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  • MOSFET Follies Clairification

    This is about the MOSFET power tube buffer application.

    I'm building an amp that originally uses a cathodyne phase inverter (12ax7), AC coupled cathode followers (12AT7), and 4 x KT-88. After playing with the setup for a while and reading a bit, I realize the AC coupled cathode followers are kinda pointless and can be dangerous since the amp uses global NFB. I would like to convert this setup to use DC coupled MOSFETs (IRF820) as described in R.G.'s MOSFET Follies but have a few questions:

    Second application down, titled "MOSFET FOLLIES - POWER TUBE DRIVER".
    http://www.geofex.com/Article_Folder...osfetfolly.htm

    1) What is -Vs? If I'm reading the text correctly, it's supposed to be approximately 2x the bias voltage. Is this just a reference voltage, or does this need to pass appreciable amounts of current (I think pass current)? Does it have to be variable (like move up/down with the bias) (I think no)? What effect does it have on the operation of the circuit (I think limit the amount + the grid can be driven)? Can this be part of the bias circuit, say I took a -120v bias source and split it somehow to generate -40v for the bias and -80v for the MOSFET (I have no clue)?

    2) How are the power tubes biased? I'm not seeing how the bias voltage being applied "in front" of the MOSFET allows the tube to be biased.

    3) Is there a recommended +Vs for this application? I'm thinking in the 400-450v range, but that's simply because that's what'll be available at that point in the circuit. If I had to knock it down I'd just use a drain resistor, yes?

    4) Will the MOSFET require a heat sink in this application?

    I tried searching this forum and the Ampage forum for this info but came up empty handed. If there's an idiot's guide out there, or even a full on schematic of implementation, I'd be interested in checking it out.
    Last edited by defaced; 12-30-2009, 09:56 PM. Reason: Clarification
    -Mike

  • #2
    1) What is -Vs? If I'm reading the text correctly, it's supposed to be approximately 2x the bias voltage.
    That sounds reasonable to me.
    +Vs might be between +40 and +80V, clamping the Mos gate to them, or simply the full voltage available to the previous tube..
    With a 50Vac winding you can get +70 and -70 "raw" volts. The low current needed eases the filtering.
    If the -70 are not enough, a voltage doubler can give you -140 raw volts, meaning well filtered -80 or -100.
    2)
    the bias voltage being applied "in front" of the MOSFET
    that means its gate, faithfully appears in its source, minus around 3 volts, biasing the following tube.
    3)
    Is there a recommended +Vs for this application?
    I guess you can use whatever +B feeds the PI.
    4) Very probably not, we are talking a couple milliamperes here; at most a very small "U" one.
    Juan Manuel Fahey

    Comment


    • #3
      Originally posted by defaced View Post
      I'm building an amp that originally uses a cathodyne phase inverter (12ax7), AC coupled cathode followers (12AT7), and 4 x KT-88. After playing with the setup for a while and reading a bit, I realize the AC coupled cathode followers are kinda pointless and can be dangerous since the amp uses global NFB.
      Yep. And they'll make any grid blocking much worse.

      1) What is -Vs? If I'm reading the text correctly, it's supposed to be approximately 2x the bias voltage. Is this just a reference voltage, or does this need to pass appreciable amounts of current (I think pass current)?
      -Vs is the power supply for the sources of the MOSFET followers. It must be more negative than the intended bias voltage because the sources sit at -Vbias normally, and must swing up to and slightly more positive than 0V, and an equal amount down. This means the total power supply available to the sources must allow it to go as much as 2xVbias negative. A bit more is helpful. It must supply only the DC current needed to get the MOSFET to swing that much, so on average it's -Vbias/Rsource. The current to the grid load is negligible.

      Does it have to be variable (like move up/down with the bias) (I think no)?
      No.

      What effect does it have on the operation of the circuit (I think limit the amount + the grid can be driven)?
      It limits the amount the grid can be driven *negative*. The lowest Vbias amount of this the grid is essentially cut off in any event for a Class AB amp.

      Can this be part of the bias circuit, say I took a -120v bias source and split it somehow to generate -40v for the bias and -80v for the MOSFET (I have no clue)?
      It makes sense to only generate one -120V DC voltage, and to run the MOSFET sources from that, as well as generating the bias for the output tubes from it. I guess that is saying "yes, they can be integrated by design." However, the MOSFET sources need much more current than the few microamps a typical grid bias needs, so the integrated supply (if done that way) needs more current capability than a normal bias supply.

      2) How are the power tubes biased? I'm not seeing how the bias voltage being applied "in front" of the MOSFET allows the tube to be biased.
      Enhancement mode MOSFETs, like the ones suggested, don't conduct at all until the gate is about 3-4V more positive than the source. Then they conduct at about 1A per extra volt of Vgs. So a MOSFET might conduct zero up to 3.463V between the gate and source. By the time it was at 4.463V, it would be trying to conduct 1A/1000ma. Since this circuit never needs more than a few milliamperes, the source never gets more than the threshold voltage plus a few millivolts; we can usually ignore the millivolts, and to a first approximation call the voltage between gate and source constant at Vthreshold.

      With that in mind, the source "follows" the gate, but Vthreshold lower. This is how the bias ahead of the MOSFET can bias the output tube. You put a variable voltage on the MOSFET gate from -Vbias (as needed for a normal output tube) to a few volts more positive. The source follows the gate Vthreshold lower. So the tube, when properly biased, is still getting the exact same voltage to make its plate current right. It's just getting it from the MOSFET source. The gate of the MOSFET sets the source voltage from the bias control. So adjusting the gate of the MOSFET adjusts the grid of the tube.

      3) Is there a recommended +Vs for this application? I'm thinking in the 400-450v range, but that's simply because that's what'll be available at that point in the circuit. If I had to knock it down I'd just use a drain resistor, yes?
      Yes, kinda. The MOSFET power rating and the amount of power you want to waste as heat gets into it. The whole point of MOSFET followers is to be able to drag the output tube grid higher than ground, if only by a few volts. A drain voltage (+Vs) for the MOSFET of +20 to +50 is adequate. But it's a PITA to generate that in a tube amp. So we try to be clever and get it somehow from B+, which we do have. A drain resistor to B+ will work. It's a kind of a waste, but will work.

      4) Will the MOSFET require a heat sink in this application?
      It depends on how much voltage difference appears across it (that is, sum of the +Vs and -Vs), how much bigger (negative) -Vs is than -Vbias for the tubes, and how much current you flow through that resistor from source to -Vs. Power is *always* voltage times current, so by setting the total voltage across it and the current through it, you set the power. Once the power is known, you can see whether the MOSFET needs a heat sink. A TO220 can dissipate up to about 2-3W by itself in open air. More than that and you need a heat sink to keep it from getting too hot.
      I tried searching this forum and the Ampage forum for this info but came up empty handed. If there's an idiot's guide out there, or even a full on schematic of implementation, I'd be interested in checking it out.
      A full schematic of an implementation would be nice, and I always intended to do that one day. These are not idiot questions, by the way. They're the real questions one needs answered to do a good implementation.

      Here's how I'd go about a design.
      1. What output tubes and class of operation do I want? This sets the bias voltage needed to get the output tube grids at the right amount negative from the cathodes, which are presumably grounded or so close as makes no difference. For this example, let's assume it's some 6L6s or EL34s and needs perhaps -40 to -50V for bias.
      2. Figure out how much current needs to flow in the source resistors. We know that the output tubes will have their sources grounded and with no signal, their grids will sit at -50V (just picking a bias voltage; it will actually be somewhat less than this). The tube specs give a maximum resistor to negative bias voltage, usually in the 100K to 470K range. Let's say that's 220K. So the pull down to -Vs needs to be 220K or less. Make it 100K for grins. And we need -Vs to be about twice the bias voltage, or -100V so the source can swing negative as far as it swings positive (about) from the negative bias voltage for the tube.

      So if we use a 100K source resistor and -100V for the -Vs, 1/2 ma flows in the source resistor. If we use a 10K, 5ma flows. The grid current can be neglected in this. Something between 100K and 10K is OK. The higher the current, the "tighter grip" the MOSFET has on the grid of the output tube.

      3. Figure the +Vs and the MOSFET dissipation. It is very convenient to derive the MOSFET drain supply from B+. All you need is a resistor from B+ to the drain. However, the B+ supply has to be able to supply another 2ma to 20ma without strain, and the resistor and MOSFET have to live with the voltage and dissipation. You want to pull the grid of the output tube up to maybe +10V; that means the drain resistor has to be able to let through B+ minus perhaps 15V at the desired current. When the grid is pulled all the way up, the source will be at about +10V, the source resistor will be letting through 110V/100K = 1.1ma if it's 100K and 11ma if it's 10K. So the drain resistor has to be something like (420-15)V/0.0011A = 368K to 36.8K for the high current case to let enough voltage get to the MOSFET.

      At zero signal, the current is half that, so the drain resistor sees 0.5 to 5ma for the two cases, and that means its voltage is 184V in both cases. This leaves 420-184V = 238V from the MOSFET drain to ground. So the MOSFET sees 238V plus -50V = 288V from drain to source, and from 0.5 to 5.5ma.

      The low current case gives a MOSFET power of 288V * 0.005 = 0.144W, and the high current case gives a power of ten times as much, 1.44W. I would bolt a small aluminum tab of maybe 1" by 2" or a small stamped heat sink to the MOSFET for the high-current case.

      A better choice might be to make a small +/- 100V or +/-`120Vdc power supply with a small transformer. This can be quite small, perhaps only a 5W to 10W transformer; a small isolation transformer would give +/-160Vdc. This would cut the dissipation on the MOSFET because now the drain resistor would only go to the lower +160V supply. You might be able to leave out the drain resistor on each MOSFET entirely and still get away with no heat sink on a TO220 MOSFET.

      Did that help?

      Or did I just muddy it up?
      Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

      Oh, wait! That sounds familiar, somehow.

      Comment


      • #4
        Originally posted by defaced View Post
        This is about the MOSFET power tube buffer application.

        I'm building an amp that originally uses a cathodyne phase inverter (12ax7), AC coupled cathode followers (12AT7), and 4 x KT-88. After playing with the setup for a while and reading a bit, I realize the AC coupled cathode followers are kinda pointless and can be dangerous since the amp uses global NFB.:

        Reading up a bit huh, or just taking good advice?
        Last edited by Joey Voltage; 12-31-2009, 03:13 AM. Reason: clarification

        Comment


        • #5
          Thank you! Now if I get snowed in this weekend and run out of painting chores to do I can play!

          Did that help?

          Or did I just muddy it up?
          That cleared up a whole lot. I'll have to play with it to really *see* everything, but just reading through it, it made a whole lotta sense. Thank you for the full-on explanation; your help is indispensable.

          Reading up a bit huh, or just taking good advice?
          Both. Between reading the same thing from you, Steve Conner, Merlin, R.G. and a hand full of other people, and what my ears were telling me on my build, I decided to open my pallet a bit more and start down this road. I figure the worst I can do is learn something.
          -Mike

          Comment


          • #6
            Originally posted by defaced View Post
            Thank you! Now if I get snowed in this weekend and run out of painting chores to do I can play!

            That cleared up a whole lot. I'll have to play with it to really *see* everything, but just reading through it, it made a whole lotta sense. Thank you for the full-on explanation; your help is indispensable.

            Both. Between reading the same thing from you, Steve Conner, Merlin, R.G. and a hand full of other people, and what my ears were telling me on my build, I decided to open my pallet a bit more and start down this road. I figure the worst I can do is learn something.
            Rg did put it quite beautifully didn't he, and you can see why I kinda unleashed on that particular circuit. It is doing absolutely nothing besides wasting a tube, taking (although probably minimally) HT, and heater current, and introducing a couple more phase shifts within the -fb loop. part of the reason you would use a CF/SF/EF circuit in the first place is to isolate the coupling caps from the output valves, which can't happen if there are two very large ones coupling the follower to the output tube grids. kinda boneheaded and redundant in that circuit, and I don't know why they did it that way exactly.

            Comment


            • #7
              Oh yeah!, and say hey to big Ilsa for me

              Comment


              • #8
                So, here's what I'm thinking:

                Use this transformer wired with the primary and secondaries in parallel. Connect the 6.3v secondary to the heater winding. This will give me primary voltage which I can turn into a bi-polar power supply. FS12-500 Triad Magnetics Transformers I've got the heater winding current capacity for this, and figuring 80% efficiency for the transformer, this should pull about half an amp if I remember my numbers right. But, if someone knows a source for a small isolation transformer with a similar footprint, I'd really like to go that route. No reason to tax the mains transformer unless I have to (though at 400VA, it'll be ok).

                Now, this is my first time doing a bipolar supply. I pulled this schematic from GGG, but if I'm thinking right, I'm just going to get half wave rectification. Is that correct? Is 47uf going to be enough filtering? I haven't looked too hard, but I might be able to get up to 100uf within the physical size I can afford.

                The rest is just the R.G.'s schem expanded for individual bias and a couple of extra taps for grounding that's specific to my amp's layout and -V for my bias supply.

                The board is included for comment. Ignore the MOSFET outlines. They are reversed because I am going to stand them up and put a heat sink over the resistors. There's interference with a power tube socket if I put the heat sink out beyond the board.
                Attached Files
                Last edited by defaced; 01-11-2010, 02:39 PM.
                -Mike

                Comment


                • #9
                  Originally posted by defaced View Post
                  Is 47uf going to be enough filtering? I haven't looked too hard, but I might be able to get up to 100uf within the physical size I can afford.
                  I worry about that, but I have not done the numbers yet. Sorry I didn't think of this earlier to warn you about.

                  Here's the issue. A source follower doesn't have much power supply rejection on the source side. Generally these are running to ground, so it's not such an issue. In this one, the source supply is directly from a capacitor with ripple, so the source power supply ripple comes through the source resistor and is attenuated by the ratio of the source resistor to the dynamic impedance of the source itself. That's low, generally, but keep it in mind.

                  Also, you're using a half-wave rectifier to get your +/- voltages. I'd feel better if you would put the two primaries in series and full wave rectify the two of them. You'd get the same output voltages, but both would be full wave rectified. That's important because it cuts the capacitor size for a given current/ripple in half. It doubles the load on the transformer, too, so the available output current is about halved. I don't think that's a problem, but smaller capacitors or lower ripple for the same capacitor is a big deal.

                  For a capacitor C charged by pulses at intervals of T seconds, discharged by a DC load of I amps, the voltage runs down by an amount of V = IT/C . The charge pulse presumably restores it to the peak voltage each time.

                  Doubling T (half wave versus full wave rectifier) means that the voltage declines by twice as much between charge pulses for the same I and C, or means that you need twice the C for the same ripple (charge/discharge) voltage.
                  Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

                  Oh, wait! That sounds familiar, somehow.

                  Comment


                  • #10
                    Running FW and series primary shouldn't be an issue. A DB107-1, or -S is small enough I can find a home for it. Is the attached schematic what you were envisioning?
                    Attached Files
                    -Mike

                    Comment


                    • #11
                      Originally posted by defaced View Post
                      Running FW and series primary shouldn't be an issue. A DB107-1, or -S is small enough I can find a home for it. Is the attached schematic what you were envisioning?
                      Yeah, that's it. You get the same output voltages, but full wave rectified instead of half wave. It's easier on the caps, makes them twice as effective as the half wave version. Be prepared to put in something like a zener regulator on the minus supply if needed to keep ripple out. The + supply is kind of an afterthought in this design. It's there to have headroom, and the high impedance at the drain means that ripple on the positive side doesn't get through to the source.

                      For diodes, you just have to have enough voltage. In this case, use a 500V or more diode bridge.
                      Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

                      Oh, wait! That sounds familiar, somehow.

                      Comment


                      • #12
                        I'm not familiar with using a zener for ripple rejection. Is it done like this:
                        http://www.reuk.co.uk/OtherImages/ze...-regulator.gif
                        -Mike

                        Comment


                        • #13
                          Originally posted by defaced View Post
                          I'm not familiar with using a zener for ripple rejection. Is it done like this:
                          http://www.reuk.co.uk/OtherImages/ze...-regulator.gif
                          yes, just a simple Zener shunt regulator. If you happen to take a look at some newer hi-fi designs that incorporate this, you will see that some use Linear IC regulators, although I don't know how truly necessary it is.

                          Comment


                          • #14
                            Was thinking about this last night and before I get too far ahead in buying a power transformer, want to make sure I have something straight. Ok, so the purpose of direct coupled followers is to drive the grid of the power tubes into conduction. Where does the current from this conduction come from/where does it go? Meaning, when I choose a power transformer for this setup, should I also be accounting for the current the grids will be conducting when driven into AB2? And if so, about how much current will that be per grid?
                            -Mike

                            Comment


                            • #15
                              More questions. About the zener ripple rejection. The current rating for the zener is the current that's going to be dumped to ground as part of the regulation, correct? I'm making sure the transformer I'm looking at is up to the job before I get too far ahead of myself. I'm not used to counting milliamps, in my day job, 5 amps is the smallest increment I care about.
                              -Mike

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