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Help with 12AQ5 class A low gain and distortion

AbaddonD

Member
2015-08-20 1:49 pm
Hello!
OK, I made this class A 3x12AQ5 amplifier with a PCL805 triode as the pre-amp.
The tubes are all in parallel with 5.1k grid stoppers for each, 220Ohm screen stopper for each and a 1MOhm leakage resistor for all of them. The PSU voltage is 170V. The audio input is 2V max P-P.

The problems I currently have are:
Really low gain compared to the pentode of one PCL805.
Distortion at high volume.

Can anyone point out the problem?

Thanks!
 
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Sorento

Member
2008-03-12 9:59 pm
Assuming it is the simplest of single ended setups which got you 4 watts before clipping.
Assuming the single output tube had a standing dc current of say 50mA.
To benefit from triple paralleling tubes you have to run 3 times the current, 50mA per tube for a total of 150.
This means you either have to use a single cathode resistor which now has to be one third of what you had before or a separate cathode resistor per tube, a total of three with the same value as before.
Now that you run 3 times the current through your output transformer you need one which can take triple the dc current without saturating. If your transformer was just o.k. for 50mA it will possibly no longer be o.k. with three times more. It has to be much bigger now.
Furthermore this transformer needs to have a lower winding ratio. If you previously had 18:1 it needs to be something like 6:1 now.
 

AbaddonD

Member
2015-08-20 1:49 pm
way too many facts missing from the description.

I'm picking there is not one problem, but several. Hard to know what they are when there are no details.

At the risk of getting too involved, I'm going to turn this on its head - what do YOU think is (are) the problem(s)? And more importantly, WHY do you think this?

I am not sure. However, when I increase the cathode resistor from 0 or 96Ohm to 500, the volume starts creeping down and eventually goes silent.
Also, when I decrease the grid leakage resistor to 100K, the power decreases and starts clipping earlier.
 
nope.

The cathode resistor sets the standing DC current through the tube and so develops the bias voltage. As you increase the resistance, you decrease the DC current through the tube, and increase the bias. You need to find out about cathode biasing and get to grips with what it is and how it works.

You can do without it entirely by setting the bias voltage of the tube's grid directly. In a paralleled tube setup like the one you are playing with, its probably best to bias each tube individually from a separate -ve supply and put a very small resistance in the cathodes just so you can measure current - typically a resistor of 2 to 10 ohms.

You seem to be designing by throwing bits together and hoping they work. You would learn more and get better results by following a good design process.

If you are designing a tube amp from parts you already have, its best to start from the OPT - what are its characteristics and what is the best way to drive it. Everything follows backward through the circuit from there.
 

AbaddonD

Member
2015-08-20 1:49 pm
nope.

The cathode resistor sets the standing DC current through the tube and so develops the bias voltage. As you increase the resistance, you decrease the DC current through the tube, and increase the bias. You need to find out about cathode biasing and get to grips with what it is and how it works.

You can do without it entirely by setting the bias voltage of the tube's grid directly. In a paralleled tube setup like the one you are playing with, its probably best to bias each tube individually from a separate -ve supply and put a very small resistance in the cathodes just so you can measure current - typically a resistor of 2 to 10 ohms.

You seem to be designing by throwing bits together and hoping they work. You would learn more and get better results by following a good design process.

If you are designing a tube amp from parts you already have, its best to start from the OPT - what are its characteristics and what is the best way to drive it. Everything follows backward through the circuit from there.
Yeah, kinda.
I have all sorts of high power and 1/4W resistors. I also don't have a shortage of caps.
The thing which I am limited to is this transformer which is a handmade power transformer with a ratio of 18:1 (220 to 12 at 3A or so). It works nicely BTW. I have tried other power transformers but they don't perform nearly as good.
I am also limited to a few tubes. I currently have these: 2x PCL805, 3x 12AQ5 and a 2C39A (which kinda has been proven to be trash for audio). I might get a 6L6GC in about 20 days.

I have looked at many DIY tube amplifiers and a few commercial amplifier schematics.
I am fairly new to this. I got 2x PCL805s about a month ago. So, I'm learning about tubes and stuff.
I am somewhat knowledgeable in electronics though.

I will try to add a few Ohm resistors to each tube, but I remember that even one 12AQ5 sounds the same.
 
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AbaddonD

Member
2015-08-20 1:49 pm
Do that. Include specs on the components.

Here it is.
20191221_120743.jpg
20191221_122434.jpg
The powersupply and the amplifier.
Oh jeez. The amplifier schematic is upside down.
But I sent the rotated one. I don't get it.
Sorry about that. Lemme see if I can fix it.
It can't be fixed :|
It is in the correct orientation on my phone, but not when I upload it.
 
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You need a coupling capacitor between the 'preamp' and 'finals' stage. But, once that's done, then the output tube G1 grids are 'floating'. They would be tied to ground via a 220 kΩ resistor. Given the 100 kΩ, and your likely desire to have frequency response down to 20 Hz, using
C (µF) = 1,000,000 / (2πFR)
F = 20 Hz
R = 100,000 Ω
C ≈ 0.1 µF at 350 V​
This would substantially open the dynamic range, also, of the amplifier.

The 96 Ω, 10 W cathode-bias resistor needs a bypass capacitor. FOR SURE, again to secure some reasonable output power.
C = 1,000,000 / (2πFR)
F = 20 Hz
R = 96 Ω
C = 86 µF (call it 100 µF or 150 µF to be sure), at 35 V​
Now the output stage will produce some power, and it'll be cleaner.

Good luck.
-= GoatGuy ✓ =-

PS: one of the reasons I write
Z = 1,000,000 / (2πFC) … instead of
Z = 159,155/FC​
is because the (2πF) part is “officially” in charge of turning hertz Hz into radians-per-second, the mathematically proper statement of oscillatory rate. One could (as was popular in the Great Tube Days of the 1950s and 1960s) just remember to estimate
1,000,000 ÷ 2π ≈ 160,000​
and be done with it; the estimate is within 1% of accurate, which is by far, good enough. -= GoatGuy ✓ =-