I am currently trying to understand much better the maths when using anode chokes or IT...and found some articles and tools about it, but clearly it is not a topic which seems to be as popular than explaining RC coupling...so I trust in the knowledge of this forum to get some more light and knowledge around this topic.
I am considering two use cases:
- Line Stage with OPT-Output (fully differential, LTP, current source, 4p1l or 10y)
- Driver stage of PP: Either Anode Choke or IT (6n6p into el34 currently, later maybe 6n6p/801a into 300b, all LTP with current sink)
What I found so far:
http://www.vt52.com/diy/tips/platechoke.xls
...a nice tool to calculate the frequency behavior for a selected tube/inductance.
My questions popped up yesterday when I read about differential driver stages:
basic-tube-5
I asked myself:
- What of the normal RC-Coupling-Calculation like for instance Calculating RLa with RLa=Ra/Rg are still true when using Anode-Loads (still C-coupled) or directly IT ?
- In case of an Anode Choke: What is the Ra for the purpose of calculation the output restistance of the driver stage ? I understand the formulars for AC where the chokes makes this a very big/infinitive impedance for the AC with raising frequency...but how do I understand RLa when the bridging relationship of 1:10 output/input impedance is met ? Somehow I am guessing that the output impedance of the driver stage will not become infinite ?
- Example: two pairs of El34 need to be driven by a 6n6p which is feed through an 150H Anode choke, but still C-coupled. The El34 got 39k grid resistors each. So, how bad is the situation ?
- What changes in terms of binding the driver to the output stage using an IT ? I guess we do not need a gridresistor anymore, the secondary of the IT becomes the grid choke as well...but how to calculate RLa ?
What I would like to understand much better is: When I would try to do the same calcs as Mr. Turner explained on LTP-Drivers (RC-coupled) but for Anode-Choke/IT: How do I calculate this so I can evaluate which options for tubes, bias-points, IT-step up or step-Down ratios I should consider.
Last example: I read from Kevin Carter and others that they like the sound of their line preamps much better with IT/OPT which have a step-down-ratio of 4:1...ok, I guess that may be empirical observations, but obviously it raises the question why this is needed when you work already with tubes like the 4p1l (rp around 1500 ohms) and driving a tube-input stage of 100k input-resistance...?...plus destroying the gain of your line-amp ? Not such an obvious choice...
I am considering two use cases:
- Line Stage with OPT-Output (fully differential, LTP, current source, 4p1l or 10y)
- Driver stage of PP: Either Anode Choke or IT (6n6p into el34 currently, later maybe 6n6p/801a into 300b, all LTP with current sink)
What I found so far:
http://www.vt52.com/diy/tips/platechoke.xls
...a nice tool to calculate the frequency behavior for a selected tube/inductance.
My questions popped up yesterday when I read about differential driver stages:
basic-tube-5
I asked myself:
- What of the normal RC-Coupling-Calculation like for instance Calculating RLa with RLa=Ra/Rg are still true when using Anode-Loads (still C-coupled) or directly IT ?
- In case of an Anode Choke: What is the Ra for the purpose of calculation the output restistance of the driver stage ? I understand the formulars for AC where the chokes makes this a very big/infinitive impedance for the AC with raising frequency...but how do I understand RLa when the bridging relationship of 1:10 output/input impedance is met ? Somehow I am guessing that the output impedance of the driver stage will not become infinite ?
- Example: two pairs of El34 need to be driven by a 6n6p which is feed through an 150H Anode choke, but still C-coupled. The El34 got 39k grid resistors each. So, how bad is the situation ?
- What changes in terms of binding the driver to the output stage using an IT ? I guess we do not need a gridresistor anymore, the secondary of the IT becomes the grid choke as well...but how to calculate RLa ?
What I would like to understand much better is: When I would try to do the same calcs as Mr. Turner explained on LTP-Drivers (RC-coupled) but for Anode-Choke/IT: How do I calculate this so I can evaluate which options for tubes, bias-points, IT-step up or step-Down ratios I should consider.
Last example: I read from Kevin Carter and others that they like the sound of their line preamps much better with IT/OPT which have a step-down-ratio of 4:1...ok, I guess that may be empirical observations, but obviously it raises the question why this is needed when you work already with tubes like the 4p1l (rp around 1500 ohms) and driving a tube-input stage of 100k input-resistance...?...plus destroying the gain of your line-amp ? Not such an obvious choice...
If you want us to comment on or explain a formula then you need to give the formula. We do not know what you consider to be "the normal RC-Coupling-Calculation", for example. What do you mean by "Calculating RLa with RLa=Ra/Rg"? What is RLa?
Too many questions; not enough information.
Bear in mind that most people using coupling transformers are aiming at a particular sound (or sufferering from nostalgia); they are not aiming at engineering excellence, as CR coupling is far more 'ideal' in its behaviour than a transformer. Hence if you try to understand what they are doing by applying science you may be left confused. An exception is when a Class AB2 output stage has to be driven, although even there it could be argued that a sold-state follower is a better solution than a transformer.
Too many questions; not enough information.
Bear in mind that most people using coupling transformers are aiming at a particular sound (or sufferering from nostalgia); they are not aiming at engineering excellence, as CR coupling is far more 'ideal' in its behaviour than a transformer. Hence if you try to understand what they are doing by applying science you may be left confused. An exception is when a Class AB2 output stage has to be driven, although even there it could be argued that a sold-state follower is a better solution than a transformer.
Sorry if this was not clear enough. Please have a look on Mr. Turner's site (link provided). He describe in mathematical detail the correct calculations for a setup of a differential pair driver with "normal RC-coupling" and showed as well what differential does to the numbers...but maths and actual measurements.
My question was: What would be the numbers when using anode chokes or IT respectively ?
the schematic:
His calcs (sorry, long post):
"These two driver stage schematics are based on the driver stage shown at Reformed RCA 30W amp. On the left hand side there is the exact stage used for the reformed RCA amp, the right hand side shows what may be used if the input stage was a single ended input stage. For these stages, I estimated the 6CG7 used had µ = 20.9 and Ra = 12k0.
The left schematic has cathode current taken to a -45Vdc supply rail via 5k0. Because the there is a well balanced input to both grids, each tube operates with the same RLa total of 30.5k, from 41k//120k. The signal voltage at cathodes is virtually zero. 2H harmonic currents with same phase flow in both loads and triodes, and a 2H harmonic voltage appears at cathodes. Some 2H voltage appears at each anode but it is same phase, common mode, and a very small % of signal and is rejected by following output stage so its presence has negligible effect on overall THD or IMD production. But the common mode gain of the stage is not low with the Rk R19 5k0 shown. If there was 0.1V of common mode signal to each grid, then 0.319V would appear at each anode with same phase. The stage above is preceded by an input LTP which produces almost zero common mode signal or distortion so high common mode rejection in the driver stage is not needed and hence no need for an active constant current sink, MJE340. Although the MJE340 is not needed, you are free to include it if you wish. The balance of output without the CCS was found to be excellent because resistances used were matched closer than +/-1%.
Output resistance at each anode is a high ohm value and poor when considering each anode separately. This is a property of all balanced amps with commoned cathodes and high value cathode resistances or CCS taken to a negative or 0V rail. However, there is no need for the Rout at each anode to be low because there is no intention or need to drive output tube grids into positive grid voltage region where the grid input resistance suddenly reduces from megohms to say 1.5kohms due to grid current onset. Therefore the LTP stage cannot easily produce drive voltages beyond the output tube grid current region and it causes voltage clipping at one LTP positive going anode while producing a peaked negative going voltage at the other anode. The negative going anode does not experience any grid current. The result of LTP overload is entirely benign and recovery is instant and there is symmetrical waveform clipping at the amp output. In most amplifiers, the onset of grid current in output tubes indicates onset of output tube overload and exponential rise of THD and IMD and the amp has reached its useful high fidelity power ceiling. The LTP driver should be designed to be able to produce at least twice the drive voltage required by output tubes for their clipping level. The above LTP may be found to generate 70Vac at each anode, at less than 1% THD, without output tubes present.
The Rout at each LTP anode where the loading of each anode is equal is much lower than where one anode only is considered. The Rout measured from anode to anode is the sum of loads on each anode in parallel with the sum of the Ra for each triode. In this case, the Rout a-a = ( 2 x 30.5k ) in parallel with ( 2 x 12k0 ) = 2 x 8k7 = 17k4. There are 2k2 "grid stopper" resistors in series with each KT88 grid so total driver g-g resistance = 17k4 + 4k4 = 21k8. This is a quite low enough resistance to drive the grid to grid input impedance including the Miller capacitance of a pair of KT88 tubes with 20% UL taps. The KT88 Cg-g may be estimated at approximately 40pF, and the resultant HF cut off pole would be at 180kHz."
My question was: What would be the numbers when using anode chokes or IT respectively ?
the schematic:
An externally hosted image should be here but it was not working when we last tested it.
His calcs (sorry, long post):
"These two driver stage schematics are based on the driver stage shown at Reformed RCA 30W amp. On the left hand side there is the exact stage used for the reformed RCA amp, the right hand side shows what may be used if the input stage was a single ended input stage. For these stages, I estimated the 6CG7 used had µ = 20.9 and Ra = 12k0.
The left schematic has cathode current taken to a -45Vdc supply rail via 5k0. Because the there is a well balanced input to both grids, each tube operates with the same RLa total of 30.5k, from 41k//120k. The signal voltage at cathodes is virtually zero. 2H harmonic currents with same phase flow in both loads and triodes, and a 2H harmonic voltage appears at cathodes. Some 2H voltage appears at each anode but it is same phase, common mode, and a very small % of signal and is rejected by following output stage so its presence has negligible effect on overall THD or IMD production. But the common mode gain of the stage is not low with the Rk R19 5k0 shown. If there was 0.1V of common mode signal to each grid, then 0.319V would appear at each anode with same phase. The stage above is preceded by an input LTP which produces almost zero common mode signal or distortion so high common mode rejection in the driver stage is not needed and hence no need for an active constant current sink, MJE340. Although the MJE340 is not needed, you are free to include it if you wish. The balance of output without the CCS was found to be excellent because resistances used were matched closer than +/-1%.
Output resistance at each anode is a high ohm value and poor when considering each anode separately. This is a property of all balanced amps with commoned cathodes and high value cathode resistances or CCS taken to a negative or 0V rail. However, there is no need for the Rout at each anode to be low because there is no intention or need to drive output tube grids into positive grid voltage region where the grid input resistance suddenly reduces from megohms to say 1.5kohms due to grid current onset. Therefore the LTP stage cannot easily produce drive voltages beyond the output tube grid current region and it causes voltage clipping at one LTP positive going anode while producing a peaked negative going voltage at the other anode. The negative going anode does not experience any grid current. The result of LTP overload is entirely benign and recovery is instant and there is symmetrical waveform clipping at the amp output. In most amplifiers, the onset of grid current in output tubes indicates onset of output tube overload and exponential rise of THD and IMD and the amp has reached its useful high fidelity power ceiling. The LTP driver should be designed to be able to produce at least twice the drive voltage required by output tubes for their clipping level. The above LTP may be found to generate 70Vac at each anode, at less than 1% THD, without output tubes present.
The Rout at each LTP anode where the loading of each anode is equal is much lower than where one anode only is considered. The Rout measured from anode to anode is the sum of loads on each anode in parallel with the sum of the Ra for each triode. In this case, the Rout a-a = ( 2 x 30.5k ) in parallel with ( 2 x 12k0 ) = 2 x 8k7 = 17k4. There are 2k2 "grid stopper" resistors in series with each KT88 grid so total driver g-g resistance = 17k4 + 4k4 = 21k8. This is a quite low enough resistance to drive the grid to grid input impedance including the Miller capacitance of a pair of KT88 tubes with 20% UL taps. The KT88 Cg-g may be estimated at approximately 40pF, and the resultant HF cut off pole would be at 180kHz."
Still not sure what you want. I am guessing you want to use chokes or transformers as the plate loads, not resistors.
If you are trying to calculate the inductance necessary for low frequency response, then use the tube plate resistance from the tube data sheet (if you run the tube at those voltages and currents). Otherwise you have to determine the plate resistance from the tube curves at the operating point you are going to be using (you will have to read up on this one, I do not have the time now to draw on a set of curves, and give an example).
The frequency response is -3 dB when the plate resistance, rp, (of the tube, not a plate load resistor, RL) is equal to the inductive reactance, Xl, of the choke or transformer primary .
Xl = rp
Where Xl = 2 * pi * f * L
re-arranging, we get
L = Xl/(2 * pi * f)
Suppose the plate resistance is 2000 Ohms, and f = 20Hz.
L = 2000 Ohms/(2 * 3.14 * 20Hz)
L = 15.9Henry
Now, I do not like to have -3dB response from only the inductor or transformer, there are other frequency rolloff points in the amp. Lets say we want -3dB at 20Hz for the entire amp circuit, and we have 3 low frequency poles in the amp.
If each pole is -1 dB at 20Hz, we have it (-1dB & -1dB & -1dB = -3dB).
For the single pole of the choke or transformer the -1dB point is 1 octave above the -3dB point (2 times the -3dB frequency).
We had -3 dB at 20Hz with 15.9Henry.
Xl = 2 * pi * f * L
If we want -1dB at 20Hz, we need to be -3dB at only 10Hz.
We get 2 * 15.9Henry. We need 31.8Henry (-3dB at 10Hz, is -1dB at 20Hz).
Be careful, if you use a push pull transformer, the turns of each half of the secondary is only 1/4 of the plate to plate inductance (root of turns ratio, 1/2 turns = 1/4 inductance).
So, with push and pull plate resistances (rp), the effective inductance in class A operation is 1/2 of the plate to plate transformer inductance.
If you do AB operation, during the time only one of the push pull tubes is on, a single plate rp is driving only 1/4 of the plate to plate inductance. Now the -3dB point has shifted upward by one octave, f * 2. What was -3dB at 10Hz now becomes -3dB at 20Hz.
There is more to the story, because using an interstage transformers to drive grids of the next stage is entirely different than an output transformer driving an 8 Ohm real resistor.
And driving a real world loudspeaker is different yet, than driving an 8 Ohm real resistor.
It is not as simple as some are led to believe.
But have fun, keep going, build, and listen to the music.
If you are trying to calculate the inductance necessary for low frequency response, then use the tube plate resistance from the tube data sheet (if you run the tube at those voltages and currents). Otherwise you have to determine the plate resistance from the tube curves at the operating point you are going to be using (you will have to read up on this one, I do not have the time now to draw on a set of curves, and give an example).
The frequency response is -3 dB when the plate resistance, rp, (of the tube, not a plate load resistor, RL) is equal to the inductive reactance, Xl, of the choke or transformer primary .
Xl = rp
Where Xl = 2 * pi * f * L
re-arranging, we get
L = Xl/(2 * pi * f)
Suppose the plate resistance is 2000 Ohms, and f = 20Hz.
L = 2000 Ohms/(2 * 3.14 * 20Hz)
L = 15.9Henry
Now, I do not like to have -3dB response from only the inductor or transformer, there are other frequency rolloff points in the amp. Lets say we want -3dB at 20Hz for the entire amp circuit, and we have 3 low frequency poles in the amp.
If each pole is -1 dB at 20Hz, we have it (-1dB & -1dB & -1dB = -3dB).
For the single pole of the choke or transformer the -1dB point is 1 octave above the -3dB point (2 times the -3dB frequency).
We had -3 dB at 20Hz with 15.9Henry.
Xl = 2 * pi * f * L
If we want -1dB at 20Hz, we need to be -3dB at only 10Hz.
We get 2 * 15.9Henry. We need 31.8Henry (-3dB at 10Hz, is -1dB at 20Hz).
Be careful, if you use a push pull transformer, the turns of each half of the secondary is only 1/4 of the plate to plate inductance (root of turns ratio, 1/2 turns = 1/4 inductance).
So, with push and pull plate resistances (rp), the effective inductance in class A operation is 1/2 of the plate to plate transformer inductance.
If you do AB operation, during the time only one of the push pull tubes is on, a single plate rp is driving only 1/4 of the plate to plate inductance. Now the -3dB point has shifted upward by one octave, f * 2. What was -3dB at 10Hz now becomes -3dB at 20Hz.
There is more to the story, because using an interstage transformers to drive grids of the next stage is entirely different than an output transformer driving an 8 Ohm real resistor.
And driving a real world loudspeaker is different yet, than driving an 8 Ohm real resistor.
It is not as simple as some are led to believe.
But have fun, keep going, build, and listen to the music.
I recently did a SET with a choke loaded input tube and ended up creating a simple worksheet for it. It is a lot like the one you linked, but it shows all the math and takes into account Rg and Miller Effect for the following stage, and the output transformer's effects on the overall frequency response.
You can download it here:
https://wtfamps.files.wordpress.com/2017/02/two-stage-set.xlsx
It's obviously assuming perfect devices and operating points, but it might help make sense of things with the way it's broken down.
You can download it here:
https://wtfamps.files.wordpress.com/2017/02/two-stage-set.xlsx
It's obviously assuming perfect devices and operating points, but it might help make sense of things with the way it's broken down.
That is extremely helpful, thanks a lot...I will play around with it...I understand that the spreadsheet is for a capacitor coupled driver/output-stage with an anode choke for the input tube ? Did you not feel like integrating the coupling cap size is worthwhile as lowering Rg will require upping the cap-value quiet a bit ?
6A3sUMMER, you summarized it well...using an IT is different from OPT into speaker and that is why I asked the question about more knowledge...on the grid side of the output tubes I would have assumed that the secondary is like a grid choke, so a high inductance for AC /Driver-Stage, but a low DCR for the output tube/grid current.
So, I believe what Kevin did here is pretty much what I want to use ... but maybe with different tubes in the different stages as well and for the line stage as well...so that is why the maths would be nice to understand:
A question is for me for example: How important is the choice of the source impedance of the IT, which are for the 1692A 10K (http://www.lundahl.se/wp-content/uploads/datasheets/1692A.pdf)...?
So, if the driver would be the 801A with rp of 5K, should the 1692A still be used ? Or better the LL1660 with its 15k (butreduced bandwith) http://www.lundahl.se/wp-content/uploads/datasheets/1660.pdf ?
6A3sUMMER, you summarized it well...using an IT is different from OPT into speaker and that is why I asked the question about more knowledge...on the grid side of the output tubes I would have assumed that the secondary is like a grid choke, so a high inductance for AC /Driver-Stage, but a low DCR for the output tube/grid current.
So, I believe what Kevin did here is pretty much what I want to use ... but maybe with different tubes in the different stages as well and for the line stage as well...so that is why the maths would be nice to understand:
An externally hosted image should be here but it was not working when we last tested it.
A question is for me for example: How important is the choice of the source impedance of the IT, which are for the 1692A 10K (http://www.lundahl.se/wp-content/uploads/datasheets/1692A.pdf)...?
So, if the driver would be the 801A with rp of 5K, should the 1692A still be used ? Or better the LL1660 with its 15k (butreduced bandwith) http://www.lundahl.se/wp-content/uploads/datasheets/1660.pdf ?
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It is hard to say which would be better. But my judgement call is that the 1692A would be better.
You may have to use RC networks on the secondaries of either transformer, just as mentioned in the data sheets.
Keep in mind that if you are worried about secondary DCR, that means you must be planning on drawing output tube grid current for large signal swings.
But nothing comes for free, the grid current lowers the impedance that the 801 plates see on the interstage primaries during those large swings.
Plan on there being more 3rd harmonic distortion when the output stage grids draw current.
If you draw large grid currents, the reflections back to the secondary can easily be lower impedance than the inductive reactance at 20Hz (those impedances are in parallel).
You may have to use RC networks on the secondaries of either transformer, just as mentioned in the data sheets.
Keep in mind that if you are worried about secondary DCR, that means you must be planning on drawing output tube grid current for large signal swings.
But nothing comes for free, the grid current lowers the impedance that the 801 plates see on the interstage primaries during those large swings.
Plan on there being more 3rd harmonic distortion when the output stage grids draw current.
If you draw large grid currents, the reflections back to the secondary can easily be lower impedance than the inductive reactance at 20Hz (those impedances are in parallel).
Ok...I digged a bitdeeper and found a less mathematical, but empiric answer from Kevin Carter in a different forum:
"Lundahl makes basically 3 transformers that are normally used for interstage service in tube circuits, LL1660, LL1692A, and LL1671. They are designed to provide somewhat overlapping application ranges and cover most medium to low Rp tubes.
For 1:1 (approximately) service I use the following Rp ranges:
4K-15K LL1660
1.5K-5K LL1692A
<1K-2.5K LL1671
How are these different when they all use the same core? Number of turns of wire is the answer. More turns means higher primary inductance, which allows higher Rp tubes to be accommodated without loss of bass performance.
You might ask why you need anything but the LL1660, because it has adequate primary inductance for any tube from 15K on down (assuming 1:1 use). Well, it turns out that stray capacitance and inductance in the LL1660 interacts to cause peaks and dips (resonances) in the response at high frequencies with lower Rp tubes, so that transformers with fewer turns actually perform better with these tubes.
"
So, the 801A could be used with both, but given the resonance issue of the LL1660, which translates in the data sheets with a significant lower bandwidth in PP, I will go withthe LL1692A and try...espescially since I want to try as wellthe 4p1l in both, a line stage and as a driver stage...
...by the way: I do not plan with grid current/a2, but understood that some people reported with grid chokes that their output tubes sounded "relieved" and their explaination was around grid current...and my guesstimate has been that the secondary of an IT acts as well for the Output tube like a grid choke as DC-Resistance is low.
"Lundahl makes basically 3 transformers that are normally used for interstage service in tube circuits, LL1660, LL1692A, and LL1671. They are designed to provide somewhat overlapping application ranges and cover most medium to low Rp tubes.
For 1:1 (approximately) service I use the following Rp ranges:
4K-15K LL1660
1.5K-5K LL1692A
<1K-2.5K LL1671
How are these different when they all use the same core? Number of turns of wire is the answer. More turns means higher primary inductance, which allows higher Rp tubes to be accommodated without loss of bass performance.
You might ask why you need anything but the LL1660, because it has adequate primary inductance for any tube from 15K on down (assuming 1:1 use). Well, it turns out that stray capacitance and inductance in the LL1660 interacts to cause peaks and dips (resonances) in the response at high frequencies with lower Rp tubes, so that transformers with fewer turns actually perform better with these tubes.
"
So, the 801A could be used with both, but given the resonance issue of the LL1660, which translates in the data sheets with a significant lower bandwidth in PP, I will go withthe LL1692A and try...espescially since I want to try as wellthe 4p1l in both, a line stage and as a driver stage...
...by the way: I do not plan with grid current/a2, but understood that some people reported with grid chokes that their output tubes sounded "relieved" and their explaination was around grid current...and my guesstimate has been that the secondary of an IT acts as well for the Output tube like a grid choke as DC-Resistance is low.
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Blitz what do you want to drive? That's the first question. Then you can determine what driver + IT you need.
If you want to use pentodes with loop feedback I do not recommend IT or chokes.
If you want to use pentodes and IT I would first get the output stage working with local feedback. The most convenient way to run the EL34's output stage is UL + cathode feeback.
So first thing get an UL output transformer with cathode feedback windings (or symmetrical secondaries that can be arranged to employ cathode feedback). 10% cathode fb is enough to get equal or better Zout than triodes and lower distortion. For best quality you could look at the Plitrons (that are sold by Amplimo in Europe).
If you don't want to buy another output transformer and still want to use IT to drive EL34's then I would connect them in triode mode. From one pair you will be still able to get 18-20 clean watts.
If you want to use pentodes with loop feedback I do not recommend IT or chokes.
If you want to use pentodes and IT I would first get the output stage working with local feedback. The most convenient way to run the EL34's output stage is UL + cathode feeback.
So first thing get an UL output transformer with cathode feedback windings (or symmetrical secondaries that can be arranged to employ cathode feedback). 10% cathode fb is enough to get equal or better Zout than triodes and lower distortion. For best quality you could look at the Plitrons (that are sold by Amplimo in Europe).
If you don't want to buy another output transformer and still want to use IT to drive EL34's then I would connect them in triode mode. From one pair you will be still able to get 18-20 clean watts.
45, my amp is right now:
http://www.triodedick.com/monobill_2/monobill_2_schema_versterker.GIF
So, no global feedback at all.
What I want to do next is:
- replace driver Anode Resistors with chokes (and lower HV)
- replace RC with IT coupling completely....just to listen and learn.
- then I want to build a new Dac-Output/Linestage which drives above amplifier, DHT&Lineout-OPT
- finally build a PP amp without global feedback based on 10y driving 300BXLS, my speakers are large line sources and need PP, SE did not work well.
So, these are many projects and I want to learn much more about IT/Anode-Chokes while staying always fully differential with LTPs and CCS...or similar like Karna of L. Olson...
So, it is about learning about the concepts which I have not used sofar. I never used IT or Anode Chokes...my current setup is completely based on Indirect heated tubes and in the power amp not even pured triodes, but UL etc. I have build a RC-Coupled 300bXls SE Amp, but my speaker sounded very soft with them...not enough air and transparency etc. 12 Ribbons and 18 4" Eton chassis want something stronger than 15W SE. The 30-40 Watt of the UL PP El34 comtrols it nicely though. I think it is more than just Watts and damping factor why it did not work with SE.
http://www.triodedick.com/monobill_2/monobill_2_schema_versterker.GIF
So, no global feedback at all.
What I want to do next is:
- replace driver Anode Resistors with chokes (and lower HV)
- replace RC with IT coupling completely....just to listen and learn.
- then I want to build a new Dac-Output/Linestage which drives above amplifier, DHT&Lineout-OPT
- finally build a PP amp without global feedback based on 10y driving 300BXLS, my speakers are large line sources and need PP, SE did not work well.
So, these are many projects and I want to learn much more about IT/Anode-Chokes while staying always fully differential with LTPs and CCS...or similar like Karna of L. Olson...
So, it is about learning about the concepts which I have not used sofar. I never used IT or Anode Chokes...my current setup is completely based on Indirect heated tubes and in the power amp not even pured triodes, but UL etc. I have build a RC-Coupled 300bXls SE Amp, but my speaker sounded very soft with them...not enough air and transparency etc. 12 Ribbons and 18 4" Eton chassis want something stronger than 15W SE. The 30-40 Watt of the UL PP El34 comtrols it nicely though. I think it is more than just Watts and damping factor why it did not work with SE.
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300B and EL34 are two different animals. With EL34 you have more options (if you use no feedback or just local feedback at the output stage).
One option is to use a SE driver with SE-to-PP IT. The LL1660 is ok. In such case I would pick a better driver than the 6n6p that IME is not as linear as other solutions.
With the LL1660/10mA in Alt V connection 2.25:2+2 the 4P1L triode connected is a better choice. You can run it at about 200-230V anode voltage and 18-20mA. 20 mA is more than the specified 18 mA but it is fine if you don't need all that signal headroom with EL34's.
The maximun swing a driver can provide is Vrms=4.44 x i x L x f in sinusoidal regime. I is the anode current, L inductance and f the frequency. So once you make sure the i x L is enough at 20 Hz you will be fine at higher frequency. Of course such max V rms should be significantly more than you need otherwise the driver will distort getting close to cut-off (assuming plate voltage is enough on the other side) despite the fact the transformer is fine. With the LL1660, 18 mA anode current and 42H the driver could swing at the primary 67Vrms @20Hz which is singificantly more than you need for the EL34's. You can expect low distortion from the driver at all levels across the audio range.
The SE-to-PP trnasformer is the most difficult to make among interstages but when the driver hasn't got a huge job it makes things simpler and better (especially soundwise).
Don't put anything across the secondaries of the IT unless you are certain. At high frequency you might guess some peaking from datasheet in unloaded conditions and for this reason Lundahl suggests some RC networks but in reality the input capacitance of power tubes + other parasitic circuit capacitance might be enough to get nice and smooth response. You will will know when you try....
One option is to use a SE driver with SE-to-PP IT. The LL1660 is ok. In such case I would pick a better driver than the 6n6p that IME is not as linear as other solutions.
With the LL1660/10mA in Alt V connection 2.25:2+2 the 4P1L triode connected is a better choice. You can run it at about 200-230V anode voltage and 18-20mA. 20 mA is more than the specified 18 mA but it is fine if you don't need all that signal headroom with EL34's.
The maximun swing a driver can provide is Vrms=4.44 x i x L x f in sinusoidal regime. I is the anode current, L inductance and f the frequency. So once you make sure the i x L is enough at 20 Hz you will be fine at higher frequency. Of course such max V rms should be significantly more than you need otherwise the driver will distort getting close to cut-off (assuming plate voltage is enough on the other side) despite the fact the transformer is fine. With the LL1660, 18 mA anode current and 42H the driver could swing at the primary 67Vrms @20Hz which is singificantly more than you need for the EL34's. You can expect low distortion from the driver at all levels across the audio range.
The SE-to-PP trnasformer is the most difficult to make among interstages but when the driver hasn't got a huge job it makes things simpler and better (especially soundwise).
Don't put anything across the secondaries of the IT unless you are certain. At high frequency you might guess some peaking from datasheet in unloaded conditions and for this reason Lundahl suggests some RC networks but in reality the input capacitance of power tubes + other parasitic circuit capacitance might be enough to get nice and smooth response. You will will know when you try....
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Hi Blitz,
I'm glad you found it useful! Because iron and tubes come in fewer configurations than caps, I used only those variables in the spreadsheet. The coupling cap will affect LF roll off, but I didn't apply it in the spreadsheet because it's fairly easy to calculate a value that won't much affect the given numbers.
45...I am still trying to understand this formular and its meaning
Vrms=4.44 x i x L x f
If I got you right, this is a limiting condition, right ?
So, let's say we want to drive a Push-Pull 300BXLS amp With a Long-Tail Pair and the inpit is not SE, but balanced...
With the 1692A we have in PP 210H/2=105H. If a 801A is used and run at 30mA, the formula would give 4.44*0.03*105*20=280Vrms ...
So, this would mean enough headroom I guess...now the issue is more, how to get this drive/gain...which raises the question if the IT should be used 1:1 or more like a 1:2 ...?
Vrms=4.44 x i x L x f
If I got you right, this is a limiting condition, right ?
So, let's say we want to drive a Push-Pull 300BXLS amp With a Long-Tail Pair and the inpit is not SE, but balanced...
With the 1692A we have in PP 210H/2=105H. If a 801A is used and run at 30mA, the formula would give 4.44*0.03*105*20=280Vrms ...
So, this would mean enough headroom I guess...now the issue is more, how to get this drive/gain...which raises the question if the IT should be used 1:1 or more like a 1:2 ...?
It comes from the transformer formulas with a DC current that generates a DC field and signal that generates an AC field. At the origin it is expressed in terms of turns, induction, core cross-section and form fator for the AC. However it can be derived using relations among those variables with the condition that BAC cannot be bigger than BDC (i.e. no zero crossing) in SE transformer. So it is a limiting condition for both transformer or tube (the first to break that rule). In your case has the LL1660 has good headroom it is the tube setting the limit at 18 mA plate current.
In principle the LL1660 in 2.25:2+2 config. would be happy with 82.5Vrms @20Hz at the primary with 18mA DC however the tube just can't provide that. You can increase the tube capability by increasing the anode current but this will reduce the headroom for the signal. So you can play with anode current until you maximize the output.
If you drive the IT in PP it is different. At worst the driver can work in class AB. You can have zero crossing.
However it is a better idea if you make it work in class A for better linearity.
If the DC currents are perfectly balanced you have the specified available voltage.
The headroom will be less if you have some imbalance in DC currents of the drivers or grid current at the power tubes.
The LL1692a PP should sustain some little imbalance of 4-5 mA. This means that 5 mA current will generate about 0.9T DC induction (according to Lundahl's way of specifying permissible DC current) and only 0.7T will be availabe for the signal.
If the configuration is 1.75+1.75:2+2 the headroom at the primary in absence of DC imbalance will be 420Vmrs (210+210 or 240+240 at the secondary) @30Hz. At 20Hz the signal you can apply is inversely proportional to frequency.
If you have 5 mA imbalance and only 0.7T left for the signal the headroom at the primary in 1.75+1.75: 2+2 will be about 175V rms @ 30Hz and about 116 Vrms (58+58) @20Hz.
However I would rather use the transformer the other way around with some step-down. Primary inductance will be even higher. Maybe you could buy the 8-10mA version. Inductance will still be high but you don't need to worry about headroom in presence of little imbalance.
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Blitz,
Not exactly.
When the driver tubes are both conducting, the plate resistance of each drives its corresponding 1/2 winding. The tubes are out of phase, driving opposite phase windings; that is good, they aid each other. As to the effective impedance that drives the transformer, they are 'effectively' in parallel across a 1/2 winding. In class A that works good. Effectively rp/2 drives 1/2 primary winding (not really that way, but impedance wise it is).
But as soon as the drivers go into the 'B' portion of AB, then one of the tubes is cut off.
We have one plate resistance, rp driving 1/2 turns of the primary, and nothing driving the other 1/2 winding.
Inductance is according to the square of the turns ratio.
1000 turns center tapped has 500 turns per 1/2 winding.
The turns ratio is 2:1, but the inductance is the square of 2:1; that is 4:1 (or going the opposite way, 1:4).
Or more simply, the 1/2 winding has 1/4 the inductance of the whole winding.
We have 210 Henrys/4 = 52.5 Henrys in 1/2 winding.
One rp is now driving 52.5 Henrys (whenever one of the tubes is cut off, the 'B' in AB).
The effect would be to cause 3rd harmonic distortion at very low frequencies, because each tube will be cut off, but at opposite extremes of the signal swing.
Therefore, the way to aim at such a driver design is to have enough drive voltage in class A operation, even enough to cause grid current in the output tubes (which at that point will also cause 3rd harmonic distortion).
Yes, at some point, properly balanced push pull amplifiers eventually head to a region where the signal is foreshortened (clipped) at both the top and bottom of the waveform.
Of course there is another principle or adage. It is especially applicable to non-feedback amplifiers with triode output tubes (or triode wired pentode / triode wired beam power tubes).
Do not let there ever be enough drive to the output stage in a push pull amplifier to cause it to draw grid current, or to ever cause one of the output tubes to cut off (keep it in class A). The reason for that is so that the rp and the rp of the 2 output tubes are always there, driving their corresponding 1/2 windings. That has the effect of keeping the output always present the same driving impedance to the loudspeaker. Loudspeakers like that.
Without feedback, properly designed push pull amps exhibit primarily 3rd harmonic distortion, with little or no 2nd harmonic distortion, especially if they are push pull or balanced through all the tube stages.
Foreshortening at only the top or only the bottom, but not both, is 2nd harmonic distortion.
Without feedback, a single ended 2 stage amplifier that does not have 'driver and output stage distortion cancellation' will have primarily 2nd harmonic distortion, and little or no 3rd harmonic distortion.
With some 'driver and output stage distortion cancellation', the 3rd harmonic distortion may be as large as, or larger than, the 2nd harmonic distortion.
Now that I have said all that, remember:
All generalizations have exceptions
Not exactly.
When the driver tubes are both conducting, the plate resistance of each drives its corresponding 1/2 winding. The tubes are out of phase, driving opposite phase windings; that is good, they aid each other. As to the effective impedance that drives the transformer, they are 'effectively' in parallel across a 1/2 winding. In class A that works good. Effectively rp/2 drives 1/2 primary winding (not really that way, but impedance wise it is).
But as soon as the drivers go into the 'B' portion of AB, then one of the tubes is cut off.
We have one plate resistance, rp driving 1/2 turns of the primary, and nothing driving the other 1/2 winding.
Inductance is according to the square of the turns ratio.
1000 turns center tapped has 500 turns per 1/2 winding.
The turns ratio is 2:1, but the inductance is the square of 2:1; that is 4:1 (or going the opposite way, 1:4).
Or more simply, the 1/2 winding has 1/4 the inductance of the whole winding.
We have 210 Henrys/4 = 52.5 Henrys in 1/2 winding.
One rp is now driving 52.5 Henrys (whenever one of the tubes is cut off, the 'B' in AB).
The effect would be to cause 3rd harmonic distortion at very low frequencies, because each tube will be cut off, but at opposite extremes of the signal swing.
Therefore, the way to aim at such a driver design is to have enough drive voltage in class A operation, even enough to cause grid current in the output tubes (which at that point will also cause 3rd harmonic distortion).
Yes, at some point, properly balanced push pull amplifiers eventually head to a region where the signal is foreshortened (clipped) at both the top and bottom of the waveform.
Of course there is another principle or adage. It is especially applicable to non-feedback amplifiers with triode output tubes (or triode wired pentode / triode wired beam power tubes).
Do not let there ever be enough drive to the output stage in a push pull amplifier to cause it to draw grid current, or to ever cause one of the output tubes to cut off (keep it in class A). The reason for that is so that the rp and the rp of the 2 output tubes are always there, driving their corresponding 1/2 windings. That has the effect of keeping the output always present the same driving impedance to the loudspeaker. Loudspeakers like that.
Without feedback, properly designed push pull amps exhibit primarily 3rd harmonic distortion, with little or no 2nd harmonic distortion, especially if they are push pull or balanced through all the tube stages.
Foreshortening at only the top or only the bottom, but not both, is 2nd harmonic distortion.
Without feedback, a single ended 2 stage amplifier that does not have 'driver and output stage distortion cancellation' will have primarily 2nd harmonic distortion, and little or no 3rd harmonic distortion.
With some 'driver and output stage distortion cancellation', the 3rd harmonic distortion may be as large as, or larger than, the 2nd harmonic distortion.
Now that I have said all that, remember:
All generalizations have exceptions
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6A3sUMMER, thanks again for all the good advice. It took a while to get the transformers ordered etc.
I tried to measure their inductance http://www.lundahl.se/wp-content/uploads/datasheets/1692A.pdf ...my LCR-Meter gave me 140H across the complete primary /(ALT-M Config) plate-to plate...but only 34H when measuring from Anode to CT...
...I continued to check the inductance by utilizing a reistor of 10k in series with a sin-generator and my Oszi and figure that the frequences response was like the specified 210H plate-to-plate...but I do not want to use it in SE-mode, but as differential stage...and with a 801A with 5000ohm...
To be precise: I want to use it for a DAC-Output Stage LTP with a current sink at the cathode trying different tubes, like as well the 801A (https://frank.pocnet.net/sheets/049/8/801A.pdf), so we talk an rp of 5000 Ohm or more (6sn7) here.
The LL1692A will be the output-transformer of the line-stage, so will drive XLR-Cable to the DIff-Pair of the Power-amp.
So, let me understand if I got you right:
1. Target: The 801A needs to stay in class A.
I guess this can be achieved by pulling enough current through the two 801A and controlled through the current sink. I planned to use them with 30-35mA each, so 60-70mA in total.
What does this now mean for the frequency behavior ? If I got it right from you, two effects will happen:
A. Both tubes conduct all the time, so Inductance is half of the 210H, not a quarter.
B. rp is becoming as well half: 5000/2=2500 Ohm, correct ?
So, if the -1db-Point for 5000 Ohm in SE over the whole winding (210H) is 15 Hz(I plugged it into the anode choke-formular of the vt52.com-sheet), the 105H would give at 2500 Ohm as well 15 Hz...correct ?
2. Target: "Therefore, the way to aim at such a driver design is to have enough drive voltage in class A operation, even enough to cause grid current in the output tubes"
Here I need a bit more detail: What is "enough" in my case ? If I am running the 801A at 300V/35mA, I should have -10V on its grid. My DAC would always give max 2Vpk as an output, so plenty of headroom. Good enough ?
As stated in my initial post I have no experience whatsoever with IT. but want to learn their utilization...so please help me to see if I have the right understanding here and the right design choices. THX a lot !
I tried to measure their inductance http://www.lundahl.se/wp-content/uploads/datasheets/1692A.pdf ...my LCR-Meter gave me 140H across the complete primary /(ALT-M Config) plate-to plate...but only 34H when measuring from Anode to CT...
...I continued to check the inductance by utilizing a reistor of 10k in series with a sin-generator and my Oszi and figure that the frequences response was like the specified 210H plate-to-plate...but I do not want to use it in SE-mode, but as differential stage...and with a 801A with 5000ohm...
To be precise: I want to use it for a DAC-Output Stage LTP with a current sink at the cathode trying different tubes, like as well the 801A (https://frank.pocnet.net/sheets/049/8/801A.pdf), so we talk an rp of 5000 Ohm or more (6sn7) here.
The LL1692A will be the output-transformer of the line-stage, so will drive XLR-Cable to the DIff-Pair of the Power-amp.
So, let me understand if I got you right:
1. Target: The 801A needs to stay in class A.
I guess this can be achieved by pulling enough current through the two 801A and controlled through the current sink. I planned to use them with 30-35mA each, so 60-70mA in total.
What does this now mean for the frequency behavior ? If I got it right from you, two effects will happen:
A. Both tubes conduct all the time, so Inductance is half of the 210H, not a quarter.
B. rp is becoming as well half: 5000/2=2500 Ohm, correct ?
So, if the -1db-Point for 5000 Ohm in SE over the whole winding (210H) is 15 Hz(I plugged it into the anode choke-formular of the vt52.com-sheet), the 105H would give at 2500 Ohm as well 15 Hz...correct ?
2. Target: "Therefore, the way to aim at such a driver design is to have enough drive voltage in class A operation, even enough to cause grid current in the output tubes"
Here I need a bit more detail: What is "enough" in my case ? If I am running the 801A at 300V/35mA, I should have -10V on its grid. My DAC would always give max 2Vpk as an output, so plenty of headroom. Good enough ?
As stated in my initial post I have no experience whatsoever with IT. but want to learn their utilization...so please help me to see if I have the right understanding here and the right design choices. THX a lot !
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Blitz,
We have covered a lot of different questions and ideas (thanks to all the others responding in this thread).
When you measure inductance of a single winding, versus a pair of identical windings in series, you will get a 4:1 inductance ratio. That is because inductance is proportional to the number of turns squared (L prop. to N squared). 2 x the turns, then inductance = 2 x 2 = 4 x the inductance.
A push pull interstage and/or output transformer has 1/4 inductance for only 1/2 the primary winding.
As you know this is a determining factor for low frequency response, along with the plate resistance, rp.
The impedance is also determined by the square of the turns ratio. Load the 8 Ohm tap with 8 Ohms. Then check the impedance of a 5k Ohm plate to plate primary. Then check the impedance of the center tap to one plate side of the primary (1/2 of the windings). The Z will be 1/4 of 5k Ohms = 1250 Ohms.
Again, impedance just like inductance is proportional to the square of the turns.
However, for 2 tubes in push pull, as long as they are each conducting a fair amount of current, they are effectively in parallel (not in phase, but in the effective impedance that drives the 1/2 winding inductance and impedances. So, 1250 Ohms driven by 2 1250 Ohm rp tubes (625 Ohms parallel ups) gives a 2:1 ratio of drive to load impedances.
Class A will have both tubes in conduction. Suppose as a sine wave goes positive, and one tube goes from 30 mA to 50 mA, the other tube will go from 30 mA to 10 mA; then as the sine wave goes negative, they reverse the trend of which tube is conducting more current.
The plate resistance will be going higher on the tube that is only conducting 10mA.
So, with larger and larger drive signals, you will get a shortening effect on both alternations (third harmonic distortion). By the time you drive grid current (A2 or AB2), one of the tubes will either be cut off, or at least will have a very high plate resistance.
This higher rp is no longer effectively driving 1/2 of the primary, so only 1 tube is driving the primary (the other 1/2). Now the effect is not a 2 parallel tube drive.
If you are using 2 801s with -10V biases, your driver ought to be able to drive each grid + and - 10V in a linear fashion. A typical CD player that is powered off the mains, has 2Vrms out (2.83V peak).
It will not drive an 801 directly.
And music CDs do not all go to 0 dB (full scale) on the DACs. Some are "quieter", so there will not be 2Vrms out.
A long XLR cable will have too much capacitance for some interstage transformers, and for some tubes driving that interstage. There can be resonances, poor square wave response, and high frequency fall off, and peaking.
Some interstage have various sub models, that allow DC, a medium DC imbalance, and a very small DC imbalance. These are the same except for the air gap spacing.
Single ended have to have the larger gap, and the push pull can use the medium or small gap, if the current imbalance of the driving tubes is small or very small.
We have covered a lot of different questions and ideas (thanks to all the others responding in this thread).
When you measure inductance of a single winding, versus a pair of identical windings in series, you will get a 4:1 inductance ratio. That is because inductance is proportional to the number of turns squared (L prop. to N squared). 2 x the turns, then inductance = 2 x 2 = 4 x the inductance.
A push pull interstage and/or output transformer has 1/4 inductance for only 1/2 the primary winding.
As you know this is a determining factor for low frequency response, along with the plate resistance, rp.
The impedance is also determined by the square of the turns ratio. Load the 8 Ohm tap with 8 Ohms. Then check the impedance of a 5k Ohm plate to plate primary. Then check the impedance of the center tap to one plate side of the primary (1/2 of the windings). The Z will be 1/4 of 5k Ohms = 1250 Ohms.
Again, impedance just like inductance is proportional to the square of the turns.
However, for 2 tubes in push pull, as long as they are each conducting a fair amount of current, they are effectively in parallel (not in phase, but in the effective impedance that drives the 1/2 winding inductance and impedances. So, 1250 Ohms driven by 2 1250 Ohm rp tubes (625 Ohms parallel ups) gives a 2:1 ratio of drive to load impedances.
Class A will have both tubes in conduction. Suppose as a sine wave goes positive, and one tube goes from 30 mA to 50 mA, the other tube will go from 30 mA to 10 mA; then as the sine wave goes negative, they reverse the trend of which tube is conducting more current.
The plate resistance will be going higher on the tube that is only conducting 10mA.
So, with larger and larger drive signals, you will get a shortening effect on both alternations (third harmonic distortion). By the time you drive grid current (A2 or AB2), one of the tubes will either be cut off, or at least will have a very high plate resistance.
This higher rp is no longer effectively driving 1/2 of the primary, so only 1 tube is driving the primary (the other 1/2). Now the effect is not a 2 parallel tube drive.
If you are using 2 801s with -10V biases, your driver ought to be able to drive each grid + and - 10V in a linear fashion. A typical CD player that is powered off the mains, has 2Vrms out (2.83V peak).
It will not drive an 801 directly.
And music CDs do not all go to 0 dB (full scale) on the DACs. Some are "quieter", so there will not be 2Vrms out.
A long XLR cable will have too much capacitance for some interstage transformers, and for some tubes driving that interstage. There can be resonances, poor square wave response, and high frequency fall off, and peaking.
Some interstage have various sub models, that allow DC, a medium DC imbalance, and a very small DC imbalance. These are the same except for the air gap spacing.
Single ended have to have the larger gap, and the push pull can use the medium or small gap, if the current imbalance of the driving tubes is small or very small.
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