Page 1 of 2 12 LastLast
Results 1 to 35 of 57

Thread: LTPI balance

  1. #1
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10

    LTPI balance

    I'm posting this as a query to see if anyone has seen this idea as a means of balancing an LTPI before. This forum audience has seen more amps than I'll likely ever see. I've looked around the web and in my Morgan book. I'm guessing it may have turned up in the hifi domain, but I have not found it. All my searches just find pointers to active current sources to help balance a PI. I have not seen this solution yet. The circuit is pretty simple...

    Click image for larger version. 

Name:	LTPI 4.jpg 
Views:	101 
Size:	114.8 KB 
ID:	52638

    It works by adding about 25% more of the NFB signal at the base of the tail than at the inverting input. That turns out to be just about right for the tail base to track the cathode voltage exactly, thereby making the tail an almost perfect current source. I've Spice'd this PI (in a Deluxe Reverb output stage) and it seems to work really well, especially given the added parts count of only 2 resistors. The outputs are damn near perfectly balanced with two 100K plate load resistors.

    I don't really want to debate whether a balanced PI sounds good or violates some secret amp juju. This is just a design question. If you wanted to build a balanced PI, this seems a way to get one without resorting to active sources or changing load resistors. I realize the 82k/100k load resistors are the standard way to balance the outputs by compensating for tube current differences. This is an alternative that actually balances the tube currents. Has anyone seen or used this idea before? Are there any downsides to the circuit behavior that I've missed?

    Given the way it works, it also seems to be a way to make a lower value tail resistor perform like a really high-impedance current source, so you can reduce the cathode elevation voltage and add headroom to the outputs without sacrificing balance.

    5 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  2. #2
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    If we ignore NFB and just concentrate on the virtual current source, then using the speaker signal is not going to give you as precise a result as, say, a constant current source. The reason being tolerance variations together with the frequency and phase response of the succeeding stages. How important these are depends on you design goals, of course.

    Afterthought: The whole idea falls flat when you start to clip. Whether that is a good or a bad thing is a whole new subject.

    0 Not allowed! Not allowed!
    Last edited by nickb; 02-23-2019 at 05:22 PM.
    Experience is something you get, just after you really needed it.

  3. #3
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by nickb View Post
    If we ignore NFB and just concentrate on the virtual current source, then using the speaker signal is not going to give you as precise a result as, say, a constant current source. The reason being tolerance variations together with the frequency and phase response of the succeeding stages. How important these are depends on you design goals, of course.
    Thanks - I see the point about spkr impedance will change the signal level at varied freqs. Fortunately, the whole NFB purpose is to reduce those variations.
    I suspect an active source would do better. I'll just have to build/test one, Or maybe I can get the sim to show some of those effects.

    Edit - I ran a sim and with small resistor tweaks, I show 25-30 db lower tail current vs plate current over 200-5kHz.
    That's pretty good IMO. Basically ua vs ma. Of course it's just a sim, but still promising. I also sim'd an active current source (mostfet) and it does no better at 1KHz. The performance goes down at low and high freq's as you suggested.

    1 Not allowed! Not allowed!
    Last edited by uneumann; 02-23-2019 at 12:53 AM.
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  4. #4
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    I would think that is we did a transient (large signal) analysis including transformer and speaker effects things would look worse. But I do agree that the differences are probably too small to worry about.

    One curiosity, I tried a CCS using two BC547B bipolars and got -89dB at 10Khz. Any FETs I tried were much worse by 40dB or so.

    0 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  5. #5
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    How is PI output signal symmetry with global NFB disconnected?
    In this case it seems that your arrangement somewhat decreases the grid to cathode input signal of V2, actually slightly increasing output asymmetry.

    But with global NFB connected, the voltage divider R9/R10 makes V2 receive less feedback signal than V1 and thus its output is increased.

    Am I missing something?
    Where does arrow "C" connect to?

    0 Not allowed! Not allowed!
    Last edited by Helmholtz; 02-23-2019 at 03:36 PM.
    - Own Opinions Only -

  6. #6
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by nickb View Post
    I would think that is we did a transient (large signal) analysis including transformer and speaker effects things would look worse. But I do agree that the differences are probably too small to worry about.

    One curiosity, I tried a CCS using two BC547B bipolars and got -89dB at 10Khz. Any FETs I tried were much worse by 40dB or so.
    Interesting, I only did a quick test with a mosfet to compare these two circuits.
    At 1Khz the tail currents are basically the same (in ua range). The CCS holds that level while the upper circuit slowly degrades at higher and lower freqs.

    Comparing the upper circuit to the "normal" LTPI (without the added tail amplitude) shows a much bigger difference.
    The signal traces below show how balanced the upper circuit outputs are at 500Hz with both plate load resistors at 100K.

    Click image for larger version. 

Name:	Capture.JPG 
Views:	45 
Size:	148.6 KB 
ID:	52649

    Click image for larger version. 

Name:	Capture3.JPG 
Views:	37 
Size:	326.6 KB 
ID:	52648

    0 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  7. #7
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by Helmholtz View Post
    How is PI output signal symmetry with global NFB disconnected?
    In this case it seems that your arrangement somewhat decreases the grid to cathode input signal of V2, actually slightly increasing output asymmetry.

    But with global NFB connected, the voltage divider R9/R10 makes V2 receive less feedback signal than V1 and thus its output is increased.

    Am I missing something?
    Where does arrow "C" connect to?
    Sorry for the confusion, A and C are just reference points. A is the input and C is just a reference for the cathode signal - it doesn't go anywhere or connect to anything.

    Yes - the V2 grid signal is smaller than the tail signal - which is exactly the point. I used the Deluxe Reverb as my example case. That LTPI uses a 820/47 ohm NFB network. In this LTPI, the tail signal is made bigger by increasing R8 to 75 ohms. That bigger tail signal now tracks the cathode signal at C more accurately - making the tail resistor look like a very high impedance. That bigger tail signal is is then attenuated by R9 and R10 to get back to the original NFB level at the input of V2. So, the V2 level is the same in a normal Deluxe Reverb and in this "enhanced-tail" LTPI.

    If you don't use NFB into the V2 grid - like in Marshall 18 watt amps, you can still use this idea. Just tap some feedback from the spkr signal and run it to the PI tail with a divider network (R7 and R8). That makes the tail resistor a very high impedance, improves the PI balance, and you can even reduce the tail resistor value to get more headroom out of the PI without sacrificing balance.

    0 Not allowed! Not allowed!
    Last edited by uneumann; 02-23-2019 at 06:10 PM. Reason: can't spell *&^^%
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  8. #8
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    That LTPI uses a 820/47 ohm NFB network. In this LTPI, the tail signal is made bigger by increasing R8 to 75 ohms. That bigger tail signal now tracks the cathode signal at C more accurately - making the tail resistor look like a very high impedance.
    Does that mean that PI output symmetry improves even without global NFB connected?

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  9. #9
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by Helmholtz View Post
    Does that mean that PI output symmetry improves even without global NFB connected?
    Yup - that's a key point that seems true. The NFB tail connection can be used to balance the PI whether you use the NFB connection to V2 or not.
    These are two independent connections that perform different functions in the circuit. (There is a more detailed description/analysis on my web site if you're interested.)

    It seems too simple (like a free lunch) and sort of obvious in hindsight, which is why I was looking for a prior article or example somewhere.
    I figure this must be something that other people have noticed or played with in the past. It's really just bootstrapping the tail resistor.
    A lack of references to something like this makes me wonder if there's a fatal flaw that leads to discarding the idea.
    So far I can not find a reference or a fatal flaw...

    0 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  10. #10
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    Quote Originally Posted by uneumann View Post
    It seems too simple (like a free lunch) and sort of obvious in hindsight, which is why I was looking for a prior article or example somewhere.
    I figure this must be something that other people have noticed or played with in the past. It's really just bootstrapping the tail resistor.
    A lack of references to something like this makes me wonder if there's a fatal flaw that leads to discarding the idea.
    So far I can not find a reference or a fatal flaw...
    But isn't what you are doing here exactly what has been done for decades? The only minor variation is a different feedback ratio due to R9 and R10. What am I missing?

    1 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  11. #11
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by nickb View Post
    But isn't what you are doing here exactly what has been done for decades? The only minor variation is a different feedback ratio due to R9 and R10. What am I missing?
    Exactly right, Nick. That's why I say it's sort of obvious. The key point is simply to make the tail signal about 25-30% bigger than the V2 grid signal. The exact resistors/ratios can be found experimentally (I did it by iterative simulation). I'm sure there is a formula that can be derived for this, but I just have not taken the time/effort to do that yet - anyone want to take a stab? The target is simply to make the tail signal match the cathode signal.

    The fact that the standard PI has been around so long and doesn't optimize the tail signal voltage is sort of surprising. It's like leaving money on the ground. Add two resistors and you get a balanced PI - then you can forget all this different load resistor stuff and it even works for non-NFB PI circuits.

    0 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  12. #12
    Supporting Member loudthud's Avatar
    Join Date
    May 2006
    Location
    Near Dallas Texas
    Posts
    3,480
    Thumbs Up/Down
    Received: 152/0
    Given: 92/1
    Rep Power
    18
    As an experiment, I'd like install a pot or switch to see how it sounds.

    0 Not allowed! Not allowed!
    WARNING! Musical Instrument amplifiers contain lethal voltages and can retain them even when unplugged. Refer service to qualified personnel.
    REMEMBER: Everybody knows that smokin' ain't allowed in school !

  13. #13
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by uneumann View Post
    So far I can not find a reference or a fatal flaw...
    Well - I found one flaw - maybe not fatal, but worth noting.
    In non-NFB amps, sometimes the PI is used as a mixer.
    In that case the V2 input produces positive feedback to the tail, making the balance worse not better.
    The bootstrap only works if the V2 input is either grounded or tied to NFB.

    0 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  14. #14
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    The signal at the bottom of the tail is ideally half the sum of the two inputs. If the open loop gain is a and the feedback ratio is b, then you can calculate the ratio of the output to bottom of tail resistor as

    1/(2a) + b

    So, for example taking a as 14 and b= 0.058 from around what you have for the design shown, with a guess for the OPT thrown in.

    Then k = 0.5/14 + 0.058 = 0.094

    In your design you have k= 82/(82+820) = 0.91 so that looks about right.

    Also, the ratio of feedback to tail resistor is

    0.058/0.094 ~= 0.63 i.e 37% lees than the tail signal.

    Still, the current source is the superior option as it's a no-brainer and works for the mixer too

    1 Not allowed! Not allowed!
    Last edited by nickb; 02-24-2019 at 07:47 PM. Reason: Typo
    Experience is something you get, just after you really needed it.

  15. #15
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    The signal at the bottom of the tail is ideally half the sum of the two inputs.
    Sorry, this I don't understand (assuming that "bottom of the tail" means the junction between R6 and R8)
    To my knowledge, its the junction of the 2 cathodes where half the input signal appears. The signal at the top of R8 (= bottom of the tail?) must be much smaller as the value of R8 is only a small fraction of the total tail resistance. For this same reason I can't see much local feedback or bootstrap effect from doubling the value of R8. The balance of the circuit still seems to be dominated by the value of R6.

    Also there seems to be some typing error in the formula(s) for k above. E.g the result of 1/(2a) + b would be 0.616. And the result of k = 0.5*14 + 0.058 = 7.06 (not 0.094).

    0 Not allowed! Not allowed!
    Last edited by Helmholtz; 02-24-2019 at 07:29 PM.
    - Own Opinions Only -

  16. #16
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    Quote Originally Posted by Helmholtz View Post
    Sorry, this I don't understand (assuming that "bottom of the tail" means the junction between R6 and R8)
    To my knowledge, its the junction of the 2 cathodes where half the input signal appears. The signal at the top of R8 (= bottom of the tail?) must be much smaller as the value of R8 is only a small fraction of the total tail resistance. For this same reason I can't see much local feedback or bootstrap effect from doubling the value of R8. The balance of the circuit still seems to be dominated by the value of R6.
    In order for R6 to appear infinite the voltage at both ends is made the same. So if half the input voltage is at the top then the same is at the bottom.
    Also there seems to be some typing error in the formula(s) for k above. E.g the result of 1/(2a) + b would be 0.616. And the result of k = 0.5*14 + 0.058 = 7.06 (not
    0.094).
    Yes there was a typo. I fixed it.

    0 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  17. #17
    Supporting Member
    Join Date
    Sep 2007
    Posts
    3,137
    Thumbs Up/Down
    Received: 181/0
    Given: 0/0
    Rep Power
    16
    Given how cheap Ds and ECs are these days, just add a negative supply and use a large resistor from the cathodes. Then it is easy to add in two channels, or use the second input for negative feedback.

    2 Not allowed! Not allowed!

  18. #18
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    [QUOTE]In order for R6 to appear infinite the voltage at both ends is made the same. So if half the input voltage is at the top then the same is at the bottom./QUOTE]

    Yes, but what would make R6 appear infinite in the proposed circuit (if you don't replace it with a real current source)? I don't see a measure that applies the same signal voltage at the bottom of R6 as on its top to achieve such bootstrapping effect in post #1. What am I missing?

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  19. #19
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    Quote Originally Posted by Helmholtz View Post
    What am I missing?
    The input from the speaker, I would guess.

    1 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  20. #20
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by nickb View Post
    The input from the speaker, I would guess.
    Did anybody verify that this actually makes (all) signal voltages the same on both sides of R6?


    Still having problems with your k formula. You say b = 0.58 but are using b = 0.058.

    0 Not allowed! Not allowed!
    Last edited by Helmholtz; 02-24-2019 at 07:45 PM.
    - Own Opinions Only -

  21. #21
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    Quote Originally Posted by Helmholtz View Post
    Did anybody verify that this actually makes (all) signal voltages the same on both sides of R6?


    Still having problems with your k formula. You say b = 0.58 but are using b = 0.058.
    It works in my simulation.

    It's 0.058 (typo again). If you look as the circuit you can see that for yourself.

    0 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  22. #22
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by Helmholtz View Post
    Did anybody verify that this actually makes (all) signal voltages the same on both sides of R6?
    Yes - This pic shows sim traces for the cathodes, V1 input and V2 input. The middle signal is the cathode signal and it's almost exactly the average of the two grid signals, as expected.

    Click image for larger version. 

Name:	Capture1.JPG 
Views:	20 
Size:	305.8 KB 
ID:	52661

    The next pic shows sim traces for the cathode and base of R6. They are almost identical, so the net signal voltage across R6 is virtually zero for the entire signal.

    Click image for larger version. 

Name:	Capture2.JPG 
Views:	17 
Size:	332.1 KB 
ID:	52662

    0 Not allowed! Not allowed!
    Last edited by uneumann; 02-24-2019 at 08:11 PM. Reason: typos
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  23. #23
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by uneumann View Post
    Yes - This pic shows sim traces for the cathodes, V1 input and V2 input. The middle signal is the cathode signal and it's almost exactly the average of the two grid signals, as expected.

    Click image for larger version. 

Name:	Capture1.JPG 
Views:	20 
Size:	305.8 KB 
ID:	52661

    The next pic shows sim traces for the cathode and base of R6. They are almost identical, so the net signal voltage across R6 is virtually zero for the entire signal.

    Click image for larger version. 

Name:	Capture2.JPG 
Views:	17 
Size:	332.1 KB 
ID:	52662
    Thanks!

    I just don't see how the circuit would improve balance for the forward signal especially in open loop situation - as claimed.

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  24. #24
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    And here are my sim results. Used b = 0.1 for no particularly good reason at all.

    Click image for larger version. 

Name:	ltpi_sim_2.JPG 
Views:	29 
Size:	466.6 KB 
ID:	52665

    Thanks for the idea uneumann. It's certainly the cheapest way to do this, just two $0.01 resistors.

    1 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  25. #25
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by nickb View Post
    And here are my sim results. Used b = 0.1 for no particularly good reason at all.

    Click image for larger version. 

Name:	ltpi_sim_2.JPG 
Views:	29 
Size:	466.6 KB 
ID:	52665

    Thanks for the idea uneumann. It's certainly the cheapest way to do this, just two $0.01 resistors.
    Thanks.
    Just to make sure: The sim results are for the open loop circuit as shown? Just wondering why you mention a feedback ratio b =0.1. Or are all "outs" connected?

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  26. #26
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by Helmholtz View Post
    Thanks!

    I just don't see how the circuit would improve balance for the forward signal especially in open loop situation - as claimed.
    So maybe this will help. Here is a basic non-NFB output stage. The input is 0.1v and the spkr signal is ~1.5v (both peak). The nfb input to the PI is grounded by a cap (C2 in this drawing), there is no NFB applied to the system. However the tail 10K resistor has an added 29 ohm resistor added (R2) which gets some of the spkr signal through the 820 ohm resistor.

    Click image for larger version. 

Name:	Capture1.JPG 
Views:	21 
Size:	88.3 KB 
ID:	52666

    The R2 signal is set to exactly match (as close as I can get with integer resistors) the cathode signal. The image below shows those two signals at R58 and R54. These are AC signals and both referenced to ground so they line up nicely in the trace image. Note they are almost identical.

    Click image for larger version. 

Name:	Capture2.JPG 
Views:	15 
Size:	241.0 KB 
ID:	52667

    The next image shows the two plate outputs. Note how nicely balanced they are for what is a pretty small tail resistor (10K).

    Click image for larger version. 

Name:	Capture3.JPG 
Views:	16 
Size:	278.2 KB 
ID:	52668

    The last image is what the plate signals look like if R2 is replaced by a wire. This is the normally imbalanced output for this 10K tail PI.

    Click image for larger version. 

Name:	Capture4.JPG 
Views:	17 
Size:	276.7 KB 
ID:	52669

    In fact you can do ridiculous things like reduce the tail resistor to 100 ohms. You'd never do that, right?
    Here are the plate outputs for the same circuit with 29 ohm R2 and a 100 ohm tail resistor. Looks pretty balanced to me.

    Click image for larger version. 

Name:	Capture5.JPG 
Views:	18 
Size:	253.1 KB 
ID:	52670

    3 Not allowed! Not allowed!
    Last edited by uneumann; 02-24-2019 at 10:05 PM. Reason: typo
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  27. #27
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    Quote Originally Posted by Helmholtz View Post
    Thanks.
    Just to make sure: The sim results are for the open loop circuit as shown? Just wondering why you mention a feedback ratio b =0.1. Or are all "outs" connected?
    All 'out' labels are the same node.

    For open loop k = 1/(2a) ( pretty obvious!!!). I see already uneumann gave you a sim for this case so I won't duplicate.

    0 Not allowed! Not allowed!
    Experience is something you get, just after you really needed it.

  28. #28
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    The high resistance R54 essentially prevents grounding of the global (speaker) NFB signal. So we still have a closed loop situation with global NFB. Why don't you just open the feedbck loop and maybe correct for the DC resistance?

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  29. #29
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by nickb View Post
    All 'out' labels are the same node.

    For open loop k = 1/(2a) ( pretty obvious!!!). I see already uneumann gave you a sim for this case so I won't duplicate.
    Sorry, but since the very beginning of this thread I have been asking about the open loop balance of the circuit.

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  30. #30
    "Thermionic Apocalypse" -JT nickb's Avatar
    Join Date
    Dec 2009
    Location
    Devon, UK
    Posts
    3,406
    Thumbs Up/Down
    Received: 632/1
    Given: 539/1
    Rep Power
    13
    Quote Originally Posted by Helmholtz View Post
    Sorry, but since the very beginning of this thread I have been asking about the open loop balance of the circuit.
    In the example given the loop is open. There is no NFB. Here is another example. I hope it helps clarify things

    Click image for larger version. 

Name:	ltpi_sim_3.JPG 
Views:	19 
Size:	264.9 KB 
ID:	52676

    0 Not allowed! Not allowed!
    Last edited by nickb; 02-24-2019 at 11:02 PM.
    Experience is something you get, just after you really needed it.

  31. #31
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by Helmholtz View Post
    Sorry, but since the very beginning of this thread I have been asking about the open loop balance of the circuit.
    Perhaps we're using some terms differently. In post #26 I'm showing a NON-NFB example. Maybe there's confusion since the spkr signal still comes back to create the tail signal. Yes, that's a fraction of the spkr signal getting fed into the tail, but it's not negative feedback (NFB) since that signal only goes to the tail and it has no impact on the gain of the circuit. The circuit is essentially open loop. To have negative feedback (NFB) you have to connect some version of the spkr signal into the V2 input. That's grounded in this example, so there is no global negative feedback in that circuit.

    As for the open loop uncorrected balance, that's shown when R2 is replaced by a wire (in post 26). That's grounding the tail as a standard PI would.

    0 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  32. #32
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by uneumann View Post
    Perhaps we're using some terms differently. In post #26 I'm showing a NON-NFB example. Maybe there's confusion since the spkr signal still comes back to create the tail signal. Yes, that's a fraction of the spkr signal getting fed into the tail, but it's not negative feedback (NFB) since that signal only goes to the tail and it has no impact on the gain of the circuit. The circuit is essentially open loop. To have negative feedback (NFB) you have to connect some version of the spkr signal into the V2 input. That's grounded in this example, so there is no global negative feedback in that circuit.

    As for the open loop uncorrected balance, that's shown when R2 is replaced by a wire (in post 26). That's grounding the tail as a standard PI would.
    OK. I see its not NFB. So the circuit needs a common mode speaker signal at the bottom of the tail resistor to improve balance, right?
    And it wouldn't have a benefit without the speaker signal?

    0 Not allowed! Not allowed!
    - Own Opinions Only -

  33. #33
    Member uneumann's Avatar
    Join Date
    Jun 2010
    Location
    Los Angeles
    Posts
    249
    Thumbs Up/Down
    Received: 51/0
    Given: 18/0
    Rep Power
    10
    Quote Originally Posted by Helmholtz View Post
    OK. I see its not NFB. So the circuit needs a common mode speaker signal to improve balance, right?
    Bingo - you got it - that's a good way to phrase it.
    Technically, the common mode signal could come from other sources too, but the spkr signal is pretty handy and works well for this purpose.

    0 Not allowed! Not allowed!
    Old Tele man: Equations provide theoretical values, SPICE provides approximate values; but, the ears provide exact values.
    Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

    https://sites.google.com/site/stringsandfrets/

  34. #34
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by uneumann View Post
    Bingo - you got it - that's a good way to phrase it.
    Technically, the common mode signal could come from other sources too, but the spkr signal is pretty handy and works well for this purpose.
    Thanks a lot.
    And congratulations on your circuit idea!

    3 Not allowed! Not allowed!
    - Own Opinions Only -

  35. #35
    Senior Member
    Join Date
    Mar 2018
    Location
    Germany
    Posts
    1,507
    Thumbs Up/Down
    Received: 1,110/1
    Given: 683/2
    Rep Power
    4
    Quote Originally Posted by uneumann View Post
    Bingo - you got it - that's a good way to phrase it.
    Technically, the common mode signal could come from other sources too, but the spkr signal is pretty handy and works well for this purpose.
    Just one more remark:
    While the common mode feedback signal produces no NFB, its phase still matters.

    0 Not allowed! Not allowed!
    - Own Opinions Only -

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. MOSFET LTPI
    By Zozobra in forum Theory & Design
    Replies: 11
    Last Post: 03-14-2017, 06:09 PM
  2. LTPI on a 6G2?
    By ron vogel in forum Mods & Tweaks
    Replies: 9
    Last Post: 11-12-2014, 01:22 PM
  3. LTPI ?
    By catnine in forum Theory & Design
    Replies: 26
    Last Post: 09-27-2013, 03:54 PM
  4. Adding NFB to LTPI
    By EFK in forum Theory & Design
    Replies: 12
    Last Post: 06-12-2010, 06:15 PM
  5. 5E3 w/ LTPI
    By hasserl in forum 5 E 3
    Replies: 16
    Last Post: 02-21-2007, 06:20 PM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •