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How does anode voltage swing to twice HT voltage?
How does anode voltage swing to twice HT voltage?
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Old 19th January 2011, 05:57 AM   #1
cricha5 is offline cricha5  United States
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Default How does anode voltage swing to twice HT voltage?

I'm trying to design my first SE tube amp. I understand just about everything except for part of the power stage. The power stage is supplied an anode voltage of say 300V, but I keep reading that the voltage it is expected to rise to twice that, 600V, and people draw their load lines as if the voltage will swing that far even though only 300V is supplied. I thought I understood how transformer coupling worked, but this is really throwing me. As soon as I get these questions answered I can finish my design and start ordering parts. Any help would be greatly appreciated.

1) What physically is going on here?

2) Why is the induced voltage exactly twice that of the supply voltage?

3) The anode voltage/current should be controlled by the grid. Won't changing the anode voltage from the transformer really mess up the signal?
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Old 19th January 2011, 06:10 AM   #2
aardvarkash10 is offline aardvarkash10  New Zealand
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the opt is an inductor - as the magnetic field associated with the primary collapses (causing the signal to transfer to the secondary) it also induces a voltage in the primary and this rises... Depends on the inductance of the opt etc etc but reasonably predictable.
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Old 19th January 2011, 07:17 AM   #3
singa is offline singa  Singapore
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Hi, I think you are confusing yourself.I suggest you download Radiotron
Engineer's handbook edition 3 and 4 from Pete Millett's website. All will be
explained.
Singa.
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Old 19th January 2011, 12:11 PM   #4
DF96 is offline DF96  England
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An inductance is not like a resistor. In a resistor the current is always in the same direction as the voltage, and proportional to it.

For an inductance the current is proportional to the integral of the voltage (for all time!). Looking at it the other way, the voltage across an inductance is proportional to the rate of change of current. This means that the current and voltage can, for a while, be in opposite directions.
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Old 19th January 2011, 03:16 PM   #5
cricha5 is offline cricha5  United States
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Singa, I have no doubt I am confusing myself. I'll check out that link, thanks.

I understand how an inductor works and I understand that a transformer is just two inductors side by side. I also understand that with a high enough change in current over a small enough time, a very large voltage can develop. What I'm not understanding is this.

1) Why have I read that the voltage swings to exactly twice the HT voltage? Why doesn't it go higher or lower? Are the transformers designed this way? I don't see how you can design a transformer this way for all tubes.

2) Doesn't this kickback voltage affect the operation of the tube? If you apply a grid voltage to a tube with a known anode voltage, you expect a certain known current to flow. If that anode voltage is changed by the transformer, how do you know how much current is flowing for a given grid voltage?
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Old 19th January 2011, 03:29 PM   #6
Wavebourn is offline Wavebourn  United States
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How does anode voltage swing to twice HT voltage?
Like capacitor saves voltage, coil saves current. When you short capacitor it tries to keep voltage, generating huge current. Similarly, when you open coil it tries to save current generating huge voltage.
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Old 19th January 2011, 03:59 PM   #7
Sheldon is offline Sheldon  United States
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Quote:
Originally Posted by cricha5 View Post
1) Why have I read that the voltage swings to exactly twice the HT voltage? Why doesn't it go higher or lower? Are the transformers designed this way? I don't see how you can design a transformer this way for all tubes.

2) Doesn't this kickback voltage affect the operation of the tube? If you apply a grid voltage to a tube with a known anode voltage, you expect a certain known current to flow. If that anode voltage is changed by the transformer, how do you know how much current is flowing for a given grid voltage?
Actually, it doesn't swing to exactly twice the HT voltage. It's usually somewhat less, which results in harmonic distortion.

In principle, it could swing higher than twice the voltage - much higher. But that assumes a very sharp rise to very high resistance. This does not happen under normal operating conditions for normal audio circuits. The tube never has infinite resistance - look at the load line. If you get much peaking, it's a bad design. A little peaking, you'd use a zobel to shunt it.

Sheldon
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Old 19th January 2011, 04:27 PM   #8
DF96 is offline DF96  England
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Assuming a resistive load attached to a perfect transformer, and an ideal valve which can pull the anode voltage down to zero volts on negative peaks and has no harmonic distortion, then the positive peak will be twice the supply voltage. In real life things are more complicated, but twice the supply is a useful estimate when assessing things.
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Old 19th January 2011, 04:30 PM   #9
Michael Koster is offline Michael Koster  United States
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There's a little confusion here. OPT inductance is not part of the formula for calculating voltage swing.

Using Ohm's law one multiplies the change in tube current from idle by the reflected load resistance. This works both ways, positive and negative-swinging signal.

The reason Sheldon that the voltage swing is unequal is due to the tube current swing being unequal (+) vs. (-), not anything about the OPT.

Tube current swings above and below the idle current as signal is applied, and the anode voltage is a function of this current and the reflected load, just the same as with a push-pull or even a parafeed stage. It's all the same, based on the load line.

The REASON the anode voltage can swing above B+ is because energy is stored in the OPT core gap due to the idle current.

Cheers,

Michael
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Old 19th January 2011, 04:41 PM   #10
Sheldon is offline Sheldon  United States
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Quote:
Originally Posted by Michael Koster View Post
The reason Sheldon that the voltage swing is unequal is due to the tube current swing being unequal (+) vs. (-), not anything about the OPT.
Yes, didn't mean to confuse the issue with a partial explanation. Just pointing out that the swing is not exactly double the B+. Can be less (or more - as with a square wave). Should have been more specific, in citing the single ended case for my particular example, where the swing is typically less on the cut off side.
Sheldon
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