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Thread: Bootstrapped Gain Stage Theory

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    Quote Originally Posted by SoulFetish View Post
    I signed up on diyAudio, but I can't download attachments yet. Could you post the 6111 model here, Dave? (unless it was written out as txt in a post?)
    Here's the 6111 model. I copied the text from post #848 - here

    6111.txt

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    I had a good hunch that the input impedance could be increased
    But the cathode input impedance of the complete cascode will still be low. I only expect an increase from about 190 Ohm to 240 Ohm.

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    I'd approach any design aimed at wringing more gain out of a tube, by thinking about how it compares to an efficient loadline in a common cathode amplifier, and then work backwards. So the starting point is to follow the old formula of aiming for a plate idle voltage that is 2/3 HT . With HT = 200, this is 133.2V. The 6111 Pmax = 1.1W. seeing as how rp = 4k, then Rp should be about 8k*. I ran a load line using the Philips datasheet, and this looks about right (see attached).

    *This follows the other old formula that ideal load resistance is rough equal to 2 x rp. ;-)

    Splitting this into a cathodyne, that's 4k for each load resistor, with the plate sitting at roughly 166V w.r.t. HT and the cathode sitting at 34V w.r.t. HT.

    If you want to anchor the grid, then aim for the grid to sit at -4V w.r.t the cathode i.e. 30V w.r.t. HT (or -70V in your dual rail supply). There are several ways of accomplishing that.

    To implement your simplified circuit, the split load for cathode load resistor would use a pair of 2k2.

    But if you want the grid to sit at 0V in your dual rail supply, then you need to unbalance the load considerably in order to get the tube to bias adequately, which means reducing the plate load resistance and increasing the cathode load resistance . My hunch is you won't get the cathode to sit where you want it without losing the plate load altogether, at which point there is no point in the feedback mechanism.

    So... you need to boost the current somehow in order to allow you to progressively increase the (unbalanced) load resistances whilst maintaining enough current to keep the tube 'on'. So maybe you could turn your other triode into current source? (e.g. by substituting it for the plate load resistor in a totem pole arrangement?) It still probably won't enable you to get the grid to sit at 0V though. Sorry for adding to the madness...

    Edit: with a 42k load and HT=200V, and -1v bias, the plate is sitting at at 50V w.r.t. HT (-50v in your dual rail supply) with s tube current of about 3.5mA. Which still leaves the issue of how to get the grid at 0v in your dual rail supply. And you wouldn’t have much signal headroom
    Attached Thumbnails Attached Thumbnails Click image for larger version. 

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    Did you measure output and input impedances of the cascode variants in the meantime?

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    Quote Originally Posted by Helmholtz View Post
    Did you measure output and input impedances of the cascode variants in the meantime?
    I set up a workable prototype board to wire this up, but got distracted by a project within the project and lost focus.
    I was going to build myself a Wein Bridge oscillator, because the Velleman signal generator I'm borrowing kind of sucks. But I wasn't sure of a suitable lamp I should use as a ptc stabilizer, and drifted farther and farther into the ether of attention deficit disorder..... But I back now! Thanks for snapping me out of it.
    I'm going to do some wiring tonight and see how far I can get. In lieu of a proper bench power supply, I'll probably tap of the supply from my amp.

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    The prototype board is all set to go.
    I picked up a Heathkit IG-18 signal generator for free and have been working on it to calibrate it. There are know issues caused by the loading of panel meter connection, so I disconnected the panel from the circuit, and turned the symmetry control fully counter clockwise to kill any signal feeding the square wave generator. I also put pretty much exact spec. cap values in the frequency multiplier switch using tektronix polycarbonate capacitors. The from 1kHz to into the +100kHz range, the sine wave look really clean with a very stable amplitude. (although it take a few cycles to bounce into stability.). The biggest issue left is as i decrease in frequency down from 1kHz, I get some pretty heavy power supply modulation of the output. I'll put a proper regulator circuit and probably replace modify the filtering when I get the chance, and hopefully that will clear things up.

    I needed to get the IG-18s output impedance to make our circuits input impedance measurements.
    So, I adjusted the unloaded output voltage to 1.002V RMS (2.833V P-P).
    Then connected to a load resistor measuring 1.005kΩ, I measured a voltage of 512mV RMS (1.447V P-P).
    Using the formula I posted earlier for Zout, the output impedance works out to me 961.82Ω.

    I think it's probably fine to round up and call this a driving impedance of 1kΩ.

    Here is the exact circuit under test.
    (note: anticipating a low input impedance possibility, I wanted to extend the bandwidth so the roll off was well below 1kHz into 100Ω. Therefor a 10F polypropylene coupling cap was used)

    Click image for larger version. 

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    I'm going to power this circuit of my amps existing supply. I'm heading out to the garage now, so hopefully I'll be able to post some results a little later on.

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    okay. the results are in. stay tuned to find out later on this week.......

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    Seems to me there is a lengthy thread about that Heath generator over at DIYAudio forum. People renovating them and updating them, and tweaking them, and making them more accurate, and so on. Might be worth a look.

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    I'm kidding.

    With the circuit connected as drawn above, the B+ where I tapped the supply of my amp dropped under load down to about +215V (which is a little closer to 200V I originally designed the circuit for).

    I connected my meter to the unloaded output of the signal generator and adjusted the voltage to 1.002V RMS.
    I connected the negative lead to the ground terminal of the 20F capacitor where the ground return connection was made, and the center conductor to the 10F coupling capacitor. Bringing up the voltage slowly with my variac, I watched the current draw (nervously) to insure I hadn't inadvertently miswired something, or installed a bum tube.
    But the DC operating point at all the terminals settled to their expected levels.

    As the tube began to draw current I watched as the loaded voltage at the cathode input drop to a measured 850mV RMS!
    That works out to an input impedance of around 5.42MΩ!
    The input impedance is way higher than anything I expected, so I ran the test again and got the same results.


    I had so many test leads flying around the bench to test all the voltages and monitoring the current draw to watch for any faults, I wasn't able to check monitor the input and output signals with a scope yet. Plus, I AC ground the grid of the second triode, and measure the results.

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  11. #46
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    Did I miss something? 1V in with 1K ohm source impedance and 850mV out? Doesn't that imply a input resistance of 0.85/(1.0-0.85) * 1k= 5.66K ohms?

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    Quote Originally Posted by nickb View Post
    Did I miss something? 1V in with 1K ohm source impedance and 850mV out? Doesn't that imply a input resistance of 0.85/(1.0-0.85) * 1k= 5.66K ohms?
    Ah Ha! I was just testing you to see if you were still payin' attention.... you passed. No, you're totally right. I just did the math and completely dropped out a decimal point or something.
    That's what I get for using a figure like "1.002V" and "961.82Ω". I'm sure we can all go through the rest of our day just fine without me having to provide an accuracy down to two hundredths of an RMS volt!
    But since I've already done it, technically, I got an input impedance of 5.27kΩ. But who's counting

    Thanks for catching that quickly. It's a good thing you guys are here to pick up the pieces.

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    I am trying to figure out why especially the cathode impedance of the left tube seems to be much higher than with a normal CF. I think one of the reasons is that the left 470k grid resistor couples most of the cathode signal to the left (input) grid.
    So I would like to ask you to repeat the cathode input impedance measurement with the left grid AC grounded via a 0.1 cap to eliminate (positive!) cathode to grid feedback.
    If - what I am expecting - cathode impedance drops, this will show that cathode input impedance depends on the source impedance at the grid.

    BTW, what is a 6C67? I thought you would use the 6111?

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    @HH

    Some observations using the 6111 model above, I see the input Z as
    2.5K for the circuit shown
    If you AC ground the grid of the second stage the Zin goes down to 600 ohms. Not a surprise since it's now got a much bigger signal going into it.
    If you remove the second stage altogether, the Zin is 2.6K
    If you also remove the AC ground (i.e the 20uf to ground) from the plate of the first stage the Zin goes up to 6.1k
    If you now replace the second stage the Zin is 4.9K i.e roughly double where we started (the two tubes are in effect in parallel).

    @SF if the objective is to get a high input Z then drive the grid instead of the cathode. Then you'll get more like 5meg. If you drive a common cathode stage from the plate of the first the gain go from a very sad gain of 1dB to a happy 24dB. It seems to be that the circuit as it stands doesn't do anything very useful at all and you'd get only slightly less gain from a bit of wire.

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    If you remove the second stage altogether, the Zin is 2.6K
    Thanks, but what is the cathode input impedance of the first stage alone with its grid AC grounded? I would expect something like 250 Ohm IIRC.

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    Quote Originally Posted by Helmholtz View Post
    Thanks, but what is the cathode input impedance of the first stage alone with its grid AC grounded? I would expect something like 250 Ohm IIRC.
    About 820 ohms. Remember the model might be poor....

    PS: Ignore that. I should have AC bypassed the 10k resistor so the DC conditions were changed to be closer the the answer you needed. Zin by this method is 116 ohms. The plate current is 4.8mA, the gm at 4.8mA is 2.9mA/V

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    Quote Originally Posted by nickb View Post
    About 820 ohms. Remember the model might be poor.
    Thanks a lot!

    Even though the sim results seem high (indicating a too low model gm) , they help me to get a better understanding of the interactions.

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    Quote Originally Posted by Helmholtz View Post
    Tanks a lot!

    Even though the sim results seem high (indicating a too low model gm) , they help me to get a better understanding of the interactions.
    Please note my revision above. I made a boo boo...

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    I should have AC bypassed the 10k resistor so the DC conditions were changed to be closer the the answer you needed. Zin by this method is 116 ohms. The plate current is 4.8mA, the gm at 4.8mA is 2.9mA/V
    No, I actually meant AC shorting the grid only, second triode disconnected. The 6111 datasheet gives me gm = 3.15mA/V @ 4.8mA/Vp = 150V. As actual plate to cathode voltage would be more like 200V, gm = 2.9mA/V seems reasonable (the 6111 is a rather non-linear tube meaning that tube parameters strongly depend on plate voltage and current). From this I calculate an internal cathode impedance of 345 Ohm. this would be shunted by 10.47K and 470k, resulting in 334 Ohm.

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    Quote Originally Posted by Helmholtz View Post
    BTW, what is a 6C67? I thought you would use the 6111?
    I'm glad you asked. The 6C67 is the tooth fairy's favorite tube, in that it doesn't exist.
    I'm sorry, that should be 6CG7. Of course, I was going to tell you that a 6CG7 has very similar characteristics to the 6111 and 6SN7 tubes; that it was designed to be a 9-pin equivalent to the 6SN7. But I suspect that actually know all this, now that it's clear that I really need to be wearing glasses more often.

    Quote Originally Posted by nickb View Post
    @SF if the objective is to get a high input Z then drive the grid instead of the cathode. Then you'll get more like 5meg. If you drive a common cathode stage from the plate of the first the gain go from a very sad gain of 1dB to a happy 24dB. It seems to be that the circuit as it stands doesn't do anything very useful at all and you'd get only slightly less gain from a bit of wire.
    Hang on a second mr 24dB. First, the circuit was an a hypothetical experiment to better understand terminal impedance characteristics, -- more specifically, how and if non traditional use of bootstrapping Techniques could be used to limit losses that due to impedance loading.
    Second, I disagree that its a useless circuit. The purpose of this circuit is to serve as a mixing stage for out of phase signals without phase cancellation, so I am using the grid as an input. Secondly, I suspect that this also offers good channel isolation without having to pad the shit out of the inputs.
    Since my application specifically does not require any voltage gain, I wasn't concerned with it. Thirdly(?), setting up this experiment and running tests is highly beneficial to me. Plus, I got this Heathkit up and running quite nicely now. New pamona banana jacks, precision capacitors in the multiplier, new filter caps, and a sweet violet indicator LED to replace the burned up neon.
    Click image for larger version. 

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    Pretty sweet, eh? (bad photo, but you get the idea.)
    Lastly, this is an evolving idea. I haven't ironed out all the kinks yet... Having said all that, I've been thinking that maybe driving the plate of the first triode is really a much better option. Your observations seem to indicate this too.

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    that maybe driving the plate of the first triode
    Plate input??

    Why don't you reveal what kind of signals you're going to mix? Might help to suggest a solution.

    Maybe you just need a differential amplifier (like a LTPI)? Reducing input impedance is easy.

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    Quote Originally Posted by Helmholtz View Post
    No, I actually meant AC shorting the grid only, second triode disconnected. The 6111 datasheet gives me gm = 3.15mA/V @ 4.8mA/Vp = 150V. As actual plate to cathode voltage would be more like 200V, gm = 2.9mA/V seems reasonable (the 6111 is a rather non-linear tube meaning that tube parameters strongly depend on plate voltage and current). From this I calculate an internal cathode impedance of 345 Ohm. this would be shunted by 10.47K and 470k, resulting in 334 Ohm.
    Ah! Okay doke. Done like that Zin is 220 ohms.

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    Quote Originally Posted by SoulFetish View Post
    I'm glad you asked. The 6C67 is the tooth fairy's favorite tube, in that it doesn't exist.
    I'm sorry, that should be 6CG7. Of course, I was going to tell you that a 6CG7 has very similar characteristics to the 6111 and 6SN7 tubes; that it was designed to be a 9-pin equivalent to the 6SN7. But I suspect that actually know all this, now that it's clear that I really need to be wearing glasses more often.



    Hang on a second mr 24dB. First, the circuit was an a hypothetical experiment to better understand terminal impedance characteristics, -- more specifically, how and if non traditional use of bootstrapping Techniques could be used to limit losses that due to impedance loading.
    Second, I disagree that its a useless circuit. The purpose of this circuit is to serve as a mixing stage for out of phase signals without phase cancellation, so I am using the grid as an input. Secondly, I suspect that this also offers good channel isolation without having to pad the shit out of the inputs.
    Since my application specifically does not require any voltage gain, I wasn't concerned with it. Thirdly(?), setting up this experiment and running tests is highly beneficial to me. Plus, I got this Heathkit up and running quite nicely now. – New pamona banana jacks, precision capacitors in the multiplier, new filter caps, and a sweet violet indicator LED to replace the burned up neon.
    Click image for larger version. 

Name:	Heathkit IG-18.jpg 
Views:	12 
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ID:	54914
    Pretty sweet, eh? (bad photo, but you get the idea.)
    Lastly, this is an evolving idea. I haven't ironed out all the kinks yet... Having said all that, I've been thinking that maybe driving the plate of the first triode is really a much better option. Your observations seem to indicate this too.
    Let's wait and see it in the final context. It seems that this is just a part of the picture.

    BTW, I'm actually more envious of the drill press.

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    Quote Originally Posted by Helmholtz View Post
    Plate???
    Yeah, instead of a grounded plate input stage. For instance... if you are mixing two out of phase signals -- one feeding the inverting input (grid), and the other feeding the non-inverting input (cathode), might it be better to take the output of the plate of that stage?
    Or maybe something like this? Just thinking aloud here, so everybody calm down if it sucks.
    Click image for larger version. 

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    Quote Originally Posted by nickb View Post
    Let's wait and see it in the final context. It seems that this is just a part of the picture.

    BTW, I'm actually more envious of the drill press.
    Oh, dude. The drill press is really nice (Jet). Picked it up second hand in great shape. I've got a set of precision bearings lined up for the next real maintenance cleaning and teardown. I'm looking a getting a LLambrich keyless chuck. Or an Albrecht if I can find on used in really nice shape, so I'll probably get into all that when I readjust and measure the runout.

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    You wrote "driving the plate of the first triode" which I interpreted as using the plate as input.

    Did you consider using a LTPI, out-of-phase signals applied to opposite grids and using either of the plates as output?
    Are you familiar with differential amplifiers?

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    Quote Originally Posted by SoulFetish View Post
    Yeah, instead of a grounded plate input stage. For instance... if you are mixing two out of phase signals -- one feeding the inverting input (grid), and the other feeding the non-inverting input (cathode), might it be better to take the output of the plate of that stage?
    Or maybe something like this? Just thinking aloud here, so everybody calm down if it sucks.
    Click image for larger version. 

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    But one input is 5Meg and the other is 5K! Why not just use a differential amplifier? You'll get about 20db of gain too.

    Click image for larger version. 

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    But anyway. One signal is dry and the other is wet from the reverb right? The reverb signal won't be out of phase but the phase will vary wildly with frequency. There's no need to be concerned about the mixing phase.

    PS: Oh.. I see HH suggested the same idea. Simulpost I guess

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    Quote Originally Posted by nickb View Post
    But anyway. One signal is dry and the other is wet from the reverb right? The reverb signal won't be out of phase but the phase will vary wildly with frequency. There's no need to be concerned about the mixing phase.
    I can dig that.

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    One signal is dry and the other is wet from the reverb right?
    If it's actually about mixing reverb and dry signals (didn't know up to now), these signals are not coherent (have no definite phase relation).

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    Quote Originally Posted by nickb View Post
    Ah! Okay doke. Done like that Zin is 220 ohms.
    If you remove the second stage altogether, the Zin is 2.6K
    What I learned from this discussion and especially from the sim results (thanks Dave and Nick) is that bootstrapping the grid leak resistor of a CF not only increases grid input impedance but can also considerably increase cathode impedance. Such increase of output impedance may not be desirable. Resulting cathode impedance obviously depends on the grid circuit (source) impedance.

    Now I wonder what bootstrapping does to a common cathode circuit.

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    Quote Originally Posted by Helmholtz View Post
    What I learned from this discussion and especially from the sim results (thanks Dave and Nick) is that bootstrapping the grid leak resistor of a CF not only increases grid input impedance but can also considerably increase cathode impedance. Such increase of output impedance may not be desirable. Resulting cathode impedance obviously depends on the grid circuit (source) impedance.

    Now I wonder what bootstrapping does to a common cathode circuit.

    Wait, didn’t we have to effectively short the grid to the “ground” point at the top of the bootstrap resistor to bootstrap the cathode? In a normal cathode biased cathode follower, where the grid is bootstrapped, how does that increase the internal cathode impedance in a meaningful way?

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    Wait, didn’t we have to effectively short the grid to the “ground” point at the top of the bootstrap resistor to bootstrap the cathode?
    Other than with a LTP, the top of the tail resistor in a single CF stage is no "ground" point. It actually carries the major part of the cathode signal (Vtail = Rtail x Icathode) and that is what is required for effective bootstrapping of the "grid leak" resistor.

    The cathode impedance values for the CF alone (provided by Nick) show a variation of a factor 12 between open grid input (2.6k) and AC shorted grid input (220 Ohm). The latter value is about what I calculated for the CF without bootstrapping. For an intermediate grid circuit impedance in the 100k range I thus expect increased cathode impedance with bootstrapping.

    As CFs typically don't need grid bootstrapping, the results may not seem relevant.
    But while in most cases increased output impedance would be considered a drawback, it might be desirable to increase the low cathode output impedance of a split-load PI. With a 12AX7 a 10k cathode impedance might be feasible.

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    Quote Originally Posted by Helmholtz View Post
    Other than with a LTP, the top of the tail resistor in a single CF stage is no "ground" point. It actually carries the major part of the cathode signal (Vtail = Rtail x Icathode) and that is what is required for effective bootstrapping of the "grid leak" resistor.
    I suppose. But it serves as the negative voltage reference point in the same way any cathode biased common cathode stage does where the grid leak is tied to ground. This was the only comparison I was trying to make.

    But while in most cases increased output impedance would be considered a drawback, it might be desirable to increase the low cathode output impedance of a split-load PI. With a 12AX7 a 10k cathode impedance might be feasible.
    I absolutely agree! (and need to do some testing) But now I have to reconsider how I set the bias in the drivers of my DC coupled output stage

    An CF output impedance of 2.6k sucks.

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    But it serves as the negative voltage reference point in the same way any cathode biased common cathode stage does where the grid leak is tied to ground.
    Obviously it makes a significant difference if the far end of the grid leak is connected to ground or to a signal.

    The interesting aspect to me is that bootstrapping seems to counteract cathode degeneration/negative feedback. Maybe not surprizing because bootstrapping is actually positive feedback.

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    Quote Originally Posted by Helmholtz View Post
    Obviously it makes a significant difference if the far end of the grid leak is connected to ground or to a signal.
    Indeed

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