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Can anyone explain me this Loftin White amp?

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Hello everybody!

I've been struck by this circuit in which there are some things I can't understand. I must premit that I'm not absolutely going to build this amp nor prototype it, I'd just like to understand how it works.

Here it is:
http://tinpan.fortunecity.com/saints/668/primer/2a3-se.gif

The question is: What is the purpose of the 33k resistor between B+ and the cathode of the 2A3?
After some hours of math yesterday evening I concluded that IT DOES NOTHING, and that's not even a faliure-protection in case of the 6SL7 missing from the socket.

Can someone enlighten me? :xeye: I've also seen circuits very similar to this where that resistor is not there.

Thanks in advance! ;)
 
Hmm....

I bet that resistor is for "bootstrapping", it ensures that there is negative grid bias during warm-up?

Do DHT's take longer to warm up?

Giaime... I have seen your picture... you are young... you live in Venice... you should not be home, studying strange SE schematics on Friday night! Make the girls pay...



:smash:
:cool:
 
I'm trying to understand, but:

1) Hum reduction shouldn't be possibile since 33k makes a too strong voltage divider, and obviously the cathode is bypassed to ground, so no injection of PSU hum at the cathode is possible.

2) DC feedback: when the amp is operating properly, Vak for the upper triode of the totem is fixed and independent of tube aging and other things (or not?), so Vg for the 2A3 is fixed. If we make the assumption that B+ doesn't vary, a decrease in cathode current of the 2A3 (tube aging) lead to the usual decrease of Vgk for the 2A3, like EVERY cathode biased triode of this planet and others. (the 390k grid resistor is TOO large to provide some variation to the Vg, that is mainly imposed by the totem).

3) protection: I've also thought about protection for the 2A3 when, accidentaly, the 6SL7 isn't in its socket. We will have at power up 0V at the 2A3 grid, but since there's the 33k in there we'll have something like 400V/85390ohm = 4.6mA (where the ohm value is the equivalent of the 2 parallel 100k in the PSU, the 33k resistor and the cathode resistors of the 2A3). Maybe more because at startup B+ could be much higher (if the PSU has a resistance of 200ohm, B+ could be somewhere between 420 and 430V).

So let's assume 5mA flowing in the 33k resistor: V = IR = 415, so cathode sits very very close to ground, like I guessed, just like the 33k resistor ISN'T there.

The 2A3 has now 420Vak with 0Vgk, it will start heavy conduction that will obviously counteracted by the cathode resistors, my guess is that even without the 6SL7 in the socket the voltages will be the same, since the grid is connected to the series of 820ohm resistors so it will have at 60mA something like 98V, just like if the 6SL7 was there.

I'm starting to think that the resistor is there to make a very light DC feedback to fix the voltages of the totem with those of the power stage, and the opposite (since the actual current in the 2A3 *could* influence totem voltages and so current in it).

But I'm not sure... :smash:

Edit: another thing crossed my mind. Maybe the resistor is just a leftover from a version of this amp that DIDN'T have a totem input stage (in which voltages are rougtly independent of the variations in tube parameters), but a standard grounded cathode stage (R loaded), where actual plate voltage is heavily influenced by the parameters of the input tube...
 
poobah said:
Hmm....

I bet that resistor is for "bootstrapping", it ensures that there is negative grid bias during warm-up?

Hello poobah, as I have calculated

So let's assume 5mA flowing in the 33k resistor: V = IR = 415, so cathode sits very very close to ground, like I guessed, just like the 33k resistor ISN'T there.

there WON'T be any positive voltage at the cathode during warm up, because the 33k resistor will drop all the voltage.

This however is valid if the second cap bank isn't charged yet. After it has charged, we could assume that they sit at 400 or more volts, so the voltage across the 33k resistor is 165V, effectively raising the cathode voltage of the 2A3 up to 250V.

But caps and the 50k resistor (100k parallel in the PSU) have a definite time constant: R*C is 10s here, so after 10s from the startup (the 2A3 should already conduce some mA) the voltage will be at 63% of the maximum value, or 250V. The current on the 33k resistor at t = 10s is 7mA, so cathode voltage will be only 19V.

A bit low. And, as you all know, single ended cathode biased tube amps have always tolerated high Vak voltages with zero bias at startup.

Maybe 400V is too much for the 2A3 and can cause arcing at startup, but if is this the case, I would have considered putting a 100k from the MAIN B+ to the cathode of the 2A3, NOT from the charging second cap bank (that won't have any voltage for LONG time).

Does it makes any sense?
 
Yes interesting read.

So it has been pointed out that the resistor that implements some sort of "fixed bias" at the output tube, thus reducing the need for a larger cathode resistor:

that is obviously wrong, because:

1) the *added* positive voltage from the resistor is very, very small, only 3V.

2) is it safer actually to use a LARGER cathode resistor:
a) so you can split them and lower dissipation, so temperature drift
b) higher resistor means BETTER control over tube current (by means of Vgk), Ohm's law.

As already mentioned, if that resistor is there to protect the 2A3 during startup, it should *really* be removed, and replaced with a 100k right from MAIN B+, not from the second filter cap bank (that have HIGH time constant).

Another thing that crossed my mind is that those resistors in the cathode of the 2A3 are pretty much standard values, and we all know that in DC coupled designs it's hard to make the PERFECT Vgk that we want, because with large valued resistors the PERFECT value is more difficult to obtain.
Maybe those 3V (Vgk would have been 42V without that mystery resistor) are ABSOLUTELY needed to obtain the operating point that was in the mind of the designer? It could be.

But I think that is a problem for tubes that have a limited bias voltage, 3V for a 12AX7 is pretty much full modulation, for a 2A3 is 6.6% difference.

Mah... :whazzat:
 
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