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Old 19th July 2005, 04:24 PM   #1
plovati is offline plovati  Italy
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Question Power SRPP

I've searched the forum without result so I started a new thread.

I'm trying to design a power amp usying two 6EM7. I want to employ the mains transformer I already have, so I'm thinking of a power SRPP for the output stage. I need 5W in class A.

What is the best load impedence for maximum power transfer for SRPP? The same that applies in the SE case, ie equal to Rp. Or Rp/2 or Rp/mu ?

There is a particular load condition for SRPP with no distortion. It is compatible with a fair efficiency or I need some different tubes?

Thanks in advance,
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Old 19th July 2005, 04:48 PM   #2
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You need... a different idea altogether. For one, you have two tubes in series driving a load for power, and not only that, you are using triodes! Triodes are NOT known for their low voltage capability. You'll need at least 400V to do it, 500-600V if mu-stage or White CF style.

Best load is still probably around 3*Rp of the top (cathode follower) tube.

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Old 19th July 2005, 08:02 PM   #3
plovati is offline plovati  Italy
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6EM7 second section works with 150V plate voltages. Gate voltages is about 15V. I have a 250V secondary so the compliance of 340/350V shold be enough for an SRPP.

3*Rp is the rule for best distortion in SE stage. I think SRPP should better because of the push pull action it should work at lower load with less distortion.

The question is : what's the (mathematically) optimum load for max power? And what's the best for lower distortion at a given power out?
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Old 20th July 2005, 01:49 AM   #4
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Quote:
Originally posted by plovati
3*Rp is the rule for best distortion in SE stage. I think SRPP should better because of the push pull action it should work at lower load with less distortion.
No, infinite load is almost always ideal load for a triode (I don't know if there are any exceptions to this, so I'm leaving it at "almost always"). You can't get any power out of this, so distortion is basically left out of the equation when deciding load. On the other hand, maximum power output is seen around 3*Rp, which typically finds distortion in the 5% range, which is also acceptable.

Quote:

The question is : what's the (mathematically) optimum load for max power? And what's the best for lower distortion at a given power out?
You haven't even told me what kind of stage it is. SRPP can be anything from a bootstrapped cathode follower to a full fledged SEPP power stage ala Futterman et al.

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Old 20th July 2005, 06:04 AM   #5
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> What is the best load impedence for maximum power transfer for SRPP? The same that applies in the SE case, ie equal to Rp. Or Rp/2 or Rp/mu ?

For SE, for a reasonably linear tube, distortion not considered, Rl = 2*Rp, but the peak is very broad and wide variation does not give much lower Power.

As Tim says, THD declines as the impedance rises. It falls pretty fast from Rl=Rp to Rl=3Rp, slower out to Rl=10*Rp at which point power may be so small we should consider another tube and start all over.

Rl=3*Rp is an excellent point. You can hardly tell the power is lower than the Rl=2*Rp condition, but the distortion has fallen enough to notice.

Rl=Rp is true for a DC coupled load. But we usually don't want the DC in the load, which leads to the factor of 2.

Rl=Rp/2 has somewhat more THD and significantly less power than Rl=Rp or Rl=2*Rp. The nice thing about a triode is that you can overload it like this without gross effect. But Rl=Rp/2 is not a nice place to be if you can do a higher Rl.

Rp/Mu is the same as 1/Gm, the cathode impedance, which is never a good load for a power amp since the supply power has to flow through Rp. (This assumes Mu is greater than 1 which is almost always true; inverted operation would be different.)

> Triodes are NOT known for their low voltage capability.

Pentodes are a lot better because they lose the electrostatic feedback from plate to grid that keeps triodes linear. But 2A3 is not too bad, and the big side of 6EM7 is a lot like a 2A3 but a bit better at low plate voltages.
http://frank.pocnet.net/sheets/049/6/6EM7.pdf

I don't see what supply voltage Piergiorgio has???? {later} ah,,, 340V.

It turns out to be quite critical. 2A3 is designed for good work at higher load impedances, 6EM7 is designed for TV pulses, their plate dissipations will not support huge currents. As a Class A plan, I think he is doomed to get less than the 5 watts he "needs". Even in AB, it looks tough to stay inside the plate dissipation line.

> There is a particular load condition for SRPP with no distortion.

No, the SRPP must be designed for a particular load impedance or it has no benefit over a simpler one-tube design. SRPP is vastly over-used today.

Read Broskie's essays on the SRPP: Decoded and Deconstructed.

My opinion: a speaker is not a constant-enough load impedance for the SRPP to work optimally. You can do better with other plans.

If Tim had not mentioned it, I would: the White Cathode Follower is worth a look. It is fairly un-critical about load, the same way that a SRPP is critical. And you have enough spare hi-Mu triodes to slap its low gain around. It is subject to the same criticisms that I have voiced in the current Cathode Follower threads: the low gain requires a gigantic driver. But not as bad as the single-ended transformer-coupled plans, because the input swing stays inside the rails, does not kick to twice the rail voltage.

The maximum power load for the WCF seems to be half the Rp of a single tube. Up to a point, the two tubes work in parallel, so half the SE optimum seems right. The DC is cancelled so a factor of 2 drops out. For tubes like 2A3 and 6EM7, about 400 ohms, though variations from 200 to 800 don't make much difference to power. Different simulations give different THD numbers: I think that when Rl is less than one Rp the THD really does not change, but for high Rl you have to re-balance the circuit because it changes from push-pull to push-push. If Rl is greater than Rp you will probably be better with some other plan.

Fooling with simulation, I get about 200V per tube (400V total), roughly -35V bias to give about 50mA idle current (this is very close to a 6EM7's heat rating). Here, 400 ohms gives almost 5 watts near clipping, but not quite 2 watts in Class A. 800 ohms gives 4 watts max, over 3 watts in Class A. But changing to 220V per tube (440V total) puts us over the 6EM7 plate dissipation rating, unless we reduce the idle current and strangle the Class A power; changing to 180V each (360V total) runs the tubes cool which really reduces the Class A power (and the AB power). We are just too close to the tubes' ratings to be making this much power with normal tube, load, and utility variations. Not a Good Design.

Output impedance is around 110 ohms, from which you can estimate the gain and damping.

The optimum top-resistor for the WCF will be about 160 to 220 ohms. The optimum input grid bias will be about 45% of the total supply voltage, and there is a null for even-order THD at half-power (and a different optimum for maximum power). Both of these adjustments are load-sensitive, and "errors" have to be judged by ear (nulling the even-order THD is probably not the most important goal).

The WCF has little ripple rejection, 6dB-10dB. With efficient speakers, you will need a LOT of supply filtering. (I don't think the SRPP is any better.)

You know.... 350V is NOT "too high" for 6EM7 with a conventional center-tapped output transformer. RCA's Design Maximum rating is 330V. Use cathode bias, you need about 55V-60V to stay inside the 10 watt plate rating. Now the tube only sees 290V-295V, well inside rating.

Staying inside Class A, idle 30mA, peaks of 60mA, you can swing 240V. That's 7 Watts output, 16K load plate to plate. With a more common 10KCT winding you approach 10 Watts in AB. And tubes like this do not have a hard line between A and AB. And you will really get less THD working the tube at idle current about 1/4 of peak current, about 25mA each for 10K load. That idles the tube comfortably below the plate dissipation rating. The 6EM7 is cheap, but they will never make any more of them, so I hate to run one AT the ratings.
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Old 20th July 2005, 06:46 AM   #6
PRR is offline PRR  United States
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Working with White Cathode Follower, 325V total supply (340V-15V for filtering), self-bias, I get this approximate optimum:

140V across each tube, 20V across each cathode resistor, ~5V across the top resistor, 50mA idle current:

2 to 2.5 Watts into 200 to 800 ohms

Full power (~2.2W-2W) in true Class A from 470 to 800 ohms

Output impedance ~70 ohms (try 100 ohm top resistor)

THD for even-order null falls from 0.4% to 0.2% as load impedance rises from 200 to 800 ohms (driver distortion WILL be higher and will dominate total THD)

Total efficiency (plate and heater) 8%.

Higher bias currents reduce power. (We slam into the zero grid bias line sooner.) Anyway, 140V 50mA is close enough to the 6EM7's 10 watt rating.

Biasing down to 35mA or 20mA gives marginally more power, like 2.7 Watts, especially for the higher impedances, but the AB distortion is starting to show.

Some of this analysis will apply to an SRPP topology, but I'm not even sure the SRPP will work well. (Personally, I do not like SRPP. It can "work", but is rarely the Best Plan for the number of tubes used.)

Using a conventional stock output transformer and topology gives three times the power, and probably similar total THD (more output stage THD but much lower driver THD).
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Old 20th July 2005, 07:39 AM   #7
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I use a pair of 13EM7 in a White CF headphone amp. Sound pretty good, but couldn't drive a low ohm speaker.

Find a few dozen 6/13EM7 and make the coolest looking monster 5W amp!
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Old 20th July 2005, 07:44 AM   #8
plovati is offline plovati  Italy
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Great advices Tim and PRR.

Just few clarification about where I started:

I supposed to have a too high voltage for conventional Push-Pull topology for this tube (PRR shows that's not true with appropriate choosing of bias, thanks), so I think a something different.

The first choice that came to my mind is White cathode follower but gain less than 1 and low PSRR of this stage pushed my to think to a conventional SRPP (two tube, two resistor). Load is trasformer coupled and with a cupling cap (no DC in trafo).
I should have better PSRR (for appropriate choice of resistor as per Broskie paper, I have constant Idc flowing thru the branch) and some gain at the expenses of slighty increase of Rout.
Furthermore, in aporopriate condition with the tubes identical, the output distortion approaches 0 at a certain load impedance.

Now, what I was not able to find out is a way to make a composite load line for SRPP to evaluate the power performances of this stage. There is lot of analisys but all valid for small signal.

The conventional rule for SE stage, that applies also to cathode follower is that max power transfer (matematically) is when load=Rp, no mention to distortion. I do not undestand PRR observation that this holds only for DC coupled load, why?
A good compromise between power out and distortion often use is as You both said 3*Rp.

But what about SRPP? The mathematically or graphical max power (no distortion matters), given tube bias and Rk, at what load is reached?

PRR, the values You reported are found by means of spice simulation or grapghical analisys?
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Old 20th July 2005, 02:52 PM   #9
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Quote:
Originally posted by plovati
Now, what I was not able to find out is a way to make a composite load line for SRPP to evaluate the power performances of this stage. There is lot of analisys but all valid for small signal.
Let's see, assuming the power supply is solid and the output is coupled to it solidly, and the tubes are driven in phase grid-to-cathode (note top at least will need a trasformer), it looks exactly like conventional PP: the load is shared between the tubes.

You get in trouble putting this into practice: the top tube, when not transformer driven, is a cathode follower with dramatically lower gain than the lower triode.

This says nothing of other SRPP arrangements such as white CF or mu stage. The latter, due to the large current-detecting resistor, has no drive capacity from the bottom tube.

Quote:

But what about SRPP? The mathematically or graphical max power (no distortion matters), given tube bias and Rk, at what load is reached?
This depends on the tube itself. It could come at any relation of Rp to RL. Screen-grid tubes and transitors have a high "Rp" and RL comes many times *lower*, as opposed to triodes. Cathode followers are even more dramatic than triodes, Zo might be 200 ohms yet best load would be say, 5k. It depends, but in general Zo is just a parameter and doesn't have much to do with anything. It does happen to be convienient for triodes, though.

So anyway, using the same triodes in an SRPP where both tubes are able to drive output, RL = 1.5 * Rp, if they are driven grid-cathode (grid-ground, this means extra drive to the top tube at least).

Tim
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Old 20th July 2005, 11:08 PM   #10
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The really-optimum loadline depends entirely on tube ratings, available supply voltage, available signal voltage, and often on distortion.

Langford-Smith defines 4 specific estimated optimum cases in Radiotron Designers Handbook 4th, chapter 13. The four cases start on page 555 (12th page of the chapter) but you should review the preceding general background to understand the specific cases.

> found by means of spice simulation or grapghical analisys?

Several different techniques including GlassWare and decades of experimentation and thinking.

I admit that I did not set up an SRPP: I don't like the idea, and so I do not know how to optimize it.

In general: if a plan is any good, the resistor losses will be low and the drives will be balanced between the two tubes. In that case, the exact placement of resistors or the scheme used to drive the tubes is not very important once you get it right (that is the hard part). So simulating the general totem-pole topology will give good answers for good designs. It may not be possible to do even this good with a specific drive scheme. (Since I did assume resistors, you could do a little better with transformer drive and separate grid bias supplies, but for few-Watt ampliers simplicity and stability is best.)

The difference between the ~2.5 watts in totem-pole and ~7W-10W in conventional topology is pretty much all about splitting the ~300V across one tube or two in series. For a tube with significant voltage drop (such as most triodes), you always get the most power by taking B+ as high as it will go. For the 6EM7, that means 300V, not half of 300V.

The totem-pole has the tubes in series with the power supply but "in parallel" to the load. That gives much lower output impedance. This is one of the few triode amps I have seen that can make a damping factor close to 10 and still have some power left. However I think if you could mis-match the conventionl center-tapped OPT plan to give a DF of 10, it would make a little more power than the WCF. Yeah, about 3 or 3.5 watts instead of 2W or 2.7W. On paper you are not gaining anything. In practice, the 10KCT winding is dead-standard while 600-800 ohm whindings are rare. However, a 600 ohm winding is likely to have less trouble with leakage inductance and winding capacitance (especially if custom-wound by a good winding designer!). But I'd guess if you use 6EM7, you are doing low-buck work, not custom-winding work. I think the 6EM7 is under-rated, but I would not pay a lot for custom windings to make the most of it.

Anyway as Gregg says: 6EM7 and its 13EM7 brother are SO cheap, that if two don't do your job, buy a six-pack or a dozen. And a dozen 6EM7 in WCF will drive 33 ohms with 12 Watts: wire an array of drivers up to 32 ohms and avoid a transformer completely.
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