Bottlehead Forum
General Category => Technical topics => Topic started by: dbishopbliss on January 03, 2011, 10:31:17 AM
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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.
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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.
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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
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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.
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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
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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.
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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?
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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?
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I don't know about raising the voltage but the math on power is right. I got 0.18W.
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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.
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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.
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David,
I believe your calculations and it looks like you will end up with ~69V on the heaters. Just what you want.
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Yes you have it. Actually 330K, 100K is usually what end up with, if drawing extra current from the supply is an issue...John
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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?
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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!
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I calculated the Vg = -2.7 by interpolating where the operating point landed between the -2 and -3 grid lines on the tube datasheet.
If the cathode to heater voltage is basically the positive value of the Vg, then it seems hard to imagine that you could exceed the Vkh rating - at least with this tube where the value is 100V. Would this value be higher in a cathode follower where the load resistor is on the cathode? If the plate voltage is 300V and the load is 30K + 540R, then the voltage from the cathode to ground would be around 150V. So in, this case, I would need to raise the potential, but if this was simple grounded cathode circuit I don't have to bother.
Do I have that right?
P.S. Thanks to everyone that has been contributing to my questions lately. I hope that there are others that find this stuff useful.
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Well, imagine a circuit where that grid is direct-coupled from the plate of a previous amp; the grid then may be at 100+ Vdc, let's say. Yet, you still might also want it to be 2-3 Vdc negative with respect to its own cathode. Which would mean that it's own cathode, using a cathode resistor of a much higher value, could be brought up to, say, 103 Vdc positive with respect to its grid. And, also, 103 Vdc positive with respect to its heater, if the heater were grounded. In that case, biasing the heater more positive with respect to the circuit ground would be indicated to keep the heater-to-cathode potential within the limit.
Now, in most Voltage Amp circuits, the grid is held at ground potential through the grid resistor. There will be no Voltage drop across the grid resistor because there will be no current flowing through it. So, with respect to ground, the grid is also at ground potential. If you measure between ground and the grid pin on the tube, your result will be "0". By making the cathode positive with respect to ground, with the grid at ground, you create the situation where the potential from grid to cathode is negative. If you use your meter, black lead on the cathode, red lead on the grid, it will read negative Voltage. Remember, the meter only knows the world between its probes. So, if "this" is more positive than "that", then "that" is also more negative than "this", depending on how you are measuring.
The other way to make the grid negative with respect to the cathode, assuming the cathode is grounded, would be to have a separate negative bias Voltage with respect to ground, and apply it to the grid. This is often done in power amps, since power tubes have more current going through them and also usually require the grid to be more negative with respect to the cathode in order to hit a useful operating point. The cathode resistor can start to get unwieldy in size and power consumption to use cathode-bias.
The Voltage on the plate or anode really doesn't figure into this discussion directly, other than it's another operating point. What you will see in your circuit, though, is that the Voltage between plate and ground will be higher than the Voltage between plate and cathode, by the amount of the positive Voltage on the cathode. IOW, if there is +150 Vdc on the plate with respect to ground, and +2.7 Vdc on the cathode with respect to ground, the Voltage from cathode to plate (black on cathode pin, red on plate pin) will be +147.3 Vdc. That is the amount of Voltage said to be being dropped across the tube, cathode to anode, at this particular point.
Note, though, how the curves approximated the drop across the cathode resistor so closely.
The other take-away is to always keep in mind that potential Voltage values are always relative to what you are considering to be the reference point. A Voltage that is +x Volts with respect to ground, may also be -y Volts with respect to some other reference point in the circuit.
Keep asking questions, if I have been unclear about any of this. I've been up a bit longer than I'm used to, and I hope I'm making sense, but I'm not sure I'm a good judge of that!
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We are actually talking about two different things. In your case, where your cathodes are just a few volt above ground. The heater/cathode difference is not the issue and you can ground the heater with no problems. The reason to bias it up 20-60V or whatever is it usually makes your circuit quieter. But not always.
Now in the case when say you have one tube on top of another like a SSRP or Aikido, or you direct couple two stages. Now you have a situation were the cathode of the top tube section is sitting on the plate of the other and at it's voltage. Say that cathode is at 150V and the bottom one is at your 2.3V. If you grounded your heater that 150V H/C is probably too much but if you where to raise the heater to say 80V then both sections will be within 100V H/C, which is usually within the limits of most tubes...John