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ray_moth 22nd February 2005 06:41 AM

Effect of cross-coupled "local" FB on plate load of previous stage
 
2 Attachment(s)
I have implemented a version of KYW's cross-coupled FB, but with FB resistor going to the grid of the driver stage rather than the plate of the preceding stage. For the purpose of this question, however, that doesn't matter because it concerns the effective "AC" plate load, not the "DC" load. Please see attached schematic for clarity.

Please correct me if I am wrong. It occurs to me that the 1M feedback resistor, which acts in parallel with the 120K plate load resistor of each half of the 6SL7 splitter, has an effective value much smaller than 1M. It is affected by the closed loop gain (about 17) of the driver and OP stage that are included in the FB loop. I think its effective value = 1M/17, about 58K. This gives an effective plate load for the 6SN7 of 58K || 120K || 330K or approx. 38K. This is much smaller than what would normally considered suitable for a triode such as the 6SL7, especially with 1K unbypassed in the cathode.

Such a low load might not matter if the previous stage were a pentode (as in the HK Citation 2, which uses a similar arrangement for one of its FB loops),since pentodesare reputed to work well with small loads. However, for a triode it seems a bad idea. Maybe this is why such cross-coupled local FB arrangements are more usually taken to the cathode rather than the grid of the driver (e.g. one of Crowhurst's designs)?

gingertube 23rd February 2005 02:56 AM

Yep - spot ON
 
Ray_moth,
Your analysis is correct. Load on the 6SL7 is about 38K as you calculated. This is way too low.

I followed your balanced shunt feedback / partial feedback thread earlier. I've been struggling with a similar design and running into the same problems.

The feedback to the driver stage cathode is an obvious solution BUT there are two draw backs.
First the driver stage can't be run as a diff amp, you have to arrange it as individual common cathode stages - so you loose the self balancing properties.
Second you loose the benefit of cross coupling - common mode rejection and odd order harmonic distortion reduction.

A couple of circuit variations I am looking at for inspiration at this time:
1) Get a hold of the Alan Kimmel Mu Stage paper and look at his Van Soyoc - ish Mu Stage Diff Amp. If you use this arrangement the Source Follower MOSFET version is the way to go (on the last or 2nd last page of the paper).
2) Go to Turner Audio AUSTRALIA site (There are two different Turner Audio s) and look at his "Sophisticated Balanced Shunt Feedback" schematics - Caution - the note on the schematics states that it was drawn up as an intellectual exercise and he has'nt actually built and tested it although the simpler balanced shunt feedback circuit has been . Patrick Turner is one of the "seniors" of the OZ valve scene and I respect his knowledge and opinions.

One other thing which would be worthwhile is have another look at KYWs posted circuit to see how he handled this. He seems to be a knowlegable fellow and so would have addressed this issue..

If I have any major brain storms in the next week or so I'll get back to you.

Cheers,
Ian

ray_moth 23rd February 2005 03:55 AM

Hi Gingertube,

Thanks for confirming my suspicion re: the load for the 6SL7. I think I'll end up using multiple local FB loops plus, maybe, some global FB, to get a reasonable OP impedance with my EL34s as pentodes. The cross-coupled loop, if I keep it, will have to have a much higher FB resistor - say 10M - so as not to foul up the load for the 6SL7. The easy way out, of course, would be to revert to triode connection for the OP tubes, so I wouldn't need much FB if any. However, I don't want to go there again yet.

You'll have noticed that Patrick Turner uses 6AU6 pentodes as the splitter/driver, in a sort of mu stage arrangement so he can get a rerasonable plate current and voltage for them without having to shunt the plate-plate FB resistor with too low a resistance. Also, he only has one stage in his FB loop (the OP stage), so he doesn't have the problem of too low an effective plate load resistance due to too much gain in the FB loop (which is my problem). Load resistance shouldn't be a problem for him.

Alan Kimmel's splitter seems to give the best of both worlds, doesn't it? I think it should have similar advantages to the Van Scoyoc (good balance, good PSRR, good gain) without the disadvantage of limited "headroom".

Of course, Kimmel's splitter still "wastes" two triodes, just like the Van Scoyoc, but who cares? All tube-based phase splitters waste at least one tube, if you think about it: the concertina has no gain; the LTP has only half the gain of a diff amplifier with balanced input (and, in my case, there's another pentode as a CCS in the tail); the floating parrotface - sorry, paraphase - can only use the gain of one of the pair of tubes, to keep the amplitude of the two phases equal.

gingertube 23rd February 2005 05:31 AM

Another Idea
 
Ray_moth,
How is your circuit for gain? If you can afford to lose a bit of gain then plate to grid resistors (with DC blocking caps) on the 6SL7 will reduce the Zout of that stage enough to drive the 38K. You need to get Zout of that stage down to about 10 to 12K to do this. This is based upon the "Rule of Thumb" which says that the load should be about 3 times the Zout for a "euphonic" blend of 2nd and 3rd harmonic distortion. You've got enough feedback in other places to clean this up a bit. Of course to some degree this is just shifting the same problem to a new place so it needs thinking through.
Cheers,
Ian

ray_moth 23rd February 2005 06:01 AM

Gingertube,

Gain is not all that much. I really need to back off some of the FB but the effect of that if I'm not careful will be to reduce the load of the 6SL7 still further. I calculate gain as follows:

6SN7 open loop gain
A2 = RL2 x mu2 / (RL2 + RP2 + RK2*(mu2+1))
= 47000 x 20 / (47000 + 7700 + 1200x21)
= 11.76

EL34 open loop gain (from Mullard's figures giving 54w across 3.5k load p-p for an input of 50vrms g-g)
A3 = (sqrt(54 x 3500))/50
= 8.7

Combined open loop gain 6SN7 + EL34
A23 = A2 x A3 = 102

OP impedance of 6SL7
RO1 = (RK1 x (mu1+1) + RP1) || RL1 || RG2 (of 6SN7)
=(1000 x 71 + 44000) || 120000 || 330000
=49852 ohm

Closed loop gain 6SN7 + EL34
ACL23 = (RO1 + RFB) x A23 / (RO1 + RFB + RO3 + RO1 x A23)
= (49852+1000000) x 102 / (49852 + 1000000 + 15000 + 49852 x 102)
= 17.4

Effective value of RFB as shunt to 6SL7 load = RFB / ACL23
= 57571 ohm
Effective load of 6SL7
RLeff1 = 57471 || 120000 || 330000
= 34802 ohm

Gain of 6SL7 (per side)
A1 = 0.5 x RLeff1 x mu1 / (RLeff1 + RP1 + Rk1 x (mu1 + 1))
= 0.5 x 34802 x 70 / (34802 + 44000 + 1000 x 71)
= 8

This gives an overall gain from IP to OP tube plates of 8 x 17.4 = 141 approx. per side. So input needs to be about 1.5v rms to get 54 watts out - not much spare gain!

gingertube 23rd February 2005 06:35 AM

Ray_moth,
BUMMER - well back to the drawing board.
If it was easy it wouldn't be fun.
We could all just build yet another bloody Williamson clone and be done with it.
Cheers,
Ian

ThorstenL 23rd February 2005 02:22 PM

Re: Effect of cross-coupled "local" FB on plate load of previous stage
 
Konnichiwa,

Quote:

Originally posted by ray_moth
I have implemented a version of KYW's cross-coupled FB, but with FB resistor going to the grid of the driver stage rather than the plate of the preceding stage.
That is a pretty bad choice, if I may say, we will see why soon enough.

Quote:

Originally posted by ray_moth
For the purpose of this question, however, that doesn't matter because it concerns the effective "AC" plate load, not the "DC" load.
True enough, but it also loads the CR Highpass coupling between the stages. That is why I used directcoupling troughout my whole Amp, with the exceptions of the output stage grids.

Quote:

Originally posted by ray_moth
Please correct me if I am wrong. It occurs to me that the 1M feedback resistor, which acts in parallel with the 120K plate load resistor of each half of the 6SL7 splitter, has an effective value much smaller than 1M. It is affected by the closed loop gain (about 17) of the driver and OP stage that are included in the FB loop. I think its effective value = 1M/17, about 58K. This gives an effective plate load for the 6SN7 of 58K || 120K || 330K or approx. 38K.

It should actually much lower than that. You are looking at a theoretical "virtual ground" at both summing junctions, in reality some resistance remains, the input impedance does not completely become zero. I find that operating a well degenerated triode into a zero load impedance is an extremely linear way of doing things, low load is GOOD.

Quote:

Originally posted by ray_moth
This is much smaller than what would normally considered suitable for a triode such as the 6SL7, especially with 1K unbypassed in the cathode.
That is if you consider conventional circuits. But what is the load seen by the lower triode of a Cascode?

The way you apply this principle shows that you have not really considered how this actually works. If you implement it so it can work as intended the results are great.

Sayonara

ray_moth 24th February 2005 03:00 AM

KYW, I was hoping you'd read this thread, thanks for commenting. You've probably gathered, by now, that I'm bot completely sure about the concepts involved, what I'm trying to do or the way I'm doing it.

I would really like to understand what is wrong with taking the FB resistors to the grids of the 6SN7 instead of the plates of the 6SL7. I would especially like to understand the following statements:

Quote:

.True enough, but it also loads the CR Highpass coupling between the stages. . . . . You are looking at a theoretical "virtual ground" at both summing junctions, in reality some resistance remains, the input impedance does not completely become zero. I find that operating a well degenerated triode into a zero load impedance is an extremely linear way of doing things, low load is GOOD.
As I thought I understood it, the voltage at end of the 1M resistor (in the FB loop) that is connected to the plate of the EL34 is not fixed but moves up and down in opposition to the signal at the 6SN7 grid. The effect, from an AC load point of view, is to make its effective resistance smaller, to an extent determined by the combined gain of the stages in between. I figured that this will be true whether the FB is taken to the 6SN7 grid or to the 6SL7 plate.

I didn't use direct coupling because of voltage limitations - didn't want to starve either the 6SL7 or 6SN7 or both, in terms of voltage. What I have sounds OK but it is somewhat lacking in gain. I also found the bass to be a bit anaemic, so I increased the coupling capacitors between the 6SL7 and 6SN7 from 0.1uF to 1.0uF. It now sounds better but still not enough gain.

My concern was that this amp might sound better still if the load presented to the 6SL7 were higher, but your statement that a low load is good interests me. Any further comments/explanations you wish to make will be gratefully received (i.e. HELP!)

Best regards, Ray

ThorstenL 24th February 2005 11:46 AM

Konnichiwa.

Quote:

Originally posted by ray_moth
I would really like to understand what is wrong with taking the FB resistors to the grids of the 6SN7 instead of the plates of the 6SL7.
You place a very low input impedance AFTER a small value coupling capacitor.

Quote:

Originally posted by ray_moth
I would especially like to understand the following statements:

"True enough, but it also loads the CR Highpass coupling between the stages. . . . . You are looking at a theoretical "virtual ground" at both summing junctions, in reality some resistance remains, the input impedance does not completely become zero. I find that operating a well degenerated triode into a zero load impedance is an extremely linear way of doing things, low load is GOOD."

Hmm, I would have though all this would be plainly obvious.

I'll note it again.

The 6SN7 Differential stage coupled to the Output Pentode stage with the crosscoupled feedback applied becomes in effect a current input amplifier with two opposite polarity inputs that are very low in impedance, in theory, if we had an ideal Amplifier with unlimited gain the input impedance would become zero.

This mean all thinking must leave the "voltage" domain and go "current".

Let us assume a 1M pair of feedback resistors. We now inject a current of 0.1mA into each of the two inputs, with opposite polarity. The feedback loop will adjust the anodes of the output valves untill in effect the input is returned to the same voltage it was at before (in reality of course some voltage will appear). Now 0.1mA will deflect that output anodes each by around 100V, so ideally we have 200V peak Voltage across our output transformer from a balanced 0.2mA balanced current input.

Our input stage now operates in effect in the same way it would in a cascode, namely as current to voltage converter.The voltage applied between both grids is converted into a current modulation. In theory you can design the circuit such that all the anode current for the input valve is supplied through the feedback rsistors, in reality you often need a bit more current than you can get that way.

As the input stage does (ideally) not see any voltage modulation on the anode it can be operated at relatively high current and relatively low voltage, giving high transconductance.

The actual transconductance of the circuit is adjusted by the resistance between the cathodes of the input stage, if it is made zero the transconductance is at datasheet values.

If we took a 6SL7 Differential Stage with no cathode resistors and operate it with 50V Anode voltage and around 0.3-0.4mA anode current we get a transconductance of around 1mA/V. I personally run something like 75V/0.75mA, which bumps the transconductance a small amount above 1mA/V BTW.

So a differential input of 0.1V into the two grids will give rise to a current modulation of around 0.1mA in the anodes which in turn would translate into around 200V theoretical across the primary of the output transformer and thus into around 10V across the speaker.

Clearly way too much gain (40db), even once we account for the difference in gain with limited open loop gain of the 6SN7 and EL34 (which is still quite high at an estimated >>300 together though). So we introduce some cathode degeneration in the 6SL7 differential Amp to limit the gain.

Does that clear things up a little?

Sayonara

ray_moth 25th February 2005 04:14 AM

Quote:

You place a very low input impedance AFTER a small value coupling capacitor.
That explains why I had to change the coupling cap from 0.1uF to 1.0uF.

Quote:

Hmm, I would have though all this would be plainly obvious.
It wasn't to me, so I'm grateful to you for taking the trouble to explain it further.

Quote:

If we took a 6SL7 Differential Stage with no cathode resistors and operate it with 50V Anode voltage and around 0.3-0.4mA anode current we get a transconductance of around 1mA/V. I personally run something like 75V/0.75mA, which bumps the transconductance a small amount above 1mA/V BTW.
Since you have been successfully running the 6SL7 at such a low plate voltage, I'm encouraged to try direct coupling again.

Quote:

Does that clear things up a little?
Yes, it does, KYW. I now realize I must think in terms of current instead of voltage for the output of the 6SL7. Thank you so much.


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