Time for my stupid question of the day. I've been studying up on push-pull tube amp designs. I get at a high level how they work with their OTs in a push-pull. This may be based on my simple understanding of push-pull transistor amplifiers from my learning 20+ years ago.
With transistors, the push-pull design (besides the biased overlap, so AB) was that one transistor would not conduct while the other one was. My understanding was that one transistor turned off simply because the base and emitter potentials were reversed and this would cut off the current, up to a point.
With valves however, treating them in the same way would/could put a 0V to positive voltage on the grid, which would cause grid current and do all sorts of bad things to the cathode, so they don't have that same reverse diode effect.
At a high level, how do transistors and valves differ in push-pull? I could see how valves are always biased negative and only dealing with an inverted signal, so essentially it is two "lighter" class A setups in opposition to each other, delivering double the power to the OT while enjoying relative linearity on both ends. It's nothing like transistors where the grid goes the wrong way and the valve simply "shuts off", right?
Looking at some schematics at a glance, I don't see a mechanism to keep the grid from going in a way that would be destructive to the valve. The two books suggested to me here don't seem to really describe the actual mechanics in detail. Valve Amplifiers by Morgan Jones illustrates that the valve will shut off when it's on the "wrong side" of the wave, but how does that not damage the valve? All it focuses on is how to balance the crossover on both sides.
With transistors, the push-pull design (besides the biased overlap, so AB) was that one transistor would not conduct while the other one was. My understanding was that one transistor turned off simply because the base and emitter potentials were reversed and this would cut off the current, up to a point.
With valves however, treating them in the same way would/could put a 0V to positive voltage on the grid, which would cause grid current and do all sorts of bad things to the cathode, so they don't have that same reverse diode effect.
At a high level, how do transistors and valves differ in push-pull? I could see how valves are always biased negative and only dealing with an inverted signal, so essentially it is two "lighter" class A setups in opposition to each other, delivering double the power to the OT while enjoying relative linearity on both ends. It's nothing like transistors where the grid goes the wrong way and the valve simply "shuts off", right?
Looking at some schematics at a glance, I don't see a mechanism to keep the grid from going in a way that would be destructive to the valve. The two books suggested to me here don't seem to really describe the actual mechanics in detail. Valve Amplifiers by Morgan Jones illustrates that the valve will shut off when it's on the "wrong side" of the wave, but how does that not damage the valve? All it focuses on is how to balance the crossover on both sides.
Maybe this will make things clearer: Fundamentals Of Radio Valve Technique : Deketh, J : Free Download, Borrow, and Streaming : Internet Archive
You are quite the wealth of information, sir. Thank you for all you have taught me so far. If I'm reading this correctly, the reverse voltage of the transformer swinging the other way on the plate basically neutralizes the "inactive" valve. Is that basically accurate?
The don't differ all that much. Transistor push pull stages typically do not use a phase inverter, unlike tubes.
If you really want to compare the behaviour of the actual devices you should choose similar topologies. Some early transistor circuits are very similar to a typical tube PP.
They are similar indeed. At low audio levels both works in Class A, then one side cuts. In a tube push-pull there can be positive grid voltage drive (with grid current), depending on the negative bias and the desired output power.
The basic difference is that a transistor output stage works as an emitter follower, while a tube output stage works as a grounded cathode circuit, having considerably higher output impedance. For this reason (and for galvanic separation) a step-down transformer is needed.
The basic difference is that a transistor output stage works as an emitter follower, while a tube output stage works as a grounded cathode circuit, having considerably higher output impedance. For this reason (and for galvanic separation) a step-down transformer is needed.
Not necessarily. And this is the reason for my advice towards the OP: compare similar topologies before making sweeping conclusions.
The vast majority of tube PP amps use the common cathode transformer coupled approach here because it best suits common tubes. Trying to make something that resembles a typical solid state amp requires exotic tubes, runs them outside h-k voltage ratings, or something equally foolish. The vast majority of solid state amps use complementary emitter followers and no transformer, because it is the most practical straightforward way of using common devices. Other topologies certainly exist in solid state amps. Complementary common emitter outputs are more common than you might think - it is a very common Pro amplifier topology. But it still doesn’t look anything like a tube amp. Common emitter transformer coupled solid state amps certainly did exist - they just aren’t used much anymore because of the cost of the output transformer. And even then, most of them used a step UP output transformer rather than step down due to differences in the operating voltages. You could certainly *build* a Common source trafo-coupled high voltage mosfet amp, but you will find no commercial implementations, and you WILL run into problems to solve. Transformer *driven* common emitter (Totem pole) outputs were once common too, especially in the days of germanium, but I’ve have also been abandoned. All of these “transformer” approaches all but disappeared when complementary silicon devices became widely available and much cheaper than iron.
I think I know where your doubt/problem lies.
I won´t "send you to a book and tell you to deal with it yourself" but straight answer.
A Tube with 0V at grid will pass maximum current possible, so normally we NEED negative voltage there to keep it under control.
We need strong negative voltage so current at idle is small.
A transistor (let´s talk NPN; PNP is the same with polarities reversed) with 0V at base passes zero current, period.
Negative voltage will do nothing, period.
"0V and reverse V are the same" so put reverse voltage out of the picture.
A transistor will need *at least* some positive 0.7V just to be at idle and then higher voltage to pass significant current although here the significant parameter is base current.
It will need (positive) base current to pass collector current, but as far as voltages are concerned, they are always positive.
In fact at high current , base voltage will raise up to, say, 3V !
And again: tubes and transistors are different.
Justb don´t mix them in the same thought block.
Tube amps are overdriven all the time so grids get slightly positive on signal peaks , nothing bad happens because phase inverters typically can provide upto 2 mA or so, while tubes are happy with tens (not tenths) of mA, go figure.
And that if amp is not overdriven , it can easily reach -150V or more.
With no damage whatsoever, the tube is built to stand that and more.
I won´t "send you to a book and tell you to deal with it yourself" but straight answer.
To simplify explanation, and focus , let´s forget Class AB for a moment , after all the Class A fraction is minimal compared to the Class B one.
No, you are "thinking transistors as tubes", they are very different.
A Tube with 0V at grid will pass maximum current possible, so normally we NEED negative voltage there to keep it under control.
We need strong negative voltage so current at idle is small.
A transistor (let´s talk NPN; PNP is the same with polarities reversed) with 0V at base passes zero current, period.
Negative voltage will do nothing, period.
"0V and reverse V are the same" so put reverse voltage out of the picture.
A transistor will need *at least* some positive 0.7V just to be at idle and then higher voltage to pass significant current although here the significant parameter is base current.
It will need (positive) base current to pass collector current, but as far as voltages are concerned, they are always positive.
In fact at high current , base voltage will raise up to, say, 3V !
Forget this analysis which is wrong.
And again: tubes and transistors are different.
Forget all this, all wrong, tubes and transistors work as I explained above, much simpler.
Justb don´t mix them in the same thought block.
Forget this, see above.
Unless you pass a ton of current which the phase inverter can not supply anyway, you will NOT damage the tube, so forget that idea.
Tube amps are overdriven all the time so grids get slightly positive on signal peaks , nothing bad happens because phase inverters typically can provide upto 2 mA or so, while tubes are happy with tens (not tenths) of mA, go figure.
Most of the time tubes are negative biased, signal drive moves them around that starting point and yes, say a grid is biased at idle -52V (6L6 in a Fender amplifier, but similar mechanism in Hi Fi amps), drive signal will sweep that voltage up to 0V at which point it will saturate passing maximum current possible, and then on the negative swing it will reach same peak value but the other way, so it will reach -104V
And that if amp is not overdriven
, it can easily reach -150V or more.With no damage whatsoever, the tube is built to stand that and more.
with the added detail that tube sound quality justifies the effort and cost.
A small germanium transistor amp?
Not so much.
Why would it damage the valve?? Any more than I damage a light bulb by giving it insufficient voltage?
Take a grid well negative, the plate current goes way low, the tube hardly know it has power on it.
OTOH, take the grid positive, you get more current, more power. And yes, burnt grid. So you build a sturdier grid (at higher price). Not worth it for wall-power amps under 40 Watts, but over 1,000 Watts it becomes worthwhile to apply 100++ Watts to the grid to slam current through the plate and load.
I think I know where your doubt/problem lies.
I won´t "send you to a book and tell you to deal with it yourself" but straight answer.
To simplify explanation, and focus , let´s forget Class AB for a moment , after all the Class A fraction is minimal compared to the Class B one.
No, you are "thinking transistors as tubes", they are very different.
A Tube with 0V at grid will pass maximum current possible, so normally we NEED negative voltage there to keep it under control.
We need strong negative voltage so current at idle is small.
A transistor (let´s talk NPN; PNP is the same with polarities reversed) with 0V at base passes zero current, period.
Negative voltage will do nothing, period.
"0V and reverse V are the same" so put reverse voltage out of the picture.
A transistor will need *at least* some positive 0.7V just to be at idle and then higher voltage to pass significant current although here the significant parameter is base current.
It will need (positive) base current to pass collector current, but as far as voltages are concerned, they are always positive.
In fact at high current , base voltage will raise up to, say, 3V !
Forget this analysis which is wrong.
And again: tubes and transistors are different.
Forget all this, all wrong, tubes and transistors work as I explained above, much simpler.
Justb don´t mix them in the same thought block.
Forget this, see above.
Unless you pass a ton of current which the phase inverter can not supply anyway, you will NOT damage the tube, so forget that idea.
Tube amps are overdriven all the time so grids get slightly positive on signal peaks , nothing bad happens because phase inverters typically can provide upto 2 mA or so, while tubes are happy with tens (not tenths) of mA, go figure.
Most of the time tubes are negative biased, signal drive moves them around that starting point and yes, say a grid is biased at idle -52V (6L6 in a Fender amplifier, but similar mechanism in Hi Fi amps), drive signal will sweep that voltage up to 0V at which point it will saturate passing maximum current possible, and then on the negative swing it will reach same peak value but the other way, so it will reach -104V
And that if amp is not overdriven
With no damage whatsoever, the tube is built to stand that and more.
I think I asked the question improperly. I definitely get they are different topologies, they are biased differently, and just generally operate completely different. Maybe I should ask in a different fashion.
My assumptions were as follows, maybe they can be corrected individually:
1) A valve must have a negative grid-to-cathode bias in order to ensure that the grid is repelling electrons to prevent them from getting from the cathode to the plate.
2) If a valve has a positive voltage(or even near zero), some or many of the electrons being emitted from the cathode will make their way into the grid and allowing a current, which under normal operation should be almost nothing. Merlin's book described such a scenario as allowing(in reverse) positive ions to possibly reach the cathode and cause cathode poisoning/sputtering/stripping, which, in his words "emissive properties will be degraded". Have I completely misunderstood this?
3) There doesn't seem to be a "mechanism" in push-pull valve schematics(that are obvious to me anyway) to prevent the grid/cathode voltage from swinging unfavorably in the #2 scenario.
4) New: From the above responses, it sounds like a phase inverter would not be able to provide a sustainable enough current for #3 to actually occur in a push-pull, is that correct?
5) Back to transistors, and I have done no design on these in 20 years and am starting from scratch. My understanding was that, in absence of a phase inverter, the NP junction between the base/emitter in the reverse polarity (assuming it doesn't break down) would be great enough to "automatically" stop conduction of any current when the wave is on the opposing side. The inactive side would be completely high impendance and the active side would take over the amplification. Because of this, a phase inverter isn't needed, because with the correct bias and parts, the large PN junction would completely turn off the transistor.
6) Back to tubes again, "Most of the time tubes are negative biased". In the context of a Hi Fi PP design, can you define "most of the time"? Do you mean in most designs, or are you talking much more granular, like amplifying a signal(some parts of the wave will not be in negative)?
Thanks for not flaming me too badly. I don't mind being sent to the books. The two originally suggested to me were Valve Amplifiers(4th edition) by Jones and Designing High-Fidelity Tube Preamps by Blencowe. Both excellent books, but I feel like(unless I totally missed it) they've skimmed over little details like this. I've gone through them, and I know with a lot of lab time and simulations I can probably figure some things out. At least with SE designs they seem to be a lot better covered without passing over some need-to-know details. If maybe there is another book that covers this better, I'm interested.
I may have another followup based on how the prior questions are answered.
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I'll give it a try.
1) What you describe is the situation at cut-off (the according grid voltage is called the "cut-off voltage"); the control grid is biased so negative that without input signal no current will flow. But for audio purposes, you never bias tubes like that. Even in class B topologies the bias is adjusted so that without input signal still some current will flow. This is done to prevent nasty high order crossover distortion caused by the switching on and off of the tubes when one of the push-pull pair takes over from the other (at very low currents tubes are very non-linear).
2) You understood this correctly, although there are tubes (mostly transmitting power tubes) that are constructed so that the control grid can stand (a lot of) control grid current (class B and C operation).
3 and 4) Overdriving an amplifier will allways give some sort of problem. But in most cases the phase inverter in an audio amplifier can only deliver little current so the amount of grid current can only be little beacuse it has to be provided by the preceding stage.
5) I’ll pass on this one because my knowledge on transistors is very poor.
6) Some transmitting power tubes (for example: the 838 power triode) work best with zero, or even positive bias voltage. For transmitting purposes, power tubes are often run with large amount of control grid current (class C operation). They were made to take this (by using proper materials/coatings for the control grid structure and by cooling the control grid structure sufficiently).
Maybe you don’t see that when a tube is biased at let’s say - 10 V, than the zero volt point of the input signal corresponds with that - 10 V point. In this situation the input signal can swing about 17 Vtt without causing control grid current in the tube beacuse the grid of one of the pair of power tubes is than swinging between - 1.5 V and - 18.5 V.
1) What you describe is the situation at cut-off (the according grid voltage is called the "cut-off voltage"); the control grid is biased so negative that without input signal no current will flow. But for audio purposes, you never bias tubes like that. Even in class B topologies the bias is adjusted so that without input signal still some current will flow. This is done to prevent nasty high order crossover distortion caused by the switching on and off of the tubes when one of the push-pull pair takes over from the other (at very low currents tubes are very non-linear).
2) You understood this correctly, although there are tubes (mostly transmitting power tubes) that are constructed so that the control grid can stand (a lot of) control grid current (class B and C operation).
3 and 4) Overdriving an amplifier will allways give some sort of problem. But in most cases the phase inverter in an audio amplifier can only deliver little current so the amount of grid current can only be little beacuse it has to be provided by the preceding stage.
5) I’ll pass on this one because my knowledge on transistors is very poor.
6) Some transmitting power tubes (for example: the 838 power triode) work best with zero, or even positive bias voltage. For transmitting purposes, power tubes are often run with large amount of control grid current (class C operation). They were made to take this (by using proper materials/coatings for the control grid structure and by cooling the control grid structure sufficiently).
Maybe you don’t see that when a tube is biased at let’s say - 10 V, than the zero volt point of the input signal corresponds with that - 10 V point. In this situation the input signal can swing about 17 Vtt without causing control grid current in the tube beacuse the grid of one of the pair of power tubes is than swinging between - 1.5 V and - 18.5 V.
1. Yes, that's correct.
2. Positive where? On the grid? You're talking about flowing grid current, which is a thing but don't worry about it until you understand the basics. You've probably misunderstood this.
3. The mechanism is called bias. For the scope of the discussion, always negative in tubes (others reading this see 2). Whether that be due to fixed bias because you actually pull the grid negative with DC or because you raise the cathode with a resistor, the result is a negative bias on the grid.
4. Grids don't pull current in your current understanding.
5. That's relatively correct, though the "stop conduction" part isn't 100% and there's crossover distortion. Biasing into class AB generally solves this with some application of feedback.
6. Look up depletion mode. That'll answer most of your questions. A tube (pentode) is basically a depletion mode NMOS. This goes back to the statement about figuring out tube operation before you consider a grid that pulls current.
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I believe this is what I was originally asking, but stated it in a different way.
By "class A setup", my intent was to mean that both valves are always conducting all 360º of the signal, even in a push-pull. So in your example, even in silence, in a perfect circuit, both tubes would be conducting at the current appropriate for the -10V grid voltage. When the signal swings one way, one valve moves down the load line to -18.5V while the other moves up toward -1.5V, but they are still both "ON" the entire time, just complementary.
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