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Transfer function of a gain stage

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  • Transfer function of a gain stage

    Is there anyone here that can help me derive the transfer function of a grounded cathode gain stage, bypassed? Or if anyone knows a link to some site that shows how to do it.

    I've tinkered some my self but I'm not sure I got it right.
    In this forum everyone is entitled to my opinion.

  • #2
    Quick and dirty.

    The cathode is grounded (bypassed) so the valve generates a voltage of uVin across RL+Ra in series. Vout is the voltage across RL so using the potential divider equation Vout = uVin x RL/(RL+Ra) or Vout = Vin x uRL/(RL+Ra)

    Ra = Valve's internal plate resistance
    RL = Plate resistor // Load resistor
    u = Valve mu
    Last edited by Dave H; 04-11-2013, 08:55 PM.

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    • #3
      Seem you missed out on the bypass capacitor... Further, the grid resistor should also be considered, right?
      In this forum everyone is entitled to my opinion.

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      • #4
        Originally posted by Dave H View Post
        Quick and dirty.

        The cathode is grounded (bypassed) so the valve generates a voltage of uVin across RL+Ra in series. Vout is the voltage across RL so using the potential divider equation Vout = uVin x RL/(RL+Ra) or Vout = Vin x uRL/(RL+Ra)

        Ra = Valve's internal plate resistance
        RL = Plate resistor // Load resistor
        u = Valve mu
        ...and of course it depends on your application. For the sound designer who wants to model the waveform, the dynamic resistances need to be considered. I've done some transfer function-based signal processing in Csound, and the basis is that the vout term approaches the electrical limit asymptotically, and the knee of the curve is what non-linearly 'colors' the sound. I don't recall off the top of my head how I generated the transfer, I'm curious now about it, and I'll have to go look through my notes to see what I did.
        If it still won't get loud enough, it's probably broken. - Steve Conner
        If the thing works, stop fixing it. - Enzo
        We need more chaos in music, in art... I'm here to make it. - Justin Thomas
        MANY things in human experience can be easily differentiated, yet *impossible* to express as a measurement. - Juan Fahey

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        • #5
          Originally posted by überfuzz View Post
          Seem you missed out on the bypass capacitor... Further, the grid resistor should also be considered, right?
          Sorry I thought you just wanted the simple equation with the cathode resistor fully bypassed at all frequencies (very large capacitor). Do you want the transfer function for a finite capacitor? I'd have to jw my way through that and I'm rusty so it could take a while. I didn't consider the grid resistor because I assumed the grid was driven from a voltage generator.

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          • #6
            I only remember some of it vaguely. The impedance Z of a resistor is Z = R. The impedance of a capacitor is Z = 1/(iwC). The thing getting me stuck is the topology, I think it's called Thevinin or something. With the proper topology it shouldn't be such a hustle deriving the transfer function. The transfer function is Y/V, where Y is the laplace tranformation of y(t) which is the in signal. Likewise for x(t), but out signal. Well this is kind of all I figured out...
            In this forum everyone is entitled to my opinion.

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            • #7
              To clarify what I mean by bypassed grounded cathode gain stage:
              Click image for larger version

Name:	grounded cathode gain stage.png
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              In this forum everyone is entitled to my opinion.

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              • #8
                I'd derive (or look up) the gain equation for a common cathode gain stage with the cathode resistor not bypassed by a capacitor then you could substitute the cathode resistor Rk in that equation with the complex impedance of a resistor in parallel with a capacitor Z = Rk // 1/jwCk

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                • #9
                  The attached photo is an X-Y plot of a 12AX7, B+ 300V, grid stopper 33K, cathode resistor 1.5K bypassed by 100uF. Generator frequency was around 200Hz tweeked to minimize looping. Vertical axis is 50V per division with zero volts being the bottom line. Horizontal axis is 2V per division with the center graticule line being zero volts.
                  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 !

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                  • #10
                    just to make sure i undestand what the OP is talking about: when you say transfer function, you're talking about analysis in the digital domain, not the time domain, right? fourier analysis?
                    "Stand back, I'm holding a calculator." - chinrest

                    "I happen to have an original 1955 Stratocaster! The neck and body have been replaced with top quality Warmoth parts, I upgraded the hardware and put in custom, hand wound pickups. It's fabulous. There's nothing like that vintage tone or owning an original." - Chuck H

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                    • #11
                      Originally posted by bob p View Post
                      just to make sure i undestand what the OP is talking about: when you say transfer function, you're talking about analysis in the digital domain, not the time domain, right? fourier analysis?
                      You're probably right. I was thinking Transfer Function = Vout(t)/Vin(t). You could derive that by the method I posted above (use s=jw if you like) but it would be tedious.

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                      • #12
                        bob and Dave Yes the Transfer Function, you could use continuous Fourier transform to determine y(t) from Y(jw). In this case we could also use the Laplace transform, Y(s) ~ y(t), since iw = s here.
                        loadthud - Cool did you scope it?
                        In this forum everyone is entitled to my opinion.

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                        • #13
                          By the way, here's the Bode plot of the amplifier I thinker with, or better the circuit I tinker with.

                          Click image for larger version

Name:	bode plot.png
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                          In this forum everyone is entitled to my opinion.

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                          • #14
                            See if you can find a copy of Burkhard Vogel's "How to Gain Gain", the transfer function of the common cathode stage is covered in the first few chapters. If you do not need to include all of the tube's internal parameters, then Merlin's site and book have the basic formulas.

                            Jaz

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                            • #15
                              To derive a frequency-domain transfer function, there are three terms to worry about:
                              The bypassed cathode resistor together with the tube's rk gives a pole-zero pair, the same as a lead-lag compensator. Lead?lag compensator - Wikipedia, the free encyclopedia

                              The Miller capacitance together with the source resistance adds a high-frequency pole, same transfer function as a RC lowpass filter. http://en.wikipedia.org/wiki/RC_circuit

                              I would go about it by writing out a generic Laplace transfer function with two poles and a zero, writing out the equation for the frequency response, taking s=j*omega, and solving as a set of equations in s to find the pole positions.

                              The Miller capacitance pole is so much higher in frequency that you could safely assume it didn't interact with the other poles. This allows you to solve the lead-lag and lowpass parts of the equation separately and just multiply them together.

                              Most Spice simulators I've used have a "laplace" block that would allow you to enter the transfer function you derived and make a Bode plot of it alongside your actual circuit.

                              The resulting transfer function captures the frequency-domain behaviour (and therefore the time-domain behaviour too) but it's a linear approximation that ignores the non-linear behaviour. You can write another kind of transfer function that captures the non-linear behaviour but not the frequency-domain behavour. This is a more complicated affair, basically done by curve fitting. Korg have a patent that describes how they did it.

                              If you want to capture both aspects in a single function, the math gets horrible. One way of doing it uses a "Volterra kernel". Volterra series - Wikipedia, the free encyclopedia I've seen this used in DSP modelling plugins.
                              Last edited by Steve Conner; 04-12-2013, 10:35 AM.
                              "Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"

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