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Old 20th April 2005, 03:55 AM   #1
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Default 6W Self Inverting 6BQ5 Amp

Has anyone built this simple amp from Triodeels website? I'm in the middle of prototyping one channel to drive my TQWTL with TB W4-1052SA drivers. What happens if a bypass capacitor is added to the cathode resistor? What about RH circuit being added to driver tube? Thanks from a newbie...
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Old 20th April 2005, 05:21 AM   #2
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Default Re: 6W Self Inverting 6BQ5 Amp

Quote:
Originally posted by Sparky OR
Has anyone built this simple amp from Triodeels website? I'm in the middle of prototyping one channel to drive my TQWTL with TB W4-1052SA drivers. What happens if a bypass capacitor is added to the cathode resistor? What about RH circuit being added to driver tube? Thanks from a newbie...
The circuit's bad enough as it is. Bypassing the output cathode resistor will make it into an SE amp with half the load resistance = low power output (probably around a quarter what it is with both tubes functioning properly) and high distortion (probably in the 15% range at full output).

I mean, look at the schematic. If you ground the cathodes to AC... no signal *at all* can get to the second 6BQ5...so wtf is it doing? Nothing!

To make this circuit actually work right, you need a CCS - they didn't have squalid state in those days, but nowadays a short negative supply (derived from the 6.3V heater winding) and quick IRF640-ish (or your favorite removed-from-scrap transistor or FET) current sink will make it run like the wind.

If you meant the 6C4 cathode...that can be bypassed without problem. Speaking of it, you can use a 6AV6 if you need more gain (a wise decision since this amp also has no NFB; in fact a pentode mode 6AU6 would be best).

Tim
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Old 20th April 2005, 08:07 AM   #3
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Have a look at the Simple EL84 at http://diyparadise.com/simpleel84.html

This is an update of this kind of amp with a lot more info. It really does need a CCS on the cathodes of the outputs to get the phase splitting to work properly. (althou i still haven't figured how he is getting his bias with the CCS tail at ground).

John Swenson also has a version of this on AA, but with a trafo out front instead of a driver tube.

http://db.AudioAsylum.com/cgi/search...&forum=tubediy

Me, i still think i'd use a CCSed long tailed pair up front.

dave
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Old 20th April 2005, 09:12 AM   #4
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Quote:
Originally posted by planet10
(althou i still haven't figured how he is getting his bias with the CCS tail at ground).
The bias is derived by the voltage drop across the CCS, since the CCS tail and the grids are at ground potential. If the CCS is programmed to draw, say, 80mA, the CCS will reach an equilibrium such that the voltage across the CCS will be such that Vgk allows 40mA through each valve for the given HT voltage (provided the valves are reasonably matched, if not, one will be slightly less and one will be slightly more). A negative supply is only needed if the CCS won't be happy operating with such little voltage across it (often the case with pentode sinks).
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Old 20th April 2005, 10:03 AM   #5
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Quote:
Originally posted by audiousername
The bias is derived by the voltage drop across the CCS, since the CCS tail and the grids are at ground potential. If the CCS is programmed to draw, say, 80mA, the CCS will reach an equilibrium such that the voltage across the CCS will be such that Vgk allows 40mA through each valve for the given HT voltage (provided the valves are reasonably matched, if not, one will be slightly less and one will be slightly more). A negative supply is only needed if the CCS won't be happy operating with such little voltage across it (often the case with pentode sinks).
I understand the current part, but how is the grid bias set at 10.5 V? (i'm still lost with SS)... we drop 40 mA x 1.2R = 48 mV across the sense resisitor means that the CCS needs to look like 10.45 V/80mA ~ 131 R... how is this magic done? Is it just where things fall?

dave
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Old 20th April 2005, 10:21 AM   #6
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The 1.2W resistors have nothing to do with the CCS, they are just there as a convenient way to measure the cathode current of each valve individually (by measuring the voltage drop across them). The actual current set resistor in this case is R1 in the circuit below:
Click the image to open in full size.
http://diyparadise.com/yhlmccs.html
(you probably already knew this, but just in case... )

Yes, at DC, the CCS may be equivalent to a 131W resistor, but to a signal, it appears to be a very much larger resistor.

You may be familiar with the fact that calculating Va/Ia for a valve is not a good approximation for the output impedance for a valve stage (particularly for pentodes). A similar thing happens here - it's just that Va/Ia is much smaller than the output Z of the circuit.

There's no magic here, SS or valve
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Old 20th April 2005, 05:26 PM   #7
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Quote:
Originally posted by audiousername
The 1.2W resistors have nothing to do with the CCS,....is much smaller than the output Z of the circuit.
I got all that, i'm just having trouble with how the grid gets 10.5 V negative to the cathode. In this case the CCS would have to convieniently look like a 130 ohm resisitor wouldn't it? What would you do if you needed, say 20V? set the CCS tail at -9.5 V? (i can see this working if the CCS is dropping V like a 130 ohm resisitor)

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Old 20th April 2005, 06:45 PM   #8
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Quote:
Originally posted by planet10


I got all that, i'm just having trouble with how the grid gets 10.5 V negative to the cathode. In this case the CCS would have to convieniently look like a 130 ohm resisitor wouldn't it? What would you do if you needed, say 20V? set the CCS tail at -9.5 V? (i can see this working if the CCS is dropping V like a 130 ohm resisitor)

dave
Ahem ! I beleive that the name "current source" is somewhat disturbing. Current regulator seems more appropriate, but ...
A current regulator is a two terminals unit in wich the current remain constant what could be the voltage accross it.
With two practical limitations, the first being that a minimum of voltage exists (no voltage, no current) and that it does not destroy the regulator.

A "true" source may then be seen as a voltage source in serie with a current regulator, the value of the voltage source does not matter (within some limits as explained above).

Now, put a current source between the cathode and the ground of a tube, and tie its grid at ground. The current in the tube will remain unchanged WHATEVER the voltage you apply to the plate because the voltage accross the current source will increase (or decrease) to maintain the current constant, and since the current source is tied between the grid and the cathode, it adjusts the bias automatically.

Yves.
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Old 20th April 2005, 08:48 PM   #9
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hi dave

well, look at it this way.

we want to do this. we want to set the ccs to 80ma with the idea that each tube conducts 40ma. that's why the 1.2ohm (1ohm is just fine) is useful for measuring voltage then calculating current.

after adjusting with a pot for 40ma each tube, we now hv 40ma flowing each tube! wonderful! you can now measure the plate voltage as well right?

if you look at el84 curves, with the current plate voltage and 40ma, you hv the grid-cathode voltage. ta da!

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Old 20th April 2005, 09:14 PM   #10
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Quote:
Originally posted by planet10


I understand the current part, but how is the grid bias set at 10.5 V? (i'm still lost with SS)... we drop 40 mA x 1.2R = 48 mV across the sense resisitor means that the CCS needs to look like 10.45 V/80mA ~ 131 R... how is this magic done? Is it just where things fall?

dave

Simple: current sinks (there's your word, Yves ) are non-ohmic devices. Just like zeners, which have a constant-voltage characteristic, and tubes which have a 3/2 power or constant current characteristic.

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