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Concertina phase inverter with LED bias - is this completely wrong?

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Must be a long time since I last logged in here. My interests spread over so many areas which makes DIY hifi just a fraction of everything else.
My last project, which is not yet finished, when it comes to hifi is to be seen here: Link

I have now decided to build my first tube amp. I will not present all the plans here, because I think I'll put up a project thread later on. Roughly it consists of a split load (concertina) phase splitter (Siemens c3g), a driver stage with two more c3g tubes, and a power stage with two triode strapped Electro-Harmonix KT88 in push-pull configuration.

From what I can see, the combination of tubes should not be an issue. Maybe I have overlooked something?

The KT88:s and the driver stage will run with fixed bias from a separate supply.
I was thinking about biasing the phase inverter. Have a look at the schematic. I have not seen this solution with LED biasing in a split load inverter before, but wouldn't this work?

Edit: Forgot to draw the in-phase output at the top of the bottom 3k resistor.
 

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Yes, I'm taking the in phase from the top of the bottom resistor. Forgot to draw that.

The grid should be a couple of volts below the cathode with the 3k resistor between cathode and ground that would be somewhere around 50 volts depending on the current flowing through it!?

The transconductance is 14000 µOhms
 
Well, let's look at the curves. You have a 300V rail. If we choose a 15mA current, there's a total of 90V of drop across the two 3k resistors, 45V across each, leaving 210V across the tube- you need to think about how much swing your output stage will need and resize those resistors if necessary. But continuing along this path, with Vak = 210V, it looks like 3.4V or so Vgk will get you the target 15mA current. That's two red LED in series.

Let's say you need more than 20VRMS of swing. We increase the resistors to 10k each and run 10mA through the tube. So there's roughly 100V left for Vak. You'll need a little more than 1V Vgk to get 10mA, so a single IR LED should work well.

In either case, the grid leak is returned to the cathode of the LED, where it meets the 3k or 10k resistor. No bias pot needed.
 
I don't get that about cathodynes having less than unity gain.

3k might be a bit low. If I increase the anode and cathode resistors to 3k4 each, biasing the amp at ~15mA a voltage swing of 2 volts in the grid would result in a current swing of ~9mA.
9mA into 3k4 ohms gives me a Vpp of 30.6 volts. That would give me a gain of 30.6/2=~15.

I am obviously missing something important here. Please don't forget I'm a newbie to tubes...
 
Rule of thumb...I aim for all voltages in a cathode follower to be a third each of the B+. My schematic I commonly use.

.....in concertina I don't bother with split cathode resistor.
solong it's DC bias strapped to preceeding stage which is in control. I've always assumed since the stage operates with fantastic neg feedback, the concertina distortion is already extremely low or an be ignored if Correctly operated..

The B+ is approx 360V which is off the pic.
richy
 

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Rule of thumb...I aim for all voltages in a cathode follower to be a third each of the B+. My schematic I commonly use.

Well, that's not exactly the case with my current values. A B+ of 300V and anode/cathode resistors on 3.4k each.
At 15mA, that would result in a Vak (voltage drop anode to cathode) of 198V.

I don't know if I misinterpreted the values in the datasheet.
Ua is set to 200V for triode mode.
 
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