Pre Planning For Modifying A Fender Super Reverb Amp

[TomCarlos]

Thanks Dave!!

I started to create a similar sheet based on my lab tests. But this is helpful - especially the examples with the choke.

Much appreciated.

[Dave H]

Have you had a good look through that page? There’s a fair bit of nonsense on it

Care to point out the nonsense for us?

[pdf64]

Care to point out the nonsense for us?

All the way through it references a V (Avg) D.C. value, without defining what it is.
eg for the two phase cap input example, it states

V (Peak) D.C. = 0.71 X Sec. V A.C.
V (Avg) D.C. = 0.45 X Sec. V A.C.
I D.C. = 1.00 X Sec. I A.C.

What is V (Avg) D.C. and why is it different to V (Peak) D.C.?
Bear in mind that it states at the top that losses aren't accounted for, so it can't be fudging in some notional degree of ripple loss.
And, as per Helmholtz's post, the Iac ratios are wonky (yes I know I stated Iac=Idc for the two phase, but that just goes to show how pernicious that page is!)

[Helmholtz]

I think V(Peak) means the no-load voltage. V (Avg) D.C. = 0.45 X Sec. V A.C. is correct acc. to my literature.

[pdf64]

I think V(Peak) means the no-load voltage...

No load is the only condition in the document's scope.

...V (Avg) D.C. = 0.45 X Sec. V A.C. is correct acc.to my literature.

In the case of a two phase cap input, for that to be true, the cap would have to be hopelessly inadequate for the loading. And in any case, losses (from loading or whatever) are excluded from the scope of the document.

[Helmholtz]

No load is the only condition in the document's scope.

I don't see that. It would make no sense, especially as the diagrams show a load resistor and they specify a DC load current.

You're right about the 0.45 voltage factor. It only holds for resistive loads (no cap), sorry. Will look up my books.

[Dave H]

What is V (Avg) D.C. and why is it different to V (Peak) D.C.?

And, as per Helmholtz's post, the Iac ratios are wonky (yes I know I stated Iac=Idc for the two phase, but that just goes to show how pernicious that page is! )

V (Avg) D.C. is the average value of the rectified waveform without filter capacitor so it shouldn't be listed in any of the examples with a capacitor.

It is possible for Iac to equal Idc for the two phase. It depends on the conduction angle.

[TomCarlos]

Thanks for the continued discussion on this. I am taking notes. I must admit, for me there is lots of confusion around Vpk, Vrms, Vdc, that Vpk = Vdc, etc. So this is interesting conversation. I have studied electronics and have my books. But it is goofy when guys use terms interchangeably. If nothing else, the Hammond paper is a reference sheet - one with pictures and allows for me to capture notes.

[Helmholtz]

It is possible for Iac to equal Idc for the two phase. It depends on the conduction angle.

As the AC currents with bridge and two-phase rectification should differ by a factor of 1.414, only one of the formulas/factors for the DC current can be correct assuming the same conduction angle.
I think it should read IDC = 0.88 X Sec. IAC for two-phase cap input. This would lead to:

Iac = 1.14*Idc for two-phase and
Iac = 1.61*Idc for bridge.

Both these conversion factors lie within the ranges I gave and would apply for large conduction angle. Meanwhile I could confirm my numbers with other literature.

[Dave H]

I think it should read IDC = 0.88 X Sec. IAC for two-phase cap input. This would lead to:

Iac = 1.14*Idc for two-phase and
Iac = 1.61*Idc for bridge.

I'm sure you are correct (assuming the same conduction angle)
I wonder why Hammond chose the value the did?
I'm sure I've measured Iac to be greater than 2*Idc on a few occasions in the past for the bridge.

I've just run up a sim for the two-phase. It shows that it has to be a really lossy power supply (small cap, high series R) for Iac to be equal to Idc as I said above. Sorry about that.

[Helmholtz]

The ranges for the AC/DC current conversion factors I posted cover conduction angles between 60° and 90°. This is what the authors call the "desirable" range.
There are more general formulas which allow the the calculation of the conversion factor from a known conduction angle.

[Dave H]

The ranges for the AC/DC current conversion factors I posted cover conduction angles between 60° and 90°. This is what the authors call the "desirable" range.

I've edited my post. It wasn't clear that I meant the value Hammond chose.

[Helmholtz]

Regarding voltages things get complicated as it strongly depends on load current.

The only thing I can safely state is that DCV must lie between the 2 extremes given as V(Peak)DC and V(Avg)DC* in the Hammond chart .

The "peak" value would apply when the amp is in standby.

* V(Avg)DC as used in the Hammond Design Guide is the averaged output DCV with resistive load (no reservoir cap). It represents the theoretical (but rather useless) lower bound for the averaged output DCV with capacitor input load.
It still depends on load current and PT source impedance.

[Dave H]

The green trace is the load voltage.

[Helmholtz]

V (Peak) DC and V AC (RMS) are fixed (ignoring PT source impedance). V (Avg) DC will drop with increasing load current and/or lower reservoir capacitance - associated with higher ripple voltage.