Ad Widget

Collapse

Announcement

Collapse
No announcement yet.

25W from 2x EL34 Class-AB Output Stage

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • 25W from 2x EL34 Class-AB Output Stage

    Searching for a 20W~25W, EL34 push-pull guitar amp info/ schematic; couldn’t find much information on it.
    Most comments were from people trying to understand why to use the more powerful EL34 to produce only 20W of output, if this can be achieved with a pair of the small 9-pin EL84 pentode or the popular 6V6 tube.
    The answer to the above question is simple – RELIABILITY


    To produce 20W output with a pair of EL84s or 6V6s, it is necessary to push these valves beyond their design limits, and thus shorten their useful life. Perhaps, good ones, at affordable prices were abundantly available back in the 60s, but not nowadays.
    A pair of EL34s should easily produce this output and last longer than these smaller tubes.

    Considering this, I decided to do a design exercise using the EL34 to get a lower-power,
    Class-AB amplifier stage, using currently available, off-the-shelf parts…



    Note: For a more detailed explanation on valve output stage design, please refer to
    “The Valve Wizard” Single-Ended and Push-Pull Output Stage tutorials.



    Initial Criteria:
    One way to get 25W output out of a EL34 pair, is to operate the valves at anode dissipation lower than the maximum value, indicate in the valve’s datasheet [orange line in the graph].
    After examining the EL34 datasheet and some experimentation, 300V was selected for Va (anode voltage), 300V for Vg2 (screen-grid) and 6,000 ohms (6K) for anode-to-anode OT primary impedance.


    Note: Va (and Vg2) = 300V is from anode to cathode and NOT from anode to ground !
    Thus, HT is the sum of Va, bias resistor drop and a few volts for the OT prim. winding.



    Class-AB load-line:
    In Class-A, both valves are load with half of the Za-a impedance, which in this design equals to 3K ohm.
    However, as soon as one valve cuts off, that half of the transformer's primary is no longer part of the circuit. Because the impedance ratio is the square of the turns-ratio, the load presented to the remaining 'ON' valve is only a quarter of Za-a (1K5 ohm).
    So, at higher signal levels, the output stage transitions to Class-B operation.


    Drawing the Class-AB load-line:
    The first step, is to draw a load line corresponding to a quarter of Za-a. In this case, a quarter of 6K = 1K5 ohm [blue line], from 300V Va. This is the Class-B part of the load.


    Note: If the blue line crosses the zero volts control grid voltage trace (Vg1 = 0V) far below its knee, then the screen voltage (Vg2) must be lowered, or a very sizeable screen-grid
    stopper resistor must be used to prevent from screen grid damage, due to excessive
    screen grid current when anode voltage approaches zero volts
    .



    The next step, is to repeat the same procedure and draw a load line corresponding to half of Za-a. In this case, half of 6K = 3K [purple line]. This is pure Class-A.
    To make the stage operate in Class-AB, the 3K (half of Za-a) line, must be moved upwards (while maintaining its gradient), towards to the anode dissipation curve, but without exceeding it [orange line].
    This is the same procedure used for the Single Ended Output Stage design.


    Quiescent anode current:
    The OT transformer load to the output valves is mostly reactive and it has very little DC resistance. As result, the voltage-drop across it is very small and generally ignored in the calculations.
    Because of this, the quiescent anode voltage is assumed to be the SAME as Va, independently of the quiescent anode current values.


    So, imagine a vertical line drawn through the Va voltage of 300V. The quiescent anode current is determined by the intersection of this vertical line and the Class-A part of the load [purple line].
    In this example, the quiescent anode current is 40mA, indicated on the chart by the ‘green dot’.


    Click image for larger version  Name:	dataurl647764.png Views:	21 Size:	714.2 KB ID:	974573

    Bias voltage:
    The last piece of information needed, is the control grid voltage (Vg1), required to set the quiescent anode current to the desired value, in this example, 40mA as previously determined.
    The major controller of the anode current, is control grid voltage (Vg1), but also, to a lesser extent, the screen voltage (Vg2). In other words, if Vg2 is reduced, then Vg1 needs to be less negative to maintain the anode current the same for a given anode voltage.
    In the “Initial Criteria” section, Vg2 was specified to be 300V. However, the anode characteristics datasheet available is for Vg2 = 360V.
    Despite that, correlating the datasheet information, we estimated that for Vg2 = 300V and everything else the same, the control grid voltages need to be 6.0V less negative [tan numbers on the graph]. So, to keep the quiescent anode current at 40mA with Vg2 = 300V, Vg1 needs to be -24.0 volts.


    Screen Grid Resistor:
    This is another topic I see a lot of debate without reaching a clear conclusion.
    Generally speaking, the higher the screen grid voltage, the higher the screen grid resistor value, in order to bring the screen voltage down to a safe value (and prevent damage) when the valve is overdriven.
    As stated previously, IF the load line crosses far below the knee of the zero volts control grid trace (Vg1 = 0V), then a sizeable screen-grid stopper resistor must be used to prevent damage to the screen grid. This condition is evident when the screen grid voltage is very high, let’s say, around 460V or more.
    However, in this design, the screen grid voltage was reduced to 300V (as shown on the anode characteristics traces plot), causing the control grid traces to slid down. Because of this, the intersection of the load line [blue line] with the zero volts control grid trace, happens just below its knee [red dot].
    So, in this case, no screen grid resistor is required.
    Nevertheless, to prevent the output tubes from developing parasitic oscillations, it is recommended the installation of individual 100 ohm resistors mounted right on each of the tubes’ sockets. 1W or 2W resistors should suffice there.

    Is noteworthy to mention, that the Philips EL34 datasheet, recommends non-decoupled, shared screen grids resistor for Class-AB push-pull operation. This is used more often in Hi-Fi designs and provide a better balance between the output tubes (as long as the amp is not clipping!).
    For anode and screen grid voltages close to the ones used in this design, one 470 ohms shared resistor is indicated. In this design, 5W to 7W resistor should suffice.
    If the reader wants to experiment and try this arrangement, first install the 100 ohm screen grid stoppers and then connect one 470 ohm resistor going to the power supply.


    Bias Resistor:
    The quiescent cathode current is the sum of the quiescent anode current (Ia) plus the quiescent screen current (Ig2).
    The data sheet gives an example where the anode current is 70mA and screen current 10mA; a ratio of 70/10 = 7:1. We already know the quiescent anode current will be 40mA [green dot], and estimating an anode to screen grid current ratio of 7:1, we can expect:
    40/7 = 5.7mA of quiescent screen current.

    So, the total cathode current will be equal to 46mA.
    Using Ohm's law to calculate the value of cathode resistor (Rk):
    24 / 0.046 = 522 ohms for each tube (like the Matchless Chieftain) or 261 ohms for shared bias resistor.

    The two closest standard values are 560 ohms for individual bias resistor OR 270 ohms shared resistor.
    The bias resistor power dissipation is:

    (0.046 * 0.046) * 560 = 1.2W
    So, we would use a 5W resistor (4.2x derate factor).
    If a shared bias resistor is used, then a 10W / 270 ohms resistor should be used.



    Cathode bypass capacitor:
    Fully decoupling the cathode resistor will prevent internal feedback and maximize input sensitivity, making the valves easier to overdrive.
    There are different approaches on how to calculate the decoupling capacitor value and some good tutorials showing how to do it are available. For this reason, I am just going to suggest values to use and leave to the reader to investigate this further.
    For guitar amps, use a bypass cap of 50uF for individual bias resistor and 100uF for shared bias resistor.
    For Hi-Fi amps, which require lower roll-off bass frequency, use a bypass cap of 150uF for individual bias resistor and 300uF for shared bias resistor.
    The bypass capacitors voltage rating must be at least three times the expected cathode voltage. This is especially necessary if you are silly and use a standby switch, as there will be a brief surge of grid current at startup, which will raise the voltage across the cathode resistor quite alarmingly!
    For this design use a 75V rated capacitor or higher.


    Output Power:
    The total output power can be closely estimated from the load line (again, we only need to look at one half of the circuit to do this).
    Simply note the peak current Ipeak (i.e., where the load line crosses the 0V grid curve); in this case it is about 183mA [red dot]. Also note the minimum anode voltage Vmin, which is about 26V in this case.
    The total (rms) output power is then approximately:
    Po = (HT-Vmin) * Ipeak / 2
    Po = (300-26) * 0.183 / 2
    Po = 25.07W



    Final Comments:
    Since the EL34 valve has more gain than the 6L6, we need less voltage swing out of the phase inverter to drive the output stage to full power.
    For this design, drive for max. power is:
    Vi = Vg1_bias * 2 * 1.414
    Vi = 24 * 2 * 1.414
    Vi ~ 68Vp-p (each input)

    It is possible to get this voltage excursion from a 12AX7 Cathodyne inverter fed at approx. 320 ~ 330 VDC.
    I would recommend looking at the Peavey Classic 20 preamp & PI circuit as an example.


    Hope this helps!
    Last edited by K Teacher; 12-09-2022, 05:54 PM.

  • #2
    Nice compilation/exercise!


    The next step, is to repeat the same procedure and draw a load line corresponding to half of Za-a. In this case, half of 6K = 3K [purple line]. This is pure Class-B.
    Should read Class A here.
    - Own Opinions Only -

    Comment


    • #3
      Helmholtz,
      Thanks for your comments.

      I meant; it would be pure Class-B - BEFORE - moving the load line upwards.
      That was poorly wrote…

      Comment


      • #4
        This is very interesting, but would also beg the question: How does it sound? Is it good for hi-fi but lacking for guitar? Since some people really love EL-84s running super hot it would be interesting to know how EL-34s giving the same power out would compare.

        Comment


        • #5
          Originally posted by K Teacher View Post
          I meant; it would be pure Class-B - BEFORE - moving the load line upwards.
          That was poorly wrote…

          In the passage I'm referring to you're talking about the lower part of the loadline having a slope correspondig to Zaa/2 (3k).
          That means class A operation.
          Moving the loadline upwards doesn't change the load impedance.

          - Own Opinions Only -

          Comment


          • #6
            Helmholtz,
            That is correct, my bad…

            glebert,
            The shared screen grid resistor is used mostly on pentode Hi-Fi output designs. I do not think it will be detrimental for guitar amps.
            The only way to find out how it sounds, is to build one. I can speculate that it will sound close to a similar amp using EL84s.
            Currently, Marshall is offering a couple of 20W guitar amps with 2x EL34. I believe you can find some sound demos on YouTube.

            Comment


            • #7
              If Zaa has already been determined I actually think it's more realistic to start the combined loadline construction with the class A loadline runnning through the operating point.
              Then join the class B part where the class B loadline meets the class A loadline at twice the idle current as that's where class B operation starts.

              Doesn't really make a difference for design.
              And with cathode bias the whole loadline will shift downward at sustained large signal level.
              Last edited by Helmholtz; 12-10-2022, 12:31 AM.
              - Own Opinions Only -

              Comment

              Working...
              X