Concertina phase inverter with LED bias - is this completely wrong? - diyAudio
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Old 15th September 2012, 04:19 PM   #1
nlinus is offline nlinus  Sweden
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Default Concertina phase inverter with LED bias - is this completely wrong?

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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Old 15th September 2012, 04:55 PM   #2
SY is offline SY  United States
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What's the transconductance of that tube? Are you taking the in-phase output from the top of the lower 3k resistor? Why do you need a bias adjustment?
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Old 15th September 2012, 05:20 PM   #3
nlinus is offline nlinus  Sweden
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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
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Old 15th September 2012, 05:45 PM   #4
SY is offline SY  United States
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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.
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Old 15th September 2012, 05:52 PM   #5
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Resistor is cheaper, and adds stability. But it does not glow, though...
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Old 15th September 2012, 05:53 PM   #6
nlinus is offline nlinus  Sweden
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Well, I just put the pot there to be able to do some experimenting. It would not do any harm, would it?

I actually don't need much gain from this stage since I have another stage (two c3g in single end) after it.
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Old 15th September 2012, 06:19 PM   #7
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Good thing you don't because a cathodyne is somewhat less than unity gain.
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Old 15th September 2012, 06:43 PM   #8
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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...
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Old 15th September 2012, 07:42 PM   #9
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Think of a cathodyne as a modified cathode follower. Exactly the same thing can be done with bipolars or FETs, although I guess it is not often used.
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Old 15th September 2012, 07:49 PM   #10
nlinus is offline nlinus  Sweden
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Of course...
The cathode resistor adds a lot of feedback. I'm learning slowly
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