you the value. In the case of the CRT filament

(and I really think you might mean the HV rectifier filament) power is supplied by a waveform having a high peak value but an RMS value that is low. Again you would need to specify the waveform exactly.



There is a very simple way to prove that my calculations are right. Get two incandescent light bulbs of the same type and wattage. Wire a diode of adequate voltage and current rating in series with one bulb and connect to the wall outlet. Use a variac to feed the second bulb. With the bulbs side by side, adjust the variac until both bulbs are the same brightness. Measure the variac output voltage. You will find that it is not 50% of the line voltage but instead very close to 70.7% of the line voltage.



In fact, the most accurate RMS voltmeters operate in a similar fashion. They let the unknown voltage heat a resistor. Then they use an adjustable DC voltage to heat an identical resistor. When both resistors are at the same temperature, the DC voltage is the RMS value of the unknown voltage. Consult the Linear Technology LT1088

datasheet for a description of one method of doing this. http://www.datasheetcatalog.org/data...ogy/lt1088.pdf



The same company also offers several other RMS-to-DC converters that operate on a delta-sigma computation technique.

http://www.linear.com/pc/downloadDoc...54,C1086,D4486



A general discussion of RMS measurements may be found at:

http://www.edn.com/contents/images/46857.pdf.



This last reference shows typical errors for different types of RMS measuring meters. I strongly suggest to anyone still confused to read it. And I really hope Phil tries the experiment I suggested.



73, Barry L. Ornitz, PhD WA4VZQ

Howdy,


A 6L6 immersed in a can of peanut oil is an old timey final, smells
good too.


73,
Grumpy
(Who can't get news express to work withg AIOE anymore.)

ken scharf wrote:
Fred McKenzie wrote:
In article
,
JIMMIE wrote:

I am considering an 807 as a replacement

Jimmie-

Just how simple will this transmitter be? My first novice transmitter
was built from an ARRL publication back in the 50s. The specification
was for a single 6V6, used as a crystal oscillator with output coupled
to the antenna.

I found I could get more power with a 6L6 plugged into the same socket.
Along the way to getting more power, I found that crystal current would
increase to the point where the crystal would fracture!

If you use a 807 or 1625, I hope you will have a lower power oscillator
driving it. Something like a 6AG7 would make a nice crystal oscillator,
and should have sufficient output to drive an 807.

73,
Fred
K4DII
The 6AG7 - 807 combo is a classic. E. F. Johnson sold a 50 Watt Novice
transmitter in the '50's with this tube line up (also a 5R4GY rectifier
IIRC). Other tubes that have been used as the oscillator/driver were
the 6CL6, 12BY7, 6GK6, and 5763.

On Dec 10, 9:27*pm, "
wrote:
Howdy,

A 6L6 immersed in a can of peanut oil is an old timey final, smells
good too.

73,
Grumpy
(Who can't get news express to work withg AIOE anymore.)



I tried doing that with a 4cx250 once. Never got it to work. Looking
back on it I should have mounted the base of the tube to the paint can
lid and hung it in the bucket upside down.
Hmmm, might be something fun to try this winter.

Jimmie

Tim Wescott wrote:
ken scharf wrote:
Tim Wescott wrote:
On Thu, 04 Dec 2008 06:23:48 -0800, JIMMIE wrote:

After years of avoiding morse code I am finally getting into it. the
ideal of operating a very simple transmitter appeals to my junkbox/
trashcan construction mentality.
I am looking for a substitue for a 6L6 that has the plate brought out
the top. I was trying to build a little Glowbug transmitter but was
having all kinds of problems neutralizing the the thing. I think I
would
have better luck if I can separate the grid and plate circuits from
each
other. I am considering an 807 as a replacement Suggestions would be
appreciated.

Jimmie

Dunno if it's been mentioned yet -- 2E26. The 807 is _not_ a 6L6 in
a different envelope -- it's quit arguably the 6L6's big brother, but
it's got different ratings; it could probably be shoe-horned into a
circuit designed for the 6L6, but you'd be missing out on about 6dB
of final output power.

The 807 IS a 6L6G with a 5 pin base and the plate connected to the top.
The published ratings of the 6L6G look different than the 807 because of

-- snip --

The type 1614 is a metal tube based on the 6L6. It is a transmitting
version, and probably has additional shielding. Otherwise its internal
structure is the same as the 6L6 metal type.

How is the 6L6 metal type different internally from the 6L6G?

I've never xrayed or took a 6L6M apart. However the 6L6M would have
shorter leads and different inter-electrode capacitance specs due to the
metal envelope and the shorter leads. Also the metal envelope might
lead to a lower max plate voltage rating due to the danger of arc-over
to the envelope (which is connected to pin 1 for ground/shield). The
glass envelope version of the tube has more electrical insulation in
this regard. The 1614 is a transmitting version of the 6L6M. Other
than being spec'ed for ICAS operation and *maybe* having additional
internal shielding it's identical to the 6L6M.

JIMMIE wrote:
On Dec 10, 9:27 pm, "
wrote:
Howdy,

A 6L6 immersed in a can of peanut oil is an old timey final, smells
good too.

73,
Grumpy
(Who can't get news express to work withg AIOE anymore.)



I tried doing that with a 4cx250 once. Never got it to work. Looking
back on it I should have mounted the base of the tube to the paint can
lid and hung it in the bucket upside down.
Hmmm, might be something fun to try this winter.

Jimmie
I recall an article on using a 4cx250 with water cooling. Actually it
sat upside down in a pool of water which was allowed to boil off from
the heat of the tube. Which means the tube didn't get hotter than 100
degrees C (until all the water boiled away). This is known as 'vapor
phase cooling'. Believe it or not you can cool a tube with STEAM!

Tim Wescott wrote:
ken scharf wrote:
Tim Wescott wrote:
On Thu, 04 Dec 2008 06:23:48 -0800, JIMMIE wrote:

After years of avoiding morse code I am finally getting into it. the
ideal of operating a very simple transmitter appeals to my junkbox/
trashcan construction mentality.
I am looking for a substitue for a 6L6 that has the plate brought out
the top. I was trying to build a little Glowbug transmitter but was
having all kinds of problems neutralizing the the thing. I think I
would
have better luck if I can separate the grid and plate circuits from
each
other. I am considering an 807 as a replacement Suggestions would be
appreciated.

Jimmie

Dunno if it's been mentioned yet -- 2E26. The 807 is _not_ a 6L6 in
a different envelope -- it's quit arguably the 6L6's big brother, but
it's got different ratings; it could probably be shoe-horned into a
circuit designed for the 6L6, but you'd be missing out on about 6dB
of final output power.

The 807 IS a 6L6G with a 5 pin base and the plate connected to the top.
The published ratings of the 6L6G look different than the 807 because
of their intended use. ICAS ratings for the 6L6G were never published.
Also the 807 has additional shielding and insulation over the 6L6G
that make it usable at higher frequencies and voltages than the 6L6G.
Most 807's have ceramic spaces to support the plate which are lacking
in the 6L6G. But make no mistake about it, the two tubes share the
IDENTICAL cathode, grids, and plate structures.

The 6BG6G tv sweep tube IS an 807 with an octal base. It has the same
ceramic plate supports, but lacks the extra rf shielding.

The 1625 is an 807 with a 12.6 volt heater and a large (same as type
'59) 7 pin base.

The type 1614 is a metal tube based on the 6L6. It is a transmitting
version, and probably has additional shielding. Otherwise its internal
structure is the same as the 6L6 metal type.

The type 1619 is sorta kinda a 6L6 with a directly heated 2.5v cathode.
Same metal bulb as the 6L6. Specs' are different due to the different
element spacing thanks to the filament cathode. This tube is often
triode connected to replace 45's and 2A3's in old radios with a socket
adapter.

As a result the of the construction differences regarding shielding,
the 807 often would NOT need neutralization while 6L6G's and 6BG6G
tubes used as rf power amps do.


The 6L6GA was identical to the 6L6G except that the bulb shrunk from
an ST16 to an ST14 size. The 6L6GB was identical except for the bulb
changing again to a T14. The 6L6GC is a totally different bottle with
higher plate and screen dissipation and plate voltage ratings. The
6L6GC was said to be a plug in replacement for the older 6L6 tubes,
but RCA kept the 6L6GB around for a while anyway. In fact a bias
re-adjustment was a good idea when replacing an older 6L6 with the 'GC
version if a cathode bias resistor wasn't used.

I bow to your superior knowledge.

How do you know? Time spent busting tubes and looking inside? Old-time
service and/or circuit design experience? Experience building tubes?

I'd love to see a detailed family tree for some of the more popular tube
types.

I've done a lot of reading on this subject. Old QST's (from the '30s),
books about tubes, and I've smashed a few open (dead ones of course).
I also have a small collection of old tube manuals (in print and pdf off
the 'web).
BTW the 6L6 family tree is quite large and includes bottles such as the
7027/A (really a 6L6 with a bigger plate and TWO base pin connections
for G1 and G2). Most sweep tubes are descendants of the 6L6 (the 6BG6G
directly and others indirectly). Other 'hifi' output BP tubes such as
the 6550, and the 'KT' series are based on the 6L6 with larger elements
for handling more power. The British have in interesting twist on the
language. KT stands for 'Kinkless Tetrode' which refers to the lack of
the dip in plate current as the screen voltage reaches the critical
point of secondary emission in tetrodes. BP tubes are considered
tetrodes since they lack a third grid, but those BP tubes with beam
deflection elements should be considered 5 element tubes. (Not all BP
tubes have the deflection plates. The critical parts of the BP formula
are aligned turns on G1 and G2, critical spacing between G2 and the
plate, and optionally deflection elements to focus the electrons into
sheets. Having a plate with 'catch pockets' such as on the 4-400 works
as well as the BD elements.)

JIMMIE wrote:
After years of avoiding morse code I am finally getting into it. the
ideal of operating a very simple transmitter appeals to my junkbox/
trashcan construction mentality.
I am looking for a substitue for a 6L6 that has the plate brought out
the top. I was trying to build a little Glowbug transmitter but was
having all kinds of problems neutralizing the the thing. I think I
would have better luck if I can separate the grid and plate circuits
from each other. I am considering an 807 as a replacement Suggestions
would be appreciated.

Jimmie
To Hijack an old thread...

The way tubes were numbered sometimes....
The 25L6 and 50L6 are NOT 6L6's with different heater voltages.
Actually both of these tubes are IDENTICAL with the 6W6GT except for the
heater ratings.