I have a cheapo push-pull stereo amp which had a little too much hum in both channels, which I could tell by ear was 60-Hz -- so it had to be coming from the heater ckt. (The amp has a full-wave rectifier, so it the hum were coming from the B+, it would be 120Hz.) The 12-V heaters were supplied by a 12V-CT winding off the power xfmr. The center-tap was grounded to the chassis.
Here's the quick fix: disconnect the 12-V CT from ground. Re-connect the CT to ground through a cap (I used a 10-uF, 150-V electrolytic).
Rig up two resistors in a simple voltage divider between B+ and ground. Size the resistors to give about 50V to ground. Now, connect the xfmr's heater winding's CT to the 50V. You're done.
It was explained that the DC bias on the heater ckt causes the tube cathodes to "repel" the 60-Hz hum. I don't understand that terminology. Assuming that the hum comes from the capacitive coupling between the heaters and the cathodes, I can't visualize why the heater bias would make a difference.
But, my 60-Hz hum virtually disappeared.
This is sometimes done with pre amplifier stages. When the filament is at or below cathode it can emit electrons. It will be seen as a negative tube element. If the filament is tied to a positive voltage now cathode is the most negative.
Norm
:To reduce 60-Hz hum originating from the heater circuit, bias the heater ckt with DC so that the voltage-to-ground stays above the tube cathode voltages. I got this idea from an audiophile website, and it worked for me.
:
:I have a cheapo push-pull stereo amp which had a little too much hum in both channels, which I could tell by ear was 60-Hz -- so it had to be coming from the heater ckt. (The amp has a full-wave rectifier, so it the hum were coming from the B+, it would be 120Hz.) The 12-V heaters were supplied by a 12V-CT winding off the power xfmr. The center-tap was grounded to the chassis.
:
:Here's the quick fix: disconnect the 12-V CT from ground. Re-connect the CT to ground through a cap (I used a 10-uF, 150-V electrolytic).
:
:Rig up two resistors in a simple voltage divider between B+ and ground. Size the resistors to give about 50V to ground. Now, connect the xfmr's heater winding's CT to the 50V. You're done.
:
:It was explained that the DC bias on the heater ckt causes the tube cathodes to "repel" the 60-Hz hum. I don't understand that terminology. Assuming that the hum comes from the capacitive coupling between the heaters and the cathodes, I can't visualize why the heater bias would make a difference.
:
:But, my 60-Hz hum virtually disappeared.
:
:
:Hi Doug
:
: This is sometimes done with pre amplifier stages. When the filament is at or below cathode it can emit electrons. It will be seen as a negative tube element. If the filament is tied to a positive voltage now cathode is the most negative.
:
:Norm
:
::To reduce 60-Hz hum originating from the heater circuit, bias the heater ckt with DC so that the voltage-to-ground stays above the tube cathode voltages. I got this idea from an audiophile website, and it worked for me.
::
::I have a cheapo push-pull stereo amp which had a little too much hum in both channels, which I could tell by ear was 60-Hz -- so it had to be coming from the heater ckt. (The amp has a full-wave rectifier, so it the hum were coming from the B+, it would be 120Hz.) The 12-V heaters were supplied by a 12V-CT winding off the power xfmr. The center-tap was grounded to the chassis.
::
::Here's the quick fix: disconnect the 12-V CT from ground. Re-connect the CT to ground through a cap (I used a 10-uF, 150-V electrolytic).
::
::Rig up two resistors in a simple voltage divider between B+ and ground. Size the resistors to give about 50V to ground. Now, connect the xfmr's heater winding's CT to the 50V. You're done.
::
::It was explained that the DC bias on the heater ckt causes the tube cathodes to "repel" the 60-Hz hum. I don't understand that terminology. Assuming that the hum comes from the capacitive coupling between the heaters and the cathodes, I can't visualize why the heater bias would make a difference.
::
::But, my 60-Hz hum virtually disappeared.
::
::
Yes, the cap's polarity will be with negative to ground, positive to the heater ckt.
For the voltage divider, you will need two resistors, say R1 and R2, in series. Yes, you want the pick-off point between the two resistors to be about 50V to ground. It doesn't have to be precise, and you can find standard sized resistors that will give a bias voltage that is close enough.
R1 is wired from the B+ line to R2. The remaining lead of R2 goes to ground. The 50V is picked off at the connection between R1 and R2, and is connected to the heater winding center tap. (If, by chance, the heater winding doesn't have a CT, let me know.)
Here's how to select the values of R1 and R2. The ratio [R2 / (R1 + R2)] = (50V / B+ voltage) (it doesn't have to be exact).
In addition, you want (R1 + R2) large enough to keep the power dissipation low enough to get by with 1/4-W resistors -- say, about 300K - 500K ohms total.
Example: my amplifier's B+ was 200V. I selected R1 = 220K and R2 = 75K. The bias voltage would then be (200V x 75K)/(220K + 75K) = 51V. (I would have been satisfied with anything between about 40V and 55V.)
Now check the power dissipation, which needs to be less than half of the resistor rating. Most of the power will be dissipated in the larger resistor, R1. Total series current, I = 200V / (R1 + R2) = 200 / 295K = 0.68 mA. Power dissipated in R1 = I^2 x R1 =(0.68mA)^2 x 220K = 0.09W. Since this is less than half of 0.25W, 1/4W resistors are fine.
Check: R2 x I = 75K x 0.68mA = 51V. Check.
Note: many radios and amps have a separate heater winding for the rectifier. Don't worry about biasing that heater.
: Doug, could you clarify two points:
: 1. When hooking up 10mfd elect., does positive goes to CT and neg. to ground?
: 2. Simple voltage divider? Assume using two resistors in series from B+ to ground at values that would give 50V (approximately) at point between the two resistors and that point goes to ground?
: I'm obviously not that experienced in electronics so you have to speak slowly. PL
:Hi, PL.
:
:Yes, the cap's polarity will be with negative to ground, positive to the heater ckt.
:
:For the voltage divider, you will need two resistors, say R1 and R2, in series. Yes, you want the pick-off point between the two resistors to be about 50V to ground. It doesn't have to be precise, and you can find standard sized resistors that will give a bias voltage that is close enough.
:
:R1 is wired from the B+ line to R2. The remaining lead of R2 goes to ground. The 50V is picked off at the connection between R1 and R2, and is connected to the heater winding center tap. (If, by chance, the heater winding doesn't have a CT, let me know.)
:
:Here's how to select the values of R1 and R2. The ratio [R2 / (R1 + R2)] = (50V / B+ voltage) (it doesn't have to be exact).
:
:In addition, you want (R1 + R2) large enough to keep the power dissipation low enough to get by with 1/4-W resistors -- say, about 300K - 500K ohms total.
:
:Example: my amplifier's B+ was 200V. I selected R1 = 220K and R2 = 75K. The bias voltage would then be (200V x 75K)/(220K + 75K) = 51V. (I would have been satisfied with anything between about 40V and 55V.)
:
:Now check the power dissipation, which needs to be less than half of the resistor rating. Most of the power will be dissipated in the larger resistor, R1. Total series current, I = 200V / (R1 + R2) = 200 / 295K = 0.68 mA. Power dissipated in R1 = I^2 x R1 =(0.68mA)^2 x 220K = 0.09W. Since this is less than half of 0.25W, 1/4W resistors are fine.
:
:Check: R2 x I = 75K x 0.68mA = 51V. Check.
:
:Note: many radios and amps have a separate heater winding for the rectifier. Don't worry about biasing that heater.
:
:
:: Doug, could you clarify two points:
:: 1. When hooking up 10mfd elect., does positive goes to CT and neg. to ground?
:: 2. Simple voltage divider? Assume using two resistors in series from B+ to ground at values that would give 50V (approximately) at point between the two resistors and that point goes to ground?
:: I'm obviously not that experienced in electronics so you have to speak slowly. PL
Regarding powering the filaments from DC, this does work well. I have a Dynaco amplifier that uses this method. It has absolutely no hum. However, to filter the DC, you need enormous condensers. This shouldn't be too hard with modern condensers. 2000 MFD condensers aren't that large anymore.
Thomas
Here is an easy way to get DC for a preamp filament. This works when the output tubes are 6L6 or similar and you want to power a 12AX7 or other tube with 150 ma filament.
Connect the cathodes through the pre amp tube filament. Current drawn by the 6L6's will operate the filament and supply DC. This also can eliminate the need for a cathode resistor.
Norm
:Marvelous! Now I know what to do to my Webster Chicago wire recorder. It hums too much, though the hum seems to come from the line cord.
:
:Regarding powering the filaments from DC, this does work well. I have a Dynaco amplifier that uses this method. It has absolutely no hum. However, to filter the DC, you need enormous condensers. This shouldn't be too hard with modern condensers. 2000 MFD condensers aren't that large anymore.
:
:Thomas
Thomas
Regarding powering the filaments from DC, this does work well. I have a Dynaco amplifier that uses this method. It has absolutely no hum. However, to filter the DC, you need enormous condensers. This shouldn't be too hard with modern condensers. 2000 MFD condensers aren't that large anymore.
Thomas
You will need the following: a power transformer (the existing filament winding won't have enough head room), a rectifier (preferably a diode bridge), a 3-terminal linear voltage regulator, a heat sink for the regulator, a resistor to set the regulator's voltage, a couple of filter caps (2200 uF would be good), and a power resistor or choke for the pi filter. If the DC supply isn't reasonably well filtered, you're likely to go from a 60-Hz hum to a 120-Hz hum, I fear.
If you try to use an unregulated supply, an one tube burns out, the voltage will increase and blow out all your tubes.
Realistically, I think DC filaments are best implemented from scratch in an amplifier design.
Biasing the heater ckt by 50V takes two resistors and one non-descript e-cap. "Mid-air" wiring can be used within in your existing chassis. My implementatation took about an hour, start to finish, and it works like a champ.
:Wouldn't rectifying the filiment line to DC work as well? All you need is a solid state rectifier and a cap.
:
:
:
::Hi, PL.
::
::Yes, the cap's polarity will be with negative to ground, positive to the heater ckt.
::
::For the voltage divider, you will need two resistors, say R1 and R2, in series. Yes, you want the pick-off point between the two resistors to be about 50V to ground. It doesn't have to be precise, and you can find standard sized resistors that will give a bias voltage that is close enough.
::
::R1 is wired from the B+ line to R2. The remaining lead of R2 goes to ground. The 50V is picked off at the connection between R1 and R2, and is connected to the heater winding center tap. (If, by chance, the heater winding doesn't have a CT, let me know.)
::
::Here's how to select the values of R1 and R2. The ratio [R2 / (R1 + R2)] = (50V / B+ voltage) (it doesn't have to be exact).
::
::In addition, you want (R1 + R2) large enough to keep the power dissipation low enough to get by with 1/4-W resistors -- say, about 300K - 500K ohms total.
::
::Example: my amplifier's B+ was 200V. I selected R1 = 220K and R2 = 75K. The bias voltage would then be (200V x 75K)/(220K + 75K) = 51V. (I would have been satisfied with anything between about 40V and 55V.)
::
::Now check the power dissipation, which needs to be less than half of the resistor rating. Most of the power will be dissipated in the larger resistor, R1. Total series current, I = 200V / (R1 + R2) = 200 / 295K = 0.68 mA. Power dissipated in R1 = I^2 x R1 =(0.68mA)^2 x 220K = 0.09W. Since this is less than half of 0.25W, 1/4W resistors are fine.
::
::Check: R2 x I = 75K x 0.68mA = 51V. Check.
::
::Note: many radios and amps have a separate heater winding for the rectifier. Don't worry about biasing that heater.
::
::
::: Doug, could you clarify two points:
::: 1. When hooking up 10mfd elect., does positive goes to CT and neg. to ground?
::: 2. Simple voltage divider? Assume using two resistors in series from B+ to ground at values that would give 50V (approximately) at point between the two resistors and that point goes to ground?
::: I'm obviously not that experienced in electronics so you have to speak slowly. PL
The DC from the rectified supply if you use a large enough cap will read approx 1.4 X the ac rating.
6.3 X 1.4 = 8.82 volts max. It can only go down with full loading, however using a dropping resistor before
the diode will get you back to 6.3. Do this by adding up the current used by all the filiments (I).
The diode will drop the voltage by aroung 0.6V.
Now we're down to 8.22. we need the resistor to drop
8.22-6.3= 1.92V.
Using ohm's law: 1.92/I = R, the size of the resistor is I^2 * R = watts.
On the other hand, you mentioned a regulated supply. Brilliant! This is even simpler! A rectifier into an IC
three pin regulator of approprite size! Maybe a couple in parallel for those heavy loads. No pi network required! You can get them in 6 volts too! If you want to boost it a coducting diode in the ground path will bump it up to 6.6V. The 6 volts should suffice though.
Now for the ripple issue. Any way you cut it, the amplitude of the DC ripple will be miniscule compared to full blown AC! In fact with those three pin regulators in line, the DC will be close to pure!
The LM7806 is good for 1.5 A DC (I thinK), so if you had 5 tubes that draw .3a each that makes 1.5A. You might want to find a larger regulator or parallell two of them. Also you need to attach them to the chassis for heat sinking. If anyone would like to try this, I might be able to find the part#'s of beefier types to do the job larger demands.
:JIMM: Yes, many low-level-preamps do use DC for the filament supply, as Thomas says. But, to switch an existing amp with AC filaments to DC is going to take quite a few additional components, probably mounted on a separate chassis or printed circuit board.
:
:You will need the following: a power transformer (the existing filament winding won't have enough head room), a rectifier (preferably a diode bridge), a 3-terminal linear voltage regulator, a heat sink for the regulator, a resistor to set the regulator's voltage, a couple of filter caps (2200 uF would be good), and a power resistor or choke for the pi filter. If the DC supply isn't reasonably well filtered, you're likely to go from a 60-Hz hum to a 120-Hz hum, I fear.
:
:If you try to use an unregulated supply, an one tube burns out, the voltage will increase and blow out all your tubes.
:
:Realistically, I think DC filaments are best implemented from scratch in an amplifier design.
:
:Biasing the heater ckt by 50V takes two resistors and one non-descript e-cap. "Mid-air" wiring can be used within in your existing chassis. My implementatation took about an hour, start to finish, and it works like a champ.
:
::Wouldn't rectifying the filiment line to DC work as well? All you need is a solid state rectifier and a cap.
::
::
::
:::Hi, PL.
:::
:::Yes, the cap's polarity will be with negative to ground, positive to the heater ckt.
:::
:::For the voltage divider, you will need two resistors, say R1 and R2, in series. Yes, you want the pick-off point between the two resistors to be about 50V to ground. It doesn't have to be precise, and you can find standard sized resistors that will give a bias voltage that is close enough.
:::
:::R1 is wired from the B+ line to R2. The remaining lead of R2 goes to ground. The 50V is picked off at the connection between R1 and R2, and is connected to the heater winding center tap. (If, by chance, the heater winding doesn't have a CT, let me know.)
:::
:::Here's how to select the values of R1 and R2. The ratio [R2 / (R1 + R2)] = (50V / B+ voltage) (it doesn't have to be exact).
:::
:::In addition, you want (R1 + R2) large enough to keep the power dissipation low enough to get by with 1/4-W resistors -- say, about 300K - 500K ohms total.
:::
:::Example: my amplifier's B+ was 200V. I selected R1 = 220K and R2 = 75K. The bias voltage would then be (200V x 75K)/(220K + 75K) = 51V. (I would have been satisfied with anything between about 40V and 55V.)
:::
:::Now check the power dissipation, which needs to be less than half of the resistor rating. Most of the power will be dissipated in the larger resistor, R1. Total series current, I = 200V / (R1 + R2) = 200 / 295K = 0.68 mA. Power dissipated in R1 = I^2 x R1 =(0.68mA)^2 x 220K = 0.09W. Since this is less than half of 0.25W, 1/4W resistors are fine.
:::
:::Check: R2 x I = 75K x 0.68mA = 51V. Check.
:::
:::Note: many radios and amps have a separate heater winding for the rectifier. Don't worry about biasing that heater.
:::
:::
:::: Doug, could you clarify two points:
:::: 1. When hooking up 10mfd elect., does positive goes to CT and neg. to ground?
:::: 2. Simple voltage divider? Assume using two resistors in series from B+ to ground at values that would give 50V (approximately) at point between the two resistors and that point goes to ground?
:::: I'm obviously not that experienced in electronics so you have to speak slowly. PL
Placing the filament supply 50 volts above the chassis sounds like the easier way to go, though. Since this supply really isn't loaded, I suggest using the highest resistances possible that will give you 50 volts while in operation (there may be slight current flow while in operation, which would make very high resistances, say in the meg-ohms, impractical). High resistances will load the power supply minimally, and will be safer in the event of a heater to cathode short. All in all, it sounds like a pretty nice idea.
Thomas
:Well, I'd like to say that my Dynaco pre-amp simply uses a selenium rectifier and a filter condenser. It uses 12AT7s. Each channel is in push-pull. The filaments are wired in series. Also, each tube in one channel is wired in series with a tube in another channel. Strange, but it works. Perhaps this is to make a higher voltage/lower current set-up, which requires less filtering than a lower voltage/higher current set-up. I don't know why, but the remaining tubes don't get much brighter when one is removed. I think that with a larger set-up (output tubes, for instance, though these don't really need DC, since they're at the end of the amplification chain), there might be more of a voltage difference if one tube was to burn out. Also, I believe that the selenium offers good filtering that a silicon might not. This is simply because a selenium has higher internal resistance. Selenium rectification is quieter. This might also help buffer the AC spikes that might otherwise normally raise the DC voltage. Also, when the electrolytics are loaded by the tubes, I don't think that the excess voltage they accumulate will remain. This may be why the tubes don't glow too brightly. I haven't measured AC voltages, though, so I don't really know what they are and what the DC voltages are when tubes are removed. Perhaps I should explore. I do know, though, that it is possible to make a DC set-up with just diodes and condensers, though linear regulators might help.
:
:Placing the filament supply 50 volts above the chassis sounds like the easier way to go, though. Since this supply really isn't loaded, I suggest using the highest resistances possible that will give you 50 volts while in operation (there may be slight current flow while in operation, which would make very high resistances, say in the meg-ohms, impractical). High resistances will load the power supply minimally, and will be safer in the event of a heater to cathode short. All in all, it sounds like a pretty nice idea.
:
:Thomas