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    Building, troubleshooting and testing of these amplifiers should only be
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    the safety precautions around high voltages.

how to calculate lower operating point?

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As I recall, this is the "choked monkey" design. Utterly horrible as you waste half the power in that 3k 25W resistor below the 300B. Just remove that and cap couple the 6AQ8 to the 300B.

Let's see, you need more than that. First, get a new power transformer that does 400VDC or so (something around 350-0-350VAC). If it's more convienient, you can use a 5U4 instead of 6AU4's. Change the first filter caps from 200uF 450V to 40uF 450 or 500V. Replace the 47 ohm 5W resistor between the 6AU4 cathodes and 1st filter cap with a short. Same for the 3k 25W, remove it. Move this end of the 65H choke to B+ instead. You'll need slightly more resistance in the 6AQ8 cathode, probably 470 ohms. Insert a .22uF capacitor in series with the 200 ohm grid stopper on the 300B, and at the junction between them, add a 100k grid leak resistor to ground.

Or you can look for another 300B SET amp circuit, there's enough out there that you WILL find exactly what I described, only in schematic form.

Tim
 
hi tim

thanks for the detailed post.

may i ask what is the function for the 0.22uF cap before the grid stopper?

and is the 100k grid leak to be placed after the 65H choke?
(ie 6AQ8 plate - 65H choke - 100k grid leak - 0.22uF - 200R - 300B grid)

with the above 2 additions, will the design still be a "choked monkey"?

actually i am interested in building a "choked monkey" 300B set.
 
The .22 seperates the high (+400V) 6AQ8 plate voltage (which is pretty high for something like that... some resistors and capacitors should be used to bring that down so it doesn't throw sparks...) from the 0V potential of the 300B grid. In the circuit shown, the 300B floats at +200V or so by way of the big resistors below it. Its operating voltage then comes from the extra high supply voltage. Capacitors block DC but they're not perfect so you need a grid leak to ensure the 300B grid stays at 0V. Its cathode will naturally float itself up to around 60-80V due to current flow through the resistor (um 800 ohms or so is it?).

Tim
 
garbage said:
any idea what is a straight-forward method to determine the resistor values that i need to change in order to make it work with a B+ of say 480V?

It wont work very well at all with only 480V. The extra B+ is the price you pay for direct coupled designs.

If you want to do a DRD / Monkey style amp with only 480V, use a 2A3 or 45 as the output valve - not a 300B.
 
6EM7/300B DRD?

Hi All

Thanks for the response so far.

How does the circuit below look alright with the indicated resistor values?

I want to use a B+ of 480V as this would make it easier to source for caps in the power supply section.

This is my first time trying to calculate the values, can someone help verify if my calculations are more or less ok?

Method of calculation taken from Jeremy Epstein's site.

Suppose I have a 480V B+. And the following information:

300b operated as follows
Vplate : 300v, 65mA
Vgrid : 120v (bias -60v)
Vcathode : 180v

Assuming the choke is 65H and 250R, this should give me resistors of 2750R and 4000R at the cathode of 300B.

6EM7 operated as follows
Vplate : 100v, 20mA
Vgrid : 3v (bias -17v)
Vcathode : 20v

This should give me 1000R for bias at the cathode of 6EM7.
 

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Realised that i am using the wrong unit for 6EM7.
It should be unit 1 instead of unit 2.

Thus with the following conditions:
486V B+, assuming 102R output transformer resistance

300B operated as follows
Vplate : 300v, 65mA
Vgrid : 120v (bias -60v)
Vcathode : 180v

6EM7 operated as follows
Vplate : 120v, 1.8mA, bias -1v

Assuming the choke is 65H and 250R, this should give me resistors of 2848R and 55R at the cathode of 300B.

And bias of 1V at 6EM7 with 1.8mA would require resistor of 555R.

nb. I have not calculated the loadlines for driver and output tube yet.
 

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i was encouraged to use both units of the double triode in the 6EM7.

also, i had suggestions that the B+ would not be sufficient enough for a DRD design.

and i misinterpreted Jeremy Epstein's Free Lunch calculations. the Free Lunch resistor should be from the plate of the driver to the cathode of the output tube, not in series with the cathode resistor as in previous schematics that i posted.

so here goes, based on the new requirements and my limited understanding of DRD and the 6EM7 tube.

B+ of 800v.
300B operated at 60mA, 450v and -100v bias
voltage at cathode = 350v
voltage at grid = 250v

giving cathode resistor of 7085R and free lunch resistor of 9184R at the cathode of the 300b.

6EM7 low mu unit 2 operated at 10mA, 150v and -30v bias
voltage at cathode = 100v
voltage at grid = 70v

giving cathode resistor of 1000R and free lunch resistor of 50000R at the cathode of 6EM7 unit 2.

6EM7 high mu unit 1 operated at 0.6mA, 69v and -1v bias
voltage at cathode = 1v

giving cathode resistor of 1666R at cathode of 6EM7 unit 1.

the 1666R resistor here creates a 1v at the cathode.
question is, how do i apply a 1v bias at the grid?

does the schematic look correct so far?
 

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  • 6em7_300b v0_3.jpg
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