Tube theory discussion

[Omega_Void]

Continuing from 5150 mod thread in a belated attempt to mitigate OT.

So let me think out loud here...if 60mA is traveling through a 6L6 with a plate voltage of 500v, that is 30W that is being dissipated across that tube.

30 Watts is the absolute maximum dissipation of 6L6GC.

Okay, theoretically you could bias to max. diss. at quiescent. But the tube will be right on the brink of meltdown even with a resistive load. With a speaker the load is reactive, and dynamic to boot. Dissipation will be kicked above maximum unless you limit input to the one specific frequency which matches the correct load. Even then it won't stay there as the impedance will vary as the speaker moves.

The bias you've described is entirely acceptable IRL provided you don't use the head to amplify any signal.

 

[Omega_Void]

Now also from my understanding is that, active components do not theoretically dissipate any heat...

You mean reactive, but yes.

...however they do in small amounts particularly inductors, which is what a speaker is.

Yes, real-life components cannot be entirely free of resistance, and where an inductor is formed by coiling up wire the wire resistance will dissipate heat.

...but the total current through all 4 tubes should stay close to the idle current when the other two tubes are cutoff,

You mean until the other two cutoff.

...and Class AB should see a max dissipation of ~1.5 times idle depending on how hot they are biased.

Nope, this is what you've forgotten here: in class AB half of the tubes may see more than max. diss., but the other half are not conducting at all. So total dissipation across all of the tubes would be half of 1.5 times max. in this instance, until the other tubes come out of cutoff when it returns to the idle dissipation.

In class AB the dissipation is least at maximum output.

 

[Omega_Void]

Oh, my quick estimate of the power consumption in the other thread left out the class A bit of AB. if the amp's in class A for the first 40W, you could add an extra 20W to the total, giving: 275W for the valves at +3dB, and 22.8W for the filaments.

So yeah, in overdive the poweramp does consume about 300W.

I'm sort of glad you're making me think through this stuff, I've reduced it all to rule of thumb long enough ago that I've lost sight of the nitty-gritty and my theory's gotten rusty. :Spin:

 

[ahjteam]

if 60mA is traveling through a 6L6 with a plate voltage of 500v, that is 30W that is being dissipated across that tube.

Where the fuck do you get that 500v? By hooking up 45 car batteries together? The American outlets output 120V and European outlets output 230V, so thats closer to 3.6w or 6.9w (assuming you were counting with the milliamps * voltage = watts formula).

So yeah, in overdive the poweramp does consume about 300W.

...more like 60-150W at max depending on the amp and practically it's usually about 2-50W unless you want to go deaf. The solid state power amps are a totally different beast, they can consume closer to 1200W if you use like an Ampeg SVT4, but again, at deafening volumes.

 

[Omega_Void]

Where the fuck do you get that 500v? By hooking up 45 car batteries together?

No. By using a step-up transformer in the power supply.

Electricity goes in to the transformer at the mains voltage, comes out much higher. (The current out is reduced WRT current in by the same proportion.)

When the AC out of the transformer is higher than the reservoir capacitors' voltage, the capacitors draw current to charge themselves up to the peak AC voltage (about 500V in a 5150).

The rectifier diodes between the transformer and the caps prevent the caps discharging back into the transformer secondary (diodes pass current in only one direction), so the caps only discharge when the circuitry in the amp draws current.

If a tube has 500V on the plate and draws 0.06 Amps current, then the power consumed in the tube is, as TheWinterSnow says, 30W.

...more like 60-150W at max depending on the amp and practically it's usually about 2-50W unless you want to go deaf. The solid state power amps are a totally different beast, they can consume closer to 1200W if you use like an Ampeg SVT4, but again, at deafening volumes.

That's the power output you're thinking of. Since amplifiers are not 100% efficient power consumption (what we're discussing here) is greater.

 

[ahjteam]

ah, gotcha.

 

[TheWinterSnow]

Many time so far have we gone over things that questions the way that I have been taught, and I have had my instructors many times before say that my knowledge in tube theory goes way over their head, but going over the basics fundamentals of electronics, what has been discussed makes me question everything I know.

I am going to think out loud again, but something a bit more simple, looking at silicon based semiconductors, the quiescent point is the voltage to the base (or gate) of a transistor where the component is half on, so that the input signal can be fully reproduced and amplified depending on the circuit.

When the transistor is conducting 100% of the time, or, never shuts off, it is in class A. And obviously preaching to the choir, if it is conducting half the time, its Class B. Class B has crossover distortion from the lag that results from the devices turning on and off (going into and out of cutoff) where the Vce=Vcc. So class A has a q point greater than Vce=Vcc but less than Ip(sat), Class B has a q point where Vce=Vcc so that the device only conducts when the input swings positive with respect to the ground. Class AB to compensate for such crossover distortion conducts more than 50% but less than 100% of the signal, generally 70%. While that is less power efficient it ensures a clean output signal. The q point is greater than Vce=Vcc, but less than the center point between Vce=Vcc and Ic(sat). I am talking about load lines here, because that is the best way I understood the class of amplifiers.

So what is Ic(sat) or Ip(sat) for a 6L6? Well at their maximum operating voltage of 500v, considering Rp if Rp=5000, is 100mA. That means, regardless of bias that tube is never capable of supplying current greater than 100mA, this is when the tubes saturate and you get power tube distortion. Now the tubes can handle AC current at saturation, because the effect DC heating is Vac/(sqr)2. With that in mind, if you brought the q point up to 100mA, the amount of DC heating from the resulting plate current would cause it to overheat. Now off the cuff if we looked at a Winged C datasheet, their Ip(sat) at 450v would be 81mA, which is a tad bit higher than the 6L6 max dissipation of 66mA if Vp=450v. So by all means the tube can supply more than dissipation, and for the most part can handle the AC signal reaching saturation, it just will cause the plate to overheat if the q point is set to saturation, which depending on the amp, maybe equal to or great than the plate dissipation (this is another reason for lowering the screen voltage, its acting as a current limiter).

The datasheet also notes that for a class AB push-pull amplifier, that the maximum plate current is 210mA when max plate voltage equals 450v and I think it said a max idle current of 100mA.

Unless I am totally screwing something up here.

Where the fuck do you get that 500v? By hooking up 45 car batteries together? The American outlets output 120V and European outlets output 230V.

step up transformer.

 

[Omega_Void]

... looking at silicon based semiconductors...

I don't disagree with anything you said about BJTs.

Well at their maximum operating voltage of 500v, considering Rp if Rp=5000, is 100mA. That means, regardless of bias that tube is never capable of supplying current greater than 100mA, this is when the tubes saturate and you get power tube distortion.

Actually, the tube will clip on positive input swing when the grid voltage exceeds the cathode voltage by more than about a volt. Strictly speaking this is "grid current limiting" rather than saturation per sé. But most folks just call it saturation.

It's a diode (just like cathode to plate) -- when the grid is negative WRT the cathode it's reverse-biased and no current flows. When the grid is positive of the cathode, (EDIT: correction here) the cathode draws current from the grid. Therefore, as long as the grid is +ve WRT cathode, the input signal cannot increase because the current output from the PI is diverted the grid-leak resistor where it normally develops the input voltage.

So the "saturation" point is controlled by the negative grid bias rather than Rp.

Now off the cuff if we looked at a Winged C datasheet, their Ip(sat) at 450v would be 81mA, which is a tad bit higher than the 6L6 max dissipation of 66mA if Vp=450v.

Yup, but Ip(sat) might not be reached due to grid current limiting. I'm not sure if any safe bias point allows it, I think GCL always kicks in first.

So by all means the tube can supply more than dissipation...

Yes but only if it switches off on the other half of the cycle so that overall dissipation over the whole cycle is lower in total than max. diss.

The datasheet also notes that for a class AB push-pull amplifier, that the maximum plate current is 210mA when max plate voltage equals 450v and I think it said a max idle current of 100mA.

The datasheet probably also said "values for two tubes". That's 50mA per tube = 22.5W each.

 

[Omega_Void]

(this is another reason for lowering the screen voltage, its acting as a current limiter).

The big deal about screen voltage is to keep it below the plate voltage under all conditions. Otherwise the screen draws current that's meant for the plate. And they just ain't built for that much.

I think some amp manufacturers aren't too careful about this, and that may be why some amps eat tubes much faster than others?

 

[TheWinterSnow]

Actually, the tube will clip on positive input swing when the grid voltage exceeds the cathode voltage by more than about a volt. Strictly speaking this is "grid current limiting" rather than saturation per sé. But most folks just call it saturation.

It's a diode (just like cathode to plate) -- when the grid is negative WRT the cathode it's reverse-biased and no current flows. When the grid is positive of the cathode, (EDIT: correction here) the cathode draws current from the grid. Therefore, as long as the grid is +ve WRT cathode, the input signal cannot increase because the current output from the PI is diverted the grid-leak resistor where it normally develops the input voltage.

So the "saturation" point is controlled by the negative grid bias rather than Rp.

from the way you worded that, you made it off to believe that the cathode can pull current from the grid. To my knowledge this is not possible as the grid does not have a heating element like the cathode to release electrons. If we are talking about triodes, from my understanding, decreasing the grid to cathode voltage so that the grid is negative with respect to the cathode, then the tube will eventually when the voltage is low enough, will cutoff. The opposite goes fro saturation, when the grid voltage is equal to or greater than cathode voltage, then the tube is at saturation because the additional current is being pulled by the grid, this would be the tube working in what is called the constant current region.

The big deal about screen voltage is to keep it below the plate voltage under all conditions. Otherwise the screen draws current that's meant for the plate. And they just ain't built for that much.

I think some amp manufacturers aren't too careful about this, and that may be why some amps eat tubes much faster than others?

I made that reference to mean that, if you want to limit or lower the Ip(sat), then you would decrease the screen voltage, that is if designing an amp you felt that Ip(sat) was too high.

I think the manufactures aren't careful on purpose so they can get people to buy their tubes (as some amp companies make their own tubes) or charge people to do repair work which is nothing more than a lot of money for replacing tubes. That or if they burn through tubes faster, if the users continue to use the companies sponsored tubes, it makes that company more money, which allows the amp company to get their bulk tubes for cheaper, in turn making them more money as well. So either way, designing an amp to eat through tubes faster is a money scheme.

 

[Omega_Void]

...you made it off to believe that the cathode can pull current from the grid. To my knowledge this is not possible as the grid does not have a heating element like the cathode to release electrons.

Yes, but conventional current is the reverse of electron flow. When the electrons are streaming from the cathode to the grid, current flow is from the grid to the cathode.

See that I corrected that bit? I initially did the same thing as you, picturing the electron flow, then wording the current statement based on that but forgetting to reverse direction.

Damn ancients not knowing which way the particles were flowing! Does my head in sometimes.

But yes, if your statement meant to say that the additional electrons are pulled by the grid then we are saying the same thing. "Constant current region" and "grid current limiting" are two terms for the same effect.

So yeah, what I was saying is the Ip(sat) figure in the datasheet is the maximum current the tube could pass at any instant if the circuit allowed it. It's technically incorrect to use Ip(sat) to refer to any externally imposed current limit.

It is not reached in practice, 'cos in an amp the tube enters constant current region first. I don't think any practical bias point would allow Ip(sat) to be achieved without overheating the valve.

 

[Omega_Void]

I made that reference to mean that, if you want to limit or lower the Ip(sat), then you would decrease the screen voltage, that is if designing an amp you felt that Ip(sat) was too high.

Apart from you should replace Ip(sat) with Ip(max) in that statement...

Yes you can reduce maximum dissipation by lowering screen voltage, but this would also have an effect on tone, increasing the compression in the power tubes.

Alternatively, you could drop B+ on the plates.

Check out Randall Aiken's page on back-biasing. You don't need to use the negative voltage for biasing, but you can still nick it off the B+. Cheaper than a new power transformer, and less heat and sag than using a series resistor before the plate supply.

 

[TheWinterSnow]

I have to ask now because this is bugging the fuck out of me. A class B amplifier amplifies only half the signal only when the signal is more positive to the cathode (in a BJT). Now with a transformer coupled amplifier, there is bias voltage at the base or grid allowing idle current to pass. What confuses me is this, if there is idle current on both sides of the push pull amplifier, doesn't that mean that both sides will conduct when the signal is present until the signal reaches the bias voltage, making the amplifier not a class B, but a class AB?

 

[MarcusGHedwig]

Where's that Scanners exploding head GIF when you need it?

 

[AllanD]

Nyess.....Nyess....

 

[TheWinterSnow]

Omega, I am looking at the power section of the schematics trying to figure this out. When I replaced the SGR with two 2K resistors, the voltage did not drop at all (which I understand because the resistance of the screen grid is damned near infinite compared to any value for the SGR) and is still running hotter than most tubes are specified for except KT88s. Now I am understanding that the voltage of the Screens are more of the voltage drop from the 400 ohm resistor from the plate to grid and such voltage drop is equivalent to the values of the other resistors for the preamp tubes. My question is, other than totally redoing the voltage divider of the power supply, what would be the best way to drop that screen voltage without messing the with the HT of the preamp.

 

[Wolfeman]

I'm pretty sure the only way to achieve what you are asking is by using a split rail design.

 

[TheWinterSnow]

what would be the best way to implement that? The only thing I can think of is regulators or something along the lines of, split the rail, and use a voltage divider.

 

[AeternusEternus]

ggggggggggggrrrrrrrrRRRRRRRRAAAAAAAAHHHHHHHHHHHHNNNNNNNNEEEEEEEEEEEEEEEEEEEEEEEEERRRRRRRRRRRRRRRRRRRRRRDDSS

 

[Wolfeman]

what would be the best way to implement that? The only thing I can think of is regulators or something along the lines of, split the rail, and use a voltage divider.

Separate transformer, or a replacement transformer with a separate winding.