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Making an amp from A to Z - The Fellow 13"

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  • Making an amp from A to Z - The Fellow 13"

    Hi,

    I worked on a new amp project and I would like to share the way I design an study the conception of my amps.

    At first we abord the preamp design.
    I'm looking for a very dynamic preamp but in the same time which can achieve warm and smooth distortion.

    It will be a two stage preamp with two 12AX7, four triodes.

    It's the opportunity to abord most of the different ways to polarize triode valves (12A*7 / ECC8* ... )

    You will recognize in the end, the kind of amp that I'm inspired.

    Now, to forward step by step we interest of V2a valve which is the first triode of preamp second stage.
    It's a very simple and traditional triode design with a plate resistor of 100k, a cathode resistor of 1,5k supply by a DC voltage of 280V. It's a classe A design with traditional cathode BIAS.

    Take a look on the loadline.



    What can we see ?

    A is the lower point when the triode drain no current, it's a blocking point.
    That's the first point I draw, easy to find.
    B is the theorical point when the tube drain maximum current.
    The math is very simple, we have U and we have Rp and I = U/R = 280 / 100k = 2,8mA
    With these two points it's easy to draw the loadline (in red on the picture).

    Each crossing line points help to achieve the tranfer caracteristic line.

    At the point, it's easy to chose a cathode resistor value.

    I'm not looking for the most linear result, in guitar amp design, we are looking for harmonic colourings.
    However I choose a "center point" around 1,5V with a current around 1mA. (there are different ways to achieve nice harmonic colourings and distortion, we will see this point belowe)
    R = U / I = 1,5 / 0.001 = 1500 Ohms
    With the loadline we can easely see that Ua = 180V and Uk = 1,5V for a consumption of 1mA, that's the bias point of V2a.
    We can also see that the real max current value drain by our triode is around 1,85mA at Ug1 = 0V.
    Until now, we are in a very traditional way.

    now, we take a look on the V1 valve of the amp, I have to tell you that it's a double triode design which share the same plate resistor and the same cathode resistor too.



    Here are the values we need to know :
    U=280V
    Rp=220k
    Rk=1,5k

    We have to understand and to concider that each components are crossing by the current of the two triodes.
    Thus each component values can be understood as to be two time more important.
    Therefore to draw the loadline we use this two new relative values, Rp=440k and Rk=3k.

    Here is the loadline :



    The loadline seem to be very "flat" and that's true !


    The double triode design is very interesting for a V1 position because of its very good dynamic and headroom.
    We most work on voltage variation which is the dynamic spring in tube amp.
    We can easely find the gain ratio.
    For grid 1V Voltage we find anode Voltage of 95V
    For grid 2V Voltage we find anode voltage of 170V

    So I can say that for a changing of 1V on the grid that result 170-95 = 75V of amplifcation.
    The Facto the gain ratio is 75.

    We can compare it with the V2a gain (see above).
    Ug1=1V => Ua=150V
    Ug1=2V => Ua=210V
    210-150=60

    So the Double Triode design give us a dynamic of 20% better than the single triode design.
    Very good thing for the first stage of our preamp, isn't it !

    The next time we will take a look on V2b which is a DC cathode follower site between V2a and the tone stack.
    We will try to understand how it's a very interesting design for achieve smooth an warm overdrive.

    Let's talk about the DC Cathode Follower which it's the last triode of our preamp design (V2b).
    You remember that we already study the first triode of our preamp second stage (V2a see above).
    I have to precise that it's the stage which achieve good preamp distortion.
    Thus this is a good erea to use a DC cathode follower.

    Here is a vue of how it looks like.



    We can easely recognize the cathode follower, the triode without plate resistor and its only resistor on the cathode which is the point we will recover the signal.

    Here are the values we have to know :
    U=280V
    Rk=56k

    Now take a look on the very interesting loadline.



    Because of the no plate resistor design, the tube works differently.
    The plate voltage is constent but the cathode works like it was the plate.
    This why we have to reverse the voltage on the up side of the graph.
    We see that the grid is in a positive voltage and we understand that it's going to tend towards zero volt and drain current.
    On the loadline, I colored it in brown, we can see that this current will tend towards 0,5mA.
    This current will come from the plate resistor of V2a which will create a SAG and by this way low down the plate voltage of V2a.
    U = RI = 100k x 0,0005 = 50V
    Thus at first we see that our DC cathode follower will works to achieve more distortion from V2a.
    But we can also see on the loadline that the tube is "locked" on 0V therefore all positive voltage will be very nicely compress.

    What could be see as a very bad design for an audiophile user is a very good way to achieve smooth and warm distortion for a guitar player.

    Now the first part of the preamp design is done.

    I can know the preamp consumption which will be necessary for the further power supply design.
    Iv2a = 1mA
    Iv2b = 3,2mA
    Iv1 = 0,75mA
    Ipreamp = 4,95mA

    It's time to talk about the voicing.

    It's important to know the power amp tube brand.
    Each tube has its own voice and that a point who has to be take in concideration very seriously.
    My power amp will run with EL84 tubes which have a very round and medium tone.
    In the same time I'm looking for a very simple tone stack with only a bass and treble pots.
    I will adjust the tone for a Vox / Matschless voicing but with my taste.
    Vox Top Boost channel tone stack is very accurate.



    For best gain capability we use a large and usual cathode capacitors value of 22µf that give us a cut frequency of 4,8Hz.
    F= 1/(2pi x Ck x Rk)= 1/(6,28 x 0,000022 x 1500) = 4,8Hz

    In the same time, we work on the design of a Rock amp and want to achive very nice distortion tone.
    We all know that distortion doesn't like too much bass in the voicing.
    We can low down the bass level with the coupling capacitor between V1 and V2a which is going to help us in the same time to protect V2a of blocking distrotion.
    Because of the very medium tone of the EL84 and the tone stack simulation I choose a cut frequency of 700Hz.
    C = 1/(2pi x F x R) = 1/(2pi x F x R) = 1/(6.28 x 700 x 220k) = 1,034nF
    1nF will be my coupling capacitor between the first and the second preamp stage.

    Well, now my preamp is becoming almost totaly design.

    It's missing the gain pot value, and I choose a 500klog for not overload the V2a grid and adding a 180p bright cap for very nice sparkle clean tone.
    Last edited by Mikka Grytviken; 05-24-2011, 10:18 PM.
    The middle way is the way.

  • #2
    Here is the preamp schematic.



    Now we are going to study the phase inverter.
    As using in the most part of push pull guitar tube amp I will use a Long-tailed pair PI with a "Schmidt" design and no negative feddback.
    Therefore I have to use a high tail resistor value for a well balanced output.

    Let's take a look on the PI schematic.



    We have to notice that the cathode resistor (Rk) is coupling with the two triode cathode, that's a very important point.

    And now the loadline to explain my choices and understand how it works.



    Because of Rq the plates see very low relative voltage value, 60V less than the power supply value.
    It's a compromise between a good output balance and an acceptable output gain and headroom.
    I know that my preamp will compress very well and the PI input level will be well control ...

    With the loadline I see easely that the PI won't clip the signal under 6V peak to peak.
    It's a value that I have to notice to compare it further with the preamp output level.

    Now we are going to estimate the gain.
    For Ug1 = 1,5V => Ua = 165V
    For Ug1 = 0,5V => Ua = 105V
    165-105 = 60V peak to peak.

    This result must be divided by a factor of 5, because the absence of decoupling capacitor of the cathode greatly reduces the gain. I deduced this factor through my own experiences.

    60 / 5 = 12

    For an input of 6V, 72V out there which is well enough to drive an EL84 push pull power amp.

    I would like explain my input coupling capacitor value.
    It is commonly acknowledged that a large capacitor will help too broad blocking distortion because of the high input impedance.
    If C1 is small, less than 10nF with 1M in gride resistance, it will improve the response at low frequencies in the balancing between the two outputs, especially if the second capacitor (C2) is sized ten times C1.

    F = (2pi x C x Z)

    About Z : If the resistance tail is broad enough to be considered a current source, there is no negative feedback, the input impedance will be equal to twice the gride resistance associated.

    F =1/(6,28 x 4,7n x 2M) = 17Hz

    It remains to calculate the current consumption of the PI.

    I = 0,00064 x 2 = 1,28mA
    The middle way is the way.

    Comment


    • #3
      Now we work on the power stage and the power supply.

      Usually, we start the study of the power stage, starting from a voltage chosen in advance. But this time, I will do it differently because I want to use my remaining transformers in stock.So I'll use a 369GX and 125E, Hammond Transformers brand.

      You will probably tell me that the 369GX is not intended to supply a push-pull with two EL84.And I would answer that it depends on a number of factors. In addition, the transformer supplying a class A EL34 SE before coming down in my stock. It operated under 250V and 82mA for a load of 2.5 k. I know that this is not enough.To do this, I modified the 369GX tranny. The 369GX is a center tap transformer, and it is well known that transformers wired in center tap do not exploit their full capabilities.
      So I simply change the internal wiring of the transformer in order to use it without the center tap. To do this, I put the two secondary windings in parallel which allowed me to double the available current intensity.

      I will not reopen the debate here, for more information I invite you to have a look on this topic :

      double the secondary current capability of a CT Power Tranny


      Now that everything is clear, I will discuss the various information that I used for the study of the power stage.

      First here is the load line of the transformer we have drawn with my friend "French-Vitriol82" from the measurements we made.



      I get an old combo and 8 ohm speaker, basing myself on this datasheet and the output transformer, Hammond 125E, I deduce the different possible load values as follows:
      6.8k / 8.2k / 11.6k / 12.8k (plate to plate of course)

      I want a power ranging from 10 to 15W, which seems quite feasible.

      To study the power supply, I use PSU Designer II.

      Helping me with my experience and software, I went by intuition and step by step to finally reach this compromise.







      This chart is extremely important in understanding what happens in the transformer and allow us to link with the load line of the transformer described above.
      We now need to read this graph by calculating the RMS current.

      Irms = (Ix0.7)/ V^2 = (0.360x0.7) / 1.414 = 0.178A



      221 x 0.178 = 39W

      The transformer is given for 35VA on the HV winding.
      It should be able to keep the 39W required especially as we are in class A.

      Now it remains for me to finalize the schematic and then I could look at the layout and component selection.

      Before posting the schematic I would like to calculate Kathode Resistor Value, Screen Grid resistor Value and RMS power of the amp.

      To do it, we have to come back on our loadline.



      Prms

      Irms = (0.085 - 0.0375) / 1.4142 = 0.0336A
      Urms = (288 - 15) / 1.4142 = 193V
      Prms = Urms x Irms = 193 x 0.0336 = 6.5W

      We are in class A so this value is for one tube, because we have two tubes

      => Prms = 6.5 x 2 = 13W

      Rk

      For one tube :
      Ip = 0.0375A
      Ig2 = 0.00375mA
      Vg1 = 10V (because Vg2 is around 280V)

      Rk = U / I = 10 / [(0.0375 x 2) + (0.00375 x 2 )] = 121,2

      Rk = 120 Ohm.

      Rg2

      That is a more complex point because voltage changes in the gride screen and increasing of the current are dependent values of both the load, the anode voltage and screen grid voltage.

      Thus, we use the conduction angle to complete a factor that will help us to calculate more precisely the behavior of our screen grid.

      CA = 2inCOS(-0.0375 / 0.085) = 232.4°
      CC = 360 / 232.4 = 1.55

      284 - 15 = 269V

      269 / 0.085 = 3165

      Rg2 = 3165 / 1.55 = 2042 ohm

      We have to concider the all impedance et resistance before the grid supply as first CRC filter and tranny impedance.

      Rg2 = 2k - 470 - 250 = 1280
      1.2k or 1k would work as well too.

      Nota : I would like to precise that this methode I used before seems to be correct for a lot of cases but it's not really the best good way to calculate de grid stopper resistor for the screen grid, I will give an other better way later.

      Until now, We didn't talk about the cathode capacitor of the power stage, we will do it below.

      Here is the complete diagram of the amplifier.

      The middle way is the way.

      Comment


      • #4
        We have already discussed the voicing of the preamp.
        Here the frequency curve of it, all knobs at noon.



        It is time for us to look at the impact of PI on the voicing.



        I state again that all the knobs are at noon.

        The volume knob mixes the two signals PI which are 180 ° out of phase.
        Thus, more the two signals are mixed and more they self cancel.

        Cut potentiometer functions as a lowpass filter.
        It is partly influenced by the volume knob.
        The fact that it works between the two parallel line of Phase Inverter, will have an impact on the calculation of its action.
        I am mindful that the potentiometers are logarithmic.

        Cut frequency

        Rc = 250k x 0.2 /0.5 = 100k

        Rm = 1M x 0.2 = 200k

        R = (200k x 100k) / (200k + 100k ) = 67k

        F = 1/(2pi x 67k x 2.2n) = 1080Hz

        Verification by simulation.



        We can also calculate the cut frequency of C10 and C11 (0.1µ coupling caps)

        F = 1/(2pi x 220k x 0.1µ) = 7,2Hz

        And now it's time to calculate the cathode BIAS cap.

        I'm looking for a low cut frequency for better gain and bass breathing.

        I know the lowest audible frequency is 20Hz

        C = 1/(2pi x 120 x 20) = 66µ

        And I choose 100µF ... hint of sub bass makes the sound more joyful and more alive.

        Hej ... I saw that I forgot to calculate the grid resistor !

        It's a very important thing to prevent high frequency auto-oscillation and noise.

        I use to choose a cut frequency of 10kHz ... for a guitar amp is largely enough !
        EL84 grid/cathode cap value is 10pF

        R = 1/(2pi x 10k x 10p) = 1592

        R = 1.5k ohm.

        The amp is working since November 2011.

        You can find HERE the PDF files, a slide show and a video ... some better clips coming soon.

        All voltage measurements are correlated with my calculations and my loadlines to 20% ... all is good !

        This clip I recorded with the Fellow 13, Blue Marvel Speaker and an Ibanez AXD82P with Seymour Duncan pups.

        It's here.



        PS : To day the amp has got an Eminence Legend GB128 speaker and sounds much better !
        The middle way is the way.

        Comment


        • #5
          Originally posted by Mikka Grytviken View Post
          We now need to read this graph by calculating the RMS current.

          Irms = (Ix0.7)/ V^2 = (0.360x0.7) / 1.414 = 0.178A
          I question the validity of this equasion. Why are you dividing by 1.414? Those numbers are only valid for perfect sine waves. The current waveform is not a sine wave! Duncan's PSUD2 will calculate the RMS current for you. Look at the guidelines in this thread: http://music-electronics-forum.com/t20730/ where I show how to get accurate numbers from PSUD2.
          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 !

          Comment


          • #6
            Hi,

            RMS values ​​which are given by psud II are not all correct.

            Take a look to my formula, in addition with 1.414, I use the factor of 0.7 which reflects the proportion with no signal on the chart.

            If I stick to the RMS values of 194,7mA ​​given by psud, I find the next factor: 360 / 194.7 = 1.849.

            How can there be more than a factor of square root of 2 while the graph shows us that we consume less than in the case of a complete and perfect sinusoid.

            It is therefore evident that RMS values ​​given by psud is totally false in this case!

            Besides, if you look at the results given by PSUDII for RMS current and compared to the usual powers of our transformers, we deduce that they are all very much under-sized, which I think is partly the case but anything approaching this scale, otherwise none of them could hold over an hour.

            I design for a very long time power supply and so far I have not encountered any problems.

            If I had relied on the RMS value given by psud, I will never mounted the Fellow 13 with the transformer in question. And if that were true, the transformer would have burnt too long. Ask a secondary winding of a transformer a power of 42.5VA while it set to roll just 35VA ensure some toast with friends, right?
            The middle way is the way.

            Comment


            • #7
              Can you describe an algorithm for computing the RMS value of an arbitrary digitized waveform? It goes like this:

              First, limit the time domain of the waveform to an intergal number of cycles. Next, take the value of each point and square it (the S), store that in a new array. Next, add all the squares and divide by the number of samples in the array (the M). Lastly, take the square root (the R). You have computed the Root of the Average Square.

              Look at the attached diagram. In the top graph, the red plot is a rectified sine wave. The green plot is computed from each point on the red graph and squaring it. Note how it looks like an inverted Cosine wave twice the frequency of the red plot? It is! It's average value is 0.5 and the square root of that is 1/(square root of 2) or about 0.707 .

              In the lower plot, the black points are a half wave rectified sine wave and the blue points are the squares of the black points. The average of the blue points is 0.25 and the square root of that is 0.5 .

              I had trouble with PSUD2 when I tried to duplicate your circuit. Some of the numbers were smeared by jpeg conversion and I could not read them. I don't think you have the correct numbers for the lumped impedance of your transformer. You need to know the unloaded secondary voltage (not the advertised voltage) and the resistance of the primary and secondary. I assume the line frequency is 50Hz in your country, that is the PSUD2 default.

              However, I believe PSUD2 is accurate within it's limitations and computes RMS values to better than 1% when the time domain is correctly limited.
              Attached Files
              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 !

              Comment


              • #8
                Some of the numbers were smeared by jpeg conversion and I could not read them.
                Bizarre, I can read them very easely.
                Click left on the picture for a larger view.
                Regards
                The middle way is the way.

                Comment

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