Announcement

**Kramusha** · 10-11-2007, 08:46 PM

Can you post the schematic? I don't understand your question clear :-\

Maybe this links are interessting for you:
http://aikenamps.com/GlobalNegativeFeedback.htm
http://aikenamps.com/LongTailPairDesign.htm
http://aikenamps.com/LongTailPair.htm

Stefan

**Earl** · 10-11-2007, 11:57 PM

Thanks for the response..but I figgered it out.

If you look at a typical Fender LT PI, like in the DR ( http://www.ampwares.com/ffg/schem/de...b763_schem.gif ) , the "top" tube's grid is connected to the preamp output, and the "bottom" tube's grid is connected to the top of the NFB shunt resistor. That node is also connected to the bottom of the PI tail resistor.

In the 'normal' sense the PI tail resistor would be tied to ground. I was wondering why they tied the normally grounded tail to the NFB shunt resistor.

I read that this is used to help balance the stage, as the typical 22K ohm tail resistor is comparatively small to what would be expected. The NFB connection at the tail resistor aids in balancing the stage's two plate signal.

I PSpiced the whole mess, and it turns out that is the case for the most part.

**MWJB** · 10-12-2007, 11:42 AM

"In the 'normal' sense the PI tail resistor would be tied to ground."...only in a design without global NFB. Many, many LTPI designs do have global NFB and the tail resistor is not grounded, so I can't concur wuith the idea that it is "normal" to ground the tail resistor.

"I read that this is used to help balance the stage, as the typical 22K ohm tail resistor is comparatively small to what would be expected. The NFB connection at the tail resistor aids in balancing the stage's two plate signal." ...22K is in fact the largest value Fender used in this location (from memory), 10K & 6.8K were also used (late Tweeds & Brown/Blonde tolex amps). Smaller values raise gain, larger values reduce gain but balance the PI better. The tap for the global NFB must be taken from the "bottom" tube/triode in order that it be 180deg out of phase with the signal generated by the "top" triode. If this were not so the amp would howl like a banshee and would be unusable (unless you swapped round power tube grid wires, or OT primary wires to resolve the issue - a common fix in builds/repairs when the OT wiring legend doesn't match the original part - which puts you back to square 1).

**Earl** · 10-14-2007, 04:47 PM

Thanks for the reply MWJB. The crux of the bisquit was why do 'they' tie the NFB shunt resistor to the cathode and the grid #2 input. The cathode as an input is not a differential input, and was wondering what the purpose/benefit of doing this was.

As mentioned, you can eliminate that cathode connection, and tie the tail resistor (22K) to ground. It still works pretty much the same with the exception of some imbalance.

All three of the signals, g #1, g #2, and cathode are the same phase...so it's a little hard to get your mind around what's happening in a dynamic state.

**MWJB** · 10-15-2007, 08:17 PM

"All three of the signals, g #1, g #2, and cathode are the same phase..." , evidently they are not. If they were how would phase inversion be achieved?

**Earl** · 10-16-2007, 01:10 AM

That statement may have been unclear.

What I meant was that the preamp signal on Grid #1 and the signal from the negative feedback divider at Grid #2 are the same phase. Since Grid #2 is also tied to the cathode(s), it is also the same phase.

**MWJB** · 10-16-2007, 08:26 AM

Maybe I'm being a bit slow & not getting your drift...the signal at the 2nd triode's grid must be out of phase with the signal at the first triode's grid for phase inversion to take place. Are we agreed on that?

**Earl** · 10-16-2007, 02:37 PM

A signal at grid #1 (I'll call this the preamp input) causes the plate signal of triode #1 to be 180 out of phase with the input at grid #1.

The signal at the plate of tridoe #2 will be in phase with the signal at grid #1, and 180 out of phase with the plate of triode #1.

A signal at grid #2 (I'll call this the NFB input) causes the plate signal of triode #2 to be 180 out of phase with the input at grid #2.

The signal at the plate of tridoe #1 will be in phase with the signal at grid #2, and 180 out of phase with the plate of triode #2.

If you put the same phase/amplitude signal on grid #! and grid #2, the in phase plate signal at plate #1 generated from grid #2 will cancel the out of phase signal at plate #1 generated by the input at grid #1....and visa versa.

In this case, there would be no plate signals at either tube, and 100% negative feedback.

When I modeled this circuit, I used one AC source that represented the P-P voltage at the speaker terminals at 22W.

I hooked the 820/47 ohm NFB voltage divider to that source. I also hooked another voltage divider to that source to use as a signal to represent the preamp to PI input that would get 35V peak PI plate signals (to match the -35V output tube grid bias).

**MWJB** · 10-16-2007, 08:29 PM

"The signal at the plate of tridoe #1 will be in phase with the signal at grid #2, and 180 out of phase with the plate of triode #2." This would be so if grid #1 & grid #2 were 180 out of phase with each other (if they weren't your 820/47ohm voltage divider would no longer be NFB and your plates would be in phase with each other).

But then you say, ..."If you put the same phase/amplitude signal on grid #! and grid #2," why would you do this?

"the in phase plate signal at plate #1 generated from grid #2 will cancel the out of phase signal at plate #1 generated by the input at grid #1....and visa versa." Plate #1 is out of phase with grid #1, but in phase with grid #2.

You do have an amp in front of you don't you?

**loudthud** · 10-17-2007, 12:07 AM

The long-tail phase inverter is in it's essence, a differential amplifier or diff-pair.

The IDEAL diff-pair has a constant current source where the tail resistor is, and equal plate resistors. If the difference between the two inputs is zero, there is no difference in plate voltage. If both inputs move up or down together, this is a common mode signal and there is no output. If there is an input condition that causes the current in one side to increase, the other side must decrease by the same amount of current because of the constant current source on the cathodes.

(+delta I) * (plate R) = voltage swing P#1
(-delta I) * (plate R) = voltage swing P#2

The two signals are equal but opposite !

In the 1950's world of Leo Fender (et al), things weren't quite so ideal. If the tail resistor was just grounded, the tail current would increase when a positive signal is applied to G#1. This would cause the signal at P#1 to be bigger than that of P#2. The way to fix this would be to decrease the plate resistor at P#1. There are many HiFi amps that have a pot to adjust the imbalance. A clever way to reduce the imbalance is to apply an in-phase signal to the bottom of the tail resistor. If the voltage across the resistor remains constant, the current through it must be constant. The circuit does not perform as well as the ideal, but it is an improvement (close enough for rock and roll).

Note that while the signals at G#1 and G#2 are in phase, the signal on G#2 is smaller than the signal on G#1 therefore: the signal on G#2 RELATIVE to G#1 is 180 degrees out of phase.

I never understood how negative feedback worked until op-amps came along. Because they have so much gain, it's easier to realize how the difference between the inputs is very small and feedback controls the gain.

Edit: And because op-amps work at DC, there is no phase to confuse things !

**Earl** · 10-17-2007, 12:52 AM

Actually, my amp is just to the right of me, in the corner. It's a 1962 Princeton. I've had it since it was brand new.

My contention is that the preamp signal feeding grid #1 (triode #1) and the NFB signal feeding grid #2 (triode #2) are the same phase.

Let's take the plate signal of triode #1, and ignore the plate of triode #2, as we know it will be the same as the plate signal of triode #1, but opposite phase.

If you apply a 1V signal to grid #1 and the stage gain is 10, the signal at the plate of triode #1 will be 10V and 180 degrees out of phase with the grid #1 signal.

Move that 1V signal to grid #2, and the signal at the plate of triode #1 will be 10V again, but in phase with the signal at grid #2.

Now we have a way of swapping the phase of the signal at the plate of triode #1 just by switching the signal from grid #1 to grid #2. The 1V signal has not changed phase.

Further, if we tied the 1V signal to grid #1 and grid #2 at the same time, the plate signal of triode #1 would be a 10V, in phase signal added to a 10V, 180 degree out of phase signal. The net result would be a 0V signal.

If this circuit is supposed to be a differential amplifier, and the inputs are the two grids, then the difference between the two grids that have the same signal on them ought to be zero.

BTW, that Princeton I have has been operating for 45 years now and has never had a malfunction.

**loudthud** · 10-17-2007, 02:34 AM

If this circuit is supposed to be a differential amplifier, and the inputs are the two grids, then the difference between the two grids that have the same signal on them ought to be zero.

If the same signal is also at the bottom of the tail resistor, the output will be zero. But the Fender implimentation isn't perfect, so common mode rejection isn't perfect. What you have described above is how the bias tremolo works. The low frequency signal is applied to both power tube grids which modulates the gain of the power tubes but cancels at the output transformer ideally.

My contention is that the preamp signal feeding grid #1 (triode #1) and the NFB signal feeding grid #2 (triode #2) are the same phase.

This is true. The subtraction takes place within the differential amplifier. If you ground the feedback (short across the 47 ohm resistor) it will only take about 0.1V at grid #1 to clip the output. With the feedback connected, it will take about 1V. The gain has been reduced by 1/10 or 20dB. The feedback signal at grid#2 will be about 0.9V. The difference between grid#1 and grid #2 is still the 0.1V it took to clip the output when the feedback was grounded.

**MWJB** · 10-17-2007, 08:04 AM

Earl, the reason I asked if you had the amp in front of you is simply because you could answer a lot of questions in the time it takes to heat up an iron...less time than it takes to read this thread. Want to see what a larger tail does, or NFB vs no NFB? In 5 mins you have the answer. I'm sure your Princeton is a great amp, it's not going to help you with queries about LTPIs though...it doesn't have one.

**Earl** · 10-18-2007, 08:26 PM

Thanks again for the comments.

"My contention .....are the same phase."

"This is true". At last! :-)

"... it will only take about 0.1V at grid #1 to clip the output."

If you need 35V peak from the PI Plates to get the output tubes grids to 0V, and the (open loop) PI stage gain is less than 20 ... ?

MWJB - I don't own any test equipment. Also, my oarticular "tone" requirement is so broad that, MPO is that almost every amp and most every design sounds fine with me. IOW, I wouldn't understand "better" or "not better" tone with a mod...just a difference.

I'm just curious about certain things I don't understand, and usually PSpice can solve the puzzle for me...hoever, in the case of the cathode and Grid #2 connection in these PI circuits, the change (with/without) was not obvious.

I have a set of "tubes" that I got from that Norm Koren's website, and can generate a set of PCC's that are almost identical to the published data...for a number of common tubes. In that case, I feel fairly confident that the models represent pretty well the real thing, electrically. I was actually surprised that a 12AX7 phase shift oscillator model I 'made' (like in Fenders) actually oscillated in a simulation.

There's a heap of pretty smart folks here, and I hope tp gain some insight into the less obvios segments of amplifiers.

Announcement

Long Tail Phase Inverter

Long Tail Phase Inverter

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment

Comment