Silvertone 1472 Early Breakup Fix

Discussion in 'Amp Tech Center' started by K Teacher, Jun 6, 2021.

  1. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    Greetings,

    I was reading a couple of threads asking how to prevent the Silvertone 1472 amp from breaking-up at very low volume settings. There were a lot of suggestions, but none of them really addressed the issue.

    What called my attention about this amp circuit design, is the fact that it has grid-leak bias in the split-load phase inverter.

    Grid-leak bias produce a very low bias voltage, and for this reason, they were mostly used on the amp's first amplification stage, where the signal voltage is low (see partial schematic below).


    SILVERTONE 1472



    upload_2021-6-6_15-4-36.jpg


    It's very unusual to have grid-leak bias in a later amplification stage and I was curious of the effect of this design on the amp's tone. The response came from these threads, reporting breakup at lower-than-expected volume settings.

    And this is why -- when the bias voltage is that low, the input voltage required to drive the amp to full power, will be such that will draw grid current. When this occurs, the grid-cathode works as clamping diode, causing output signal distortion. This will become clear, after examining the phase inverter load-line and bias voltage on the following graph.

    From the values on the schematic:

    Ebb = 220V
    Eb_k = 157V - 65V = 92V
    R_load_dc = 68K + 68K = 136K
    I_max = Ebb / R_load_dc = 220V / 136K = 1.62 mA


    Plotting these values on the 12AX7 "AVERAGE PLATE CHARACTERISTICS" graph:


    upload_2021-6-6_15-4-36.jpg


    The Silvertone 1472 schematic shows 92 V drop across the 12AX7. From the plot, to get that drop with
    68K + 68K load, it is necessary a plate current of 0.94 mA, commanded by a grid voltage bias of - 0.58V.

    Thus, the amplitude of the input signal to produce an enough output to drive the 6V6's to full power, will exceed the ZERO volts grid line.

    Now, if we insert a bias resistor in the circuit, at the point marked with the "RED X", we can modify the split-load tube bias point. In the plot's example, using a 1K8 bias resistor for the 12AX7, will generate a more centered bias voltage and allow for a larger input signal, without driving the input grid to current draw region and thus resolving the low volume breakup issue.

    Just to check, with the new bias point, each 68K resistor will drop (220V - 130V)/ 2 = 45V each… thus, the 12AX7 lower 68K resistor to ground will measure 45V, and the plate to ground voltage will be (220V - 45V) = 175V, give or take few percent due to components' tolerances.

    If the load-line drawing and bias calculations is not entirely clear, there are several good articles on the net explaining that in much more detail.

    If you implement these changes, please report the before and after results.

    Just my 2 cents… hope this helps the Silvertone 1472 and similar guitar amp owners'…
     
  2. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    Where are you seeing that? Remember that the grid voltage in the chart is relative to the cathode and in this case it's "bootstrapped" via the 1M grid-leak resistor. Also remember that because that resistor is bootstrapped, it's actually more like a 10M resistor which is why the grid-leak bias works here.
     
    ThermionicScott likes this.
  3. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    mrriggs,

    I am not sure what you are asking.

    I understand that the bootstrapping is moving the voltage in the junction cathode-1M resistor, in phase with the input (grid-to-ground) voltage. That just means that the grid-to-ground voltage needs to be larger, in order to obtain the desired control-grid-to-cathode voltage value.

    However, what controls the current through the triode tube is always the voltage between control-grid-to-cathode, no matter what circuit configuration is used.

    What I am demonstrating here, through the load-line plot is, with low bias voltage, the amplitude of the signal that needs to be applied between the control-grid-to-cathode, to produce enough output on both of the 68K resistors to drive the 6V6's to full power, will exceed the ZERO volts grid line, drawing PI grid current and thus, resulting in distortion (breakup) at low amp volume settings, as pointed out by the Silvertone 1472 owners' threads.
     
  4. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    Perhaps I missed it but I didn't see anything indicating how much voltage swing is required.

    The schematic doesn't list the bias voltage of the power stage. A quick Google search showed a similar Silvertone amp at 16 Volts. Typically the power tubes will start distorting before the grids reach 0 Volts (due to screen robbing current from plate) so let's say you need to swing ±15 Volts for max clean output.

    With 68k load resistors on the phase inverter, to swing 15 Volts would be a current change of 0.22mA. That added to the 0.94mA idle current is 1.16mA which intersects the load line below the 0V grid line.
     
  5. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    mrriggs,

    Thanks for validating my point.

    Using your numbers -- to move the tube current from 0.94 mA to 1.16 mA, in this specific circuit, the control-grid-to-cathode voltage needs to move from -0.58V to -0.11V. This value is a smidgen before it gets to ZERO volts, and just like you stated, at this point the 12AX7 is distorting due to grid current drawn from the previous amplification stage. Just keep in mind that these calculations were made using "nominal center values", which do not include neither, variations from tube-to-tube nor aging.

    Put in other words, in the Silvertone 1472 design, the phase-inverter will distort before the output tubes, which for most, is very unpleasant distortion.

    But I am still not sure what you are asking.

    This study, is offered to the people that have the Silvertone 1472 amp (or similar design) and want to change this effect of the amp breaking up at lower volumes and could not get a satisfactory explanation on how to do that. They simply need to change the 12AX7 PI bias as explained in the beginning.

    However, if you like how the original amp sounds, by all means, do not change the circuit.
     
  6. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    mrriggs,

    Thanks for validating my point.

    Using your numbers -- to move the tube current from 0.94 mA to 1.16 mA, in this specific circuit, the control-grid-to-cathode voltage needs to move from -0.58V to -0.11V. This value is a smidgen before it gets to ZERO volts, and just like you stated, at this point, the 12AX7 is distorting due to grid current draw from the previous amplification stage. Just keep in mind that these calculations were made using "nominal center values", which do not include neither, variations from tube-to-tube nor aging.

    Put in other words, in the Silvertone 1472 design, the phase-inverter will distort before the output tubes, which for most, is very unpleasant distortion.

    But I am still not sure what you are asking.

    This study, is offered to the people that have the Silvertone 1472 amp (or similar design) and want to change this effect of the amp breaking up at lower volumes and could not get a satisfactory explanation on how to do that. They simply need to change the 12AX7 PI bias as explained in the beginning.

    However, if you like how the original amp sounds, by all means, do not change the circuit.
     
  7. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    My numbers were pulled out of my ass. I don't know how much voltage swing the power stage can take with so little plate and screen voltage. Is it ±15? More? Less? Without knowing that, you can't conclusively say the phase inverter is the weakest link.

    Please don't get me wrong, I'm not trying to say one way or the other whether it is or isn't. I really appreciate your thought process and explanation. It's just that, while reading through your original post, there was a logical jump that wasn't supported. It says that it can't produce enough output, but doesn't specifically say how much it can produce or how much is required.
     
  8. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    upload_2021-6-7_22-45-15.png
     
  9. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    Thank you for bringing up this topic. I must admit that I have never messed with a grid-leak-biased cathodyne phase-inverter before. It's a fascinating concept. Out of curiosity, I bread-boarded one this evening to see what it could do.

    With 220 Volt B+, 68k load resistors, 1M grid leak, my quiescent plate voltage [in reference to ground] was 155 Volts, cathode voltage was 62 Volts. That puts the plate voltage [in reference to cathode] nearer to your chart than could be expected, 93 Volts.

    I kept cranking up the input signal amplitude and was absolutely blown away to see this coming out of the phase inverter.

    [​IMG]

    The scope is set to 10 Volts/div! That's 70 Volts peak-to-peak before it starts clipping. I never would have guessed that.

    What's happening is that when the input swings high, instead of saturating, it re-biases the tube further down the load line. At that 70Vp-p, the DC plate voltage [to ground] was 183 Volts and the DC cathode voltage was 37 volts, so the new bias point is at 146 Volts.

    What happens when you push it further? Believe it or not, it'll reach cutoff before it ever saturates.

    [​IMG]

    That's right, the increased bias voltage plus the increased voltage swing eventually adds up to B+ and it can't swing any higher than that.

    Thanks again for bringing light to this obscure design. This kind of stuff is fascinating and it's fun to mess around with it to see how it works.
     
  10. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    mrriggs,

    Thanks for taking the time to build a grid-leak-biased cathodyne phase-inverter, make the measurements and share the results. This is very interesting particularly the part of the stage re-biasing itself.

    I am curios… did you applied the input signal directly to the cathodyne phase-inverter or the signal went through the other 12AX7 half first?
     
  11. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    Ran out of time last night, had to get a couple hours sleep before work. Back at it today.

    To better illustrate how this thing is working, I DC-coupled the probes, one on plate other on cathode, scale on oscilloscope is set so the top of the grid is 220 Volts (B+) and the bottom is zero volts. (Sorry some of the pics turned out blurry)

    Idle; the two are ~90V apart, that's the point where grid current starts flowing and sets the bias.
    [​IMG]

    Increase signal; notice the minimum distance between them stays fairly constant, each trace expands outwards towards the "rails".
    [​IMG]

    [​IMG]

    [​IMG]

    Once it hits the rails, it's game over, no more room to expand.
    [​IMG]

    Of course, we still haven't answered the question of how much voltage the phase inverter can swing in a Silvertone 1472, because we've neglected to account for the load on the circuit. The 330k power-tube grid-leak resistors are each parallel to the 68k phase-inverter load resistors, so the AC load line is (68k||330k) + (68k||330k) = 113k

    [​IMG]

    I added the 0.01uF coupling caps and 330k resistors to the test circuit and the peak-to-peak signal swing to each power tube was reduced to 65 Volts. Not a huge loss.

    A shot of it not quite hitting the rails now due to the steeper load line
    [​IMG]
     
  12. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    From the [50 Ohm] signal generator I went through a 150k resistor [to simulate the output impedance of the previous stage] and a 0.01uF cap. I think the output impedance of the previous stage is actually closer to 120k but I had the 150k resistor handy and didn't feel like digging for a 120k. I also tried a 68k resistor (cause it was handy) and it didn't make any difference.
     
  13. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    mrriggs,

    Wow… thanks for your detailed testing and info.

    I omitted the AC-load-line for simplicity. I figured since the grid resistor of the 6V6 is almost 4x the load resistor of the PI, the result would not be too far from the real circuit. You graph is more representative of the real behavior.

    If you still have the scope set to 10 Volts/div, with loads closer to the real circuit as you described, looks that the output of the inverter is struggling to get to 30Vp-p per side!

    Did I read this correctly?
     
  14. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    On shots with both traces on the screen, full scale is 220 Volts, there are 8 divisions so it's 27.5V/div. Yeah, that's not a convenient number to work with but this was meant more for a big picture view than it was for nit-picking exact values. This isn't the best scope so the actual value is likely ±5 Volts from what it's indicating.

    When I zoomed in on a single trace at 10V/div for a more accurate measurement, it was ~70 Vp-p with no load, and ~65 Vp-p with load. Again, this isn't the best scope so the actual numbers are likely ±2 Volts from what it's indicating.
     
  15. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    Here it is with the 1.8k bias resistor added.

    [​IMG]

    [​IMG]

    [​IMG]

    [​IMG]

    [​IMG]

    The traces start further apart [due to the higher bias] so they don't need to move as the signal amplitude increases. Once the clipping level is reached, the traces are in pretty much the exact same spot as the grid-leak-biased version. Max peak-to-peak output is ~62 Volts.
     
    ThermionicScott likes this.
  16. NTC

    NTC Tele-Holic

    Posts:
    656
    Joined:
    Mar 17, 2003
    Nice work mrriggs. And K Teacher.

    One other thing not taken into account is the effect of the 6V6 control grid load as the input voltage approaches 0v. This may also start to distort things.

    Also, are you taking the non inverted output from the cathode when the 1.8k resistor is there, or from the junction of the two resistors? It shouldn't make much difference in this test circuit, but it might when driving actual 6V6's.
     
  17. K Teacher

    K Teacher TDPRI Member

    Age:
    51
    Posts:
    25
    Joined:
    Jun 7, 2020
    Location:
    Farmington, MI - USA
    NTC,

    Thanks for your comments.

    With regards of the non-inverting output connection, it's six of one, half a dozen of the other. Fender amps take that output directly from the cathode; Silvertone takes it exactly where is indicated on the posted diagram, that is, between the bias resistor and the load resistor. Just look at its big brother 1484 amp.

    I personally prefer the Silvertone way. Both methods will work the same under normal circumstances. However, when overdriven, the current drawn by the output tube control-grid-to-cathode "diode", will partially bypass the driver stage cathode resistor (like a bypass capacitor), causing that stage to have a sudden jump in gain and generating unpleasant distortion.

    If you take the output from the junction between the two resistors, you have some cathode isolation from the output tube grid current provided by the bias resistor, in this case, 1K8.
     
    NTC likes this.
  18. mrriggs

    mrriggs Tele-Meister

    Age:
    41
    Posts:
    192
    Joined:
    May 24, 2020
    Location:
    Vancouver, WA
    I added the power tube grids to the breadboard this morning but had to get the kids on the bus and go to work before I could get any measurements or pictures.

    Not actual power tube grids, but a 6AL5 dual-diode to simulate the overdriven power tube grids. The plates of the 6AL5 are used as the "grids" and the cathodes are hooked to a +15 Volt supply. It's obviously not an exact representation but close enough to highlight any differences in the various phase-inverter topologies.

    Stay tuned...
     
  19. ThermionicScott

    ThermionicScott Poster Extraordinaire

    Age:
    40
    Posts:
    6,467
    Joined:
    Dec 3, 2005
    Location:
    CID
    Don't take this the wrong way, but next time you should verify the "fix" before you promise it to the world. ;)

    Sounds like the phase inverter isn't the culprit after all. If a 1472 has "early" breakup at "low" volumes (both of these terms being very subjective), my money is on either the low B+ or inefficient speakers.
     
    Last edited: Jun 10, 2021 at 4:44 PM
  20. NTC

    NTC Tele-Holic

    Posts:
    656
    Joined:
    Mar 17, 2003
    I like the fact that he threw down the gauntlet and someone gave us a very nice real life test with scope shots. We wouldn't have gotten that otherwise!

    Nice explanation, I had a more simplistic view of what goes on when done the Fender way. This is a split load pi. If you take the non-inverted signal from the cathode, you haven't split the load properly. That and the tube acts as a cathode follower to some extent and may be able to drive the non-inverted output tube grid some. It probably can't.
     
IMPORTANT: Treat everyone here with respect, no matter how difficult!
No sex, drug, political, religion or hate discussion permitted here.