LTPI balance

[uneumann]

I'm posting this as a query to see if anyone has seen this idea as a means of balancing an LTPI before. This forum audience has seen more amps than I'll likely ever see. I've looked around the web and in my Morgan book. I'm guessing it may have turned up in the hifi domain, but I have not found it. All my searches just find pointers to active current sources to help balance a PI. I have not seen this solution yet. The circuit is pretty simple...



It works by adding about 25% more of the NFB signal at the base of the tail than at the inverting input. That turns out to be just about right for the tail base to track the cathode voltage exactly, thereby making the tail an almost perfect current source. I've Spice'd this PI (in a Deluxe Reverb output stage) and it seems to work really well, especially given the added parts count of only 2 resistors. The outputs are damn near perfectly balanced with two 100K plate load resistors.

I don't really want to debate whether a balanced PI sounds good or violates some secret amp juju. This is just a design question. If you wanted to build a balanced PI, this seems a way to get one without resorting to active sources or changing load resistors. I realize the 82k/100k load resistors are the standard way to balance the outputs by compensating for tube current differences. This is an alternative that actually balances the tube currents. Has anyone seen or used this idea before? Are there any downsides to the circuit behavior that I've missed?

Given the way it works, it also seems to be a way to make a lower value tail resistor perform like a really high-impedance current source, so you can reduce the cathode elevation voltage and add headroom to the outputs without sacrificing balance.

[nickb]

If we ignore NFB and just concentrate on the virtual current source, then using the speaker signal is not going to give you as precise a result as, say, a constant current source. The reason being tolerance variations together with the frequency and phase response of the succeeding stages. How important these are depends on you design goals, of course.

Afterthought: The whole idea falls flat when you start to clip. Whether that is a good or a bad thing is a whole new subject.

[uneumann]

If we ignore NFB and just concentrate on the virtual current source, then using the speaker signal is not going to give you as precise a result as, say, a constant current source. The reason being tolerance variations together with the frequency and phase response of the succeeding stages. How important these are depends on you design goals, of course.

Thanks - I see the point about spkr impedance will change the signal level at varied freqs. Fortunately, the whole NFB purpose is to reduce those variations.
I suspect an active source would do better. I'll just have to build/test one, Or maybe I can get the sim to show some of those effects.

Edit - I ran a sim and with small resistor tweaks, I show 25-30 db lower tail current vs plate current over 200-5kHz.
That's pretty good IMO. Basically ua vs ma. Of course it's just a sim, but still promising. I also sim'd an active current source (mostfet) and it does no better at 1KHz. The performance goes down at low and high freq's as you suggested.

[nickb]

I would think that is we did a transient (large signal) analysis including transformer and speaker effects things would look worse. But I do agree that the differences are probably too small to worry about.

One curiosity, I tried a CCS using two BC547B bipolars and got -89dB at 10Khz. Any FETs I tried were much worse by 40dB or so.

[Helmholtz]

How is PI output signal symmetry with global NFB disconnected?
In this case it seems that your arrangement somewhat decreases the grid to cathode input signal of V2, actually slightly increasing output asymmetry.

But with global NFB connected, the voltage divider R9/R10 makes V2 receive less feedback signal than V1 and thus its output is increased.

Am I missing something?
Where does arrow "C" connect to?

[uneumann]

I would think that is we did a transient (large signal) analysis including transformer and speaker effects things would look worse. But I do agree that the differences are probably too small to worry about.

One curiosity, I tried a CCS using two BC547B bipolars and got -89dB at 10Khz. Any FETs I tried were much worse by 40dB or so.

Interesting, I only did a quick test with a mosfet to compare these two circuits.
At 1Khz the tail currents are basically the same (in ua range). The CCS holds that level while the upper circuit slowly degrades at higher and lower freqs.

Comparing the upper circuit to the "normal" LTPI (without the added tail amplitude) shows a much bigger difference.
The signal traces below show how balanced the upper circuit outputs are at 500Hz with both plate load resistors at 100K.

[uneumann]

How is PI output signal symmetry with global NFB disconnected?
In this case it seems that your arrangement somewhat decreases the grid to cathode input signal of V2, actually slightly increasing output asymmetry.

But with global NFB connected, the voltage divider R9/R10 makes V2 receive less feedback signal than V1 and thus its output is increased.

Am I missing something?
Where does arrow "C" connect to?

Sorry for the confusion, A and C are just reference points. A is the input and C is just a reference for the cathode signal - it doesn't go anywhere or connect to anything.

Yes - the V2 grid signal is smaller than the tail signal - which is exactly the point. I used the Deluxe Reverb as my example case. That LTPI uses a 820/47 ohm NFB network. In this LTPI, the tail signal is made bigger by increasing R8 to 75 ohms. That bigger tail signal now tracks the cathode signal at C more accurately - making the tail resistor look like a very high impedance. That bigger tail signal is is then attenuated by R9 and R10 to get back to the original NFB level at the input of V2. So, the V2 level is the same in a normal Deluxe Reverb and in this "enhanced-tail" LTPI.

If you don't use NFB into the V2 grid - like in Marshall 18 watt amps, you can still use this idea. Just tap some feedback from the spkr signal and run it to the PI tail with a divider network (R7 and R8). That makes the tail resistor a very high impedance, improves the PI balance, and you can even reduce the tail resistor value to get more headroom out of the PI without sacrificing balance.

[Helmholtz]

That LTPI uses a 820/47 ohm NFB network. In this LTPI, the tail signal is made bigger by increasing R8 to 75 ohms. That bigger tail signal now tracks the cathode signal at C more accurately - making the tail resistor look like a very high impedance.

Does that mean that PI output symmetry improves even without global NFB connected?

[uneumann]

Does that mean that PI output symmetry improves even without global NFB connected?

Yup - that's a key point that seems true. The NFB tail connection can be used to balance the PI whether you use the NFB connection to V2 or not.
These are two independent connections that perform different functions in the circuit. (There is a more detailed description/analysis on my web site if you're interested.)

It seems too simple (like a free lunch) and sort of obvious in hindsight, which is why I was looking for a prior article or example somewhere.
I figure this must be something that other people have noticed or played with in the past. It's really just bootstrapping the tail resistor.
A lack of references to something like this makes me wonder if there's a fatal flaw that leads to discarding the idea.
So far I can not find a reference or a fatal flaw...

[nickb]

It seems too simple (like a free lunch) and sort of obvious in hindsight, which is why I was looking for a prior article or example somewhere.
I figure this must be something that other people have noticed or played with in the past. It's really just bootstrapping the tail resistor.
A lack of references to something like this makes me wonder if there's a fatal flaw that leads to discarding the idea.
So far I can not find a reference or a fatal flaw...

But isn't what you are doing here exactly what has been done for decades? The only minor variation is a different feedback ratio due to R9 and R10. What am I missing?

[uneumann]

But isn't what you are doing here exactly what has been done for decades? The only minor variation is a different feedback ratio due to R9 and R10. What am I missing?

Exactly right, Nick. That's why I say it's sort of obvious. The key point is simply to make the tail signal about 25-30% bigger than the V2 grid signal. The exact resistors/ratios can be found experimentally (I did it by iterative simulation). I'm sure there is a formula that can be derived for this, but I just have not taken the time/effort to do that yet - anyone want to take a stab? The target is simply to make the tail signal match the cathode signal.

The fact that the standard PI has been around so long and doesn't optimize the tail signal voltage is sort of surprising. It's like leaving money on the ground. Add two resistors and you get a balanced PI - then you can forget all this different load resistor stuff and it even works for non-NFB PI circuits.

[loudthud]

As an experiment, I'd like install a pot or switch to see how it sounds.

[uneumann]

So far I can not find a reference or a fatal flaw...

Well - I found one flaw - maybe not fatal, but worth noting.
In non-NFB amps, sometimes the PI is used as a mixer.
In that case the V2 input produces positive feedback to the tail, making the balance worse not better.
The bootstrap only works if the V2 input is either grounded or tied to NFB.

[nickb]

The signal at the bottom of the tail is ideally half the sum of the two inputs. If the open loop gain is a and the feedback ratio is b, then you can calculate the ratio of the output to bottom of tail resistor as

1/(2a) + b

So, for example taking a as 14 and b= 0.058 from around what you have for the design shown, with a guess for the OPT thrown in.

Then k = 0.5/14 + 0.058 = 0.094

In your design you have k= 82/(82+820) = 0.91 so that looks about right.

Also, the ratio of feedback to tail resistor is

0.058/0.094 ~= 0.63 i.e 37% lees than the tail signal.

Still, the current source is the superior option as it's a no-brainer and works for the mixer too

[Helmholtz]

The signal at the bottom of the tail is ideally half the sum of the two inputs.

Sorry, this I don't understand (assuming that "bottom of the tail" means the junction between R6 and R8)
To my knowledge, its the junction of the 2 cathodes where half the input signal appears. The signal at the top of R8 (= bottom of the tail?) must be much smaller as the value of R8 is only a small fraction of the total tail resistance. For this same reason I can't see much local feedback or bootstrap effect from doubling the value of R8. The balance of the circuit still seems to be dominated by the value of R6.

Also there seems to be some typing error in the formula(s) for k above. E.g the result of 1/(2a) + b would be 0.616. And the result of k = 0.5*14 + 0.058 = 7.06 (not 0.094).