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Old 14th August 2012, 05:52 PM   #21
ilimzn is offline ilimzn  Croatia
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Quote:
Originally Posted by 20to20 View Post
I'll stand corrected for the wiper on the input. If that loads the output from the prevoius stage.
Yes, at minimum, the wiper goes to ground. Also, you get typ. 100k in series with the grid which will do two things - it will shift the bias of the tube somewhat due to grid leakage current developing extra voltage on the 100k of the pot (this is in fact exactly how the grid leak test function is implemented in most tube testers - you check for this voltage!), and, worse, it forms a lo pass filter with the tube's input capacitance, so at the extreme low setting, this filter will have a high cut function inside the audio band. And - most importantly, the pot will actually not attenuate, just provide this filtering function. However, if you try this, you might be fooled that it indeed does work precisely because of the (over)loading of the previous stage. It has an output impedance so this acts as a voltage divider with the portion of the pot between wiper and ground. The attenuation curve will be very odd depending on the source, usually only a bit over minimum and you will get full output, and as I said, the source signal will be overloaded when the wiper is set to zero, i.e. the input signal will be shorted.
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Old 14th August 2012, 06:20 PM   #22
20to20 is offline 20to20  United States
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Also, you get typ. 100k in series with the grid
The pot is in parallel to the grid, just as the original 1M grid leak was. At full volume the grid leak would be 100K and at 0 volume it is still 100K.
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Old 14th August 2012, 07:02 PM   #23
20to20 is offline 20to20  United States
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Originally Posted by 20to20 View Post
The pot is in parallel to the grid, just as the original 1M grid leak was. At full volume the grid leak would be 100K and at 0 volume it is still 100K.
You could also just tie the wiper to the grounded end of the pot. Then the wiper would ground the grid at one extreme 0 volume, and be at full volume and 100K at the other extreme. Again, this is if grounding the grid does not also place an overload condition on the previous stage for some reason.
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Old 14th August 2012, 07:26 PM   #24
ilimzn is offline ilimzn  Croatia
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Originally Posted by 20to20 View Post
The pot is in parallel to the grid, just as the original 1M grid leak was. At full volume the grid leak would be 100K and at 0 volume it is still 100K.
No, at full volume it would be the output resistance (at DC) of the previous stage in parallel with 100k. If you are making an amp that connects to 'a signal source' it is not good practice to assume what the source looks like at AC or DC (for instance assume that the source has a coupling cap at the output, or that it will operate correctly with it's output shorted). You could tie the wiper and ground together and add a series resistor in front assuming you use a linear taper pot, a correctly calculated resistor will get you a usable audio-taper like attenuation curve. However, in this case the variation in series resistance seen by the tube is higher than in the classic connection and there is always some attenuation even at the highest setting of the pot.

All that being said I really see no reason to further elaborate on this, if you like you can draw the circuits themselves (not forgetting about avoiding assumptions re previous stage) and figure it all out for yourself, as I have no intention of runnig through all the circuit variants that save a whopping one resistor for the price of added disadvantages.
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Old 15th August 2012, 12:12 AM   #25
AJT is offline AJT  Philippines
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Quote:
Originally Posted by ilimzn View Post
Your diagram has obvious mistakes.
The 6AG5 is triode connected to a 'current source' that is bypassed to ground through a 1uF capacitor - the way it is drawn, the plate load for this section is the 545ohm resistor. This will provide a lot of attenuation and distortion, not gain.
The bottom half 6N1P grid must not be connected to ground, since the top half is DC coupled to ~143V if you want this to work at all. This way the bottom half of the 6N1P is WAAAAY into cut-off (-140 something volts at G1) so no phase splitting.
The output tube bias network has a common flaw, namely if the wiper of the adjustment pots go open (which may happen during adjustment as they turn even on pots that normally seem to work OK), the tubes will be left with no negative bias, i.e. maximum current will flow.
good analysis as always........i was thinking Mullard 5-20 when i saw his schematic......i was unsure about the CCS....

about the output biasing network, i also use such type, to avert disaster in case of pot failure, i do connect a 100k resistor from the wiper arm to the negative voltage source, this way tube meltdown is avoided.....
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Old 15th August 2012, 12:17 AM   #26
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Originally Posted by Tony View Post
you have direct coupling from your 6ag5 plate to the grid of the 6n1p, voltages do not add up, did you mean to do cap coupling instead?
here i was thinking that a cap coupling with grid leak to the LTP grid from the plate of the 6AG5 in order for his phase splitter to work as shown.....
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Old 15th August 2012, 01:54 PM   #27
ilimzn is offline ilimzn  Croatia
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Originally Posted by sorenj07 View Post
Any glaring errors before I start ordering parts?

Signal
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LTP Curves
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Input Curves
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It's still a nice project, so I think you should go on with it just correct the mistakes. So:

1) First stage needs to have a proper CCS as the plate load. In your drawing it is blocked to ground through the 1u cap, i.e. for AC the tube plate sees only the 545 ohm resistor. I can't read exactly what the CCS part is but in this position a good CCS can save you a lot of extra 'housekeeping' - for instance, no need to specially filter the power supply for that stage as a good CCS will provide very high ripple rejection (basically it will be the ratio of CCS dynamic resistance and plate resistance of the tube, it's quite simple to get 1000+ i.e. 60dB+ here even with a simple CCS). You are using the full mu of the tube, which might be high for your amp given there is no global feedback. The output tubes need some 100Vpp clean swing at least, and the gain of the input times LTP is well over 600, so about 120mV will get you to full power. If this is too low, using a simple resistor load for the first stage may be a good idea, but then you do need extra filtering for it's power supply.

2) LTP stage has two problems - not enough lean swing and there is an error in connecting the grid of the bottom half of the 6N1P. This should be connected to the top half grid via a resistor of 470k to 1meg, and also to ground via a capacitor - the 1u/630V part you have in the first stage on your original drawing, will do the trick. The purpose of this connection is to give both grids the same DC voltage, but prevent any of the AC output of the first stage go to the bottom half of the 6N1P's grid.
Getting more swing out of a DC coupled LTP means in essence maximizing the available voltage across the tubes + their respective plate loads. Since the amplification factor is not too high, you might consider lowering the plate current in the first stage (this will give you bit less mu there but as i said it may be too high to begin with), which will also lower the plate voltage. This in turn lowers the grid voltage on the 6N1P, and as a consequence the common cathode voltage must go down nearly by the same amount, giving more voltage headroom. Secondly, the power supply for this stage needs to be kept as high as possible. One thing that goes hand in hand with this requirement is that the power supply ripple for this stage does not need to be the apsolute lowest possible because it is common mode and hence rejected by this stage and the output stage. This means that a simple RC filter from the main supply will be more than enough, calculated so that minimum voltage is lost across the R - say about 20V. Both of these measures should give you some 50V more to play with in this stage. Also, you can lower the current somewhat. 15-18k resistors in the plate circuit should do fine, and the output tube grid leaks (100l) should still not present a big enough load to significantly reduce the output swing.

3) Output biassing correction - there is a number of ways you can do this but in general I have found the best way is to make a voltage divider with a resistor from the bias supply, and the trimer (one side connected to wiper to form a variable resistor) to ground. This way, if the wiper goes open, the voltage divider automatically gets the maximum negative voltage, thus saving the tubes. You could also keep it as it is now just add a 100k from the wiper to the bias supply (and slightly correct the value of the resistors on the side of the pot that goes towards ground), but that will need more parts. I have also found that a bias network that has a bias and balance pot is much more practical and easyer to adjust. but this is more a thing of personal preference.

4) A few cosmetic touch-ups, probably drawing errors - the 8 ohm speaker needs to be connected between 0 and 8ohm taps, the input stage security grid leak (1 meg) should be on the tube side of the shielded cable. It may be a good idea to reduce the cathode resistors in the output stage to 1 ohm (make these small, so they also act as fuses) - especially if you want to try triode connection. 1 ohm gives you a direct 1mV = 1mA reading of the cathode current. You also already have the resistance of the output winding in series with the cathodes. In pentode and UL this does make a small but hardly signifficant difference, in triode it increases the apparent plate resistance of the triode connected tube by (Rk + Rwindig) * mu, which can get up to 5-10% extra apparent Rp, and a few W of power lost. Also, perhaps I have missed some other smaller points already mentioned.

In any case I wish you good luck with the project, it looks very promising!
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Old 15th August 2012, 11:32 PM   #28
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First, I'd just like to say that I only have the tubes, transformers, 4 10M45's and 4 LM334's. The rest is all in the air

Quote:
Originally Posted by ilimzn View Post
1) First stage needs to have a proper CCS as the plate load. In your drawing it is blocked to ground through the 1u cap, i.e. for AC the tube plate sees only the 545 ohm resistor.
I just followed the example schematic provided with the LR8's datasheet. It did look a little strange but I trusted the manufacturer..?

See p. 4:
http://www.supertex.com/pdf/datasheets/LR8.pdf

Quote:
Originally Posted by ilimzn View Post
The output tubes need some 100Vpp clean swing at least, and the gain of the input times LTP is well over 600, so about 120mV will get you to full power.
I didn't have the formulas handy to calculate the decrease in gain due to the CFB but I wanted to leave room for global FB if needed.


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Originally Posted by ilimzn View Post
LTP stage has two problems - not enough lean swing and there is an error in connecting the grid of the bottom half of the 6N1P.
Yep, the bottom half of that 6N1P is definitely not getting the right voltage, an easy fix I guess. What do you mean by lean swing though? Sure the curves are a little bunched but that looks like mostly 2nd harmonic, so the push-pull topology should neutralize that right?

Quote:
Originally Posted by ilimzn View Post
Since the amplification factor is not too high, you might consider lowering the plate current in the first stage (this will give you bit less mu there but as i said it may be too high to begin with), which will also lower the plate voltage.
Probably what I'll end up doing. It was just a bit of a juggling effort, between trying to stay within the .76 watt dissipation of the LR8, as well as finding the most linear area of the 6AG5 curves in which to operate.

Quote:
Originally Posted by ilimzn View Post
a simple RC filter from the main supply will be more than enough, calculated so that minimum voltage is lost across the R - say about 20V. Both of these measures should give you some 50V more to play with in this stage.
Well, I'd planned on using an LC filter so there's a good deal more potential voltage to be had. The Hammond 282X' 500V secondary should give over 700V if needed. I just didn't want to have to put electrolytic capacitors in series :P

Quote:
Originally Posted by ilimzn View Post
Also, you can lower the current somewhat. 15-18k resistors in the plate circuit should do fine, and the output tube grid leaks (100l) should still not present a big enough load to significantly reduce the output swing.
You think the 10M45's will be happy at 8mA? If so I'll gladly proceed with that plan. Or is there another IC that'd do a better job in this place? I can save the 10M45's for another project.

Quote:
Originally Posted by ilimzn View Post
Output biassing correction - there is a number of ways you can do this
Thanks for all the options, I still haven't quite decided but the extra resistor doesn't make too much difference to me.

Quote:
Originally Posted by ilimzn View Post
the 8 ohm speaker needs to be connected between 0 and 8ohm taps
Since we are grounding the 4 ohm tap in CFB scheme, I'm pretty sure this renders the 0-16 ohm winding equivalent to 8 ohms, yes?

Quote:
Originally Posted by ilimzn View Post
It may be a good idea to reduce the cathode resistors in the output stage to 1 ohm (make these small, so they also act as fuses) - especially if you want to try triode connection. 1 ohm gives you a direct 1mV = 1mA reading of the cathode current.
Sounds like a plan.

Quote:
Originally Posted by ilimzn View Post
In any case I wish you good luck with the project, it looks very promising!
Thanks very much, especially for the detailed explanations. I've built a few amps previously but it was a few years ago, and it's nice to have things laid out clearly.
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Old 16th August 2012, 12:22 AM   #29
20to20 is offline 20to20  United States
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Quote:
Quote:
Originally Posted by ilimzn Click the image to open in full size.
It may be a good idea to reduce the cathode resistors in the output stage to 1 ohm (make these small, so they also act as fuses) - especially if you want to try triode connection. 1 ohm gives you a direct 1mV = 1mA reading of the cathode current.
Quote:
Sounds like a plan.
Since you are using the output tranny secondary to ground your cathode through you need to consider the total resistance to ground. You'll have maybe .5 -.75 ohms additional and that is a large percentage of error with respect to a 1 ohm resistor when calculating the cathode current. Better to leave it at 10 ohms. Even then you should still read the total for accuracy.
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Old 16th August 2012, 12:56 AM   #30
AJT is offline AJT  Philippines
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Quote:
Originally Posted by sorenj07 View Post

I didn't have the formulas handy to calculate the decrease in gain due to the CFB but I wanted to leave room for global FB if needed.
not much....Cathode feedback


Quote:

Since we are grounding the 4 ohm tap in CFB scheme, I'm pretty sure this renders the 0-16 ohm winding equivalent to 8 ohms, yes?
no, the 8ohm speaker goes to the 8ohm tap, if you put your 8ohm speaker on the 16ohm tap, the reflected primary impedance is halved....

the reason to connect one cathode to 16ohm tap and the other to the 0 tap while grounding the 4ohm tap is so that the cathodes get equal voltage feedback... the 0 to 4 ohm tap and the 4ohm to 16 ohm tap have equal number of turns.....
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