ZOTL patent - convert to model for simulation?

Started by earthtonesaudio, October 17, 2011, 10:04:36 AM

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earthtonesaudio

I think the Berning/Milbert tube amp "impedance converter" is fascinating, but mysterious.  So I'm trying to understand it better, with the end goal of building a working model (but puny - maybe 1W or less).
I simplified the schematic from David Berning's patent significantly, and tried to get the results to work in Falstad's circuit simulator here.

I substituted an NPN transistor because the triode didn't seem to work correctly in the sim, and the two DPDT analog switches would have to be replaced by a real MOSFET H-bridge in a working unit, but for the sim I think that's an acceptable substitution.  Finally, I have shown the simplistic non-bridged load; for a speaker load obviously you'd need to block DC or use a BTL configuration.

Choosing the right transformer is critical, of course.  I don't really want to wind my own so I want to figure out what specs to look for in an off-the-shelf unit, then tailor the rest of the circuit to fit.


earthtonesaudio


PRR

> the triode didn't seem to work correctly in the sim

Needs standard bias.

> you'd need to block DC or use a BTL configuration.

This may be essential. That patent is too tricky for me to understand for free.

> I don't really want to wind my own so I want to figure out what specs

That's unlikely. This application is nothing like any standard switcher.

You need a switcher-guru, and that won't be cheap.
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earthtonesaudio

Thanks for the reply.  I think this circuit is just a magnetic amplifier with modified source and load, and I suspect that it may be possible to use "just any" high frequency transformer by adjusting the quiescent current of the tube (quiescent current acts like a bias winding in a mag amp).

PRR

The quiescent current and voltage of a tube Power amplifier can not be adjusted over any arbitrary range.

Taking my SE 6550 amp: below 250V 140mA it could not take full input power, over 600V 58mA exceeded a voltage rating.

> just a magnetic amplifier

In the 1950's sense of the word? No.

The H-bridge switches battery voltage. The transformer steps-up/down to tube voltage/current. The vacuum tube IS the amplifier. Its variations of voltage and current are transformed back to battery voltage and current in the load.

Winding 52 is selected for battery voltage. Windings 56..58 are selected for tube voltage. All of this of course times switching frequency and power handling capacity.

For a reasonable SE tube amp, battery voltage could be 20V and tube voltage could be 350V. PerHAPS you could do something with a small switcher with a ~~12V or 24VCT winding and a line-voltage primary.

The switching and rectifiers allow the transformer to have low inductance (small size) despite handling low frequencies.

The multiple rectifiers are not conceptually essential. It may be a detail due to fast rectifiers having low breakdown voltage, or a need for multiple windings to control breakdown or stray capacitance.

I'm merly discussing the circuit. The actual Patent coverage may be on only a portion of what is shown in the drawing. 27 Claims?? He's sliced it so thin there may be a way to motocross around the Claims. (Not your worry: anybody may construct a patent for self study, and I bet by the time you have it working good/better this patent will be expired.)
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jasperoosthoek

Interesting design.  ;D

Looking at the first page and not reading too much I think it works like this: The oscillator transformer switches the pair of MOSFETs of and on in a complimentary (C)MOS fashion. On one wave of the oscillator signal say the top left MOSFET and bottom right MOSFETs are closed and the other ones are opened. As a remark I'd say that can maybe be replicated by standard CMOS.

So the small winding of the second transformer connected to the tube get a square wave from the complimentary switching mosfets. This square wave gets rectified at the secondary leading to the tube. The resistance/transconductance of the tube determines whether current flows. More current in the tube (positive grid signal) means that an equal but transformed current flows in the smaller winding primary. This current will change direction every time though.

I'd say this might be replicated with a CMOS rectifier IC (CD4049), connecting the speaker load between the Vdd and the ground. This takes out the oscillator transformer. Two pairs get a positive square wave, two a negative (spreading the current) at say 300kHz. When nothing is connected, no current should flow to the speaker. In between them is a step-up transformer able to handle the oscillation frequency. Just simple trial and error, get something from Mouser that steps up from 9 volts to 250 volts and works at 300kHz.

I don't see right now how negative voltage swings supplied to the load work. In my head it only works with positive voltage swings. I'll think about it.

Maybe I'm overlooking something important or I oversimplify. Maybe I'm plain wrong, and PRR laughs at my explanation. It's just my 1cnt trying to understand this sucker.  :icon_mrgreen:
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earthtonesaudio



Think of it this way: the output transformer has two primary connections, A and B.  During one half-cycle, A is connected to the battery and B to the load.  During the next half-cycle, they switch so B is to the battery and A to the load.

This polarity reversal provides the square wave to the secondaries which is rectified into a DC voltage for the tube/transistor to use.  The tube/transistor pulls some current for bias, and then more or less depending on the signal.

It's impossible to tell from the patent drawings if the magnetic core is saturating, BUT the output waveform in my simulation looks very similar to the waveform of a magnetic amplifier from here

PRR

> maybe be replicated by standard CMOS.

Logic CMOS? Well, that stuff can hardly pass 0.1 Watts. "Tube power amps" generally eat 1 Watt, 10 Watts, or more.

The H-bridge is clearly Power MOSFETs. A similar scheme is in your computer power supply, though often as half-bridge form.

> if the magnetic core is saturating

No.

Works the same as modern PC switching supplies, except quite different impedance levels.
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jasperoosthoek

Quote from: PRR on October 24, 2011, 10:10:39 PM
> maybe be replicated by standard CMOS.

Logic CMOS? Well, that stuff can hardly pass 0.1 Watts. "Tube power amps" generally eat 1 Watt, 10 Watts, or more.

The H-bridge is clearly Power MOSFETs. A similar scheme is in your computer power supply, though often as half-bridge form.
True, but I was thinking about CMOS as a proof of principle. Maybe of you have 5 of them in parallel to drive a simple 12AT7 reverb.

What I found interesting is that he didn't use complimentary (power) mosfets but all N-type. But his oscillator transformer arrangement makes it very easy. But, I was wondering whether this could be simplified by a CMOS-type arrangement with two N-type and P-type (power) MOSFETs as an H-bridge. That way you will only need one transformer.
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PRR

P-type FETs have significantly worse conductance. When you work with POWER, you really want to hack your driver to use only N-type.
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