You want to drop 4 volts @ .42 amps across the resistor. Divide 4 by .42 and you get 9.5 ohms. You will need at least a 2 watt resistor. 4 x .42 = 1.68 watts.
In this situation don't remove tubes from their sockets with power on. Voltage will go up since load current goes down.
Norm
:I'm looking for the proper resistor connection to reduce a 6 volt storage battery to operate filaments of six 2 volt tubes with total current .42 amps ( 5 @ .060 amps, 1 @.12 amps). Can someone advise me?
But, anyway, here is an alternative for you to consider:
Put three of the 60-mA tubes in series. 3 x 2V = 6, so you can put them directly across the 6V battery supply.
Now put the remaining two 60-mA tubes in parallel (so that the parallel combination is rated 2V, 120mA). Now, put that parallel combination in series with the 120-mA tube. So that whole combination of three tubes is rated 4V, 120mA. Put in a dropping resistor to drop 2V at 120mA = 16.7 ohms (say 15 ohms). The power consumption is only 0.25W, so a 1/2-W resistor will work fine.
I can see several advantages to this scheme, depending upon your application.
When tubes are in parallel, and one tube blows, then the others will get higher voltage - ping, ping, ping! The three tubes I propose to be in series will be safe from this problem.
(However, the proposed two 60-mA tubes in parallel (and thence in series with the 120-mA tube) would be somewhat at risk.)
Another advantage is only needing a 1/2-ohm resistor and less wasted power drain on the battery.
Doug
:Hi Peter
:
: You want to drop 4 volts @ .42 amps across the resistor. Divide 4 by .42 and you get 9.5 ohms. You will need at least a 2 watt resistor. 4 x .42 = 1.68 watts.
:
: In this situation don't remove tubes from their sockets with power on. Voltage will go up since load current goes down.
:
:Norm
:
::I'm looking for the proper resistor connection to reduce a 6 volt storage battery to operate filaments of six 2 volt tubes with total current .42 amps ( 5 @ .060 amps, 1 @.12 amps). Can someone advise me?
:Peter, I'm curious about what you're cooking up here.
:
:But, anyway, here is an alternative for you to consider:
:
:Put three of the 60-mA tubes in series. 3 x 2V = 6, so you can put them directly across the 6V battery supply.
:
:Now put the remaining two 60-mA tubes in parallel (so that the parallel combination is rated 2V, 120mA). Now, put that parallel combination in series with the 120-mA tube. So that whole combination of three tubes is rated 4V, 120mA. Put in a dropping resistor to drop 2V at 120mA = 16.7 ohms (say 15 ohms). The power consumption is only 0.25W, so a 1/2-W resistor will work fine.
:
:I can see several advantages to this scheme, depending upon your application.
:
:When tubes are in parallel, and one tube blows, then the others will get higher voltage - ping, ping, ping! The three tubes I propose to be in series will be safe from this problem.
:
:(However, the proposed two 60-mA tubes in parallel (and thence in series with the 120-mA tube) would be somewhat at risk.)
:
:Another advantage is only needing a 1/2-ohm resistor and less wasted power drain on the battery.
:Doug
:
:
Getting out the old Ohm's law book, it seems that we wish to drop 4 Volts across the resistor, leaveing 2 Volts for the tubes. Since R=E/I, we are looking at 2/.42, or 9.52 Ohms. The power would be EI, or 4 times .42 = 1.68 Amperes. If you were to use a 10 Ohm resistor, you would have 4.2 Volts across the resistor, but since wet cell batteries are 2.2 Volts per cell, you are starting out with 6.6 volts, so a ten Ohm, 2.5 Watt resistor should work out just fine.
Lewis L.
~~~~~~~~~~~~~~~~~~ Corrected copy~~~~~~~~~~~~~~~~~~~~
Getting out the old Ohm's law book, it seems that we wish to drop 4 Volts across the resistor, leaveing 2 Volts for the tubes. Since R=E/I, we are looking at 4/.42, or 9.52 Ohms The power would be EI, or 4 times .42 = 1.68 Amperes. If you were to use a 10 Ohm resistor, you would have 4.2 Volts across the resistor, but since wet cell batteries are 2.2 Volts per cell, you are starting out with 6.6 volts, so a ten Ohm, 2.5 Watt resistor should work out just fine.
:
:Lewis L.
::::I'm looking for the proper resistor connection to reduce a 6 volt storage battery to operate filaments of six 2 volt tubes with total current .42 amps ( 5 @ .060 amps, 1 @.12 amps). Can someone advise me?
::
::
::~~~~~~~~~~~~~~~~~~ Corrected copy~~~~~~~~~~~~~~~~~~~~
::
::Getting out the old Ohm's law book, it seems that we wish to drop 4 Volts across the resistor, leaveing 2 Volts for the tubes. Since R=E/I, we are looking
at 4/.42, or 9.52 Ohms The power would be EI, or 4 times .42 = 1.68 Amperes. If you were to use a 10 Ohm resistor, you would have 4.2 Volts across the resistor, but since wet cell batteries are 2.2 Volts per cell, you are starting out with 6.6 volts, so a ten Ohm, 2.5 Watt resistor should work out just fine.
:::
:::Lewis L.
::
Resistor is in series with filaments in positive or negative side. It doesn't matter.
Norm
:That would be in series with positive side of battery? Thank you all for your help.
:
:
:
:
:::::I'm looking for the proper resistor connection to reduce a 6 volt storage battery to operate filaments of six 2 volt tubes with total current .42 amps ( 5 @ .060 amps, 1 @.12 amps). Can someone advise me?
:::
:::
:::~~~~~~~~~~~~~~~~~~ Corrected copy~~~~~~~~~~~~~~~~~~~~
:::
:::Getting out the old Ohm's law book, it seems that we wish to drop 4 Volts across the resistor, leaveing 2 Volts for the tubes. Since R=E/I, we are looking
:
:at 4/.42, or 9.52 Ohms The power would be EI, or 4 times .42 = 1.68 Amperes. If you were to use a 10 Ohm resistor, you would have 4.2 Volts across the resistor, but since wet cell batteries are 2.2 Volts per cell, you are starting out with 6.6 volts, so a ten Ohm, 2.5 Watt resistor should work out just fine.
::::
::::Lewis L.
:::
:
As far as where each resistor goes, if you put the resistor in series with the positive side, and the negative is the same as B-, then all you need to do is bias your grids with respect to B-. If you put the resistor on the negative side, you will be adding an additional positive bias to the cathode with respect to B-, and an additional negative bias to the grid, if it is biased at or below B-. If the additional bias isn't necessary, you might want to connect the other side of your C source on the filament side of the negatively positioned resistor. If you are using grid leak bias, connect the leak resistor to the filament, and not to B-. If you are using an IF or RF transformer, connect the other side of the transformer to the filament, not to B-, unless the additional negativity is desired.
In 3-way portables, where the tubes are strung in series, special attention must be paid to how each tube is biased. When done properly, you'll have one very fine performing radio. It's a bit trickier than biasing an AC set, though. Also, like I said before, putting a resistor between the tube filament and B- will also affect grid bias. If you want some extra bias, but don't need all that is provided by a resistor in the negative side of the filament (4 volts of bias developed), you can instead put a resistor on the positive and the negative side, and adjust each so that you have the right amount of bias on the negative side....say you want only 1 or 2 volts on the negative side....then you'd put in an appropriate dropping resistor there, and then put a resistor on the positive side that'd drop 3 or 2 volts, depending on what the other one was dropping, making the total drop of the two resistors 4 volts.
Thomas
Thomas
Make sure that the zeners can handle it on their own. They should be able to operate without overheating, with every tube in the set removed.
Thomas
If you are going to that much trouble, with semiconductors, how about a 2 Volt regulator made with a one Amp NPN transistor, a Zener diode, and a 100 Ohm resistor? Connect the battery positive to the collector and the Zener, the resistor between the battery/collector and the base, and the Zener between the base/resistor and the battery negative. Perhaps a 2.5 or 3 Volt Zener should give you about 2 to 2.5 Volts, regardless of load, experment and see.
Lewis
Lewis
~~~~~~~~~~~~~~~~~~~~Another correcton~~~~~~~~~~~~~~
The battery positive goes to the collector and the resistor, the resistor goes to the base and the Zener anode, the Zener cathode goes to battery negative, power is taken off of the Zerer emitter (+) to battery (-)
I don't know how many junction drops (.7V) you will have, may have to experiment with 1N914 diodes forward biased in series to get the exact Voltage you want.
Lewis
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Lewis
:::Also, if you decide to wire the tubes in parallel, and you don't want disaster from a burnout, you can wire 2 volt zener diodes across each tube. This would make sure that no more than 2 volts is ever developed across each tube.
:::
:::Make sure that the zeners can handle it on their own. They should be able to operate without overheating, with every tube in the set removed.
::
::
::
::If you are going to that much trouble, with semiconductors, how about a 2 Volt regulator made with a one Amp NPN transistor, a Zener diode, and a 100 Ohm resistor? Connect the battery positive to the collector and the Zener, the resistor between the battery/collector and the base, and the Zener between the base/resistor and the battery negative. Perhaps a 2.5 or 3 Volt Zener should give you about 2 to 2.5 Volts, regardless of load, experment and see.
::Lewis
:
:~~~~~~~~~~~~~~~~~~~~Another correcton~~~~~~~~~~~~~~
:
:The battery positive goes to the collector and the resistor, the resistor goes to the base and the Zener anode, the Zener cathode goes to battery negative, power is taken off of the Zerer emitter (+) to battery (-)
:
:I don't know how many junction drops (.7V) you will have, may have to experiment with 1N914 diodes forward biased in series to get the exact Voltage you want.
:
:Lewis
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::
Thomas
:Here's another idea----put a hefty four Volt Zener diode in series with the battery lead. You will have: 1. A four Volt drop regardless of current drawn, 2. A simple and cheap fix for your problem, and 3. No fooling around changing vintage radios. The downside, if the battery charges up to, say 7.3 volts, the filament Voltage will be 3.3 Volts. There are also ways to wire an LM 317 regulator, using a single resistor, to provide a constant current. This guy could be placed in series with the positive lead, with no connection to the negative, another cheap and quick fix that is better than the resistor.
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:Lewis
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:
:
::::Also, if you decide to wire the tubes in parallel, and you don't want disaster from a burnout, you can wire 2 volt zener diodes across each tube. This would make sure that no more than 2 volts is ever developed across each tube.
::::
::::Make sure that the zeners can handle it on their own. They should be able to operate without overheating, with every tube in the set removed.
:::
:::
:::
:::If you are going to that much trouble, with semiconductors, how about a 2 Volt regulator made with a one Amp NPN transistor, a Zener diode, and a 100 Ohm resistor? Connect the battery positive to the collector and the Zener, the resistor between the battery/collector and the base, and the Zener between the base/resistor and the battery negative. Perhaps a 2.5 or 3 Volt Zener should give you about 2 to 2.5 Volts, regardless of load, experment and see.
:::Lewis
::
::~~~~~~~~~~~~~~~~~~~~Another correcton~~~~~~~~~~~~~~
::
::The battery positive goes to the collector and the resistor, the resistor goes to the base and the Zener anode, the Zener cathode goes to battery negative, power is taken off of the Zerer emitter (+) to battery (-)
::
::I don't know how many junction drops (.7V) you will have, may have to experiment with 1N914 diodes forward biased in series to get the exact Voltage you want.
::
::Lewis
:::
:::
:::
If you connect a zener diode forward biased, then it will act like a regular diode with less than a volt of drop.
Doug
:By putting the zener diode (IN4730A) in series off the positive post of the battery, does the cathode or band on the zener point to the radio or battery terminal?
:
~~~~~~~~~~~~~~~Correction to correction~~~~~~~~~~~~~
Above power would be 1.68 Watts, not Amperes. I know the math, but have forgotten the terminology.
Lewis
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