Voltage level question

[VMS]

Hi guys,

I probably should know this by now but I'm drawing a blank here.

What is the maximum output voltage from the PI on the following circuit and is it enough to drive a pair of EL84's or ECC99 triodes on push-pull.

I know the gain in various stages but haven't quite yet figured the voltage levels and what are the limiting factors for max voltage.

[HotBluePlates]

What is the maximum output voltage from the PI on the following circuit ...

That depends on the B+ (which is not indicated).

.... is it enough to drive a pair of EL84's or ECC99 triodes on push-pull.

This schematic fragment looks like a 5F6-A Bassman, with a 5E3's split-load inverter grafted on.

A 5E3 can drive the 6V6's it came with, at the bias voltage present in the 5E3 (and with the B+ voltages present throughout the 5E3). EL84's are similar enough to 6V6's, except they generally have less bias voltage, so they're easier to drive.

I have no experience with ECC99's, but expect the bias to be even lower, and therefore requiring less drive voltage from the preamp/inverter. So this thing should work, but I don't know how well yet.

Guitarists always focus on the preamp first, and this is a mistake. An amp design always starts with the output tubes and output transformer, and has to assume a reasonably available supply voltage (which means you need to have some good idea of what power transformers are available and suitable for your output section).

Once all the details of the proposed output stage are known, the bias voltage sets the drive requirements that the phase inverter must deliver. You have to plan on the phase inverter being able to easily drive each output tube grid from its idle bias voltage to a peak of 0v. That means if the bias is -18v (like the 18vdc across a 5E3's cathode resistor; look at the 5C3 schematic to actually have voltages shown), each output of the phase inverter must be able to deliver 18v peak output.

Then you consider the power supply and a reasonable drop in voltage from the output tube plate node to output tube screen node to phase inverter plate resistor, and see if the guesstimated B+ voltage after those drops is sufficient to get the required output voltage from the phase inverter type you selected. If you find the phase inverter can't make the output required with the existing design and supply voltage, you either have to select a different inverter or a different B+ and start the design over.

If this sounds like a lot of work, it is (for a well-designed amp). I'm maybe 1/4 of the way through reading a ~50-page thread on another forum for a scratch design. There were a lot of sharp folks contributing to that, so all those pages aren't a lot of fluff. It came down to design, error-finding, discovering wrong assumptions, re-re-re-design, tweaking, considering failure modes, etc.

[VMS]

Thanks HBP.

The reason that this looks like a backwards design is that I wan't to know if I can leave out the gain stage that is usually in front of the split-load PI.

Lets say that I have enough gain stages in front of the 12ax7 driving the tone stack that I can overdrive it. Using the voltages on the picture I would like to know what is the voltage level (in ballpark) on points A, B and C. So basically how much the signal attenuates on tone stack and on PI?

 

[HotBluePlates]

I used the Duncan Tone Stack Calculator to figure the amount of attenuation in the tone stack (using the Marshall sim for the cathode follower source impedance, but with Fender parts values). With all controls at 0, everything is cut at least 15dB. With Treble/Bass at 5 and Mids at 0, midrange is -23dB but treble and bass are only -7-8dB. So we'll take -15dB as an average tone stack attenuation.

For voltage, 20 log (0.1778/1) = -15dB.

So tone stack "gain" is 0.1778, or prior voltage leveled divided by 5.6.

The phase inverter has no gain, and typical loss will be something like 0.98. You could figure unity gain for that and be fine.

In other words, whatever peak voltage is needed at the output tube grids (numerically equal to bias voltage), you'll need slightly more than that at the input to the phase inverter. An AC-30 shows 12.5v across its cathode resistor at full power, so you'll need 12.5v-13v at the input of the phase inverter for EL84's at similar operating voltage as an AC-30.

Using the voltages on the picture I would like to know what is the voltage level (in ballpark) on points A, B and C.

Good initiative. But I think you took those voltage from a 5F6-A Bassman, which will be unrealistically high for EL84's and ECC99's. I say that because the voltage of the filter cap feeding the phase inverter is not listed, and is 385v in the Bassman. We have to know that voltage to know the Vout capability of the split-load inverter.

So let's work backwards and make some guesses. You'd probably use a B+ between 300-350v for EL84's, so let's pick 350v at the EL84 plates. Voltage drop across a choke will be minimal, so the screen node will probably be around 345v.

Without recalculating everything for the preamp with assumed voltages, let's figure where the B+ nodes will probably land. In the 5F6-A and your schematic fragment, the voltage drops from 385v at the phase inverter filter cap to 325v at the preamp filter cap. 385v-325v = 60v -> 60v/10k = 6mA drawn by the preamp at the 325v level.

Looking at a Princeton Reverb schematic, the phase inverter draws about 1mA with a 240v supply (~1v is shown across the bias resistor of the inverter).

We might need 10k or so as a dropping resistor from the screen node to the phase inverter. 6mA + 1mA = 7mA, so 70v would be dropped across 10kΩ. 345v - 70v = 275v. The preamp filter cap will then likely have 275v - 60v = 215v.

Now, I could plot a loadline for the split-load inverter and figure out voltage output capability, but I'll note the Princeton Reverb has 240v as its phase inverter supply node, and can drive 6V6's with -34v of bias. So we'll almost certainly have sufficient drive for EL84's.

Double-check: a split-load inverter usually idles so that the plate is at 2/3 B+ and the cathode is at 1/3 b+. That puts the cathode around 91v and the plate around 184v. There should be sufficient voltage swing available for 10-13v peak output swing as long as the tube is not seriously mis-biased.


Still going with the 13v guess for EL84 bias (yes, I keep rounding up a bit), input to the phase inverter needs to be at least 13/.98 = ~13.25v peak.

Tonestack imposes a 15dB loss, so output of the cathode follower needs to be at least 13.25v/0.1778 = 74.5v peak.

The cathode follower itself probably has a gain around 0.98, so output of the prior stage needs to be 74.5v/0.98 = 76v peak.

By all means, figure actual stage gains using plate curves, but I'll now assume a gain of ~55 for each 12AX7.

The 12AX7 stage ahead of the cathode follower needs 76v/55 = 1.38v of input to make 76v output (whether the stage can manage this should be verified with curves and a loadline, and the 215v supply we figured earlier).

You'd like your amp to hit full clean output power at something like 5 or 7 on the Volume control, to account for weaker pickups or soft strumming. For audio taper, half-up is 10% output (although there are 20% and 30% tapers out there). So full output happens at 1.38v/0.1 = 13.8v out from the first preamp stage if we want the onset of output stage distortion around half up on the volume control.

13.8v/55 = 0.25v input at the first preamp tube. Since we figure guitars make ~100mV output, this is pretty low sensitivity. However, realistically, this means that the amp is still clean at 5 on the volume control, so you only need to turn up a bit more to get sufficient voltage. If you put this in a real amp or calculate looking at a log taper chart, you'll hit the needed voltage level between 7-8 on the Volume control.

As this shows, work from output backwards to input. You can't get direction for where you want to go without specifying the destination; you can't realistically design/check a preamp's ability to drive the output stage without defining the output stage.

[VMS]

Thanks again HBP. I have to save this info to my scrapbook.

The 12AX7 stage ahead of the cathode follower needs 76v/55 = 1.38v of input to make 76v output (whether the stage can manage this should be verified with curves and a loadline, and the 215v supply we figured earlier).

If I could ask one more favor, is there a simple way to determine what limits this 76v output other than the input signal?

Haven't quite figured out the loadlines yet.

[HotBluePlates]

... is there a simple way to determine what limits this 76v output other than the input signal?

Haven't quite figured out the loadlines yet.

Loadlines are the only way to avoid pure guesswork.

So I generated a loadline for the stage prior to the cathode follower and assumed a 215v supply voltage (as I'd figured earlier), a 100k plate load resistor and an 820Ω cathode resistor as in the original schematic.

The red line in the drawing is the 100k plate load for a 215v supply. The green line is the voltage/current across the cathode resistor. Where they cross is the point the tube will idle at with these resistors and supply voltage.

There is a middle blue line to show what voltage the plate will be at idle. The blue line to the left shows what the plate voltage will be when the grid is driven momentarily to 0v by a positive input signal. Because the tube idles at ~0.8v on the cathode, the maximum input signal in the positive direction is 0.8v peak. The blue line to the right is 0.8v peak input in the negative direction.

Output voltage is indicated by the resulting plate voltage at each of these two points (the outer blue lines). It drops by 52v with a 0.8v peak input, and increases by 40v for a -0.8v peak input. That the change in plate voltage is different in either direction indicates some amount of distortion. the total change of plate voltage divided by total change of grid voltage gives voltage gain; in this case 92v/1.6v = 57.5.

So we can't get 76v of output from this stage with the 215v supply voltage. The answer would be to raise the supply voltage some amount.

By the way, I should note I overlooked that I said we needed over 1.3v of input to this stage for 76v output earlier... The voltage indicated on the schematic at the cathode was 1.2v, which means there's no way to drive it with 1.3v+ cleanly.

But this may not matter: Remember when I calculated tone stack loss with all controls at 0? When was the last time you played your amp that way? I picked that because it was a worst-case, but now maybe I need to figure with less loss in the tone circuit.

If the tone circuit results in an 8dB loss (as I mentioned was typical for controls half-up except midrange), that equates to multiplying by about 0.4 instead of 0.1778. Our figured 13.25v peak requirement at the input of the phase inverter implies a needed output from the cathode follower of 13.25/0.4 = 33.125v peak.

Now, this is well within the capability of the stage prior to the cathode follower. What does this tell us? If you turn all tone controls to 0, you won't be able to drive the output tubes to distortion, because the 2nd gain stage can't manage the output and distorts first. But with the tone controls turned up more, the 2nd gain stage easily provides enough output.

The other lesson to be had is everything interacts with everything else.

We haven't calculated everything and checked that every stage works exactly at its optimum, but we see the design will produce a functioning amp.

[PRR]

> I want to know if I can leave out the gain stage that is usually in front of the split-load PI.

Assume the amp you are plagiarizing is "good". (Hmmm.... this seems to be a paste-up job?)

EL84 need about half the drive of 6V6.

Therefore you could remove a stage which had a gain of 2.

ECC99 can be run *many* ways. However using cusomary voltages, it will need a drive somewhere between 6V6 and EL84. You can remove a stage witg gain of 1.5 or so.

Your preamp stages all have gains of 20 to 50. Not 1.5 or 2. You can't remove a stage. You may want to pad-down signal; but a factor of 2 excess gain is usually best left in.

It is very difficult to drive a "power" bottle with just one gain stage. Triode does not give enough gain. Pentode does, but tends to overload. First Champ did it, and I think that's why 2nd Champ went to two triodes with gain pot between.

Also: the usual power bottles *were designed* to drive well from a voltage amplifier driver fed slightly lower B+. (Contrast with 2A3, which run at 250V needs more like 350V on its driver to smack the full output.) EL84 is even easier to drive. However if you put attenuation (such as a tone-stack) between driver and power bottle, you can NOT smack full output.

As a ballpark, a well-biased triode voltage amplifier can deliver peak output about 20% of its B+. So for 225V supply you can get up to 45V of output. This will directly drive the usual bottles at 300V-400V supply. However put 5:1 attenuation between driver and output, you can only get 9V peak. Which is not enough to fully smack the usual outputs.

Yes, this is commonly done to favor preamp-clipping at the cost of not-full-Watts. For bedroom amps this may be a valid path.

[HotBluePlates]

.. I wan't to know if I can leave out the gain stage that is usually in front of the split-load PI. ...

I see 5 triodes in your drawing out of the 6 available from your three 12AX7's.

Why can't we still have the gain stage in front of the split-load, since the extra triode is available? It would guarantee full output from EL84's and the ECC99, and the extra triode is just sitting around anyway...

[VMS]

Assume the amp you are plagiarizing is "good". (Hmmm.... this seems to be a paste-up job?)

I see 5 triodes in your drawing out of the 6 available from your three 12AX7's.

Why can't we still have the gain stage in front of the split-load, since the extra triode is available? It would guarantee full output from EL84's and the ECC99, and the extra triode is just sitting around anyway...

Yes, I should clarify that the drawing I put together is just an example of CF driven tonestack in to a split-load PI (fender circuits was the fastest way to do this).

...and the question was what can I expect for the PI output voltage.

And thanks to you guys now I'm a lot smarter.

Except that I think my written english getting worse. I probably should practice it by writing more on this forum.


Anyway the front end of the preamp that I'm thinkin of building is going to be different than what is on the drawing.

[HotBluePlates]

Okay, cool then.  

If you're just reaching for some general rules regarding that phase inverter, consider this rule-o-thumb:

The split-load by itself has no gain, due to the large cathode resistor needed as a load for one of the outputs. You add a pre-gain stage in front of it, and now the two triodes have the overall gain of about a single normal gain stage.

The long-tail pair has gain, but not as much as you'd think. The non-driven triode of the long-tail gets its input signal by way of cathode-coupling from the driven triode; this method of coupling causes the long-tail's triodes to need twice as big an input signal for a given output voltage as you'd normally have with a single-triode gain stage (take this fact on faith for a moment; I can explain it further if you want later). That you have 2 triodes each with half the gain they'd typically exhibit in a normal gain stage, the pair gives a total output voltage equal to a single normal stage.

Huh? What does that mean?   Plain english, please!

These two facts mean that in general the long-tail phase inverter will have roughly the same gain as a split-load when the extra triode ahead is used.

 

So why use a long-tail over the split-load?

It boils down to how big the voltage output has to be, and how much supply voltage you have available. The split-load has to develop 2 output signals and still leave voltage across the triode, all at the same time. All 3 of those voltages have to add up to equal or less than the supply voltage feeding the split-load.

While the long-tail does have its cathode at an elevated voltage which subtracts from possible output signal size, each triode only needs to develop one voltage.

Some feel that the overload characteristics of the split-load are ideal for big rock amps. There are workarounds for that. Ultimately, the split-load was generally used (*) for smaller guitar amps with small output tubes with smaller bias voltages needing smaller drive signals. The long-tail was used more often for bigger amps with bigger tubes and bigger bias voltages. Most everyone later just copied what they saw work in earlier amps.


*Funny enough, it was also used in some of the biggest, highest-powered guitar and bass amps, and a lot of hi-fi stuff. The designers of those however added other extra bits to the circuit to facilitate what they were trying to accomplish. They didn't just slap in a split-load and call it done.

[Ed_Chambley]

The long-tail pair has gain, but not as much as you'd think. The non-driven triode of the long-tail gets its input signal by way of cathode-coupling from the driven triode; this method of coupling causes the long-tail's triodes to need twice as big an input signal for a given output voltage as you'd normally have with a single-triode gain stage (take this fact on faith for a moment; I can explain it further if you want later). That you have 2 triodes each with half the gain they'd typically exhibit in a normal gain stage, the pair gives a total output voltage equal to a single normal stage.

[/quote]
Somewhere in this would explain why the LTP PI normally has different resistances on the plates.  In every Marshall I have built has a 100k and 82k plate resistors.  I have never understood this since it has a common Cathode resistor of 470 ohm.   Looking at the 1987 Marshall schematic I have never been able to determine exactly why.  I have been told it balances.

Comment?

[sluckey]

Somewhere in this would explain why the LTP PI normally has different resistances on the plates.  In every Marshall I have built has a 100k and 82k plate resistors.  I have never understood this since it has a common Cathode resistor of 470 ohm.   Looking at the 1987 Marshall schematic I have never been able to determine exactly why.  I have been told it balances.

Quoting Aiken... When only one signal input is used (ignoring feedback inputs) R1 is usually made 10% - 20% lower than R2 to compensate the unbalanced gains of the two tube sections and make the two output amplitudes equal.

For more detailed info read... http://www.aikenamps.com/LongTailPair.htm

[HotBluePlates]

... explain why the LTP PI normally has different resistances on the plates.  ...

Quoting Aiken... When only one signal input is used (ignoring feedback inputs) R1 is usually made 10% - 20% lower than R2 to compensate the unbalanced gains of the two tube sections and make the two output amplitudes equal.

Aiken's right, but he dances around why gain is unequal in the first place, which is what Ed really wants to know.

I know I'd read the reason many times, and likely PRR has told me numerous times; however, it didn't stick until an explanation by Dave Gillespie that I read on another forum. So credit for the paraphrase below goes to him.

The long-tail is based on a differential amplifier. This circuit receives two equal and opposite signals on its grids and develops equal and opposite output signals at the plates. The action of the differential amplifier is enhanced when there is a constant-current source (CCS) between cathode and ground. The CCS strives to keep the total current through it constant at all times. So when one triode receives an input which causes its current to increase, the CCS insures that the other triode's current falls by the same amount. The resulting equal and opposite tube currents, when used with equal plate load resistors, causes equal and opposite output voltages at the plates.

Good constant current sources are made with active devices. In the old days, they might have been pentodes because their high plate resistance means a changing voltage at their plate causes almost no current change through the pentode. A modern CCS would be made of at least a FET, and likely several solid-state devices or a chip to get ever more-perfect function.

You can kinda use a large-value resistor to fake a CCS if you have a very big supply voltage (or an additional negative supply). Guitar amps have neither of these, and don't really use resistors big enough to fake a "good" CCS.

Things change when you drive only one input, as in a long-tail pair. That cap from the grid of the non-driven input (often 0.1uF) runs to a point which is equivalent to ground. If you only drive one grid, where does the drive signal from the other triode come from?

It is effectively coupled from the cathode of the 1st triode to the cathode of the 2nd triode; by causing the cathode voltage to change with respect to ground, you change the grid-to-cathode voltage of the 2nd triode, and induce a current in that triode. You cause that cathode voltage change by changing the resistor through the (typically 470Ω) bias resistor.

But wait... if the tail resistor is a fake CCS, which keeps total current in the tail constant, then how does a changing current appear through the bias resistor to induce a signal in the 2nd triode?

In order to make things work, the 1st triode has to have more current flowing under signal than the 2nd triode, in order to develop the drive signal for the 2nd triode. And it's because the 1st triode's signal current is greater, it's plate resistor must be smaller to arrive at equal voltage outputs at the plate. It's ohm's law.

In the end the way to understand how it works is to figure out how it must not adhere to standard theory in order for things to allow it to work. That's why it's hard to see what's happening based on what you know from normal preamp stages.

[Ed_Chambley]

Thanks Sluckey and HBP.  It is so strange how easy it is to understand something when you get it.  One more question.  What does the pf cap across do?  I have built with and without a cap and did not notice a great difference, but I have never tried different values in the same amp to listen.

[HotBluePlates]

What does the pf cap across do?  I have built with and without a cap and did not notice a great difference, but I have never tried different values in the same amp to listen.

Reduces highs, but that cap is usually small enough to reduce highs beyond the range of hearing.

1 + (-1) = 0

The two outputs are equal and opposite, right? So the cap is a path of "least impedance" from one output to the other for high enough frequencies, and cancels them out.

Really, Fender noticed that when they built their amps, some had oscillation and some didn't. In general they were using grid stoppers. So as a production expedient, they added that cap to kill frequencies in the range of the oscillation.

If small enough, it has no audible effect but the amp doesn't oscillate. If too small, it doesn't reduce the oscillation effectively. If bigger, you'll start to hear it slightly reduce extreme highs. Bigger still and the reduction extends lower and lower.

Also, this is the basic idea behind the Vox Cut control, except they use a set frequency (set cap) and vary resistance to put the cap in or out of the circuit.