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?
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?
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.
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.
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?
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?
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
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.
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
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
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
I like to reduce the system to the minimum components that show an effect
for "swing above the rail" all you need is the tube and plate load Inductor - so explainations that de-emphasize the Inductor seem wrong headed to me
energy is stored in the Inductance - whether gap or core is a irrelevent detail for the circuit analysis
a Inductor has Zero DC V drop (ignoring parasitic winding R) - with a AC signal across an Inductor the average over a full cycle (or long term) has to be 0 V - but the stored energy can support "above the rail" AC 1/2 cycle output
the Inductance is required in the circuit analysis along with the plate R to determine the low frequency roll-off
I would claim you need to understand the above before introducing the details of calculating xmfr pri Inductance, avoiding saturation, coupling properties and load circuit impedance transformation
for "swing above the rail" all you need is the tube and plate load Inductor - so explainations that de-emphasize the Inductor seem wrong headed to me
energy is stored in the Inductance - whether gap or core is a irrelevent detail for the circuit analysis
a Inductor has Zero DC V drop (ignoring parasitic winding R) - with a AC signal across an Inductor the average over a full cycle (or long term) has to be 0 V - but the stored energy can support "above the rail" AC 1/2 cycle output
the Inductance is required in the circuit analysis along with the plate R to determine the low frequency roll-off
I would claim you need to understand the above before introducing the details of calculating xmfr pri Inductance, avoiding saturation, coupling properties and load circuit impedance transformation
It's a convenient guesstimate for designing SE loadlines.
All that's required for OPT design is to match the actual impedance of the load to that of the loadline, which will be larger since VTs are high voltage, low current (Hi-Z) devices.
That's taken care of by the loadline, and impedance matching. You can have much higher voltages in designs like horizontal deflection circuits for TVs. In that particular case, the load is nearly pure inductance (since you need a magnetic field to deflect the electrons across the CRT screen). The return stroke does induce voltages way in excess of the normal Vpp. The spec sheets for TV HD types will specify how much extra voltage, and for how long, the VT can withstand that voltage.
It's also the source of "B boost" voltages used for other subsystems (usually vertical deflection, horizontal oscillator, and perhaps the audio final) that comes from the damper diode connected across the horizontal deflection coils. That's why the HD circuit needs to be so much more robust than the vertical deflection circuit: much more is demanded of it, including providing the HT needed by the CRT itself.
(For vertical deflection, all you really need is the usual damping you'd see in an audio amp: maybe some resistors paralleling the vertical deflection coils and the low r(p) of low-u triode finals or pents used as pseudotriodes. If the vertical coils ring a bit, who cares? You'll never see it happen since the electron beam will be off-screen anyway.)
If the load is nearly pure resistance (speeks, antennae, T-line, next RF stage, etc) then the OPT voltage is limited to a theoretical 2Vpp. (In RF practice, you can also see some hellacious overvolts if you mistune and thereby cause an excessive SWR -- that can, and has, poofed RF finals.) In practice, you never see that, nor do you want to for Class A operation, since you definitely don't want a SE stage going all the way to plate current cutoff, since that will mean clipping distortion (sounds quite nasty).
Ok, this is good stuff. What I'm hearing is that the 2X HT voltage thing is just a rule of thumb and is just what actually happens with a real life signal. I think I'm going to spend a little time with pencil and paper and see what actually goes on in the tube-OPT circuit with various input signals.
The frustrating thing is that I know that spending all this time trying to understand this doesn't actually gain me any practical knowledge. It seems that you pick the HT voltage and draw the load line assuming twice the HT voltage. Simple. It's just that I've built a couple of things (theremin) without fully understanding everything, and it has always come back to bite me.
The frustrating thing is that I know that spending all this time trying to understand this doesn't actually gain me any practical knowledge. It seems that you pick the HT voltage and draw the load line assuming twice the HT voltage. Simple. It's just that I've built a couple of things (theremin) without fully understanding everything, and it has always come back to bite me.
You don't have to undertsand at a scientist or engineer level.
1. Just understand the relationships of ohm's law. V/I X R.
With it you can calculate any one quantity when the other two is known.
2.Run the tubes at 80% or lower and you are in safe territory.Whether voltage
or current.
3. The current you wish to bias at the anode is equal to the bias at the cathode.This is the most important thing you must know for a standard circuit.
4.The impedance resistor to match output will be 3 to 8 times the internal
resistance of the tube (anode),output tranny included for best performance in terms of linearity,
distortion and power.
5. Adjust power supply voltage to suit your circuit.
These are the main things that should guide you.Then you begin to enjoy
the hobby.Of course safety first. Singa.
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No, you don't draw a load line assuming twice the HT voltage. You draw a load line through the quiescent point, with a slope set by the total load impedance. Under certain circumstances this may look identical to one drawn from twice the HT.
If you want to "match" the output you need a resistor equal to the anode impedance. In almost all circuits this is not what you want. For triodes you normally have a much higher resistor, to get low distortion. For pentodes you normally have a much lower resistor, to get enough power output or appropriate matching to the next stage. I'm afraid singa is oversimplifying things.
If you understand things like a scientist or engineer you will find that they come back to bite you far less often. Remember: the opposite of knowledge is ignorance. So learn as much as you can.
If you want to "match" the output you need a resistor equal to the anode impedance. In almost all circuits this is not what you want. For triodes you normally have a much higher resistor, to get low distortion. For pentodes you normally have a much lower resistor, to get enough power output or appropriate matching to the next stage. I'm afraid singa is oversimplifying things.
If you understand things like a scientist or engineer you will find that they come back to bite you far less often. Remember: the opposite of knowledge is ignorance. So learn as much as you can.
I was struggling with the same question. I stumbled upon this on Valve Wizard. http://www.freewebs.com/valvewizard/index.html
"If you are wondering why it appears that the signal voltage can now swing higher than the HT voltage, it is because this is exactly what happens! Inductances abhor changes in current. When current through the transformer increases it stores energy, which is released when the current falls again, allowing up to twice the HT voltage to be developed. Because of this, the HT in a Class A amp must never be more than half the maximum peak anode voltage rating of the valve, given on the data sheet. For the EL34 this is 2000V so we are well within safe limits!
Now we have the load line we need to set the screen voltage before we can continue."
The complete article is under Power-amp / Output Stage....Single Ended on the left hand side of the page.
The other thing is the current that is below your idle current and the resulting HT that are defined by the load line are still controlled by the grid.
"If you are wondering why it appears that the signal voltage can now swing higher than the HT voltage, it is because this is exactly what happens! Inductances abhor changes in current. When current through the transformer increases it stores energy, which is released when the current falls again, allowing up to twice the HT voltage to be developed. Because of this, the HT in a Class A amp must never be more than half the maximum peak anode voltage rating of the valve, given on the data sheet. For the EL34 this is 2000V so we are well within safe limits!
Now we have the load line we need to set the screen voltage before we can continue."
The complete article is under Power-amp / Output Stage....Single Ended on the left hand side of the page.
The other thing is the current that is below your idle current and the resulting HT that are defined by the load line are still controlled by the grid.
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