Hello,
İ found this schematic of an el84 6n2 tube push pull amplifier
İ need some opinions on if its worth building or should i find something else
İs it decent or garbage?
And any upgrades/replacements that can be done on the board
Thanks,
İ found this schematic of an el84 6n2 tube push pull amplifier
İ need some opinions on if its worth building or should i find something else
İs it decent or garbage?
And any upgrades/replacements that can be done on the board
Thanks,
So now we know you are smart 😎
Maybe you can also find some time to answer the OP's questions?
Jan
Maybe you can also find some time to answer the OP's questions?
Jan
Is this already a PCB? Or will you hand-wire it? If so, I'd add 100ohm Screen Grid Resistors at Pin 9 of each power tube. 1K resistors at Pin 2, Grid Stoppers help prevent HF oscillation.
Use high quality film caps for C2 & C3, signal goes through them.
You could prob make C4 bigger, 22uF - 47uF.
If you get a high-pitched whine when you first turn it on, the FB wire is connected to the wrong side of the Secondary. Sometimes the easiest fix is to swap P1 and P2 on the Primary side.
Most audio pentodes have the Suppressor Grid, g3, internally wired to the Cathode, to simplify wiring in the amp. 6P15P is a TV tube and g3 is wired to Pins 1&6. EL84/6P14P have g3 wired to Pin 3. You can make this amp to use both types of tube if you add a wire from Pin 3 to Pin 6.
Looks like you'll want a b+ or 300-350V, 120mA per channel, so 250mA for a stereo amp. All caps must be rated for the full unloaded voltage of the transformer, I'd go 450v for safety.
Be careful. Safety first.
Oh, and make sure your power supply has a bleeder resistor.
Use high quality film caps for C2 & C3, signal goes through them.
You could prob make C4 bigger, 22uF - 47uF.
If you get a high-pitched whine when you first turn it on, the FB wire is connected to the wrong side of the Secondary. Sometimes the easiest fix is to swap P1 and P2 on the Primary side.
Most audio pentodes have the Suppressor Grid, g3, internally wired to the Cathode, to simplify wiring in the amp. 6P15P is a TV tube and g3 is wired to Pins 1&6. EL84/6P14P have g3 wired to Pin 3. You can make this amp to use both types of tube if you add a wire from Pin 3 to Pin 6.
Looks like you'll want a b+ or 300-350V, 120mA per channel, so 250mA for a stereo amp. All caps must be rated for the full unloaded voltage of the transformer, I'd go 450v for safety.
Be careful. Safety first.
Oh, and make sure your power supply has a bleeder resistor.
6A3sUMMER two 25 parallel? That's 12k5. Not a lot. I always use 100K. Long time ago I had the wonderful idea "a smaller bleeder will drain the caps quicker" but it drained the power supply so much there wasn't any current left for the amplifier. Can't remember which value I used, but it must have been around 10K.
Miniwatt,
No, Not two 25k in Parallel.
50k Ohms (2 x 25k in Series).
Example:
350V / 50k Ohms = 7 mA load on the B+
Suppose the input stage is 6mA, and the output stage is 50mA = 56mA.
With the bleeder, 56mA + 7mA = 63mA total B+ load
I hope your amplifier B+ can take the extra 7mA.
Turn off your tube amplifier, the B+ needs to collapse, and remove the power cord (you can get shocked from the mains wiring inside the amp).
I sometimes use a total of 600uF (200uF after the choke, 200uF after a dropping resistor, and 200uF after another dropping resistor).
I use that kind of brute force filtering, it keeps stage to stage from coupling at bass frequencies;
And . . . I want to get less than 100uV hum on a single ended amplifier that does not have negative feedback.
For B+ to collapse after the power is turned off, we need to consider that 3 x 200uF = 600uF of capacitance has to be discharged.
Calculation: Time constant of 50k Ohms x 600uF = 30 seconds.
In one Time Constant, the capacitors will discharge 63% of the voltage, so there is still 37% of B+ voltage there after 30 seconds.
350V B+ x 0.37 = 129.5V B+ remains after 30 seconds.
If you use your 100k bleeder, 129.5V B+ remains after 1 minute.
100k Bleeder, you saved 3.5mA load current, versus 50k . . .
You are in a hurry, so you remove the cover plate off of the amplifier in 30 seconds.
You slip, and OUCH, about 200V is there to "greet your finger".
The correct resistance for a bleeder depends on 3 things:
The total capacitance in your B+ filters
Your B+ Voltage
How quickly you will remove the cover plate.
Key formulas:
After 1 Time Constant, you have 37% of B+ remaining
After 2 Time Constants, you have 13.7 % of B+ remaining.
Do not count on your output tube current to discharge the B+.
That is because the output tube just Failed Open, and now you are in a hurry to find the cause of your problem.
Screw out, screw out, screw out, and screw out. Bottom plate removed, and OUCH!
Good Safety does Not rely on just one thing.
Safety requires knowledge of the complete "System": Your amplifier details, your speedy electric screwdriver, and a hasty forgetful action.
Prevent the "Surviving Spouse Syndrome".
Vacuum Tube amplifier repair persons are educated (either at the beginning of their carrier, or after having been shocked!).
They will love you when there is a proper bleeder resistor in the amplifier that they are repairing.
No, Not two 25k in Parallel.
50k Ohms (2 x 25k in Series).
Example:
350V / 50k Ohms = 7 mA load on the B+
Suppose the input stage is 6mA, and the output stage is 50mA = 56mA.
With the bleeder, 56mA + 7mA = 63mA total B+ load
I hope your amplifier B+ can take the extra 7mA.
Turn off your tube amplifier, the B+ needs to collapse, and remove the power cord (you can get shocked from the mains wiring inside the amp).
I sometimes use a total of 600uF (200uF after the choke, 200uF after a dropping resistor, and 200uF after another dropping resistor).
I use that kind of brute force filtering, it keeps stage to stage from coupling at bass frequencies;
And . . . I want to get less than 100uV hum on a single ended amplifier that does not have negative feedback.
For B+ to collapse after the power is turned off, we need to consider that 3 x 200uF = 600uF of capacitance has to be discharged.
Calculation: Time constant of 50k Ohms x 600uF = 30 seconds.
In one Time Constant, the capacitors will discharge 63% of the voltage, so there is still 37% of B+ voltage there after 30 seconds.
350V B+ x 0.37 = 129.5V B+ remains after 30 seconds.
If you use your 100k bleeder, 129.5V B+ remains after 1 minute.
100k Bleeder, you saved 3.5mA load current, versus 50k . . .
You are in a hurry, so you remove the cover plate off of the amplifier in 30 seconds.
You slip, and OUCH, about 200V is there to "greet your finger".
The correct resistance for a bleeder depends on 3 things:
The total capacitance in your B+ filters
Your B+ Voltage
How quickly you will remove the cover plate.
Key formulas:
After 1 Time Constant, you have 37% of B+ remaining
After 2 Time Constants, you have 13.7 % of B+ remaining.
Do not count on your output tube current to discharge the B+.
That is because the output tube just Failed Open, and now you are in a hurry to find the cause of your problem.
Screw out, screw out, screw out, and screw out. Bottom plate removed, and OUCH!
Good Safety does Not rely on just one thing.
Safety requires knowledge of the complete "System": Your amplifier details, your speedy electric screwdriver, and a hasty forgetful action.
Prevent the "Surviving Spouse Syndrome".
Vacuum Tube amplifier repair persons are educated (either at the beginning of their carrier, or after having been shocked!).
They will love you when there is a proper bleeder resistor in the amplifier that they are repairing.
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Osvaldo de Banfield,
Good!
1. I like your idea of using a common self bias resistor (un-bypased), but only if the output tubes are Extremely Well Matched @ the quiescent operating plate voltage and quiescent operating plate current.
The output transformer really loves exactly equal plate currents.
As an example:
low-power-balanced-mono-block-tube-amplifier.404771
You can see the un-bypassed common self bias resistor on the input stage, where exact quiescen current matching is a little less critical than the output stage; but the gain match of the input triodes is very important.
My output stage, has 10 Ohms between the tube cathodes (10 Ohms allows me to measure and verify the cathode current balance).
2, Back to the original posters schematic:
The amplifier schematic in Post # 1, is capable of Class AB1 operation.
With a large enough input signal, it will come out of Class A.
It does not have the most perfect set of resistors for the input stage phase splitter, it has un-equal amplitude of the 2 phases.
The global negative feedback reduces the distortion caused by the un-equal amplitudes.
But, the un-balanced amplitudes can be easily fixed if desired.
Good!
1. I like your idea of using a common self bias resistor (un-bypased), but only if the output tubes are Extremely Well Matched @ the quiescent operating plate voltage and quiescent operating plate current.
The output transformer really loves exactly equal plate currents.
As an example:
low-power-balanced-mono-block-tube-amplifier.404771
You can see the un-bypassed common self bias resistor on the input stage, where exact quiescen current matching is a little less critical than the output stage; but the gain match of the input triodes is very important.
My output stage, has 10 Ohms between the tube cathodes (10 Ohms allows me to measure and verify the cathode current balance).
2, Back to the original posters schematic:
The amplifier schematic in Post # 1, is capable of Class AB1 operation.
With a large enough input signal, it will come out of Class A.
It does not have the most perfect set of resistors for the input stage phase splitter, it has un-equal amplitude of the 2 phases.
The global negative feedback reduces the distortion caused by the un-equal amplitudes.
But, the un-balanced amplitudes can be easily fixed if desired.
Last edited:
Ok,I'll bite.
How do ya get the just exactly perfect set of resistors to balance amplitudes for a circuit such as this?
How do ya get the just exactly perfect set of resistors to balance amplitudes for a circuit such as this?
i am not familiar with the 6N2, but here is the principal to balance the phase splitter.
The plate load is 100k in parallel with 470k = 82.4k.
Use the 6N2 plate curve graph, and an 82.4k load line to find the gain of the 6N2 in-circuit.
We want the amplitudes of the two plates to be out of phase, and have equal amplitude.
As it is now, the closer the signal amplitudes are from the two 6SN1 plates, the more the junction of R8 and R9 will tend to cancel out those signals, so the grid will get very little or no signal.
Suppose the 6N2 with the 82.4k AC plate load has a gain of 40.
If the plate pin 1 puts out 20V, then we need 20/40 = 0.5V at pin 7, so the gain of 40 will put out 20V at pin 6.
If the 6N2 gain is 40, The divider of R8, R9, and R7 needs to be adjusted to give 1/40 more voltage to the grid, pin 7.
That will make the amplitudes much closer to equal.
The divider output to pin 7 depends on the gain of the triodes in-circuit. Which means you can not just do a calculation from the plate curves, because the triode gains may not be the same as the graph of plate curves, and may not be the same for the two triodes in the same glass envelop.
That makes it more difficult to match the gains.
I use a cathode coupled LTP phase splitter, because as long as you match the plate load resistors RL, and the next stages RG resistors, the gain match is automatic.
The plate load is 100k in parallel with 470k = 82.4k.
Use the 6N2 plate curve graph, and an 82.4k load line to find the gain of the 6N2 in-circuit.
We want the amplitudes of the two plates to be out of phase, and have equal amplitude.
As it is now, the closer the signal amplitudes are from the two 6SN1 plates, the more the junction of R8 and R9 will tend to cancel out those signals, so the grid will get very little or no signal.
Suppose the 6N2 with the 82.4k AC plate load has a gain of 40.
If the plate pin 1 puts out 20V, then we need 20/40 = 0.5V at pin 7, so the gain of 40 will put out 20V at pin 6.
If the 6N2 gain is 40, The divider of R8, R9, and R7 needs to be adjusted to give 1/40 more voltage to the grid, pin 7.
That will make the amplitudes much closer to equal.
The divider output to pin 7 depends on the gain of the triodes in-circuit. Which means you can not just do a calculation from the plate curves, because the triode gains may not be the same as the graph of plate curves, and may not be the same for the two triodes in the same glass envelop.
That makes it more difficult to match the gains.
I use a cathode coupled LTP phase splitter, because as long as you match the plate load resistors RL, and the next stages RG resistors, the gain match is automatic.
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With the same preamp tubes, I personally would rather go for a cathodyne phase inverter.
Will give a bit lower distortion, is more simple and less fiddly.
Will give a bit lower distortion, is more simple and less fiddly.
b_force,
I was editing my post # 13 to say exactly that, after you posted.
In different words than in my post . . .
i like coupled cathodes, CCS, and matched plate loads = Intrinsic Balance
In case nobody noticed, this schematic is like the Dyna 6V6 schematic.
Only the output tubes have been changed to protect the "guilty/non-guilty" of Plagiarism.
I was editing my post # 13 to say exactly that, after you posted.
In different words than in my post . . .
i like coupled cathodes, CCS, and matched plate loads = Intrinsic Balance
In case nobody noticed, this schematic is like the Dyna 6V6 schematic.
Only the output tubes have been changed to protect the "guilty/non-guilty" of Plagiarism.
Aliexpress is virtually full of amps/pcb boards following the Dynaco 6V6 schematic in post #1. Most even directly refer to it as Dynaco. You will see 6V6/EL84/6P14/6P15 variants.
Doesn't the Dyna circuit use the Floating Paraphrase splitter with the driver tube grounded thru the Fb?
Plenty of sources on the internet claim the 6N2P is an equivalent to the 6AX7. Others say it has similar behavior despite different construction. Used in a Champ 5F1 style circuit, it does behave very much like the 12AX7, however, in a Shure M65 phono pre, with only 100v on the plates, it does not behave like the 12AX7.
2¢
2¢