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Resistor Wattage?

D

Dave.H

Jan 1, 1970
0
I have a regen radio that uses one type 30 triode tube with 2.0 volt,
60 mA filament. The radio had two 33 ohm 1/2 watt resistors wired in
parallel for 16.5 ohms in series to the tube filament to cut down 3
volts from 2 D cells to 2 volts. I removed these resistors so I could
run the filament off 1 D cell. I now wish to run the filament off 2 D
cells again, problem is I put the resistors in a safe place, now I
can't remember where that "safe place" is. I need to buy a new 20 ohm
resistor but should I use 1/2 watt, or 1 watt? The filament is 2
volts, 60 mA as I previously mentioned.

Thanks,
Dave
Australia
 
J

Jonathan Kirwan

Jan 1, 1970
0
I have a regen radio that uses one type 30 triode tube with 2.0 volt,
60 mA filament. The radio had two 33 ohm 1/2 watt resistors wired in
parallel for 16.5 ohms in series to the tube filament to cut down 3
volts from 2 D cells to 2 volts. I removed these resistors so I could
run the filament off 1 D cell. I now wish to run the filament off 2 D
cells again, problem is I put the resistors in a safe place, now I
can't remember where that "safe place" is. I need to buy a new 20 ohm
resistor but should I use 1/2 watt, or 1 watt? The filament is 2
volts, 60 mA as I previously mentioned.

The basic equations to keep in mind often is E=I*R and P=E*I (power.)
You need to drop 1V at 60mA, which works out to 1V/60mA or 16 2/3 ohm,
as you pretty much agree with from the above comments. The power is
1V*60mA or 60mW. So a 1/2 watt resistor would seem to do the job.

If you use a 20 ohm resistor, it will drop the current somewhat. Your
filament was about 2V/60mA or 33 1/3 ohms. That, added to your 20
ohms at 3V will drop the current to about 56mA. So 1V*56mA or 56mW.
Same deal -- the 1/2 watt will be fine. So would 1/4 or 1/8 watt,
too.

You could still buy two 33 ohm resistors and parallel them. They
aren't expensive and often sold in pairs, anyway. (Each would share
half the power, so still no problem with low wattage resistors.)

Jon
 
D

Dave.H

Jan 1, 1970
0
The basic equations to keep in mind often is E=I*R and P=E*I (power.)
You need to drop 1V at 60mA, which works out to 1V/60mA or 16 2/3 ohm,
as you pretty much agree with from the above comments.  The power is
1V*60mA or 60mW.  So a 1/2 watt resistor would seem to do the job.

If you use a 20 ohm resistor, it will drop the current somewhat.  Your
filament was about 2V/60mA or 33 1/3 ohms.  That, added to your 20
ohms at 3V will drop the current to about 56mA.  So 1V*56mA or 56mW.
Same deal -- the 1/2 watt will be fine.  So would 1/4 or 1/8 watt,
too.

You could still buy two 33 ohm resistors and parallel them.  They
aren't expensive and often sold in pairs, anyway.  (Each would share
half the power, so still no problem with low wattage resistors.)

Jon

Is there any disadvantages to running the filament with 56 mA? The
filament is in series with a 20 ohm rheostat, which I always leave on
maximum. Audio distorts if it is set any lower.


Thanks,
Dave
Australia
 
J

Jonathan Kirwan

Jan 1, 1970
0
Is there any disadvantages to running the filament with 56 mA? The
filament is in series with a 20 ohm rheostat, which I always leave on
maximum. Audio distorts if it is set any lower.

I'd imagine that the peak plate current might be lower for some plate
voltages. The key reason for heating the cathode up more is to
encourage more electrons to boil off the cathode material. The plate
voltage attracts them across a small gap in the tube.

Diversion: Imagine inside the tube heat is removed from the filament
through two methods -- radiation and conduction (convection is
unlikely given the rough vacuum inside the tube.) The radiation part
is proportional to T^4, so it is some k1*T^4. The conduction part is
largely proportional to the difference in temperature from the
surroundings via some unknown thermal resistance, so k2*T. The sum of
the two relates to heat removal. Lowering the current will definitely
lower the temperature, which will lower the heat removal rate, and it
will find some new equilibrium temperature. I can't say what the new
temperature will be, but it will be colder. If you can see the
difference with your eyes, you can be pretty sure it is more than just
a few degrees.

Diversion coming around to point: The thermal agitation of electrons
(thermionic emission) is strongly (non-linear to Kelvins) temperature
dependent (see the Richardson- Dushman equation) and is some
k3*T^2*e^(k4/T), where k4 is negative. The derivative essentially
tells you that the exponential part dominates. For tungsten, for
example, a 20 Kelvin change might double (or halve) the electron cloud
generated. These are the available electrons for motion across to the
plate. There is another effect, a field effect, which can pull
electrons off of the cathode. But in vacuum tubes I think it is
pretty much nil for most plate voltages and tubes in use. (Sometimes,
there is something called a Wehnelt or another electrode used to
enhance this field effect along with thermionic emissions, but not in
usual tubes like these.)

Back to point: The change in filament current is about 7% and this
lowering of the filament temperature will make a change on the
electrons that are agitated from the cathode and thus lower the plate
currents somewhat. But whether or not it matters that much for your
use is another story. It might work well. Resistors are cheap, not
accounting for time and travel. If the 20 ohm doesn't do it, then you
can always go back and get the pair of 33 ohm resistors. Or buy a
selection and play around and have some fun with it.

Jon
 

neon

Oct 21, 2006
1,325
Joined
Oct 21, 2006
Messages
1,325
Jonathan Kirwan said:
On Sat, 26 Jul 2008 03:14:20 -0700 (PDT), "Dave.H"
<[email protected]> wrote:

>On 26 Jul, 17:08, Jonathan Kirwan <[email protected]> wrote:
>> On Fri, 25 Jul 2008 23:49:07 -0700 (PDT), "Dave.H"
>>
>> <[email protected]> wrote:
>> >I have a regen radio that uses one type 30 triode tube with 2.0 volt,
>> >60 mA filament. *The radio had two 33 ohm 1/2 watt resistors wired in
>> >parallel for 16.5 ohms in series to the tube filament to cut down 3
>> >volts from 2 D cells to 2 volts. I removed these resistors so I could
>> >run the filament off 1 D cell. I now wish to run the filament off 2 D
>> >cells again, problem is I put the resistors in a safe place, now I
>> >can't remember where that "safe place" is. I need to buy a new 20 ohm
>> >resistor but should I use 1/2 watt, or 1 watt? The filament is 2
>> >volts, 60 mA as I previously mentioned.

>>
>> The basic equations to keep in mind often is E=I*R and P=E*I (power.)
>> You need to drop 1V at 60mA, which works out to 1V/60mA or 16 2/3 ohm,
>> as you pretty much agree with from the above comments. *The power is
>> 1V*60mA or 60mW. *So a 1/2 watt resistor would seem to do the job.
>>
>> If you use a 20 ohm resistor, it will drop the current somewhat. *Your
>> filament was about 2V/60mA or 33 1/3 ohms. *That, added to your 20
>> ohms at 3V will drop the current to about 56mA. *So 1V*56mA or 56mW.
>> Same deal -- the 1/2 watt will be fine. *So would 1/4 or 1/8 watt,
>> too.
>>
>> You could still buy two 33 ohm resistors and parallel them. *They
>> aren't expensive and often sold in pairs, anyway. *(Each would share
>> half the power, so still no problem with low wattage resistors.)

>
>Is there any disadvantages to running the filament with 56 mA? The
>filament is in series with a 20 ohm rheostat, which I always leave on
>maximum. Audio distorts if it is set any lower.


I'd imagine that the peak plate current might be lower for some plate
voltages. The key reason for heating the cathode up more is to
encourage more electrons to boil off the cathode material. The plate
voltage attracts them across a small gap in the tube.

Diversion: Imagine inside the tube heat is removed from the filament
through two methods -- radiation and conduction (convection is
unlikely given the rough vacuum inside the tube.) The radiation part
is proportional to T^4, so it is some k1*T^4. The conduction part is
largely proportional to the difference in temperature from the
surroundings via some unknown thermal resistance, so k2*T. The sum of
the two relates to heat removal. Lowering the current will definitely
lower the temperature, which will lower the heat removal rate, and it
will find some new equilibrium temperature. I can't say what the new
temperature will be, but it will be colder. If you can see the
difference with your eyes, you can be pretty sure it is more than just
a few degrees.

Diversion coming around to point: The thermal agitation of electrons
(thermionic emission) is strongly (non-linear to Kelvins) temperature
dependent (see the Richardson- Dushman equation) and is some
k3*T^2*e^(k4/T), where k4 is negative. The derivative essentially
tells you that the exponential part dominates. For tungsten, for
example, a 20 Kelvin change might double (or halve) the electron cloud
generated. These are the available electrons for motion across to the
plate. There is another effect, a field effect, which can pull
electrons off of the cathode. But in vacuum tubes I think it is
pretty much nil for most plate voltages and tubes in use. (Sometimes,
there is something called a Wehnelt or another electrode used to
enhance this field effect along with thermionic emissions, but not in
usual tubes like these.)

Back to point: The change in filament current is about 7% and this
lowering of the filament temperature will make a change on the
electrons that are agitated from the cathode and thus lower the plate
currents somewhat. But whether or not it matters that much for your
use is another story. It might work well. Resistors are cheap, not
accounting for time and travel. If the 20 ohm doesn't do it, then you
can always go back and get the pair of 33 ohm resistors. Or buy a
selection and play around and have some fun with it.

Jon
finaly we have somebody that realy knows tubes. I complement you JON. very well done to details.
 
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