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Old 7th December 2017, 09:39 AM   #21
Kay Pirinha is offline Kay Pirinha  Germany
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Generally spoken, in order to achieve maximum signal output a Concertina should be designed in a way that the tube sees about half of B+ and both resistors see a fourth each.


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Old 8th December 2017, 02:51 PM   #22
Fenris is offline Fenris  United States
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Broskie, who originally drew the circuit in the diagram, puts the same resistor values in the concertina as in the anode of the gain stage. This is done so that the currents through each stage are equal and opposite, resulting in a constant current draw. This should reduce the distortion. A downside is that there is less current available to drive the final stage tubes. Not usually a problem with EL 84 type tubes, but for some others it could be an issue.

In this circuit, the operating point of the concertina is set by the voltage divider made up of the two 1Meg resistors. Assuming that the anode of the gain stage is at B+/2, then it should be at B+/4, give or take a few volts. Since tubes are never at exactly their datasheet values, there will be some variance. Also, since there will be only about 12v needed to drive the finals, the setpoint of the concertina technically only needs to be 15 volts less than B+/2, or about 110v for 250v B+. I wouldn't do this personally, since tubes change with age and the setpoint could drift. I would aim for between 70 and 100 volts DC at the grid of the concertina tube. It should be in that range using the values in the circuit, but can be adjusted by varying the ratio of the resistor divider between stages.
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Old 8th December 2017, 03:51 PM   #23
artosalo is offline artosalo  Finland
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Separate biasing of voltage amplifier and the cathodyne is mandatory when higher drive voltages are needed.
I don't see any reason to avoid that topology in low level constructions as well.
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Old 9th December 2017, 08:44 AM   #24
Tyimo is offline Tyimo  Hungary
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Quote:
A downside is that there is less current available to drive the final stage tubes.
That was the reason for my question. Thanks for your answer!
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Old 9th December 2017, 12:39 PM   #25
artosalo is offline artosalo  Finland
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Normal audio output tubes do not require current to be driven. The main current goes to grid resistors. Assuming we have EL84, the grid resistor is 1M and reguired drive voltage, say 12 Vpeak. Anybody can calculate the "current" needed.

There are commercial amplifiers where a 12AX7 LTP drives 4 x EL34 output stage with no lack of current.
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Old 9th December 2017, 02:10 PM   #26
kodabmx is offline kodabmx  Canada
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If you're driving the output triode connected, then you must factor in the input capacitance as well. That 12AX7 couldn't drive 4 EL34 in triode for sh*t.
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Old 9th December 2017, 03:06 PM   #27
artosalo is offline artosalo  Finland
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The purpose of my comment was to draw attention that vacuum tubes are usually "driven" by voltage, not current (A2 and AB2 classes make the exception). Many DIYers seem not to know this at all.

I agree that 12AX7 is not good tube to drive any triode connected output tubes, but that is an other discussion.
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Old 9th December 2017, 04:06 PM   #28
stocktrader200 is offline stocktrader200  Canada
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The 12AX7 driving a 500v mosfet concertina at 3ma to 5ma bias would drive those EL34's perfectly
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Old 10th December 2017, 04:39 PM   #29
kodabmx is offline kodabmx  Canada
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So would a 6SN7...
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