Adding NFB to an 18W

[Silvio55]

I want to add a NFB loop to my modded 18 Watt clone, just the loop, not a presence control. I was wondering if this is correct:

I've seen the Fender brown Deluxe schematic, it has a LTP PI and NFB, it's the same idea but it has this resistor before the PI ground, wich I don't know what does it do, do I have to add this resistor too?

Thanks!

[Dave H]

Your feedback resistor just goes to ground on the first schematic. It needs the second resistor like the Deluxe. The second resistor in combination with the feedback resistor sets the gain of the output stage.

[Silvio55]

Thanks Dave! Should I start experimenting with the Deluxe resistor values, or should I use different values as a starting point in the 18W?

[Mike Sulzer]

There is some analysis to do here, that you cannot do yourself. Wait for advice before proceeding.

Thanks Dave! Should I start experimenting with the Deluxe resistor values, or should I use different values as a starting point in the 18W?

[Chuck H]

You can use any impedance tap you want, but the circuit must be adjusted accordingly. I use the 8 ohm tap because the I don't have a 16 ohm tap. From the 8 ohm tap a 22:1 ratio is working well for me. The typical circuit is a voltage divider. The NFB loop is about voltage, not current. The feedback series resistor should be a high enough value not to interfere with the speaker load. Because the feedback shunt resistor elevates the PI tail you want to choose a value range for the circuit that won't affect the speaker load or interfere with the PI much. So... Using the KISS principal, try a 22k series and a 1k shunt. Like this:



You can experiment with the 22k value if you want to try different levels of NFB. Don't go below about 10k and anything higher than about 100k may as well be no NFB at all.

Cheers

EDIT: This schematic has been updated.

[66 Kicks]

Do you think that the NFB will work with the grid of the non-inverting stage of the PI AC grounded like that? It almost looks like positive feedback to me for that half of the PI.

[Chuck H]

Do you think that the NFB will work with the grid of the non-inverting stage of the PI AC grounded like that? It almost looks like positive feedback to me for that half of the PI.

Ah! Good catch. No, it doesn't make it a positive feedback loop. But it looks like it might nullify the loop on that PI output. As far as I've ever seen that grid bypass cap should go to the node of the PI tail and the feedback shunt resistor. Thanks for spotting that and being vigilant. I'll update the pic or post a new one if that isn't possible now.

[Mike Sulzer]

Ah! Good catch. No, it doesn't make it a positive feedback loop. But it looks like it might nullify the loop on that PI output. As far as I've ever seen that grid bypass cap should go to the node of the PI tail and the feedback shunt resistor. Thanks for spotting that and being vigilant. I'll update the pic or post a new one if that isn't possible now.

In the Marshall you have a long tail phase inverter which is not a bad differential amp when used in the conventional way. I do not see any need for the boot strapping. Why not just pull the capacitor to the right hand grid off ground and connect it directly to the junction of the resistors in a voltage divider from the output transformer?

[Dave H]

I do not see any need for the boot strapping. Why not just pull the capacitor to the right hand grid off ground and connect it directly to the junction of the resistors in a voltage divider from the output transformer?

I always wondered about that so I tried a simulation and the PI outputs were out of balance without the bootstrapping.

[Mike Sulzer]

I always wondered about that so I tried a simulation and the PI outputs were out of balance without the bootstrapping.

Even using a 56K resistor on the cathodes to ground? (The original 18 circuit does not use bootstrapping. I do not see why putting something other than ground into the right hand input would increase the need.)

[Dave H]

Even using a 56K resistor on the cathodes to ground? (The original 18 circuit does not use bootstrapping.

The original circuit is quite well balanced. It's a while since I did it but I seem to remember that adding negative feedback to the other input without bootstrapping put it out of balance. I'll do the simulation again to check.

[Mike Sulzer]

The original circuit is quite well balanced. It's a while since I did it but I seem to remember that adding negative feedback to the other input without bootstrapping put it out of balance. I'll do the simulation again to check.

Thanks, simulations are great. Having been around since computers used punched cards, I like to do approximate analyses by inspection as well. Here we see that 56K+ is large compared to the inverse of the gm; also, we have symmetry, and so which input should not matter.

[Dave H]

Here are the simulation results.

The circuit was - LTP plate resistors 100k, tail resistor 47k, bias resistor 1k
Feedback resistors 33k/1k from the 8ohm tap.

First plot is the standard circuit without -ve feedback.
Second plot is with -ve feedback and bootstrapping.
Third plot is with -ve feedback but no bootstrapping.

[Mike Sulzer]

Very interesting! I wonder where the extra asymmetry comes from in the feedback/no_bootstrapping case?

Thanks,

Mike

Here are the simulation results.

The circuit was - LTP plate resistors 100k, tail resistor 47k, bias resistor 1k
Feedback resistors 33k/1k from the 8ohm tap.

First plot is the standard circuit without -ve feedback.
Second plot is with -ve feedback and bootstrapping.
Third plot is with -ve feedback but no bootstrapping.

[ATTACH=CONFIG]37920[/ATTACH]

[ATTACH=CONFIG]37921[/ATTACH]

[ATTACH=CONFIG]37922[/ATTACH]

[Dave H]

I wonder where the extra asymmetry comes from in the feedback/no_bootstrapping case?

I'm guessing that the bootstrapped circuit has better common mode rejection because the bootstrapping keeps the tail current more constant. When I replaced the tail resistor with a constant current sink both circuits were equally well balanced.