Modifying a vintage Ampeg B25 to a vintage Ampeg V3

[seven]

Hi all,

I recently learned that the early 1970s Ampeg V3 and the B25 were very similar amps. The story is that the V3 was modified/voiced for guitar (the B25 was a bass amp).

I looked at the schematics to see the differences and I'd like to learn more about them, specifically the output/phase inverter area as it seems that it has the most significant changes between the two.

* cathode and plate on the V3: 14k + 1k and 15k, respectively (cathode and plate on the B25: 47k)
* control grid resistors on the V3: 47k (on the B25: 100k)
* cathode bypass cap on 12AX7 channel 1 doesn't exist on the V3
* no NFB on the V3
* 3M9 missing from PI in V3





There are other differences in the power supply but I don't suspect that they contribute significantly in the difference in tone between the two.

I was wondering if I were to convert my B25 head to a V3 if it would then "lose the magic" when used with bass guitar? It is a nice round tone when playing bass through it.

Thoughts?

[g1]

I would say the only noticeable difference is the removal of NFB and type of speakers used. No idea who would put an 'ultra-lo' switch on a guitar amp at that time, so it just seems like a re-purposed bass amp with very little difference.
I would try it with guitar as is, then lift the NFB and try again. If you like it better for guitar with no NFB, just put in a lift switch, and engage the NFB when you want to run bass through it.
Assume you are planning on different speaker cabs for bass vs guitar?

[seven]

I would try it with guitar as is, then lift the NFB and try again. If you like it better for guitar with no NFB, just put in a lift switch, and engage the NFB when you want to run bass through it.

For sure. Was thinking I could use the now unused polarity switch on the front panel for the lift switch (amp now has a new replacement 3-prong power cord)?

Assume you are planning on different speaker cabs for bass vs guitar?

Oh yes, most definitely. Got a cab with two 16-ohm speakers wired up in parallel to use with this bad boy!

[Helmholtz]

I looked at the schematics to see the differences and I'd like to learn more about them, specifically the output/phase inverter area as it seems that it has the most significant changes between the two.

PI differences.
B25: 47k plate and cathode loads, plate current 3.1mA, fixed bias, 100k power tube grid stoppers.
V-3: 15k plate and cathode loads, plate current 7.5mA, auto bias, 47k power tube grid stoppers.

Higher value grid stoppers may be needed for HF stability with global NFB.

[Chuck H]

The grid stops on the B25 are actually 1k. I missed it too on the first look because of how the print is arranged.

[Chuck H]

Seven,

There's something to notice about Helmholts observations. That he stipulated fixed bias for the B25 and auto bias for the V3. Just adding the 3.9M resistor to the V3 (as shown in red?) may not provide suitable bias for the circuit. Notice that the B25 also has a 1.5M resistor to ground from the PI grid. The 3.9M and 1.5M resistors form a voltage divider from the HV rail providing correct bias voltage for the PI grid.

The V3 has a 1M reference resistor from the PI cathode to the grid. The 1k/14k circuit at that cathode is also a voltage divider and provides correct bias to the PI grid.

Adding the 3.9M resistor to the V3 as shown will inject voltage at the grid AND the cathode. I didn't attempt the math to determine what sort of bias adding the 3.9M resistor as shown would be.

[Helmholtz]

The grid stops on the B25 are actually 1k. I missed it too on the first look because of how the print is arranged.

Well spotted.
Unfortunately my eyes are weak.

[Helmholtz]

The V3 has a 1M reference resistor from the PI cathode to the grid. The 1k/14k circuit at that cathode is also a voltage divider and provides correct bias to the PI grid.

No voltage dividing involved.
Grid voltage is exactly the same as the voltage at the node between R25 and R31 as there's no current through R30.
Grid bias ( more exactly Vgk = grid voltage minus cathode voltage) is the negative of the voltage drop across the 1k bias resistor.
And yes, adding the 3.9M resistor would "spoil" center bias. Thought that was clear now.

[Chuck H]

No voltage dividing involved.
Grid voltage is exactly the same as the voltage at the node between R25 and R31 as there's no current through R30.
Grid bias ( more exactly Vgk = grid voltage minus cathode voltage) is the negative of the voltage drop across the 1k bias resistor.

Which is why I said that it was the 1k/14k resistors in the cathode circuit (R25 and R31) are a voltage divider. Which they are as far as I can tell. I didn't intent to imply the 1M (R30) was in a voltage divider circuit.

[Helmholtz]

Which is why I said that it was the 1k/14k resistors in the cathode circuit (R25 and R31) are a voltage divider. Which they are as far as I can tell. I didn't intent to imply the 1M (R30) was in a voltage divider circuit.

Seems I misunderstood.
In fact R25/R31 can be seen as a voltage divider.
But regarding tube bias, the value of R25 is quite insignificant as the bias voltage (Vgk) is determined by cathode current times bias resistor value (1k).

[Chuck H]

Seems I misunderstood.
In fact R25/R31 can be seen as a voltage divider.
But regarding tube bias, the value of R25 is quite insignificant as the bias voltage (Vgk) is determined by cathode current times bias resistor value (1k).

Agreed. I almost indicated that too. The voltage standing on the cathode is dictated by Rp and Rk and the "division" of Rk is adjusted for the correct Vgk relatioship.

[Helmholtz]

Bias generation via the 1k resistor works exactly the same as with a common cathode gain stage, where the grid and the lower end of the 1k bias resistor are at ground potential.
The drop across R25 just lifts both grid and cathode by the same voltage.

[Chuck H]

Bias generation via the 1k resistor works exactly the same as with a common cathode gain stage, where the grid and the lower end of the 1k bias resistor are at ground potential.
The drop across R25 just lifts both grid and cathode by the same voltage.

Yep. Steve Conner was the first to explain that to me many years ago. I was confused by the high voltage on a particular "elevated" circuit and I think his analogy was something like: "If a man is standing on a ten foot hill that doesn't mean his pants need a twelve foot 8 inch inseam."

[Helmholtz]

A tube only sees the voltage differences between its pins.
It can't know which circuit point or potential the designer chose for common ground.