tube amp output question

[SUPERMAGOO]

the typical configuration output tube for mashall 800`s amps are class ab1 or ab2? :? :? :? :?

 

[CJ]

Remember, Einstein didn't know how to drive a car, and he always forgot to wear socks. So what is your defination of stupid?

I am so stupid I don't even know the answer to your question. No wait, that makes me ignorant, not stupid, right?
:grin:

 

[SUPERMAGOO]

cj you are a idol!!! :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb: :thumb:

 

[Consul]

I'm pretty sure it's AB Negative. :green: :green:

Sorry, I'll stop joking around now.

As far as I know, there is only class AB, no AB1, or AB2. Please someone correct me if I'm wrong.

 

[SUPERMAGOO]

I`m sure that.I know there are some diferferences.
i need calculate a output transformer for 6l6gc output.but in datasheet i see p to p load impedance class ab1=5000 ohm ab2 =3800ohm.i dont know which data use for calculate the xformers.... :? :? :?

 

[pstamler]

[quote author="Consul"]As far as I know, there is only class AB, no AB1, or AB2. Please someone correct me if I'm wrong.[/quote]

If I recall correctly, a Class AB2 amp draws grid current during part of the waveform, while a Class AB1 amp does not.

Peace,
Paul

 

[BYacey]

IIRC AB2 is more of a reference to transistor push-pull biasing to minimize crossover distortion. Tubes don't really draw grid current in a practical sense, due to the very high input impedance.

 

[VacuumVoodoo]

[quote author="pstamler"][quote author="Consul"]As far as I know, there is only class AB, no AB1, or AB2. Please someone correct me if I'm wrong.[/quote]

If I recall correctly, a Class AB2 amp draws grid current during part of the waveform, while a Class AB1 amp does not.

Peace,
Paul[/quote]

Correct. JCM800 and hundreds of other guitaramps are AB1. The ones operating in class AB2 are most often high power bass amps. Usually you'll find there cathode followers driving the grid(s) of output power tubes to provide enough grid current.

 

[NewYorkDave]

Tubes can be driven into grid current. This is done in AB2 and B output stages, and was being done long before transistors. As usual, there's more detail in Radiotron 4.

That whole "high grid impedance" thing goes out the window when the grid-cathode diode is forward biased. What happens when a diode is forward-biased? Current flows! :wink:

An AB2 or B output stage is usually precded by a driver stage that's designed for low output impedance. The typical high-impedance "Schmitt inverter" found in Fenders and Marshalls can't drive the output tubes much into the positive grid region.

 

[PRR]

> mashall 800`s amps are class ab1 or ab2?

I don't know the amp; yet I say confidently that it is AB1 and if it is 6L6 you should be using ~6K loading. 5K4 or 6K6 nominal load makes very little difference when driving speakers.

(push-pull) Audio amplifiers up to 100 watts are run AB1. The exceptions are VERY few. Even the mighty Ampeg SVT ran AB1. The largest Macintosh ran very slightly AB2. Bogen had a 15W AB2 HiFi amp and a 120W AB2 PA amp: the little one was insanely complicated and expensive for what it did, the big one was a cheap monster.

If the output stage grids are capacitor-coupled, you can NOT run AB2. The amp runs class A for very small signals, AB1 for medium signals, and only goes AB2 (grid current) for the largest signals. And when it starts drawing grid current, the grid cap charges so the average current is zero, taking you out of AB2 on peaks and screwing-up your A/AB1 bias in moderato passages following a peak.

To run AB2 you usually need a POWER driver stage AND a transformer or choke. In some old AB2 designs they used a third same-type power tube for the driver (picture three 6L6). Alternately you can try to use cathode followers to drive the output grids, but a simple cathode follower won't pull a grid up well. The cathode output impedance is similar to the positive grid input impedance. Bogen used push-pull cathode followers plus a center tap choke so both cathodes could yank the active grid. In the Macintosh the output stage itself is a half-cathode-follower so grid impedance is somewhat higher; even so they only used a 12AX7 driver so you know they were only going a few mA into AB2. And this existed probably only because Macintosh's first amp used a grid transformer into 6L6.

The AB2 rating on the 6L6 is an antique: When 6L6 was introduced, triodes ruled the world and big amps all ran transformer coupled. But the literature for the 6L6 emphasized the simplicity and fidelity of omitting the grid transformer and taking overall feedback to cancel some of the 6L6's sins. AB2 does not begin to make sense until you get to very high plate voltages and modest fidelity levels. 10,000 watt AM transmitters, etc. AB2 starts to make sense with the 807 tube, and the early 6L6 was a 807 cheapened with a lower plate voltage rating. But 6L6GC and most later 6L6 equivs can make as much power in AB1 at 450V as the 6L6 could at its 360V rating when hard-driven AB2. Finding another 100V and another couple-K of load is a lot easier than building the driver for an AB2 output.

For historical/hysterical completeness, there were a series of "Zero-Bias Class B" triodes. These "always" ran grid current: at idle they had zero volts grid-cathode and a little grid current, on signal swing one grid sucked current while the other was cut-off. Simple and fairly powerful, and a fairly constant load on the driver. While popular for early battery radios, the crossover distortion (not much unlike a mis-biased transistor amp) was too irritating for general use.

 

[PRR]

> need calculate a output transformer for 6l6gc output

For pentodes, working reasonably optimally (not super low or high voltage):

Find your power supply voltage V (not counting cathode-bias) and your output power P.

Subtract 50 from V.

Square it.

Multiply by 2.

Divide by P.

For 450V supply, 55 watts out:
450-50= 400; 400^2= 160,000; times 2 is 320,000, divided by 55 is 5,818. The 6L6 sheet gives 5,600 load for this condition.

For 285V supply, 10.5 watts out:
285-50= 235; 235^2= 55,200; times 2 is 110,400, divided by 10.5 is 10,500. The 6K6 sheet gives 12,000 load for this condition.

For 285V supply, 14 watts out:
285-50= 235; 235^2= 55,200; times 2 is 110,400, divided by 14 is 7,900. The 6V6 sheet gives 8,000 load for this condition.

For 540V supply, 76 watts out:
540-50= 490; 235^2= 240,000; times 2 is 480,000, divided by 76 is 6,300. The 7027 sheet gives 6,500 load for this condition.

 

[bcarso]

It struck me just now that the similarities between JFETs and triodes extend to positive bias on gates/grids as well. Many simulators don't have a clue as to how to handle positive gate-source bias on an N-JFET, but a little bit doesn't screw things up too much and enhances transconductance beyond the canonical maximum at Idss. This means voltage noise should be reduced as well.

I was doubtful and irritated when I read Van der Ziel's advocacy of the use of JFETs in "floating gate" mode the first time (Noise in Measurements, Wiley, 1976). But although the noise is theoretically higher because two current noise generators are adding their power, the fact that everything else except the source (like a condenser mic capsule for instance) can be disconnected from the input circuit confers some interesting benefits.

 

[SUPERMAGOO]

thank you very very much!!!! prr :thumb: :thumb: :thumb: :thumb:

 

[Test Point]

I move to make a motion that we canonize PRR....He needs to write a book called "The Gospel According to PRR"....very few people write stuff so I can understand it....Isaac Asimov was another that could do it....that puts you in incredible company PRR. :wink: :wink:

TP

 

[scott_humphrey]

[quote author="Test Point"]I move to make a motion that we canonize PRR....[/quote]


All in favor, say "Aye."

That was a beautiful explanation of push-pull, tube output stages.

 

[SUPERMAGOO]

PRR: Can you send me a formula to calculate output transformers (AB1) because the one I have I think It´s given me bad results.

For class A push-pull, AB1 and AB2 do I have to use the same output transformers formula?

Between AB1 and AB2 the only think that change is the data sheet impendance or the circuit diagram, or both.

 

[PRR]

> Between AB1 and AB2 the only think that change is the data sheet impendance or the circuit diagram, or both.

The REAL difference is: in AB1 you drive the grids from some negative bias voltage up to zero volts, and the grids are very-very-high impedance. AB2 is exactly the same except you drive the grids positive. This is very hard work because at and above zero volts the grid impedance drops from very-very-high (or 100K grid resistor) to about 500Ω!

In short, you "NEVER" run Class AB2 in audio. It sounds awful.

And specifically: the high output of original 6L6 in AB2 can easily be beaten, with less effort and distortion, with the higher voltage 6L6GC staying in AB1.

On the 6L6 data sheets, AB2 suggests a lower load resistance than AB1 because yanking the grids positive allows more current to flow. Same voltage, more current, lower impedance, more power. However the extra power means a MUCH more elaborate and expensive driver, to make the huge grid current peaks (around 20mA, instead of ~0.5mA peak in AB1 with 100K grid resistors). The driver needs to be over 1,000 times as powerful to go into the AB2 range as to do AB1. So we never do that.

Plate lines for 6L6 working AB1 and AB2:

The lower curve (higher impedance) is AB1, as you see because it goes just to zero grid volts. Actually I extended a dotted line above zero grid volts, into AB2, to show that only a wee bit more power is available, not worth the hassle and strain of going into grid current.

The upper curve is the 47W AB2 rating. Much more current is available when you drag the grid positive, but actually a little less voltage, and you have to use a lower impedance load to take advantage of that. Up to zero grid volts, it is still in AB1. The lower load actually makes less power, because the current at zero grid volts has not changed. You can go further, to +12V on the grid, and get almost 50% more current. However it takes 24mA of grid current to reach this point, and that is a LOT of work for the driver.

 

[PRR]

If you look through the entire RCA or Sylvania Receiving Tube Handbook, 1945-1975, you only find ONE tube with a suggested operating condition for audio AB2: the 6L6/6L6GC. Receiving tubes don't run AB2: it is too much work for home audio, and really for anything less than kilowatt transmitters. The 6L6 stands alone because it is not originally a receiving tube: it was the 807 transmitting tube. A 6L6 is a 807 in a different bottle and using cheap 400V plate metal instead of good 600V plate metal. In context of a big expensive radio transmitter, it makes some sense to run 807s at 600V and AB2: you can get 80 to 120 Watts out of a pair for around 0.2-0.5 Watts of power to the grids. For 1934, this was astonishing show-off power for a tube that fit in your hand.

Back around 1930, fashion was different. Here is a transmitter audio power amp getting about 100 watts from a pair of type 800 tubes working far into Class AB2.

The driver stage is not a small tube: it is two type 45, the biggest receiving tube of the day, working in push-pull at maximum rating (about 2 Watts output if the load were linear). Transformer coupling between '45 and 800 is necessary: cap-coupling would charge-up and mis-bias the amp on signal peaks that hit the AB2 range.

Using pentodes, we can today (since 1950) get similar power from a quad of 6L6GC or a pair of 6550, working AB1, using only a small dual-triode and two caps to drive the output. Why mess with AB2?

 

[Test Point]

AHHHHHHHHHHHH....a pair of 833's modulating a pair of 833's! Now that was the day......That was the day of beautiful sounding AM radio...driving those final amp tubes into HEAVY grid current....but, that was class C. it worked because of the tank circuit but we won't go there :grin: :grin:

TP