RCA 6L6 Data Question

[Ed_Chambley]

Looking at RCA Data Sheets (others do not have the same) under cathode bias there is something I cannot understand.  The voltage listing of 360v on the 2 tube cathode bias shows 9K Plate to Plate.  What has changed in this operational conditions to increase the resistance this much?

[HotBluePlates]

... 9K Plate to Plate.  What has changed in this operational conditions to increase the resistance this much?

Nothing. Except the designer chose to use a 9kΩ plate-to-plate load. 

Let's compare the middle & right columns:
Fixed Bias
Plate                      360v
Screen                   270v
Bias                      -22.5v
Peak Grid input       22.5v (to one tube)
Idle Current            44mA (for one tube)
Max Plate current  140mA
Z, plate-to-plate    3800Ω
Power                     18w
THD                        2%

Cathode Bias
Plate                      360v
Screen                   270v
Bias                      -21.8v (88mA * 248Ω)
Peak Grid input       20.3v (to one tube)
Idle Current            44mA (for one tube)
Max Plate current  100mA
Z, plate-to-plate    9000Ω
Power                    24.5w
THD                        4%

So what happened? The designer used slightly less bias and therefore slightly less drive voltage, and got ~33% more output power for 2% more THD with the higher load impedance. Max-signal plate current dropped (as it must) because the same supply voltage is being used, but there is a higher load impedance. Just like if you raised resistance in a circuit but kept voltage the same, current goes down.

What happens in the Low-Z condition? Assume the max plate current is an RMS value; then the peak value appears across a load of the plate-to-plate impedance divided by 4 (because this is class AB and one side is shut off). 3800/4 = 950Ω and 140mA * 1.414 = ~198mA peak, so 198mA * 950Ω = 188v peak across the load impedance. That's pretty far short of the 360v available, even considering the minimum plate voltage allowable for the 6L6. Power should be calculated with the RMS numbers, so (140mA)² * 950Ω = 18.6w, close enough to what the data sheet says.

What happens in the High-Z condition? 9000/4 = 2250Ω, and 100mA * 1.414 = ~141mA. Peak voltage swing across the load is 141mA * 2250Ω = ~318v peak. Much more of the available voltage is used as output swing, though peak current draw is lowered. Power output is (100mA)² * 2250Ω = 22.5w, but that calculation assumes zero distortion and we know there is 4% THD (which raises measured output power).

Net effect: RCA showed a condition where you could get more power from a cathode-bias setup with slightly less drive, lowered peak current demand due to high load impedance. And all for a pretty measly increase in THD. The cathode bias condition also implies a relatively cheap power supply will suffice, because there's relatively low peak current and a relatively small change when the tubes move from idle to maximum output power.

The 1st column in that trio shows what you can get when the load impedance is jiggered between the other two to 6.6kΩ. Now peak current has increased to ~187mA and peak voltage swing is then 187mA * 1650Ω = 308v. Formula power output is 134mA² * 1650Ω = 134mA RMS * 221v RMS = 29.6w, which guesses the output a few watts high. Close enough unless you need data sheet accuracy, in which case you'd break out the plate curves and plot lines for a number of different loads until you maximize power output while minimizing distortion.

[HotBluePlates]

Note, all numbers should be taken with a grain of salt, and a dash of practical experience...

The screen voltage was only 270v. When that voltage is raised, the maximum peak plate current available from the tube goes up. So does idle current, and therefore idle dissipation. In response, you'd raise the bias voltage to tame idle current.

If you understand that for an output tube it's a waste to not drive the grid close to 0v (meaning a peak input to 1 tube equal to the value of bias voltage), then you'll also see the required drive voltage gets bigger. All of these conspire when designing a class AB amp towards using lower load impedances to make use of the availability of higher peak current. This is balanced against a peak plate voltage swing that big, but not so big as to touch the minimum plate voltage needed across the tube (roughly where the knee of the plate curves sits).

Which explains why you're not used to seeing numbers like "9kΩ" as a 6L6 load: you're used to looking at amps with screen voltages almost 200v higher. You could use a 9kΩ load with an amp with 400v plate & screen, you'd just get closer to the 20-25w on the sheet instead of 35-40w you'd expect with the high supply voltage.

Then there's the concept of "Power Sensitivity" or power output divided by grid driving voltage required to produce it. For a given supply voltage, some balance of bias and load impedance may result in an amp that delivers the most output with a given drive voltage. This might be a design figure of merit.

Or maybe you really want to use a certain type of phase inverter circuit, with output limitations. Maybe you're willing to accept less total power output as a tradeoff for better balance & lower distortion. Now maybe a different balance of all the factors seems most appealing.

[jjasilli]

2 questions:


1.  Is it proper to say that there is 50% more output in watts as a result of doubling the THD?


2.  ". . .there is a higher load impedance. Just like if you raised resistance in a circuit but kept voltage the same, current goes down."


I thought that plate-to-plate impedance is a definite, objective measurement, not an outside condition imposed upon the plate(s) by a downstream load.  Que passa?  (This seems similar to Ed's question which posted while I was typing.)

[HotBluePlates]

1.  Is it proper to say that there is 50% more output in watts as a result of doubling the THD?

No. The power increased 36% by using 2.37x bigger load, and 2% higher distortion (you could say 100% more distortion), but with reduced drive signal. More horsepower, less gas...

The straight mathematical equation assumed zero distortion, and predicted 18.6w for the Low-Z load (0.6w higher than the data sheet, with 2% THD) and 22.5w for the High-Z load (2w lower than the data sheet, with 4% THD).

Distortion will raise output power, and RCA probably got a hair more output as a result of increasing drive voltage somewhat until there was 4% THD in the High-Z condition. Neither of us will know for sure unless we build it ourselves and see what power output result with only enough drive to create 2% THD.

Predicting distortion is a very time-intensive process using plate curves. Regardless, the difference of power output from the straight equation (whether an increase or decrease) is a result of a tube not have perfectly straight, equidistant grid lines. Which is another way of saying "distortion".

2.  ". . .there is a higher load impedance. Just like if you raised resistance in a circuit but kept voltage the same, current goes down."


I thought that plate-to-plate impedance is a definite, objective measurement, not an outside condition imposed upon the plate(s) by a downstream load.

Plate-to-plate impedance is a definite, objective measurement of an output transformer when a rated load is attach to a secondary to reflect the rated primary impedance. If you change the secondary load, or use a different transformer, you change the plate-to-plate load impedance.

What you may have really asked was, "I thought plate-to-plate impedance was a measurement of the tube." It's not. Pentodes and beam power tubes have quite high internal plate resistances compared to the load attached to them. Power triodes might be run with load impedances more like 2x their internal plate resistance, but you also get less output power for the same supply voltage as a result.

[HotBluePlates]

I need to keep expanding the thread to deal with fundamental questions/misperceptions like these, but look at this post.

The diagonal line is the load impedance. Horizontal axis is voltage, vertical axis is current. Imagine the load line which runs corner-to-corner represents the ideal load for the condition shown, where plate voltage and current swings are maximized.

Now make the diagonal line more-vertical. This corresponds to less-impedance. Current swing can be as big as in the ideal case, but now voltage swing is reduced on both ends. Overall power is reduced (current times voltage).

Go back to the first case, but make the diagonal line more-horizontal (more impedance). Voltage swing can be as big as in the ideal case, but current swing is reduced on both ends (makes sense: more resistance/impedance + same voltage = less current). Power output is also reduced compared to the ideal case.

And so it was with the 3 conditions shown for the same plate and screen voltages (and largely similar bias) on the data sheet. Really, you got to see the effect of changing the load (by selecting a different OT) on available output power.

[HotBluePlates]

1.  Is it proper to say that there is 50% more output in watts as a result of doubling the THD?

No. The power increased 36% by using 2.37x bigger load, and 2% higher distortion (you could say 100% more distortion), but with reduced drive signal. More horsepower, less gas...

The straight mathematical equation assumed zero distortion, and predicted 18.6w for the Low-Z load (0.6w higher than the data sheet, with 2% THD) and 22.5w for the High-Z load (2w lower than the data sheet, with 4% THD).

Distortion will raise output power, and RCA probably got a hair more output as a result of increasing drive voltage somewhat until there was 4% THD in the High-Z condition. Neither of us will know for sure unless we build it ourselves and see what power output result with only enough drive to create 2% THD.

Predicting distortion is a very time-intensive process using plate curves. Regardless, the difference of power output from the straight equation (whether an increase or decrease) is a result of a tube not have perfectly straight, equidistant grid lines. Which is another way of saying "distortion".

2.  ". . .there is a higher load impedance. Just like if you raised resistance in a circuit but kept voltage the same, current goes down."


I thought that plate-to-plate impedance is a definite, objective measurement, not an outside condition imposed upon the plate(s) by a downstream load.

Plate-to-plate impedance is a definite, objective measurement of an output transformer when a rated load is attached to a secondary to reflect the rated primary impedance. If you change the secondary load, or use a different transformer, you change the plate-to-plate load impedance.

What you may have really asked was, "I thought plate-to-plate impedance was a measurement of the tube." It's not. Pentodes and beam power tubes have quite high internal plate resistances compared to the load attached to them. Power triodes might be run with load impedances more like 2x their internal plate resistance, but you also get less output power for the same supply voltage as a result.

[PRR]

>> "I thought plate-to-plate impedance was a measurement of the tube."
> It's not.

+1.

> Pentodes and beam power tubes have quite high internal plate resistances compared to the load attached to them.

When you use a pentode for Power you look at *two* internal resistances. See attached.

On 6L6: the low plate voltage line runs about 300 Ohms. Load should be much greater than this for good efficiency. The traditional "plate resistance" runs 10K to 30K. Load should be lower than this.

Taking a split-difference we get 1.5K-2K load per tube, or 6K-8K rating for the push-pull load. That is a broad maximum of plate circuit efficiency.

Efficiency always sucks and getting away from the "maximum" may help other compromises more than the lost power is worth. Going to 4KCT loading is maximum power output for the improved 6L6 (5881, 6L6GC) which can stand a high G2 voltage and has good dissipation. Going to 8K or 9K reduces full-power current and makes self-bias more practical (note only 1:1.25 shift of cathode current for the 9K condition).

[Ed_Chambley]

Nothing. Except the designer chose to use a 9kΩ plate-to-plate load. 
 
I understand this comment.  I had no better way to ask the question.  I was following along in your "Class A" thread.

Here is one of the reasons I am asking.  The CARR Rambler is a 2 6L6 28 watt amp..  I had one on loan and drew a schematic of the preamp.  I can draw the power section, but I have not been able to determine the transformers.  It has custom would Heybour trannys.

The PA section is exactly like the 5E5, not the 5E5a since it has no NFB and is cathode bias.  I do know the OT is 6K6/8ohm, but the voltage is not consistent  with the 6L6 data since the Plate and screen are almost the same.  Plate voltage read with (variac 120AC) 366vdc and screens 363. 

It is not my plan to build one anytime soon, but it is a very nice sounding amp.  The reason of interest is I have always wondered why Fender designed the 5E5 with a 6K primary.  I see now that it was not uncommon at all especially with 6L6/G/GB.

Another question.  How do you determine the necessary HT amperage for a given tube arrangement?

For instance HBP mentioned:
What happens in the High-Z condition? 9000/4 = 2250Ω, and 100mA * 1.414 = ~141mA. Peak voltage swing across the load is 141mA * 2250Ω = ~318v peak. Much more of the available voltage is used as output swing, though peak current draw is lowered. Power output is (100mA)² * 2250Ω = 22.5w, but that calculation assumes zero distortion and we know there is 4% THD (which raises measured output power).

In the above comment, does increasing the impedance lower the needed amperage for the tube?

[HotBluePlates]

... The CARR Rambler is a 2 6L6 28 watt amp..  I had one on loan and drew a schematic of the preamp.  I can draw the power section, but I have not been able to determine the transformers.  ...  I do know the OT is 6K6/8ohm ... How do you determine the necessary HT amperage ...

For the situation you described (reverse-engineering an amp-in-hand), the "right" answer is:

• Power off; disconnect the PT high voltage center-tap from ground. Insert an ammeter between the center-tap and ground.
• Attach a dummy load (8Ω replacing the speaker), attach a voltmeter across the dummy load.
• Inject a test sine wave signal wherever convenient, power on, increase the signal until you measure 14.97v RMS across the dummy load (corresponds to 28w into 8Ω).
• Read measurement on ammeter of RMS current draw.

I'll add another write-up shortly to show how to estimate this in the absence of direct measurement, and using what you already know.

[sluckey]

I use another method to determine total B+ current that may be more convenient than lifting the HT center tap if you have a STBY switch. Just tack solder a 1Ω resistor across the STBY switch and measure the voltage drop across the switch.

[Ed_Chambley]

I use another method to determine total B+ current that may be more convenient than lifting the HT center tap if you have a STBY switch. Just tack solder a 1Ω resistor across the STBY switch and measure the voltage drop across the switch.

I use that all the time.  Great if you have the amp in-hand.

BUT
How do you determine the necessary HT amperage for a given tube arrangement in the design prior to build?

[HotBluePlates]

How do you determine the necessary HT amperage for a given tube arrangement in the design prior to build?

...  The CARR Rambler is a 2 6L6 28 watt amp. ...

And according to Carr's website, Class A. You already know the amp is ~360v plate and screen. If the amp uses a 6L6GC tube (30w), idling at maximum dissipation results in 30w/360v = ~83mA per tube. If the amp is operating Class A, the operating point is on the loadline, the current (at most) swings up to twice-idle (166mA) and down to 0mA (these are peak values in opposite directions).

Also, since the amp is Class A, both sides are conducting at all times, so neither half-primary is disconnected from the circuit. That means each side of the output sees 1/2 the plate-to-plate impedance at all times. You already determined the impedance was 6.6kΩ, so the tube sees 3.3kΩ.

Converting peak current to RMS, 166mA * 0.7071 = 117mA. The RMS voltage swing is then 3.3kΩ * 117mA = 386v. That's not possible with only a 360v supply. If the supply voltage was higher, we might predict output power as 117mA² * 3300Ω = 45w, obviously not on spec. So the current draw is less than 117mA RMS. That may be enough for you to pick a PT right there, but let's keep going.

Let's try using the equation for watts and volts across resistance to determine the required plate voltage swing. Volts = √(Power * Resistance) = √(28w * 3300Ω) = 304v RMS. That would leave 360v-304v = 56v across the plate at peak output.

If you look at the graph on the top of Page 6 of the 6L6GC data sheet, you have a graph of plate current when G1 = 0v (i.e., peak positive input signal, and maximum plate current) for various G2 voltages. Follow the G2 line for 350v (close enough) and you'll see the knee is at ~75v plate and ~260mA. With this G2 voltage, the tube could be capable of a lot more output power (say, in Class AB with a smaller load impedance). But below the knee of the curve, the 6L6GC hits 166mA at well under 50v plate, so the calculated voltage swing is possible.

304v RMS / 3300Ω = 92mA RMS. So pick a 100mA high voltage winding, yielding ~360v with your chosen rectifier. 92mA is also 130mA peak current, but the RMS number is what matters for PT current rating (the peak number matters for peeping tube data sheets).

304v * 92mA = 27.968w. I'd call that dead-nuts 28w.

The slight rise from idle to full-power RMS current tells you there is a little distortion at the full 28w rating. Again, don't ask me to predict how much, because the accurate method using plate curves is very tedious. I doubt it'll matter to you anyhow.

[HotBluePlates]

Oh yeah... the 28w final calculation was a little self-serving. It dawned on me that since we calculated plate voltage swing by using the known output power and primary impedance, that of course we'd end up with the same number when multiplying the derived-current times the derived-voltage.

That doesn't make the results less-valid, because we knew 2 of 4 variables, and could readily derive the other 2.

Next go-round, I'll explain a bit about what you do when you know only 1 of the 4 variables... Hopefully, that will be educational for all.

[PRR]

> voltage is not consistent  with the 6L6 data ... Plate ... 366vdc and screens 363.

That data-sheet you posted shows _270_V_Max_ on screens, which is the traditional early-6L6 rating.

> why Fender designed the 5E5 with a 6K primary

Because 6K was much more available than 9K. 9K would generally be suitable for a BIG Triode power amp (aka Williamson), aimed at a high-class hi-fi market. 6K was more usual with early 6L6 and readily available in Fender's price/performance zone.

______________________________________________

> How do you determine the necessary HT amperage for a given tube arrangement?

The first-hack is to realize that the output power has to come through the lossy tubes from the B+. You rarely get over 50% efficient with small (<100W) tubes. So power from B+ will be "about twice" the power output.

Next: data-sheets and generalize?

There is a problem with the early 6L6 sheets. They list four conditions for contrast: 6.6K and 3.8K in AB1 and AB2. The AB1 3.8K loading is non-optimum: 6L6 at 270V G2 can't really pull 3.8K without going into grid current. In the part you cut off, it will pull 3.8K very well IF your driver can pump grid-current (99% of guitar-amps can't).

Take the "good" conditions.

6L6  360Vp  6.6K  132mA
6L6  360Vp  9K    100mA
6V6  250Vp  10K    79mA
6V6  285Vp  8K     92mA
EL84 250Vp  8K     75mA
EL84 300Vp  8K     92mA
6550 400Vp  5K    185mA
6550 400Vp  3.5K  275mA
6550 600Vp  5K    270mA
HK354 2000V 9.3K  480mA
HK354 3000V 19K   370mA

These are all different conditions. However since the Full Power Output condition probably has "something to do with" the load impedance, let us compute the "equivalent resistance" of these voltages and currents. 

6L6  360Vp  6.6K  132mA  --- 360V/132mA 2.7K
6L6  360Vp  9K    100mA    3.6K
6V6  250Vp  10K    79mA    3.2
6V6  285Vp  8K     92mA    3.1K
EL84 250Vp  8K     75mA    3.3K
EL84 300Vp  8K     92mA    3.3K
6550 400Vp  5K    185mA    2.2K
6550 400Vp  3.5K  275mA    1.5K
6550 600Vp  5K    270mA    2.2K
HK354 2000V 9.3K  480mA    4.2K
HK354 3000V 19K   370mA    8.1K

Now compare the DC B+ equivalent resistance to the load impedance:

Tube  condition  Rl    R(B+)/Rl
6L6  360Vp 6.6K  2.7K  0.41
6L6  360Vp 9K    3.6K  0.40
6V6  250Vp 10K   3.2   0.32  (*)
6V6  285Vp 8K    3.1K  0.39
EL84 250Vp 8K    3.3K  0.41
EL84 300Vp 8K    3.3K  0.41
6550 400Vp 5K    2.2K  0.44
6550 400Vp 3.5K  1.5K  0.43
6550 600Vp 5K    2.2K  0.44
HK354 2000V 9.3K 4.2K  0.45
HK354 3000V 19K  8.1K  0.43

These are "all the same" with remarkably little spread considering the wide range (250V to 3,000V, 3.5K to 19K, 10W to 810W). (*)And we may know that the 6V6 10K condition leaves some crumbs on the table (this is really a 6F6 condition when 6V6 was stealing 6F6 jobs).

Working the math with "perfect" (zero resistance tubes), Sine-wave drive, I think the answer is 0.3535. The difference from "0.4" is explained if we assume typical tubes with typical loads can only pull-down 88% of the supply voltage. Pull-down from say 450V to 54V seems entirely reasonable.

Multiply the plate-plate load impedance by "about 0.4" to get an equivalent resistance, drop that across B+, compute the current.

[HotBluePlates]

Next go-round, I'll explain a bit about what you do when you know only 1 of the 4 variables... Hopefully, that will be educational for all.

I keep forgetting to come back around & finish this...

There are a lot of variables that play into output stage operation. How much output power? What supply voltage? What B+ current? What output transformer primary impedance?

At the end of the day, only those 4 variables matter in output stage power. Sure, you care about "What tube type?" but that doesn't dictate output power directly, as you'll hopefully see below.

So let's start this as though we're designing the output stage. The first question to answer is "What output power is needed?" Let's say we need exactly 7w (just to pick a number that's not same-as a typical plan).

We now need to figure out power supply requirements and OT primary impedance. We could just pick random numbers to start for any of those, but the reality is there are only so many available power transformer voltages/currents, and only so many common OT's. Since we will be bound by these (unless you have money to burn on custom transformers), we should start by picking one of these. I'm going to use a transformer from Hoffman's Fender Transformers, starting with the power transformer.

The list includes parts which will yield a solid-state rectified B+ of:
353v (unknown current, let's say 50mA; Mojo 779)
410v @ 75mA (Mojo 759)
460v @ 70mA (125P1B)
466v @ 120mA (041316)

If you multiply the B+ voltage by the rated current, all seem absurdly high except the Mojo 779 (353v * 0.05A = ~17.7w), so we'll use that transformer.

We know that the output tube will need some amount of voltage across it at maximum current to operate properly. Let's take that voltage to be 100v for now. We could convert the resulting maximum peak voltage of 253v to an RMS value so that our calculations can include the known RMS output power. That's 253v / √2 = ~179v RMS

Let's also assume we will be using a class A push-pull output stage. This means that we can calculate based on the maximum-signal conditions for one side of the push-pull stage, and the resulting calculated load impedance will be half of the plate-to-plate load. In all cases below, I write "Resistance" in place of "Impedance" because the equations derive from Ohm's Law and the Equation for Power, and the change doesn't alter the results one bit.

Power = Voltage²/Resistance, or rearranging for the quantities we know:
Resistance = Voltage²/Power
Resistance = 179v² / 7w = ~4600Ω, or 9200Ω plate-to-plate

Looking back at Hoffman's Fender Transformer list, the closest stock OT primary is 8.5kΩ (the 022913 & 041318), so now we ought to re-calculate using the available OT impedance (8500/2 = 4250Ω per tube), this time for needed voltage swing.

Voltage = √(Power * Resistance) = √(7w * 4250Ω) = ~172v RMS at full output.

This result implies an RMS current of 172v/4250Ω = ~40mA RMS, or 57mA peak. The RMS value is what we ought to use to subtract from the PT's available B+ current rating, as it most-closely equates to d.c. draw.

Ideally, Our output tube will idle at or a little above 1/2 the peak current for maximum power, because that implies full utilization of the output tube in this case. 57mA/2 = 28.5mA, and 28.5mA * 253v = 7.2w, which just goes to show for class A push-pull, output stage efficiency is under 50% so the total dissipation of all output tubes should be more than double the needed output power.

So now we have an output stage designed, and we don't even know the output tube type... because it doesn't matter. Well, the tubes matter in that they need to be able to deliver the current swing required with the available supply volts, and within their ratings. But it's the PT and OT characteristics which set the amount of amount power possible.

So now you can pick any type and number of tubes you want to use, so long as the total dissipation rating per side exceeds 7.2w, and the minimum plate voltage required at the 0v gridline with a 4250Ω loadline and a B+ of 353v is less than 110v (172v RMS * √2 = 243v, 353v - 243v = 110v).

Depending on whether you chose a pentode or a triode (and how many per side), you'll need to follow this up with choosing screen voltage (if needed) and bias method/voltage. Depending on the obstacles you run into, you may have to return to the overall output stage design, select a different PT, OT or B+ voltage/current, rinse & repeat until you get a workable total output stage and power supply design.

[lego4040]

Those posts are where I get lost. I'm glad you guys understand all off that and I can just ask. The little I do grab and understand is thanks to you

[Ed_Chambley]

Thanks guys.  I got it, however I reserve the right to revisit this.

From what I am understanding, I have been using Power Transformers larger than needed.  I bought a few new in box Thordarson transformers.  50's Iron and very nice looking stuff.  They range in ma from 85 to 120, but usually have 3 or 4 amp heater winding and 5v as well.

Next question:
If the data sheet states a 360 plate / 270 screen voltage but in guitar amps the screen voltage is very near the same.  Why are guitar amps this way and how does it change the results, if any at all?  Also is there any significance with the screen voltage being 33% of the plate?

[HotBluePlates]

From what I am understanding, I have been using Power Transformers larger than needed. 

My example was for a 7w Class A output stage. If you need higher power output with a reasonable OT primary impedance, and available tubes, then high B+ voltage and current makes sense.

I steered away from a Class AB stage design because it is probably reasonable to start with a desired tube for the output stage, and with an available OT impedance, then calculate what the PT requirements will be. There's also a semi-painful process of figuring out d.c. power input to the tube vs. audio power output to determine in advance if the tube will red-plate in operation.

All of the above distracts from the simple fact I wanted you to see: Power Output is a product of Voltage and Current Swings, the possible Voltage Swing is pre-determined by the B+ voltage (and tube limitations), and the Current Swing is a result of the Voltage Swing acting across the Load Impedance (in this Class A case, a half-primary). This boils down to simply Ohm's Law and the Equation for Power.

The challenge is that the basic understanding presented above is then complicated by the fact tubes aren't perfect devices. The tubes allow you to use a voltage signal to allow current swings to take place through the OT primary impedance, resulting in a.c. voltage across across the primary. Once you have to stop and consider all the limitations imposed by the tubes being imperfect, those added complications distract from the fact the overall picture is pretty simple.

If the data sheet states a 360 plate / 270 screen voltage but in guitar amps the screen voltage is very near the same. 

So what?

The screen affects plate current, just as the control grid does. The screen is further from the cathode and not as tightly-wound, so it doesn't choke back plate current as easily as the control grid. But disconnect the screen from a voltage positive of the cathode and you'll nearly-kill all plate current.

If the screen voltage is reduced, the tube will have a lower maximum plate current for 0v on the grid than if the screen were at a higher voltage. More-complete data sheets show this with a set of curves where G1=0v, and G2 voltage is varied. In all tubes, higher G2 voltage results in higher plate current when G1 is 0v. Except for circuits where G1 is driven positive and draws grid current, the G1=0v curve respresents the peak current the tube can deliver at that screen voltage. This helps you determine supply voltage requirements for the screen when considering if a tube can swing the currents needed for your desired output stage (see what I mean about tube issues being "complicating factors" for a simple idea?).

... but in guitar amps the screen voltage is very near the same.  Why are guitar amps this way ...

It's easier to build a power supply where the screen voltage is very near the plate voltage.

Ideally, the screen is pinned to a tightly-regulated voltage. If screen voltage drops appreciably, plate current gets choked back. Screen current tends to rise at the moment plate current is reaching a maximum; if you derived your screen voltage through a dropping resistor, that extra current is drawn through the resistor, and supply voltage falls some amount. Great if you want sag, but not if you want to get every watt from your output stage.

Which is where a choke between the plate and screen supply nodes comes in... The choke has relatively low resistance (so low voltage drop at maximum screen current), but gives the filtering effect of a very large resistor. When an amp is big enough that the price point justifies spending for a choke instead of just a resistor, companies like Fender used a choke (which also happened to be on their big-output amps, where screen current draw could be more substantial).

The exception is very big output tubes (6550, KT88) using really high plate voltage (~600v). The tube might not be rated to stand the full plate voltage on the screen, and might deliver all the needed current at a screen voltage of half the plate voltage. Now you use a bridge rectifier on a PT winding with a center-tap, connect the CT to the mid-point of your series filter caps, and you have a convenient half-plate voltage for the screen. The filtering may not be as-good as having a dedicated supply just for the screen and preamp voltage, but it's a lot cheaper. Some circuits go whole-hog and actually have a 2nd power supply, but you usually don't see that in guitar/bass amps.

[HotBluePlates]

... how does it change the results, if any at all?  Also is there any significance with the screen voltage being 33% of the plate?

Changing the screen voltage moves all the G1 curves. With a set plate voltage, you might find a load impedance that promises the best output power (when drawing loadlines for 3 or 4 different impedances). But if you're designing a hi-fi, there might be too much distortion at the best-power load. Jiggering the screen voltage could change the amount of distortion (though it will also change the possible power output), so a hi-fi designer could find an ideal balance that gives low distortion, and then wrap that in feedback for even-lower distortion performance. The power supply complications may be worth enduring to them, because low distortion is the overriding consideration.

And hi-fi will endure a lot of hassle at times when something is perceived as intolerable. There are some who think the compromises inherent in output transformers are so evil that they'll forgo OT's altogether; see output-transformerless (OTL) amps. LOTs of big output tubes, run in a non-ideal manner from the standpoint of driving them, to get relatively low power output and the possibility of sending your speakers up in smoke. Not to mention more power supply hassles, and transferring output stage design headaches to the preamp/phase inverter/driver.

[PRR]

> 360 plate / 270 screen voltage

Both numbers are essentially datasheet MAX for the original 6L6, and mostly were not up-rated when better materials came into use.

33% is certainly not a magic number.

There is a *small* compromise designing a power pentode for Vg2 near Vp. In many other fields we use low-Mu tubes like 6146 which will make huge power with 500V on plate and 190V on G2. However the separate power supplies just annoy audio designers who already have a lot to think about.

Higher G2 is louder with less grid-blocking, so is common in guitar amps. It nominally reduces tube life, but what do we care? And all the tubes we buy today use tougher stuff.

5881 is rated for higher G2 voltage which makes 40+ Watt AB1 operation legal. Most of our "6L6" favorites are really 5881-spec designs. 6L6GC also has higher ratings (is a cheap 5881). These two types dominated the 1960s, so "all" modern 6L6 types should be up to this spec.

[Ed_Chambley]

The contributors to this thread did a great job,

Two more cents:  The png found in the first posting, containing, some wording, "Design Center Values", This is a conservative design rating. There are a couple of other less conservative ratings developed after this RCA sheet was printed.  Those less conservative ratings based on "Standards by RMA changes "pushed" the same tube published values, without changing the design of the tube.  IMO, those sheets are just as valid, but have caused confusion.  If you find an RCA 6L6 spec sheet of a later date with different design values, look at the specified rating system.  The tube didn't change, the standards used to provide design changed.  However, a 6L6a, spec sheet indicates the tube was changed, and the older design 6L6 could be a problem when replacing a 6L6a or later tube. 

It is my intent in this posting to add some useful info to this discussion.

I understand the data sheets could be updated, but not really what I am looking at.  I am actually been fascinated with Output Transformers and some of the cool things which can be done.

For instance I have a 58 Bandmaster with a Triad 1848.  If you look in the Triad catalog from that year the OT is listed as an Audio replacement.  Actually a stock item and sold for HiFi, not exclusive to guitar.  Now I hear all these stories about how cheap Leo Fender was, but question if this had the same impact as in the Golden Years of audio.  IOW, audio Output transformers were made with laminated designed for audio output.  I am reading that most winders are using the same core materials, laminates as they do with their Power Transformers, yet in the earlier days a softer cold rolled grain steel increasing the magnetic flux and reducing the magnetic saturation.

What I am currently doing is checking a circuit with different output tubes and have found something I do not understand yet.  I have a fender circuit designed for 5881.  Around 350Vdc plate.  I am glad PRR mentioned the 5881 because this is where I am hearing differences.

With USA Tung sol 5881 and RCA 6L6GB and a pair of 6L6GB philips.  While they all sound a little different, there is a consistent occurrence that does not happen with New Production tubes.  At low volumes the old tubes have what I call hair.  Sort of like a very slight fuzz coming from even order harmonics so I assume something in the preamp is doing it, but why will it not react the same as JJ6L6GC or some other newer ones I have.  At the same volume the "hair" is not present.  It is a Bandmaster so you want that unique tone.

I know these Russian tubes are some different, but how much?  Consider the JJ6V6.  I put a pair in a Super Reverb and they are running fine, but I would not consider a old 6V6.  These cannot be the same.  What I am trying to understand is if it is possible to "tweak" new production tubes to get this touch response? 

[HotBluePlates]

I am actually been fascinated with Output Transformers and some of the cool things which can be done.

For instance I have a 58 Bandmaster with a Triad 1848.  If you look in the Triad catalog from that year the OT is listed as an Audio replacement.  Actually a stock item and sold for HiFi, not exclusive to guitar.  Now I hear all these stories about how cheap Leo Fender was, but question if this had the same impact as in the Golden Years of audio.

Do you know which catalog you found the 1848? The Triad catalogs I've seen have different numbering conventions...

But to the point of Leo being cheap: Even if Fender bought one of Triad's hi-fi transformers, Triad was not king of the hill in those days. A hi-fi transformer from UTC would set you back much more, and be worth it. Jensen speakers were cheap compared to offerings from Altec and JBL. Leo's genious was in building the amp only as-good as it needed to be.

With USA Tung sol 5881 and RCA 6L6GB and a pair of 6L6GB philips.  While they all sound a little different, there is a consistent occurrence that does not happen with New Production tubes.  At low volumes the old tubes have what I call hair.  Sort of like a very slight fuzz coming from even order harmonics so I assume something in the preamp is doing it, but why will it not react the same as JJ6L6GC or some other newer ones I have.  At the same volume the "hair" is not present.  ...

I know these Russian tubes are some different, but how much? 

It would take a lot of measuring to figure out what quantifiable factor correlates with your listening experience.

[Ed_Chambley]

Do you know which catalog you found the 1848? The Triad catalogs I've seen have different numbering conventions...

I will look and follow-up.  I bought a box of old catalogs and this was one.


But to the point of Leo being cheap: Even if Fender bought one of Triad's hi-fi transformers, Triad was not king of the hill in those days. A hi-fi transformer from UTC would set you back much more, and be worth it. Jensen speakers were cheap compared to offerings from Altec and JBL. Leo's genious was in building the amp only as-good as it needed to be.

Sure, UTC is much higher grade.  When I speak of HiFi, I mean Dynaco, Heathkit.  You know, party amps.  This was told to me and I am not sure is it is not folklore, but the story is since we did begin to concentrate on High Fidelity that old guitar amp OT's were derived radio and then Hifi.  Not to spread misinformation, I am posting hoping you guys with more experience know for sure.

During the decline of tube amplification in favor of Solid State for the consumer market, the Output Transformer for the audio market became very small and the need for the materials fell as well.  Smaller production makes higher unit pricing.  Let me just ask it this way:
Is "average" workmanship and metals from the golden era superior to the majority of Audio transformers we purchase today.  Beside the obvious exceptions of some of the Audiophile equipment.

And yes, the Jensen was cost saving, but far from a discount speaker.  They are not EV's or JBL and the like.



It would take a lot of measuring to figure out what quantifiable factor correlates with your listening experience.

Understood, not simple.

Thanks for the response!

[HotBluePlates]

When I speak of HiFi, I mean Dynaco, Heathkit.  You know, party amps.  This was told to me and I am not sure is it is not folklore, but the story is since we did begin to concentrate on High Fidelity that old guitar amp OT's were derived radio and then Hifi. 

If I follow, you're speculating that guitar amps started with radio parts, then eventually moved up to hi-fi parts as those became available. Then onward to today, guitar amps stayed stagnant at early-hi-fi while audiophile stuff kept evolving.

This isn't wrong but might be an overstatement. I've got some McIntosh MC-30's from the early 50's. The OT passes 30w at full audio bandwidth (or close enough; there may be a bit of bass-shave on the bottom octave). See the pic below for a comparison of the size of the OT (on the left of the tubes) to a pair of what appear to be G.E. 6L6GC's. That OT is huge compared to any 50w guitar amp OT, though admittedly the potting can adds a bit to the perceived bulk.

By comparison, Fender used quite smallish OT's for their 2x 6L6 amps in the 50's, though they also did not need full power down to 20-30Hz. That allows the core to be smaller for the same rated output power.

Now, are the materials better/worse than the golden era? I don't know; I'd defer to folks PRR knows on other forums who wind their own transformers.

[PRR]

> same core materials, laminates as they do with their Power Transformers

There's no question. If you are going down near 50Hz, at more than a part-Watt, the ONLY practical core is power-transformer stuff.

Very-very-very rarely you will find Nickel in the mix; this improves initial permeability but may triple the cost.

PT iron has changed some in 50 years, but not much in ways that matter to us. The eddy-current loss is less. This acts as a shunt resistance. Even in 1950 this resistance was much-much higher than the load resistance.

If you can hear the difference between a vintage 6L6 and a modern 6L6, maybe you can hear differences between vintage/modern transformers. But tracing the difference to a reason may be impossible.

> McIntosh MC-30

Mac was something else. However the core-iron was just a extra-thin lam of standard power transformer iron.

> See the pic below

Lost in the innernet.

[Willabe]

Mac was something else. However the core-iron was just a extra-thin lam of standard power transformer iron.

Why did they use 'extra-thin lam'?


                Brad   

[Ed_Chambley]

> same core materials, laminates as they do with their Power Transformers

There's no question. If you are going down near 50Hz, at more than a part-Watt, the ONLY practical core is power-transformer stuff.

Very-very-very rarely you will find Nickel in the mix; this improves initial permeability but may triple the cost.

PT iron has changed some in 50 years, but not much in ways that matter to us. The eddy-current loss is less. This acts as a shunt resistance. Even in 1950 this resistance was much-much higher than the load resistance.

If you can hear the difference between a vintage 6L6 and a modern 6L6, maybe you can hear differences between vintage/modern transformers. But tracing the difference to a reason may be impossible.

> McIntosh MC-30

Mac was something else. However the core-iron was just a extra-thin lam of standard power transformer iron.

> See the pic below

Lost in the innernet.

Thanks for the information and sticking in with my insanity.  I sort of get obsessive and a few weekends ago I was in a guy's shop and he was restoring a 1960 Tweed Deluxe.  Man it looked nice.  Said someone put 6L6 in it and it took the transformer out.

Well he rewinds them and I did not know it.  I have taken them apart and fixed them, but had no understanding of what even "interleaved" meant.  The "I" and "E" of it.  To see the nylon bobbin compared to paper.  He simply had a motor with a counter.  Within about an hour he had a rewound OT.

Something about seeing it changed my thoughts on them.  Sort of demystified it.  Above all it was fun to actually begin to understand.

[HotBluePlates]

Lost in the innernet.

Yep, the inter-webs were fine... I just forgot to link the photo!