Smaller Cathode Resistor = More Gain?

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers' started by robrob, Dec 4, 2017.


  1. robrob

    robrob Member

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    It seems to be "common internet wisdom" that decreasing a preamp tube's cathode resistor will warm the bias and lead to more gain. But drawing tube load line charts doesn't seem to bare this out.

    On the charts, changing the cathode resistor value has a very minimal effect on the plate load line slope that determines the tube circuit's gain.

    1.5k cathode resistor = 250v / (100k + 1.5k) = 2.46ma
    2.7k cathode resistor = 250v / (100k + 2.7k) = 2.43ma

    These values form the left end of the red plate load line in the chart below so the slope of the line (and therefore gain) barely change.

    The main effect of altering the cathode resistor value is the operating point (intersection of plate loade line and cathode load line) is shifted along the plate load line which obviously affects headroom. A center bias allows the most voltage swing without distortion giving the max headroom. Moving the operating point away from the center bias point will give earlier distortion thus less headroom.

    The chart below shows a common 12AX7 guitar amp gain stage with 250v supply voltage, 100k plate load and 1.5k and 2.7k cathode resistors (unbypassed in this example for simplicity).
    [​IMG]

    Am I missing something or does changing the cathode resistor value have minimal effect on gain?
     
  2. darkfenriz

    darkfenriz Member

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    Not even wrong.

    Anode current is not determined by 100k anode load, but mostly by transconductance characteristics (transfer characteristics)
     
    Last edited: Dec 4, 2017
  3. HotBluePlates

    HotBluePlates Member

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    You marked out 2 operating points given 2 values of cathode resistors, but haven't calculated gain for either (which means so far, we don't know the impact of the Rk change).

    But let's think in general terms for a moment. What happened when Rk was made smaller? Bias dropped and plate current increased. What does increased plate current mean for internal plate resistance (rp)? For plate current to have risen, rp dropped.

    Check out Aiken's white paper on Common Cathode Triode Amplifiers. Scroll down to "Gain (fully bypassed cathode)", and note the formula:
    - Av = (mu*Ra)/(Ra+ra)

    Aiken uses "Ra" as total load resistance outside the tube (plate load in parallel with input resistance of the following stage), and "ra" instead of the "rp" I defined earlier (internal plate resistance). If you remove "Mu" from the equation, you get the form of an equation for a voltage divider:
    - Ra/(Ra+ra)

    In fact, you could leave the equation in that form, and Simply multiply Mu by the result and get the same answer:
    - Mu * [Ra/(Ra+ra)]

    What does that mean? If Ra is unchanged, a smaller "ra" (or rp) in the denominator yields a larger final result. Assume Ra is 100kΩ, and ra is either 40kΩ or 100kΩ (and Mu is 100 for our 12AX7 example):
    - Av = Mu * [Ra/(Ra+ra)] = 100 * [100kΩ/(100kΩ+100kΩ)] = Gain of 50 (ra = 100kΩ)
    - Av = Mu * [Ra/(Ra+ra)] = 100 * [100kΩ/(100kΩ+40kΩ)] = Gain of 71.4 (ra = 40kΩ)

    The above result shows that if internal plate resistance is smaller, the resulting gain of a stage is a larger fraction of the tube's Mu. So lower rp = higher gain.

    Page 2 of this 12AX7 data sheet shows how the tube's rp changes with plate current. Hotter bias always equal lower rp, though the relative change of rp is geater when plate current moves from being "very low" to "merely-low".

    "Gain" doesn't equal distortion, as we could get more distortion of one kind by running into curved portions of the plate characteristics you posted when using a cathode resistor bigger than 2.7kΩ. So we can draw some general conclusions, but where people create headaches for themselves is that the individual amplifier stage circuit & conditions must be evaluated for many different factors, which are often influencing the stage in contradictory ways, and which may yield a stage that at first seems to defy the general conclusions we think we know.
     
    Last edited: Dec 4, 2017
  4. HotBluePlates

    HotBluePlates Member

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    And what @darkfenriz is saying above aligns with what I'm saying. We're presenting 2 sides of the same coin because Mu = transconductance (Gm) * rp, and Gm goes up as rp goes down.

    This is shown in the data sheet I linked earlier (as well as the fact Mu isn't a constant, but is the most-closely constant tube characteristic).
     
  5. robrob

    robrob Member

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    The voltage gain of the stage can be determined with this chart using the yellow AC load line vs a 1 volt grid voltage change as seen below.

    My charts assume a 1M grid leak as the following stage input impedance so the slope of the plate load line and AC load line are very close.

    [​IMG]

    Changing the cathode resistor from 1.5k to 2.7k has only a tiny effect on the AC load line slope and therefore voltage gain.
     
    Last edited: Dec 4, 2017
  6. HotBluePlates

    HotBluePlates Member

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    But your new graph only evaluates one operating point; you need to evaluate both to know if there was a change in gain and how much.

    Note also that the spacing between grid voltage curves narrows at lower plate current. This is another indicator that gain is falling (because rp is going up). Your same 1v change of grid voltage results in a shorter span across the plate voltage axis when those grid curves are closely-spaced.

    That's not the idle current. It's is the current through the plate load resistor if the tube is replaced by a short circuit. 250v supply / 100v plate load = 2.5mA.

    The idle current is at the location where the cathode resistor line crosses the plate resistor load line. In your graph of Reply #5 above, the idle current appears to be ~0.75mA. In your graph in the initial post, the 2.7kΩ cathode resistor results in an idle current of 0.65mA.

    Looking at the graph I referenced on the data sheet earlier, there is not a huge change of rp (~71kΩ for 0.65mA and ~67kΩ for 0.75mA), but it's there. There would be a much bigger change of gain for a change of idle current from 0.3mA or 0.4mA up to either of the idle currents for the cathode resistors you're looking at. Easier to see if you don't make small changes, but evaluate an 820Ω resistor vs a 4.7kΩ resistor.

    My intent here isn't to "sharpshoot" you, but to note the general rule is always valid. However, as you & I have discussed in other threads before, sometimes the impact of the general rule in a specific situation isn't very significant. That's why it is often dangerous to oversimplify & always rely on the general rules...
     
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  7. robrob

    robrob Member

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    Here's how the 12AX7's mu fits into this graph:
    [​IMG]
     
  8. robrob

    robrob Member

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    There it is--that's what I was missing. I'll calculate the new lower gain numbers and report back.

    "The plate current change at idle only decreases from 2.46ma to 2.43ma with the cathode resistor change from 1.5k to 2.7k ..."


    I realized just after posting that that it was incorrect.

    HotBluePlates, I don't take your posts as "sharpshooting" at all. I posted this here because I know you guys know your stuff. Thanks for helping me understand this.
     
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  9. robrob

    robrob Member

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    I plotted the 2.7k cathode resistor voltage gain and it drops from 58 with 1.5k cathode resistor to 55 with the 2.7k resistor. The tighter spacing of the grid voltage curves at the bottom of the chart incorporate the changing internal plate resistance into the graph.

    Thank you HotBluePlates for helping me understand how a hotter bias leads to more voltage gain.
     
    Last edited: Dec 5, 2017
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  10. HotBluePlates

    HotBluePlates Member

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    Yep, you can approach it by recognizing the relationship graphically, or by seeing it in mathematical form. Since each is accurately reflecting the tube's behavior & characteristics, each is correct. But usually one will resonate more than the other with a single individual.

    Yes, but...

    For any evaluation of Mu, the lines you're using should extend an equal grid voltage change on either side of the operating point. This is tricky to do if your operating point is not exactly on a grid bias curve.

    The same applies for evaluating transconductance, and for in-circuit voltage gain (except the former is vertical, and the latter is along the loadline).

    The big challenge is estimating where the, say, -0.58v and -1.58v curves would be when evaluating an operating point located at -1.08v. It's a challenge because the grid lines are not equidistant and parallel (as they would be for an ideal tube), and the difference in length from center-point to either end indicates distortion (one side of the signal is not amplified the same as the other side of the signal).
     
    Last edited: Dec 4, 2017
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  11. HotBluePlates

    HotBluePlates Member

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    The below image should help you see how to approach the gain estimate. Rather than selecting particular cathode resistors & seeing what operating point they land on, I picked arbitrary operating points of -1v (blue) and -2v (green). The loadline is for 100kΩ and a supply voltage of 250v, and ignores any impact due to following-stage input impedance (a.c. loadline).

    There is an orange vertical placed at a grid voltage change of 0.5v either side of the operating points. These help evaluate gain at the operating point, and estimate distortion for that size signal (1v peak-to-peak, or 1v p-p).

    [​IMG]

    I didn't try to squint & measure to get exact figures, but a rough count for the -1v operating point looks like about 62v p-p plate voltage change for a 1v p-p grid voltage change. Gain is about 62, and implies the internal plate resistance is around 62kΩ at this point (use Aiken's page, do the math).

    The -2v operating point looks like roughly 52v p-p plate voltage change for the same 1v p-p grid voltage change. This implies the internal plate resistance is around 92kΩ. If you look back at the earlier-linked 12AX7 data sheet, the curves for Mu, Gm & rp suggest internal plate resistance should only be about 80kΩ for an idle current of 0.5mA, but for as quick & dirty as I did this that's probably not too much error.

    Now look at the plate voltage for the operating point, and count the plate voltage change for the 0.5v grid voltage change in either direction. For the -1v operating point I see ~5.5 lines plus a little bit (29-30v?) when the grid moves from -1v to -1.5v. I also see about 6.5 lines (32.5v) when the grid moves from -1v to -0.5v.

    This still tallies up to roughly 62v change at the plate, but notice how the 2 figures are not exactly the same. That means one side of the input signal is amplified more than the other side, despite assumed symmetry of the input, and is distortion (just a very little bit). Perfect linearity would happen if those plate voltage changes are exactly equal.

    Now while there is some distortion at each of these operating points (we might guess there is a bit more at the -2v operating point), if the signal is small neither one is that bad. Imagine the stage idled at -2.5v and applied a 2v p-p signal (up to -1.5v and down to -3.5v), where there will be 225v on the plate at idle. We can see plate voltage can swing downward to about 173v (~52v plate voltage change) but up only as high as 250v (our supply voltage; 25v plate voltage change). Might not actually completely cut off the tube and swing the plate all the way up to 250v (due to grid curve crowding & the difficulty of totally shutting-off a tube). We could calculate, but it's obvious the distortion is higher than our previous case, even though we know gain is lower (at (250v-173v)/2 = 38.5).

    Idle plate current at the -2.5v operating point is a bit over 0.25mA. The rp curves on that 12AX7 data sheet shows an internal plate resistance of ~120kΩ for this current; my quick & dirty math would estimate closer to 150kΩ. Either way, this is much higher than the typically-quoted data sheet figure of 62.5kΩ for a 12AX7. This internal resistance swings very high (infinity at cutoff?) in one direction of grid input, and swings much lower (dwn to around 65kΩ) in the other direction of grid input.

    There are some critical key points here:
    • Down in this low-current region, there are big swings of internal plate resistance with input signal.
    • Big changes of internal plate resistance correlate with higher distortion.
    • While gain is provably-lower at these low-current operating points, distortion is provably-higher; consider this proof of how "Gain" & "Distortion" are different things and we can have low gain with high distortion.
    • Whether there will be distortion (and how much) can greatly depend on input signal size relative to the operating point.
    Depending on your preferred way of thinking about tube characteristics, we could have just as easily said "big changes of transconductance results in increased distortion."
     
    Last edited: Dec 5, 2017
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  12. robrob

    robrob Member

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    Excellent graphic and post HotBluePlates, thank you.
     
  13. robrob

    robrob Member

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    Here's a chart showing how to calculate internal plate resistance on the plate characteristics chart:

    [​IMG]
    To calculate internal plate or anode resistance, called ra or rp we'll move +/-25 plate volts from the operating point paralleling the nearest grid voltage curve and read the plate current at +25 and -25 plate volts.

    We then divide the plate voltage change of 50 volts by the plate current change of .85ma to get the internal plate resistance:

    50V / .00085A = 58.8kΩ rp
     
    Last edited: Dec 6, 2017
  14. robrob

    robrob Member

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    Here's how transconductance is related to the characteristics chart. Transconductance expresses the relationship between voltage change on the grid to the change in plate current. We see below that moving from -1 volt on the grid to -2 volts results in plate current going from 1.39ma to .16ma or 1.23ma per volt.

    [​IMG]

    A 1 volt change on the grid (-1V to -2V) gives you a change of 1.39ma - .16ma = 1.23ma of plate current change per volt (shown in orange).

    .00123A change * 1V change = .00123 mho of transconductance.
     
  15. HotBluePlates

    HotBluePlates Member

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    Life is very much harder when you're trying to do graphical calculations to a grid voltage curve that's not plotted. It requires the person to imagine where a grid voltage curve should be for some value of bias, but doing so requires they have a solid grasp of how the tube deviates from being an ideal device (even then it's just a good guess).


    For internal plate resistance, you need a line that is tangent to the grid voltage curve, touching only at the operating point. If you use the actual grid curve, you will introduce error in your estimate. This is equivalent to the math operation of taking the derivate of a curve's equation, to find the slope of the curve at a single point.

    Make the tangent line large enough to extend all the way down to 0mA on the voltage axis, and as far up/right as possible. You need the biggest possible triangle to minimize error.

    [​IMG]


    But your operating point is not at -1.5v. Supposedly, it matters that you measure an equal change of grid voltage above/below the operating point (because you want to evaluate the tube's performance, or distortion, when applying a non-distorted or equal-swing signal). For my -1v example, this is easy to do by plotting the -0.5v and -1.5v currents. In the graph above, you can easily see that for -1v grid and 125v plate, there is a bigger change in current when moving from -1v to -0.5v than for moving from -1v to -1.5v.

    Squinting at my graph I estimate 125v plate and -0.5v grid results in a plate current of 1.65mA. 125v plate and -1.5v grid results in a plate current of 0.25mA. (1.65mA - 0.25mA)/1v = 1.4mA/V or 1400 micromhos.


    Try repeating your examples with an operating point chosen with 100v on the plate, and on a grid bias curve (either -1v or -0.5v). Compare your answers to those graphed on page 2 of that 12AX7 data sheet I linked in my first reply. You'll get a solid feel for whether you're working the process correctly, and will have an easier time plotting to curves explicitly drawn on the characteristics.
     
    Last edited: Dec 7, 2017
  16. robrob

    robrob Member

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    Thanks HotBluePlates, I re-ran the transconductance chart and came up with 1410 micromhos. I revised the chart but can't post it because this website is caching the chart jpg.
     

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