Resistor types for 60’s fender.

[BrownIsound]

My feedback resistor was too high, >1.2k, so over 40% high. The tail-ground resistor was in spec at 104.

Quite an enlightening experiment…a parallel 1k to the now 1.2k (550) gave it a Twin Reverb like tightness. Paralleled a 3.3k for now to get it closer to spec while I await a replacement resistor.

Interesting to see that the 820/100 feedback circuit is used in most of the BF/SF amps which means higher feedback in the TR (double power output at same 4ohm load) vs the other 2 speaker 40W amps. The SR has a 2 ohm output but same 820/100 feedback circuit, so half the feedback again compared to the vibrolux/Pro. DR has 820/47, but an 8ohm load, but also half the power so 1/2 feedback again.

I also wonder, with the amp cranked into distortion, is it putting more than 1/2W through the 820 resistor?

[B Ingram]

Interesting to see that the 820/100 feedback circuit is used in most of the BF/SF amps which means higher feedback in the TR (double power output at same 4ohm load) vs the other 2 speaker 40W amps. The SR has a 2 ohm output but same 820/100 feedback circuit, so half the feedback again compared to the vibrolux/Pro. DR has 820/47, but an 8ohm load, but also half the power so 1/2 feedback again.

I would recommend looking at "quantity of feedback" differently: "8Ω circuit" vs "4Ω circuit" (pretend the Super Reverb doesn't exist)

If you were reading the Radiotron Designer's Handbook, they would be calling Fender's feedback around the power section "negative voltage feedback." There is a voltage present across the speaker(s), and a sample of that is being applied to an earlier point in the amp. So how much voltage is sampled?

Pretend we have a case of "16Ω load" and "4Ω load." Let's find the voltage present for the same power at the speaker(s), and let's assume 40 watts of power:
16Ω
Voltage = √(Power x Impedance) = √(40w x 16Ω) = 25.3v RMS
- Check: Current = Volts / Impedance = 25.3v RMS / 16Ω = 1.58A RMS, and Power = Volts x Current = 25.3v RMS x 1.58A RMS = 40 watts

4Ω
Voltage = √(Power x Impedance) = √(40w x 4Ω) = 12.65v RMS
- Check: Current = Volts / Impedance = 12.65v RMS / 4Ω = 3.16A RMS, and Power = Volts x Current = 12.65v RMS x 3.16A RMS = 40 watts

- We see that with same power:
- "Impedance x 4" gives twice the voltage at the speaker(s) compared to our starting point. This is going from 4Ω to 16Ω.
- "Impedance / 4" gives half the voltage at the speaker(s) compared to our starting point. This is going from 16Ω to 4Ω.

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Look at any blackface Fender amp schematic, making a note of whether there is 1 speaker or 2 speakers. Since Fender used 8Ω speakers for most of the lineup, this is noting whether the amp has an 8Ω load or a 4Ω load.

As shown in the Showman schematic, Fender used "100Ω" in the feedback loop if the amp had an 8Ω load, but halved that to 47Ω for a 4Ω load.

Check any amp and you'll find that case generally holds.
4Ω Pro Reverb got a 100Ω resistor.
4Ω Tremolux got a 100Ω resistor.
4Ω Twin Reverb got a 100Ω resistor.

8Ω (single) Showman got a 47Ω resistor.
8Ω Deluxe Reverb got a 47Ω resistor.
8Ω Vibrolux got a 47Ω resistor.

The Champ & Princeton are special cases, so you should leave them out of this comparison (though funny enough, they get a 47Ω resistor to go with their single-speaker). The Super Reverb is also a special case.

Now Fender seems to have wanted to give "similar feedback" to amps with the same power section (making the same power output) when their speaker loads changed by halving the resistor-value for the higher speaker-voltage of an 8Ω load. Except the step from "4Ω to 8Ω" doesn't yield "Volts x 2" but rather "Volts x √2" (you can repeat the steps shown earlier to prove that to yourself).

Funny enough, if 100Ω is perfect for 4Ω loads, something more like 68Ω would be closer to "same feedback" for the 8Ω amps.
I don't think Fender was trying to "optimize every model" but instead made some quick decisions to adapt a single design to "single-speaker version" and "dual-speaker version."

[solderhead]

I have a drip edge late 1967/early 68 AA165 that is all original other than the electrolytic caps and added 3 prong power cord. Many of the resistors have drifted up out of spec. What is the current consensus these days on the best type of resistor to replace the carbon comps? Dependent on the specific part of the circuit?

I'm going to stick my neck out and disagree with some of the conventional wisdom in this thread.

You've already noted that the old CC resistors drifted upwards, often taking them out of spec. I don't agree with the opinion that you don't need to change resistors unless they exceed their 20% tolerance spec. IMO you need to focus your attention on the sound of the amp and not on specs. There are amps out there that sound wonderful because they have drifted out of spec, and there are other amps that wound wonderful because of a combination of drifting tolerances that stack up on one another. And there are other amps that sound bad because of tolerance drift. When making the decision on what parts to replace, I use my ears, my calculator, and some alligator clips.

The nice thing about CC resistors is that they drift upward in value, and it's a simple task to calculate how much parallel resistance you need to drape across them to bring that portion of the circuit back into spec. Alligator clips mounted back to back work wonders for temporarily adding components to a circuit to determine whether or not it's worth replacing the parts, and how much compensation for drift (if any) results in the best sound.

When it comes to CC resistors I would definitely consider them for use as plate load resistors if noise is not a problem, but I'm not sure that I'd use them elsewhere -- especially when the circuit calls for a large value because noise is your enemy. IMO this is where CF and MF tend to work better. I know -- it's heresy. But then I'm the guy who if I've got crackling output tubes I won't hesitate to replace the CC resistors with flameproof MO.

> I also wonder, with the amp cranked into distortion, is it putting more than 1/2W through the 820 resistor?

I would say No. But if you really want to be sure, then drive the PI into distortion with a sine wave and measure the voltage drop across the resistor.