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Old 9th June 2011, 05:56 PM   #1
rongon is offline rongon  United States
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Default Driver stage to finals newbie questions - (I hope not dumb questions)

OK, I'm going to show my ignorance here, but I have a question that's been nagging at me for a long time. I don't think I understand the concept behind its proper answer. So I'm asking for help.

I'm trying to learn how to predict if a driver stage can drive an output stage to full output with low distortion. The fundamental part of it is voltage swing -- When do I have enough?

I hope the concept is similar for a SE driver into SE output as compared to a PP driver into PP output stage.

I've read and re-read MJ's Valve Amplifiers. The output volts rms vs. pk-to-pk stuff totally confuses me.

So, to get to it... If I have 2 volts rms of input (like the output of a CD player), do I need to convert that to peak? Do I multiply 2Vrms by 2(square root of 2)? That's close to 2.83, so that would be:

2Vrms(2.82) = 5.66Vpk

If that's correct, then I will assume that the grid bias of the input of my driver stage must be at least a volt more than that, to avoid driving the driver stage triode into grid current. So let's say a grid bias of -7V.

Let's say my triode driver stage has a Mu of 15 (like a 6N6P or 5687 with a plate choke).

Now, let's say my 5687 is driving a triode wired 6L6. So the grid bias is -45V.

I have 5.66Vpk at the driver input (grid), times 15 (Mu), equals 84.9Vpk.

Looking at the plate curves for a 5687 with Va = 165V, Vg = -7V, and Ia = 15mA, it looks like the 5687 w/ choke load (flat loadline at 15mA) can swing +/- 100V (down to 65V with -1V bias, up to 265V with -14V bias).

So from that, it looks like the 5687 can swing a lot more volts than can be accepted by the 6L6-triode's grid. That means we have plenty of headroom and everything's good up to that point... Right?

Did I do that right? I'm not so sure... But one the assumption that I did:

- How do I convert the above to Push-Pull operation?

Let's say (for example) a 5687 differential phase-splitter/driver into a class A, PP pair of 6L6-triode.

__________________________________________________ _______________________________

I have an amp that I think sounds very good, but it clips in a rather harsh way. The amp's very simple:

- 6N6P differential phase-splitter/driver with split load choke in the plates, CCS in the tail, with a -30V supply for the tail.

- The 6N6P plates are DC-coupled to a pair of 2A3's.

6N6P op points are Va = 165V, Vg = -7.5V, Ia = 13mA (per triode section), Mu = approx. 15 (I'm lowballing it a little, just to be safe)

2A3 op points are Va = 275V, Vg = -40V, Ia = 55mA (per triode).

The OPT is a Tango XE-45-5. I'm using the 5kOhm primary (there are also 3.5kOhm primary taps).

I use a line stage in front of the amp with plenty of output swing for the purpose. Way overkill.

I get 6Wpc from the amp. It looks really good on the scope, but it clipped with a loud scrunch on peaks when it was trying to drive my friend's fairly inefficient speaker setup (four Jordan JX52's per side) in his rather large room.

My question is, did I goof and the 6N6P driver clips before driving the 2A3's all the way into full power? Or is 6Wpc all I'm gonna get, and the harsh clipping is just the 2A3's giving out in that big room?

Feel free to just point me to a good explanation if you know of one, or even tell me to re-read a particular part of MJ's book, if that's what I need.

Thanks in advance for any guidance on this.

-=|=-
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Old 9th June 2011, 07:12 PM   #2
tomchr is offline tomchr  United States
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I suggest posting a schematic. It's much easier to talk about circuits when you can refer to a schematic.

Anyway. Let's clear out this RMS vs peak stuff first. Wikipedia has a good write-up on sine waves. So a sine wave has an amplitude 'A' (Vpeak or Vp). That's all fine and good. So why bother with this RMS stuff? Well, it's mostly relevant when calculating or measuring average power. For a DC waveform, the power dissipated in a resistor for a given voltage is P = V^2/R. A DC waveform is constant versus time. A sine wave voltage is not constant with time (the amplitude may be constant but the absolute value of the voltage does vary). So to calculate the power dissipated in a resistor by a sine wave voltage, you need to calculate the Root-Mean-Square. A sine wave with an amplitude of, say, 2 Vrms will dissipate the same power in a resistor as a 2 V DC. I.e. Pavg = Vrms^2/R. In fact that's how some true RMS voltmeters are built - they measure the temperature rise of a resistor... It turns out that for a sine wave, the ratio between Vp and Vrms is sqrt(2) = 1.414. I.e. Vrms = Vp/sqrt(2).

In amplifiers, confusion typically arise because the input voltage is often given in Vrms and circuit designers are used to flipping back and forth between Vp and Vrms.

I forget the numbers for your amplifiers, but here is one example from a 300B amp that I'm working on:

The 300B is biased to Vgk = -80 V. As a rule of thumb, I prefer to run in class A1 (especially with 300B and the like), so I want to keep Vgk less than 0 V at all times. This limits the signal swing on the gate to 80 Vp (0-(-80) = 80).
So say, I use Vgk = -80 V bias; and a 80 Vp sine wave from the driver. Then Vgk would swing from 0 V to -160 V (the sine wave swings from +Vp to -Vp, remember). I need to make sure that the 300B is happy under those conditions.
OK, so I've checked the data sheet and load line and the 300B is happy. I now need a driver circuit that can provide 80 Vp to the 300B when provided with 2 Vrms from a CD player. What's the gain of the driver stage? Well... The gain is Vout/Vin, right. Vout = 80 Vp. Vin = 2 Vrms. 2 Vrms = 2*sqrt(2) = 2.82 Vp. So the gain of the driver should be 80/2.82 = 28.4 V/V.

Does this help?

~Tom
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Old 9th June 2011, 08:07 PM   #3
rongon is offline rongon  United States
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Thanks for that, I think it does help!

The math in that Wikipedia article escapes me, but I think I get the idea on peak to peak amplitude (that's pretty easy to visualize). RMS is harder to really grasp, but it's not too much for me to get (at least enough to work with it).

I won't be able to send a schematic of the amp until I get home.


OK, if I take your 300B example and plug in 5687 --> 2A3 numbers...


I have a 2A3 with a bias of -43.5V. I don't want to drive it into any kind of grid current, just class A1.

The Vgk of the 2A3 can swing from 0V to -87V and the load is 2500kOhms. That is equivalent to "43.5V peak to peak" -- correct?

How much gain do I need to drive that 2A3 from the 2Vrms CD player output? (Now I convert 2Vrms into the peak voltage)

output Vp/source Vrms = amplification needed, or

43.5/2.82 = 15.42


It looks like my 5687 has just *barely* enough output swing to drive the 2A3 to full power from 2Vrms at its input. If I used a 6SN7 with a mu of 20, I'd add a little headroom. Correct?

However, if I put another voltage amplifier stage before the 5687, I can then bias the 5687 so it can swing way more than 43.5Vp. Let's say my 5687 is biased so that it has -8Vgk.

What would the maximum clean swing be from my 5687 with -8Vgk?

The 5687 will likely draw grid current at -1Vgk, and it will start to distort if the grid sees over 7Vp. So the input signal at the 5687 grid could be limited to 6Vp to minimize distortion, correct?

Does that mean that the 6V peak at the 5687 grid, multiplied by the amplification factor of 15, equals a peak output voltage of 90V from the 5687?

That seems like a lot of volts. That 5687 with a Vgk = -8V should be able to drive a 2A3 with Vgk = 43.5V very cleanly. Correct?

-=|=-
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Old 9th June 2011, 11:06 PM   #4
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Yew, BUT all these assumptions are correct if you do not have a global negative feedback loop, terminating at the cathode resistor of the first stage. In that case, the feedback voltage will counteract the input voltage (how much depends on the feedback ratio) and the biasing will not need be as "deep". That is the reason you have seen schematics where the first stage bias is at -1 Volt or so. This thing had confused me greatly in the past.
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Old 9th June 2011, 11:25 PM   #5
tomchr is offline tomchr  United States
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Quote:
Originally Posted by rongon View Post
The math in that Wikipedia article escapes me, but I think I get the idea on peak to peak amplitude (that's pretty easy to visualize). RMS is harder to really grasp, but it's not too much for me to get (at least enough to work with it).
Well... Visualize a rectified sine wave with an amplitude of A = 1 Vp. Now - super-imposed on this plot - visualize a DC with an amplitude of 0.707 V (1/sqrt(2)). You'll notice that the area between the sine wave curve and the X-axis will fit nicely under the curve for the DC. That's what RMS amounts to. In visual terms.

Quote:
Originally Posted by rongon View Post
I have a 2A3 with a bias of -43.5V. I don't want to drive it into any kind of grid current, just class A1.

The Vgk of the 2A3 can swing from 0V to -87V and the load is 2500kOhms. That is equivalent to "43.5V peak to peak" -- correct?
Draw a graph of it. Then tell me the peak-to-peak voltage. It's not 43.5 V. Recall Vpp = 2*Vp.

The 2500 ohm load is on the 2A3 anode and has nothing to do with the grid swing.

Quote:
Originally Posted by rongon View Post
How much gain do I need to drive that 2A3 from the 2Vrms CD player output? (Now I convert 2Vrms into the peak voltage)

output Vp/source Vrms = amplification needed, or

43.5/2.82 = 15.42
You did the math right, but the text wrong. It should be output Vp/source Vp. Your 15.4 is correct, though. A CCS loaded 6J5/6SN7 will do that....

Quote:
Originally Posted by rongon View Post
If I used a 6SN7 with a mu of 20, I'd add a little headroom. Correct?
Didn't see that you were one step ahead of me already... Yep. With a mu of 20, a 6SN7 will deliver a gain of about 20 V/V with an infinite plate load. The closest you can get to an infinite load is a CCS. An IXYS 10M45 set to 10~12 mA works real well for this.

Quote:
Originally Posted by rongon View Post
However, if I put another voltage amplifier stage before the 5687, I can then bias the 5687 so it can swing way more than 43.5Vp.
Yep. If you really want to use that tube, you can cascade multiple stages to get the gain up. Same math applies.

Quote:
Originally Posted by rongon View Post
Does that mean that the 6V peak at the 5687 grid, multiplied by the amplification factor of 15, equals a peak output voltage of 90V from the 5687?
Sounds about right. And if the tube can deliver this cleanly, you might have a good amplifier. Although, a rather sensitive one. To lower the sensitivity (= gain) of the amp, some people employ global negative feedback or various forms of local feedback. But I think that's a topic for another day.

~Tom
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Old 10th June 2011, 09:10 AM   #6
rongon is offline rongon  United States
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Quote:
Draw a graph of it. Then tell me the peak-to-peak voltage. It's not 43.5 V. Recall Vpp = 2*Vp.
Hmmm... I guess since it's a single-ended input it's just peak voltage. Only goes from 0V to -43.5V, not -43.5V thru 0 to +43.5V (which would be 87Vp-p).

Quote:
Quote:
Originally Posted by rongon
43.5/2.82 = 15.42
You did the math right, but the text wrong. It should be output Vp/source Vp. Your 15.4 is correct, though. A CCS loaded 6J5/6SN7 will do that....
Of course. Vp needed at output tube grid divided by Vp at driver grid.

I spent a little time reading MJ about the Scrapbox Challenge SE amp; specifically how he went about designing the driver stage. He chose a pair of 6J5 in SRPP (A = 14) driving a 6528 triode with Vgk of -25V. He was aiming for full power from an input of 1.3Vrms. He kept everything in rms terms, so he converted that 6528 -25Vgk to its rms equivalent and came up with 18V. I guess that's 25(0.707) = 17.675 (close enough). So...

1.3Vrms input, times 14 (gain of 6J5 SRPP) equals approx 18Vrms at 6528 input.

If using a 2A3 as an example,

43.5Vp = approx 31Vrms

31Vrms/15 (approx amp factor of 5687) = 2.07Vrms approx input to 5687 needed

Or put another way...

31Vrms/1.5Vrms (desired input V to full power) = 20.7 (desired amp factor)

Jeez, I should've used a 6SN7! I might try a 6FQ7 there, just to see how it works. I originally chose a 5687 because I wanted to use a driver tube with a low anode resistance (Ra), figuring the 2A3 triode would have a large Miller capacitance. I later switched it to a 6N6P, because I wanted to try them (I like them). I figured I'd be using a line stage anyway, so didn't need a highly sensitive power amp section. So I have a single-ended line stage (with some excess gain) driving a low-sensitivity PP power amp. In the end, the system works so that most CDs play back at 'moderately loud' level with the volume control turned up about half-way to 60%.

Here's a schematic of the line stage and power amp (sorry about the sloppy drawing):

Click the image to open in full size.

I didn't show the power supplies (those are the most complicated parts).

Keeping everything in rms terms, if I need 31Vrms to drive the 2A3's, then I need 2.07Vrms to drive the 6N6P's (mu = approx 15). That means the 6DJ8 line amp (mu = 25?) only needs 83mV rms at its grid (input) !

Hmmm... There must be losses at work here, because the system doesn't seem overly sensitive in use... How much insertion loss occurs in the OPT? -3dB? Or maybe it works because I'm not using super-efficient horn loudspeakers or anything like that (I'm using a pair of old Tannoy T185, rated at 91dB/1w@1m).

I guess that's another topic...

Thanks for the help. I'll try to get all this to sink in!

-=|=-

Last edited by rongon; 10th June 2011 at 09:35 AM.
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Old 12th June 2011, 12:29 AM   #7
tomchr is offline tomchr  United States
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Quote:
Originally Posted by rongon View Post
Hmmm... I guess since it's a single-ended input it's just peak voltage. Only goes from 0V to -43.5V, not -43.5V thru 0 to +43.5V (which would be 87Vp-p).
If the output tube is biased to -43.5 V, the max swing in A1 would be the max sine wave that would keep Vgk negative while being centered at -43.5 V. So the max sine wave would swing from Vgk = -87 V to 0 V.

Quote:
Originally Posted by rongon View Post
1.3Vrms input, times 14 (gain of 6J5 SRPP) equals approx 18Vrms at 6528 input.
MJ tends to use resistive loads, hence, the gain of 14 vs the mu = 20 for 6J5.

Quote:
Originally Posted by rongon View Post
Here's a schematic of the line stage and power amp (sorry about the sloppy drawing)
No worries on the schematic quality. I'd probably cap couple (or transformer couple) the input. That would allow you to lose the -15 V bias on the input stage.

~Tom
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Old 12th June 2011, 12:50 PM   #8
tricomp is offline tricomp  Denmark
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I'm by no means a tube expert so mostly I have questions about your design, not answers.
First of all; aren't you wasting a lot of energy running the 2A3's essentially with a supply-voltage Anode-Cathode of 252V feed from 460V?
Then you worry about the Miller-effect. Shouldn't you drive the 2A3's from a Cathode-Follower if that's a problem?
Looking at the RCA data-sheet for the 2A3 is states that the load resistance (Plate-to-Plate) is around 3K5 with fixed bias and a Plate voltage of 300V, class AB1 P-P.
You are running at 252V but it should still be in the vicinity of 3K5. RCA expects 15Watt of output in that situation. Why not try the other tabs on your OPT?
I see that you are trying to accomplish everything in a minimalistic way. Why not simply add a triode stage in front of the LTP if you need more amplification? It will leave you with dB's to apply NFB and that's good for lowering output-Z and distortion.

Hope I'm not way off

rgds,

/tri-comp
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Old 13th June 2011, 01:45 PM   #9
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the schematic looks like a monkey.

tricomp, I don't think the excess voltage is being wasted, it is used to run the input/driver stage.
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Old 13th June 2011, 03:30 PM   #10
rongon is offline rongon  United States
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Quote:
Originally Posted by quikie22 View Post
tricomp, I don't think the excess voltage is being wasted, it is used to run the input/driver stage.
Right, the power supply for the 2A3's is "stacked" on top of the power supply for the phase splitter/driver stage. The objective was to DC-couple the driver's plates to the output tubes' grids, and retain a DC-coupled input.

A transformer input would be a good solution (that would act as the phase splitter as well as blocking DC from the input), but that kind of a PP input xformer ain't cheap. I should look to see what Jensen or Lundahl have available...

Another (maybe better) way to optimize this circuit would be to put a phase splitter LTP as the first stage, then have that drive the PP driver stage. (It's not a big deal to DC-couple the first LTP to the PP driver with both being fed from the same B+ supply.) That will be my next amp, when I scale this idea up for PP 300B's (or PPP 2A3's, if I can afford the extra power supply iron that would require).

That's the stuff I do understand. Now back to my confusion...


Quote:
Originally Posted by tomchr
If the output tube is biased to -43.5 V, the max swing in A1 would be the max sine wave that would keep Vgk negative while being centered at -43.5 V.
I had to visualize that for it to sink in. The sort of "virtual 0V point" is at the bias voltage of the tube. The input signal makes it swing -43.5V below that point and +43.5V above that point. Does that sound right?

Quote:
Originally Posted by tomchr
So the max sine wave would swing from Vgk = -87 V to 0 V.
So that driver needs to provide 87Vp to the output tube? In other words, 61.5Vrms? If so, the gain of the first stage would need to be a little more than 30 to drive the output tube to full power from 2Vrms input at the driver grid. Right?

If this is the case, then it would take 4Vrms at the input of the driver stage to drive the output stage to full power. Correct?

If that's the case, then that would agree with how the PP 2A3 amp in the schematic is actually working, and why it requires a line stage in front of it.

Given the above, I totally agree that I should have put that 6DJ8 as a voltage amp in front of the PP driver, and made a sort of all-triode Mullard 5-20-style amp, or make it a "dual-differential" amp (6DJ8 diff phase splitter -> 6N6P diff driver -> PP 2A3's). That would be basically the same as JC Morrison's "Tube-o-saurus Rex" from an early '90s Sound Practices article, which isn't that far off from MJones' "Crystal Palace" amp. (I built a Tube-o-saurus Rex years ago, but I made mistakes and it oscillated. It was then scavenged for parts, but that's a different story...)

So, my confusion persists, and my question is, how do I know how many volts need to be swung from the output of the driver stage to the grid of the output stage?

If my output tube's bias is -43.5V, then the driver needs to swing from 0V to -87V? Is that "peak voltage" then?

If my output tube's bias was -80V (as in a 300B) then the driver would need to swing from 0V to -160V (peak)?


Quote:
Originally Posted by tomchr
I'd probably cap couple (or transformer couple) the input. That would allow you to lose the -15 V bias on the input stage.
I thought the bias for the input stage was -7.5V. My thinking was that the input grid is at 0V, and the second triode's grid is grounded (so a "Schmit" inverter), while the joined cathodes are at +7.5V, so the grid is at -7.5V compared to the cathodes.

I feel like there are some very basic things I'm missing here.

Thnx again...


-=|=-

Last edited by rongon; 13th June 2011 at 03:41 PM.
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