tube, SS recto switch sanity check

[Merlinb]

The purpose of the Rs would be to drop the voltage prior to rectification to match the rectified voltage of the tube, so both options SS or Tube would result in the same HV rectified VDC.

That's what I thought you meant. As I said, this would be a silly thing to do because the amp would sound the same on both settings, but hey, guitarists hear what they want to hear.

As I understand it the Vdrop across each resistor would be:
V = I * X where I is current draw from the amp and X is impedance

I think you made a typo there somewhere. The voltage across a resistor is simply V = IR!

If i switch out the tube, why would the impedance of the tube be used to calculate the voltage drop across the AC resistor for the SS option? I thought this would be the impedance of the transformer.

The transformer impedance is in series with the diodes in both situations! It contributes to voltage drop all the time! The only thing you're switching is the type of diodes, not the transformer.
Therefore the only extra bit of impedance standing between the transformer and the reservoir capacitor is the resistance of the tube diodes. Therefore, to get the same total voltage loss with the SS diodes you need to add resistors that are equal to the internal resistance of the valve diodes. You can get that from the valve datasheet by estimating the slope of the diode curve.

I have your book but its not in there either. Hard to find info on AC voltage drop for an amp.

It is covered in my second book (power supplies, logically enough!) towards the end of chapter 2. Example valve diode resistances are on page 46.

[surfsup]

I think you made a typo there somewhere. The voltage across a resistor is simply V = IR!

That's what I thought too, until my math resulted in the wrong DC voltage, and then researching why I read this at Wikipedia:

The voltage drop in an AC circuit is the product of the current and the impedance (Z) of the circuit. Electrical impedance, like resistance, is expressed in ohms. Electrical impedance is the vector sum of electrical resistance, capacitive reactance, and inductive reactance. The voltage drop occurring in an alternating current circuit is the product of the current and impedance of the circuit. It is expressed by the formula E = IZ, analogous to Ohm's law for direct current circuits.

I meant to write IZ not IX, sorry. I was getting way low voltages with a 300-0-300 PT with two 1k5W Rs on the AC side. With two el84s and three 12ax7s I figured each side would pull 0.06A and I would get ~330VDC on the other end:

300-(1000*0,06)*1.4=300-60*1.4=240*1.4=336VDC

I was getting 274VDC instead which is way low. I wound up using 330R 5W resistors. I'm getting 340 which is close enough now but something in my estimates is wrong if V=IR on a resistor on AC.

It is covered in my second book

I missed this because it was covering a tube and not a physical resistor I guess. I'll check that out.

Hmm, I think I know what the problem is, I thought in a PP amp each EL84 would only be "on" half the time drawing current. The math works if I figure 50mA for each EL84 and they are both "on" drawing current all the time?

[Merlinb]

Hmm, I think I know what the problem is, I thought in a PP amp each EL84 would only be "on" half the time drawing current. The math works if I figure 50mA for each EL84 and they are both "on" drawing current all the time?

This has nothing to do with what the rest of the amp is doing! It doesn't matter what you attach to your power supply; if you want the DC voltages to be the same with either rectifier then the resistors must be equal to the internal resistance of the tube diodes, it really is that simple!

And still I insist that it is also pointless since all you would achieve is to make a solid-state simulation of the tube diode, so the audio amp will react exactly the same with either rectifier. The whole point of switching to SS is to get higher voltages and a stiffer supply!

FYI, the loss in DC voltage is equal to the *peak* ripple current times the source impedance (not the load current). From fig 2.22 in my second book you will see that this is around 3.5 times the DC load current. Incidentally, are you using a 5Y3?

[surfsup]

I apologize for being vague. I've slept < 4hrs in two days.

Originally discovering this thread, as a math exercise, I wanted to verify how to calc AC voltage drop matching the numbers of the tube rectified value VDC.

I have another amp with no tube recto, but it does have resistors on the AC side, hence the math I posted because I couldn't calculate the proper R value and had to waste 3 pairs of Rs testing, and I'm still a bit off. I was just clarifying the AC drop vs DC drop being impedance vs resistance. Fundamentally I just wanted to know how to calculate this value (AC V drop), tube recto included, or not.

Yes I'm using a 5Y3 in the amp with the tube recto.