More Resistor/Diode Fil Supply Musing

More Resistor/Diode Fil Supply Musing

Post by Ken » Sun, 06 May 2007 04:17:19



I took some scope readings and found the half wave peak across a 6V tube
was 12V. This will cause a 600ma current through the 300ma fil. This
makes sense, since if no current passes during the first half cycle,
then twice as much will need to pass through the fil for the same one
cycle heating. The thing that has always bothered me about this method
is hitting a tube fil with twice the current. It works great, but I
wonder if the fil life is affected. Any thoughts? Ken
 
 
 

More Resistor/Diode Fil Supply Musing

Post by Uncle Pete » Sun, 06 May 2007 04:32:49



Quote:
>I took some scope readings and found the half wave peak across a 6V tube
>was 12V. This will cause a 600ma current through the 300ma fil. This makes
>sense, since if no current passes during the first half cycle, then twice
>as much will need to pass through the fil for the same one cycle heating.
>The thing that has always bothered me about this method is hitting a tube
>fil with twice the current. It works great, but I wonder if the fil life is
>affected. Any thoughts? Ken

I think you are confusing the issue. The peak current on the
crest of the AC sinewave will also be 1.414 times the RMS
value. The diode delivers 1/2 power, or an voltage equivalent
of .707 times RMS.  The critical thing is average heat energy for
the filament.  Not what is happening at some single point
on a curve. Also consider that there is no energy being delivered
on the missing half cycle.

Also, you can't just look at the AC component coming out
of the rectifier, there is a DC bias which must also be
calculated for.

Pete

 
 
 

More Resistor/Diode Fil Supply Musing

Post by R! » Sun, 06 May 2007 05:20:51




Quote:



>>I took some scope readings and found the half wave peak across a 6V
>>tube was 12V. This will cause a 600ma current through the 300ma fil.
>>This makes sense, since if no current passes during the first half
>>cycle, then twice as much will need to pass through the fil for the
>>same one cycle heating. The thing that has always bothered me about
>>this method is hitting a tube fil with twice the current. It works
>>great, but I wonder if the fil life is affected. Any thoughts? Ken

> I think you are confusing the issue. The peak current on the
> crest of the AC sinewave will also be 1.414 times the RMS
> value. The diode delivers 1/2 power, or an voltage equivalent
> of .707 times RMS.  The critical thing is average heat energy for
> the filament.  Not what is happening at some single point
> on a curve. Also consider that there is no energy being delivered
> on the missing half cycle.

> Also, you can't just look at the AC component coming out
> of the rectifier, there is a DC bias which must also be
> calculated for.

> Pete

Zenith used this method to drop the filament voltage on some of their
1970's 12" Black and white TV's.

                      Resistor
     ---FUSE-->|----|-/\/\/\--/\-/\-/\-- GND
AC ==               --       Filaments
          Diodes    /\
     ---------------|-------------------GND

They used a safety fuse and extra diode if one diode shorts it takes out
the fuse...

Use fixed width font to see ascii diagram better...

R!

 
 
 

More Resistor/Diode Fil Supply Musing

Post by neses » Sun, 06 May 2007 05:29:04



Quote:
> I took some scope readings and found the half wave peak across a 6V tube
> was 12V. This will cause a 600ma current through the 300ma fil. This
> makes sense, since if no current passes during the first half cycle,
> then twice as much will need to pass through the fil for the same one
> cycle heating. The thing that has always bothered me about this method
> is hitting a tube fil with twice the current. It works great, but I
> wonder if the fil life is affected. Any thoughts? Ken

Hang on, Ken. If you set it up so you are providing 1/2 wave through a
diode such that you have twice the peak voltage, then the dissipation
will be double. W=ExI = (12 x .6) x 1/2 = 3.6W where as the nominal
dissipation would be 6 x .3 =1.8W. The voltage across the tube on half
wave should be 1.414 times the nominal rating, or 8.9V on a 6.3v tube.

It is a good point, though. When the diode and resistance are set
corectly it clearly does work okay. Operationally, it is the heat in
watts that is the important thing, not how it's dissipation is
distributed over a cycle. When you consider 'switch on' the current is
many times the 'normal' current because the cold resistance of the
filaments is far less than the hot resistance, so the filaments are
capable of carrying much more than the nominal 300 or 150mA. That
said, there must be a limit to how much current can be safely carried
because it would be silly to expect to put, say, 3A for 1% of the time
to get the correct heating.

Neil S.

 
 
 

More Resistor/Diode Fil Supply Musing

Post by Ken » Sun, 06 May 2007 05:50:11


Quote:


>> I took some scope readings and found the half wave peak across a 6V tube
>> was 12V. This will cause a 600ma current through the 300ma fil. This
>> makes sense, since if no current passes during the first half cycle,
>> then twice as much will need to pass through the fil for the same one
>> cycle heating. The thing that has always bothered me about this method
>> is hitting a tube fil with twice the current. It works great, but I
>> wonder if the fil life is affected. Any thoughts? Ken

> Hang on, Ken. If you set it up so you are providing 1/2 wave through a
> diode such that you have twice the peak voltage, then the dissipation
> will be double. W=ExI = (12 x .6) x 1/2 = 3.6W where as the nominal
> dissipation would be 6 x .3 =1.8W. The voltage across the tube on half
> wave should be 1.414 times the nominal rating, or 8.9V on a 6.3v tube.

> It is a good point, though. When the diode and resistance are set
> corectly it clearly does work okay. Operationally, it is the heat in
> watts that is the important thing, not how it's dissipation is
> distributed over a cycle. When you consider 'switch on' the current is
> many times the 'normal' current because the cold resistance of the
> filaments is far less than the hot resistance, so the filaments are
> capable of carrying much more than the nominal 300 or 150mA. That
> said, there must be a limit to how much current can be safely carried
> because it would be silly to expect to put, say, 3A for 1% of the time
> to get the correct heating.

> Neil S.

Yeah, that's what I was wondering about. I just didn't know how much
over current these filaments could take. Sounds like this fix is not
distructive. Thanks, Ken
 
 
 

More Resistor/Diode Fil Supply Musing

Post by Ken » Sun, 06 May 2007 05:56:40


Quote:





>>> I took some scope readings and found the half wave peak across a 6V
>>> tube was 12V. This will cause a 600ma current through the 300ma fil.
>>> This makes sense, since if no current passes during the first half
>>> cycle, then twice as much will need to pass through the fil for the
>>> same one cycle heating. The thing that has always bothered me about
>>> this method is hitting a tube fil with twice the current. It works
>>> great, but I wonder if the fil life is affected. Any thoughts? Ken
>> I think you are confusing the issue. The peak current on the
>> crest of the AC sinewave will also be 1.414 times the RMS
>> value. The diode delivers 1/2 power, or an voltage equivalent
>> of .707 times RMS.  The critical thing is average heat energy for
>> the filament.  Not what is happening at some single point
>> on a curve. Also consider that there is no energy being delivered
>> on the missing half cycle.

>> Also, you can't just look at the AC component coming out
>> of the rectifier, there is a DC bias which must also be
>> calculated for.

>> Pete

> Zenith used this method to drop the filament voltage on some of their
> 1970's 12" Black and white TV's.

>                       Resistor
>      ---FUSE-->|----|-/\/\/\--/\-/\-/\-- GND
> AC ==               --       Filaments
>           Diodes    /\
>      ---------------|-------------------GND

> They used a safety fuse and extra diode if one diode shorts it takes out
> the fuse...

> Use fixed width font to see ascii diagram better...

> R!

Neat circuit, to guarantee that fuse will blow. Thanks. Ken
 
 
 

More Resistor/Diode Fil Supply Musing

Post by neses » Sun, 06 May 2007 07:05:58



Quote:

> > I took some scope readings and found the half wave peak across a 6V tube
> > was 12V. This will cause a 600ma current through the 300ma fil. This
> > makes sense, since if no current passes during the first half cycle,
> > then twice as much will need to pass through the fil for the same one
> > cycle heating. The thing that has always bothered me about this method
> > is hitting a tube fil with twice the current. It works great, but I
> > wonder if the fil life is affected. Any thoughts? Ken

> Hang on, Ken. If you set it up so you are providing 1/2 wave through a
> diode such that you have twice the peak voltage, then the dissipation
> will be double. W=ExI = (12 x .6) x 1/2 = 3.6W where as the nominal
> dissipation would be 6 x .3 =1.8W. The voltage across the tube on half
> wave should be 1.414 times the nominal rating, or 8.9V on a 6.3v tube.

> It is a good point, though. When the diode and resistance are set
> corectly it clearly does work okay. Operationally, it is the heat in
> watts that is the important thing, not how it's dissipation is
> distributed over a cycle. When you consider 'switch on' the current is
> many times the 'normal' current because the cold resistance of the
> filaments is far less than the hot resistance, so the filaments are
> capable of carrying much more than the nominal 300 or 150mA. That
> said, there must be a limit to how much current can be safely carried
> because it would be silly to expect to put, say, 3A for 1% of the time
> to get the correct heating.

> Neil S.

Oops, I failed to note you said 'peak' measured with a 'scope. In that
case a 6V tube on a full sine wave would have 8.5V peak on it, while
on equal power 1/2 wave it would be 12V peak. I guess you never
actually measured the current waveform since that is more awkward to
do accurately.

Neil S.

 
 
 

More Resistor/Diode Fil Supply Musing

Post by maar.. » Sun, 06 May 2007 08:48:08


Quote:
>> Zenith used this method to drop the filament voltage on some of their
>> 1970's 12" Black and white TV's.

>>                       Resistor
>>      ---FUSE-->|----|-/\/\/\--/\-/\-/\-- GND
>> AC ==               --       Filaments
>>           Diodes    /\
>>      ---------------|-------------------GND

>> They used a safety fuse and extra diode if one diode shorts it takes out
>> the fuse...

Philips never added a safety diode their circuit that was used in
televisions, but it had a much better load factor:

-FUSE----|----->|-------------------- +B
         |  diodes
AC       |----|<----/\--/\--/\--|
                     filaments  |
                                |
--------------------------------|--- GND

--
Met vriendelijke groet,

Maarten Bakker.

 
 
 

More Resistor/Diode Fil Supply Musing

Post by Mark Oppa » Sun, 06 May 2007 10:30:35


Yeah, that's what I was wondering about. I just didn't know how much
over current these filaments could take. Sounds like this fix is not
distructive. Thanks, Ken

thats the bottom line, Ken, it works great.  I have done it on several sets
and not one has failed... so, I give it the thumbs up.

Mark Oppat

 
 
 

More Resistor/Diode Fil Supply Musing

Post by thisjukeboxplays33rp » Sun, 06 May 2007 10:28:22



Quote:
> Yeah, that's what I was wondering about. I just didn't know how much
> over current these filaments could take. Sounds like this fix is not
> distructive. Thanks, Ken

> thats the bottom line, Ken, it works great.  I have done it on several sets
> and not one has failed... so, I give it the thumbs up.

> Mark Oppat

Your computer is ten minutes fast, Mark.

Or maybe I'm 5 slow and you 2 ahead.

 
 
 

More Resistor/Diode Fil Supply Musing

Post by DumpsterDive » Wed, 09 May 2007 10:52:45



Quote:

>... there must be a limit to how much current can be safely carried
> because it would be silly to expect to put, say, 3A for 1% of the time
> to get the correct heating.

> Neil S.

I think the key here is the FREQUENCY of the current spikes, not their
maximum value.  As long as the temperature of the heater or filament
doesn't vary significantly over the cycle, the shape of the waveform,
peak current value, etc. shouldn't matter at all.  I bet "3A for 1% of
the time" at 60 Hz would work just fine, at least in terms of not
damaging the filament or shortening its life.

The filament has a property known as "thermal mass" or "thermal
inertia" which refers to how quickly its temperature can change in
response to changes in the instantaneous power input.  If the average
current over the time interval determined by its thermal inertia meets
the rated nominal value, it will operate the same whether the waveform
is a series of spikes, full wave, half wave, DC or whatever.

Of course, extremely spiky waveforms are likely to be problematic for
other reasons, like noise propagation, "ringing", etc., but they would
not likely harm the tube at all.  Sorta like those lamp dimmer
switches ... lotsa *** RF noise radiation, but no harm to the light
bulbs.

DD