Tube amplifier analysis

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I'm trying to understand the use of the amplifiers in some of the schematics above.

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In several of the amplifiers there are preamp sections that do not use emitter degeneration but are configured as CC. V3B is one example. I'm a bit unsure why a capacitor was not added across the emitter resistor.

I've seen similar circuitry used for the input tube where no emitter resistor was used at all such as the supro Amp or the silvertones with no emitter degeneration at all.

From what I understand the "emitter degeneration" is actually used to create a stable bias for the tubes essentially lifting the emitter up a few volts which effectively lowers the gate a few volts. This allows input singles with no DC have full swing instead of being clipped.

Even more confusing is the input stage into the power amplifiers. This looks to be class B, i.e., push pull, with the input stage being a cathode coupled paraphase amplifier. What I don't understand is how the gate's are creating the 180 out of phase signals that are driving the power stage.

Any thoughts on whats going on?

Reply to
bob.jones5400
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The resistor is used in this case as a negative feedback. It has the effect of reducing the gain while keeping the frequency response flat over a wider range.

Note that there are two inputs to these stages. One is feedback from the output and the other is the audio input It is a little confusing from the way it is drawn but this is a differential amplifier. Any signal fed into one side will appear at both plates with a 180 deg phase shift. The feedback input compensates for the transformer frequency response and makes sure the gain is the same for both phases going to the power stage.

Reply to
JK17PWGBDR

Shunting the cathode/emitter resistor of a common cathode/emitter stage with a capacitor doesn't change the static (i.e.. 'DC') working point but it reduces the dynamic (i.e. 'AC') negative feedback within the stage, and this is often done to increase the stage gain. V3B in the given example follows the effects return input and its purpose is to bring the output of an effect device up to the same level (i.e. AC voltage) as what was fed into V3A, so the same overall gain is achieved with the effect in or out of circuit. The signal output by the cathode-follower V3A can be attenuated by up to about

19 dB by the 'send level' pot (to avoid clipping in the outboard effect device, which may have been designed to take a guitar-level signal). V3B then needs to provide 'make up' gain of up to 19 dB. With its non-bypassed cathode resistor, the stage will provide a voltage gain of about 12x or 21 dB which is then reduced somewhat by the resistors in series with the 'output' pot.

The cathode-coupled 'paraphase' phase splitter operates in the same way as a long-tailed pair. The pair of triodes share a common cathode resistor so when one triode is biased further into conduction, its increased cathode current raises the voltage of both cathodes so the second triode is biased further out of conduction (assuming its grid voltage doesn't change). The given example is a bit more complicated because the 330 k grid-leak resistors are connected to the 15 k + 100 R part of the common tail resistor (probably to raise the static operating voltage of the whole stage), whilst the 470 R part between the cathodes and the junction with the grid-leak resistors is the part that will cause the long-tailed pair behaviour. The signal to be phase-split is fed to the grid of one triode, two antiphase signals appear at the pair of anodes to be passed to the push-pull output valves, and the grid of the other triode can be connected to earth or used to add another signal (with reversed phase relative to the grid of the triode first mentioned). In the given example, a signal from the loudspeaker output is fed to this other grid to provide output-stage negative feedback.

Chris

Reply to
christofire

THERE IS NO PHASE SHIFT!!!!!!!!!!!!!

There is merely a polarity inversion.

Admittedly, an inverted sine wave LOOKS EXACLTY THE SAME AS one that's phase-delayed by 180 degrees, but THEY ARE NOT THE SAME. To shift the phase, you need some reactance in the signal path. There is none here, merely a POLARITY INVERSION.

I wish people could get this abstract thought through their concrete heads.

Thanks, Rich

Reply to
Rich Grise

Nothing abstract at all: polarity inversion is 180 degree phase shift.

John

Reply to
John Larkin

I WISH YOU OMNIFICENTS WOULD LEAVE US IGNORANT MORTALS ALONE!!!!!!!!!!!!

Reply to
bob.jones5400

Tubes DO NOT have emitters or collectors or gates. You have your terminology of tubes, bipolars and FETs all mixed up. And what is this with the meaningless "with no DC have full swing" junk? No voltage across a device means no *possibility* of any "swing". Oh yea; that "Lonestar" reference is not a schematic so "cathode coupled" is an ASS-u-ME-ption along with others...

Reply to
Robert Baer

Yes, they are. They're two different ways of talking about the same thing -- one in the the time domain, the other in the frequency domain.

You seem to have a misconception about what the term "phase shift" means. It doesn't imply any kind of time delay, nor any particular physical process.

It's purely a description of how the output waveform is related to the input waveform. It says nothing about how the relationship comes about.

--
Greg
Reply to
greg

OOPS! Sorry about that..having dial-up makes one impatient.

See V5A & V5B; they are configured as what is called as a "long-tailed pair". The unequal plate resistors compensate for the lower gain of the "B" side WRT the "A" side. Signal going into the grid of "A" drives it making a same-phase signal on the cathode and an opposite (and larger) signal on its plate. Assume that the GRID of "B" is constant; the voltage variations from "A" drives "B" giving a same phase signal on its plate. Those semi-equal (remember those plate resistors) signals then drive the power output stage. Too bad the transformer does not have screen taps...

Reply to
Robert Baer

...then build us a (say) 135 degree phase shifter good from (say) 10Hz to 100KHz...

Reply to
Robert Baer

I'm not sure what you are suggesting. I didn't say anything about 135 degrees.

The thing you suggest isn't impossible, or even seriously difficult, but it's not a polarity inverter.

John

Reply to
John Larkin

It's almost trivial to make phase shifters that are quite good over an octave. I do it all the time for use in image-reject mixers.

Over a wide range, it's been done, though not as easily... IIRC, see papers by Darlington... yep the same guy (at Bell Labs, of course ;-)

...Jim Thompson

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

NO IT IS NOT! The two waveforms LOOK identical, but one has been phase shifted, the other merely iverted.

Inversion does not shift phase.

Thanks, Rich

Reply to
Rich Grise

What's an "omnificient"?

Thanks, Rich

Reply to
Rich Grise

Let's try this little demonstration with a non-sinusoidal wave, say a pulse train:

_ _ _ _ _ _____| |_____| |_____| |_____| |_____| |___

Now look at it phase- shifted 180 degrees:

_ _ _ _ _ _ _| |_____| |_____| |_____| |_____| |_____| |___

Now look at it inverted:

_____ _____ _____ _____ _____ _____ |_| |_| |_| |_| |_|

See the difference?

Thanks, Rich

Reply to
Rich Grise

How bizarre.

John

Reply to
John Larkin

In the case of a pure sine wave, inversion and a 180-degree phase shift are precisely identical, mathematically and in practice. You cannot distinguish them based solely on the signals themselves (although you can look inside the "black box" and figure out whether the *mechanism* was one of inversion or time delay / phase shift).

Mathematically, sin(x + pi) = -sin(x) - the former is a phase shift and the latter an inversion. For any repeating signal (i.e. composed of the sum of sines of different frequencies), you can exactly invert the signal by phase-shifting each frequency component by exactly 180 degrees at that frequency.

You cannot exactly invert such a signal (in general) by shifting the

*whole* signal by a time equivalent to 180 degrees at its primary component frequency. It will (in general) look different.

The same is true for non-repeating or irregular signals.

--
Dave Platt                                    AE6EO
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior
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Reply to
Dave Platt

And, as you pointed out, that's not a sine wave. It's a sum of different (harmonically-related) sines (assuming that it's a precisely regular pulse train, as your drawing suggests).

In this case, inverting the signal shifts *each* of these component sines by 180 degrees at its individual frequency.

What you describe as "phase-shifted 180 degrees" is only a 180-degree phase shift at *one* frequency - that of the fundamental. It's quite a bit more than 180 degrees at each of the other component frequencies. As a result, the component sines don't sum up to an inverted version of the original pulse train.

--
Dave Platt                                    AE6EO
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior
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Reply to
Dave Platt

driving

that's

here,

shift.

Sorry John, Rich is actually right on this one. Just consider a moderately asymmetrical waveform.

Reply to
JosephKK

driving

that's

here,

How dare you besmirch the onmificient OZ?

Reply to
JosephKK

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