Center Tapped Filaments Question

[dbishopbliss]

I have a transformer with the primary/secondaries as shown in the attachment.  I didn't realize when I bought it that the filament secondaries were center tapped. 

Assuming I am running AC filaments:

Would I use the wires labeled GRN as I would non-center tapped filament wires? 
What do I do with the GRN/YEL wire?

Thanks for your help. 

[Grainger49]

The FP series of preamps uses a DC bias on the AC heaters.  Well, the FP III uses DC with a DC offset to keep the heater to cathode voltage low, but the two previous had AC with the same DC on the heaters for the same reason.  If you put a DC bias on the heaters then no wire is grounded.  If not, then you can ground one leg of the AC heater winding.  In the ST-70 it is the CT of the heaters is grounded through a 0.02uF cap.  In my last Dynaco PAS 3X rebuild I let the DC heaters float, it hummed and I grounded one of the legs.

Was I wishy-washy enough?  I don't know that there is a definitive answer.

[2wo]

You can tape off the CT if you want. If you plan to raise (bias) the heater, this is a the best spot to add it...John

[dbishopbliss]

More questions about this...

From what I've read it sounds like I should be grounding the center-tap of my filaments to reduce hum.  Some say the wire should just be connected to ground while others recommend using a small capacitor. 

I've also come across DC Elevation with regard to heaters.  I've also read about connecting the center tap to the cathode of one of the tubes, or creating a potential divider from the HT (is this the same as B+).

I really don't understand how this works.  I've tried looking in the various tube manuals but I haven't found much on the topic (perhaps I'm looking for the wrong terms).  Can someone explain or point me to a good explanation.

[2wo]

Often grounding the CT is enough to reduce hum. Sometimes raising the heater potential 40-60V or so works better. If you place a cap from CT to ground with AC powered heaters, it will float to some dc value and raise the heater potential with it.

Most ofter I use a combination power supply bleeder/heater raise-er-upper. Say I want to bleed roughly 10ma with a 300V supply. V/I= 30K. So I might use a 5K and 25K as a voltage divider across my B+ with the 25K to B+ and 5K to ground and tie my CT to the 50V or so between them. Check my decimal places...John            

[dbishopbliss]

I never thought about this before... does a bleeder resistor contribute to the current draw on the power supply when the amplifier is operating? 

In the example John gave where there is a 30K resistor across a 300V supply, do I need to add an additional 10mA to the total current draw for the circuit?  That is the same current that my tubes are drawing... might have to get a bigger transformer.  Perhaps that is why people connect the center tap to the cathode of a tube.

[Grainger49]

I never thought about this before... does a bleeder resistor contribute to the current draw on the power supply when the amplifier is operating?   .  .  .  .   

Yes, but the value is chosen so it doesn't drain much current from the B+, some but not much.  They discharge the capacitors according to the 1/2*(Pi)*R*C time constant.  Where C is the total capacitance and R is usually the bleeder.  It kills the supply on a walk from my listening room to my work room.  (I really don't want to be shocked, I am dependent on a pacemaker)

Bottlehead typically uses 249k ohms.  The 30k ohm would be too low a value in my estimation because of the current it would take away from the supply.  How about 1/2 Meg?

[dbishopbliss]

How about 1/2 Meg?

That seems more like it.  If I were to put 500K across a 300V power supply, the current would be 0.6mA.

To use John's strategy of using bleeder/heater raiser, I would try standard value resistors of 100K and 330K.  This would give me around 0.69mA.  Therefore, my voltage would be raised around 69V (100000 * .000069).  The resistors would need to be 1/2W for safety (0.2 ~= .69mA * 300).

Is my math correct?

[Grainger49]

I don't know about raising the voltage but the math on power is right.  I got 0.18W.

[dbishopbliss]

I probably confused things by including the 500K comment above.  Here's what I was thinking... 


My B+ is 300V.  My Va is 150V.  The Vkh rating of my tube is 100V.  Therefore, I need to raise the potential of the heaters at least 50 volts.

I want to add less that 1mA additional load on my transformer so I should choose bleeder resistance value greater than 300K:

I = 300V / 300,000R
I = 0.001A
I = 1mA

Now its time to start playing with standard value resistors to determine the current drawn.

R = 330K + 100K
R = 430K

I = 300 / 430,000
I ~= 0.00069A
I ~= 0.69mA

Now that I know the current, I can calculate the voltage drop across each resistor:

V = 0.69mA * 330K
V ~= 228

V = 0.69mA * 100K
V ~= 69

As John mentioned before, I would connect the 330K to B+ and 100K to ground, then connect the CT of my heaters to both resistors.  This will not only give me the potential rise I want, but bleed the voltage from my power supply when the amp is turned off. 

Pretty cool.  I hope I have it right.

[JC]

I probably missed it, but what Voltage are you expecting/shooting for on the cathode of the tube?

"Cathode Voltage with respect to circuit ground", I should have said.

[Grainger49]

David,

I believe your calculations and it looks like you will end up with ~69V on the heaters.  Just what you want.

[2wo]

Yes you have it. Actually 330K, 100K is usually what  end up with, if drawing extra current from the supply is an issue...John   

[dbishopbliss]

I probably missed it, but what Voltage are you expecting/shooting for on the cathode of the tube?

"Cathode Voltage with respect to circuit ground", I should have said.

Just when I think I'm getting this stuff...

How do I figure that out.  Here is what I know:

Va = 150V
Ia = 5mA
Load = 30K
Rk = 540R
Vg = -2.7V

How do I calculate the Cathode Voltage with respect to ground?

[JC]

OK, assuming all Voltages are relative to circuit ground other than the grid Voltage which is usually referenced to the cathode, then you kind of already know the answer.  That negative 2.7 Vdc on the grid is actually the same as a positive Vdc on the cathode with respect to ground, providing that that -2.7 Vdc is with respect to the cathode instead of ground.

You can double-check this with Ohm's Law again.  The Voltage drop across 540 Ohms with .005 Amps running through it.

If I have all this right, then, your Heater-to-Cathode potential DC is 2.7 Vdc if both the heater circuit and the cathode are referenced to circuit ground.  IOW, well within the manufacturer's maximum spec.  And, it would still be within the spec you indicated if you add ~70 Vdc "bias" to the heater supply with respect to ground.

I know I often slip up by not always indicating what a stated Voltage is with respect to.  Unless I miss my guess, this is a great example of why it is important to do so!