A quick LTspice simulation shows:
A tube with a 20k output impedance drives a 3k to 8R output transformer. This gives us a DF of 0.15 from the impedance POV.
The transformer has the following parameters:
-50mH overall leakage inductance (quite a lot)
-100pF of overall capacitance (it's realistic and can be done
-60H of primary inductance
The result is a smooth frequency response of 5Hz to 78kHz at -3dB. Loading does most of the job.
A tube with a 20k output impedance drives a 3k to 8R output transformer. This gives us a DF of 0.15 from the impedance POV.
The transformer has the following parameters:
-50mH overall leakage inductance (quite a lot)
-100pF of overall capacitance (it's realistic and can be done
-60H of primary inductance
The result is a smooth frequency response of 5Hz to 78kHz at -3dB. Loading does most of the job.
The 6JC5 beam pentode gives you 50K Ohm output Z. (19 Watt Pdiss)
One could also put a small amount of drive signal on grid 2 (still mainly using grid 1 for conventional drive) to cancel the small plate feedback for most any beam pentode, to raise output Z.
Even more accurate, would be to use some small actual plate signal from the other P-P side, resistively added to the local screen V's (ie, a few % cross coupled R compensations to screen V) to cancel internal plate effects to get high Z out.
One could also put a small amount of drive signal on grid 2 (still mainly using grid 1 for conventional drive) to cancel the small plate feedback for most any beam pentode, to raise output Z.
Even more accurate, would be to use some small actual plate signal from the other P-P side, resistively added to the local screen V's (ie, a few % cross coupled R compensations to screen V) to cancel internal plate effects to get high Z out.
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6A3summer: I've read somewhere that coupling capacitor distortion is minimized if capacitor's impedance is much lower than input impedance of the coupled stage. But I still wouldn't take a risk. Almost all capacitors are complex electromechanical / mechanoelectric transducers with multiple intrinsic resonances. Capacitor distortion cannot be adequately measured at a single frequency. Pretty much like speaker distortion.
50AE: So, it isn't as bad as it seems! Practically, if leakage inductance is less important than winding capacitance, there are clear transformer design guidelines. Wide core with short central stem, and preferably round cross section. Short, large diameter, multiple layer windings. Silk-clad magnet wire, if possible. Funny, they were making such transformers in 1930s, I have a piece somewhere in my junk pile.
Smoking-amp: very interesting ideas. The Rp cross-coupling compensation is somewhat reminiscent of Miller capacitance compensation scheme in PP triodes. That scheme wasn't really practical though because it was prone to ultrasonic oscillation. Maybe with resistors instead of capacitors, your circuit would be more stable?
I decided to do some transformer math.
A transformer capable of 6W SE power with 60H primary inductance, 60mA max Idc, primary impedance of 3500R.
My DIY simulation excel sheet gives the following data
-Primary Rdc - 105 R with 24 layers of 0.37 effective dia magnet wire
-Secondary Rdc - 0.3R at 4R output impedance with 8 layers of 0.45 effective dia magnet wire (quite high)
-Leakage inductance goes to 90mH. Quite a lot.
Double HiB C-core, single coil, layer length of 52mm, surface area of 11.2 mm2.
Sectioning is P12-S8-P12 with 138 turns of primary per each layer.
The inner primary section will begin with the anode potential at core end and finish with 0.25 capacitance factor towards secondary. We assume the secondary to be at zero potential due to being grounded. The outer primary section will be reverse wound with the B+ connection facing the secondary, resulting in almost zero capacitance in this area.
The MLT is approximately 205mm and the surface area derived from the MLT is 10660mm2
If we assume we are going to use 0.1mm thick pressed paper of dielectric constant of 3 between primary layers, the static capacitance is equal to 2830pF asuming we are using flat wires, but for round wires it is less, although I like assuming worst cases.
Then overall primary capacitance is calculated by
Cp = ( 2830 * 1.33 * 24 / (24-1)^2 ) 2
Cp = 85.3pF
The 1.33 factor is used when layers are wound in a normal left to right, right to left direction. Cst is static capacitance. The division by 2 is because the transformer primary in divided into two sections.
Now for the primary to secondary capacitance, we can assume a factor of 0.25 * Cst. It is slightly less due to the MLT being lower in this area, but I'll skip calculating it more thoroughly due to laziness.
I will be putting a 1mm thick, low epsilon (1.3) dielectric in this area, which gives me a static capacitance of 125pF. Multiplying it by a factor of 0.25 gives a primary to secondary capacitance of 31pF
We can neglect the other P/S layer, being almost close to zero potential, with a capacitance factor of 0.00043. There we can put a thin 0.1mm sheet of pressed paper, resulting in roughly 1pF of capacitance.
So, overall capacitance is 31 + 85.3pF = 116pF
With a 50k driver impedance and a 3500Rload LTSPICE says a bandwidth of 9.2Hz to 60kHz, although there is a series (Ls+Cp) resonance at 42kHz. Such transformer will be A LOT sensitive to capacitive loads.
A transformer capable of 6W SE power with 60H primary inductance, 60mA max Idc, primary impedance of 3500R.
My DIY simulation excel sheet gives the following data
-Primary Rdc - 105 R with 24 layers of 0.37 effective dia magnet wire
-Secondary Rdc - 0.3R at 4R output impedance with 8 layers of 0.45 effective dia magnet wire (quite high)
-Leakage inductance goes to 90mH. Quite a lot.
Double HiB C-core, single coil, layer length of 52mm, surface area of 11.2 mm2.
Sectioning is P12-S8-P12 with 138 turns of primary per each layer.
The inner primary section will begin with the anode potential at core end and finish with 0.25 capacitance factor towards secondary. We assume the secondary to be at zero potential due to being grounded. The outer primary section will be reverse wound with the B+ connection facing the secondary, resulting in almost zero capacitance in this area.
The MLT is approximately 205mm and the surface area derived from the MLT is 10660mm2
If we assume we are going to use 0.1mm thick pressed paper of dielectric constant of 3 between primary layers, the static capacitance is equal to 2830pF asuming we are using flat wires, but for round wires it is less, although I like assuming worst cases.
Then overall primary capacitance is calculated by
Cp = ( 2830 * 1.33 * 24 / (24-1)^2 ) 2
Cp = 85.3pF
The 1.33 factor is used when layers are wound in a normal left to right, right to left direction. Cst is static capacitance. The division by 2 is because the transformer primary in divided into two sections.
Now for the primary to secondary capacitance, we can assume a factor of 0.25 * Cst. It is slightly less due to the MLT being lower in this area, but I'll skip calculating it more thoroughly due to laziness.
I will be putting a 1mm thick, low epsilon (1.3) dielectric in this area, which gives me a static capacitance of 125pF. Multiplying it by a factor of 0.25 gives a primary to secondary capacitance of 31pF
We can neglect the other P/S layer, being almost close to zero potential, with a capacitance factor of 0.00043. There we can put a thin 0.1mm sheet of pressed paper, resulting in roughly 1pF of capacitance.
So, overall capacitance is 31 + 85.3pF = 116pF
With a 50k driver impedance and a 3500Rload LTSPICE says a bandwidth of 9.2Hz to 60kHz, although there is a series (Ls+Cp) resonance at 42kHz. Such transformer will be A LOT sensitive to capacitive loads.
Maybe with resistors instead of capacitors, your circuit would be more stable?
Possibly. Both plate and grid 2 are near 3/2 power law, while grid 1 is closer to square law, due to grid 1 wire proximity effects. So plate to grid 2 cross compensation should remain more constant versus changing cathode current.
Some small comp C could also be tried across the comp R's to neutralize versus frequency effect from tiny internal capacitance. But at audio freq. the main plate feed-thru is going to be just ordinary "DC" transconductance effect (ie, cathode emission versus V field).
Some pentodes with an external pin connection for grid 3 can have their plate curve knees squared up with some small positive V on grid 3. This can give a higher plate Rp at the lower plate voltages.
One can also just put some degenerative R below the cathode to raise Zout. (no bypass cap) Will require more drive V then.
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Edit:
I just noticed that the Sylvania handbook lists the 6JC5 Rp at 5K Ohms, while the GE handbook and datasheet list 6JC5/6JB5 Rp at 50K Ohms. Someone has a typo.
The GE datasheet curves look more like 50K.
The 6JC5 (and 6JB5, 6EZ5, and 6HE5) appear to copy the specs for 6V6, which has a 50K Rp. However, 6MF8 appears to have the same pentode, but lists Rp at 5K. Confusion reigns.
I just noticed that the Sylvania handbook lists the 6JC5 Rp at 5K Ohms, while the GE handbook and datasheet list 6JC5/6JB5 Rp at 50K Ohms. Someone has a typo.
The GE datasheet curves look more like 50K.
The 6JC5 (and 6JB5, 6EZ5, and 6HE5) appear to copy the specs for 6V6, which has a 50K Rp. However, 6MF8 appears to have the same pentode, but lists Rp at 5K. Confusion reigns.
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I just found the original Sylvania 6MF8 datasheet from 1967. It lists Rp as 5K, BUT the curves on the datasheet show 50K for Rp. I'm convinced that all these tubes are 50K Rp now.
For a simple amp to experiment I suggest the 6CL6. At typical operative conditions it has 150K plate resistance. Make PSE to get about 6W output. Zout = 160R excluding series resistance from OPT. Moreover it can be a spud amp with high output sources. It only needs 2.1V RMS drive to get full power....
The OPT for PSE needs to have 3.75K primary impedance.
The OPT for PSE needs to have 3.75K primary impedance.
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...a current drive amplifier driving a fullrange or midrange/tweeter section of a speaker
If you drive a speaker with a current source/sink output, the speaker will see a very high source impedance, and will not get any electrical damping from the amplifier.
That is why we are looking to drive a loudspeaker with a relatively flat impedance, which is easily the case with a FR/mid/tweeter driven directly driving the unit. All the needed damping is supplied by the speaker. Speaker resonance peak not in the driven bandwidth (and aperiodic enclosures can greatly reduce that peak if needed), and copper polepiece et al can flatten the higher frequency impedance.
dave
Aperiodic mid-high speaker is not a problem...
Measured impedance of a real-world aperiodic midTL used >250 Hz.
dave
All of you are giving me too many ideas!
I mentioned the ported woofer with metal with holes in it over the port to make it aperiodic. Was that a KLH 33?
Then there was the Aperiodic Dynaco A25, with the port closed off with something like wool (sandwiched between?) cross hatched chicken wire.
But that is only the beginning.
I have always wanted to Bi-Amp.
Design the speaker; Build a passive or active crossover to 2 amplifiers (Bi-Amp it with current source amplifiers).
And just doing the midrange could make that sound really sweet.
I hope I get the time and energy.
Tri-Amp Anybody?
Thanks!
I mentioned the ported woofer with metal with holes in it over the port to make it aperiodic. Was that a KLH 33?
Then there was the Aperiodic Dynaco A25, with the port closed off with something like wool (sandwiched between?) cross hatched chicken wire.
But that is only the beginning.
I have always wanted to Bi-Amp.
Design the speaker; Build a passive or active crossover to 2 amplifiers (Bi-Amp it with current source amplifiers).
And just doing the midrange could make that sound really sweet.
I hope I get the time and energy.
Tri-Amp Anybody?
Thanks!
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Aperiodic Dynaco A25, with the port closed off with something like wool (sandwiched between?) cross hatched chicken wire.
Fiberglass insulation sandwhiched between 2 pieces of plastic gutter cover.
ScanSpeak VarioVent is a pretty version of the same thing.
dave
Tri-Amp Anybody? Thanks!
That's what I am currently working on.
Sub (20-100 Hz) aperiodic array, driven by 150W PPP super-6BG6GA transformer-coupled to fully differential 6P5G driver.
Mid (100-1000 Hz) Magnavox 12" field coil full range in open baffle. The amp is PPP 2A3 transformer-coupled to differential 56 driver.
High (1000-20,000) BG Neo-8 magnetostat. The amp is PP 1P24B. It is connected to the 2A3 driver in such a way that 2A3 Miller capacitance serves as part of mid-hi crossover network.
Sub and mid are separated at amp inputs by RL filters.
Advantages of tri-amping are: 1) Simple design of output transformers and 2) No horrendous speaker level crossovers.
sser2,
Another major advantage of Tri - Amping is much lower Intermodulation Distortion in all 3 amplifiers.
And the drivers are controlled by the amplifier design and amplifier output impedance, etc.
No crossover elements that change the impedance to each driver, and no resonance of one crossover to another crossover (like low frequency crossover LCR to mid frequency crossover LCR, etc.
Clean as a whistle.
Makes me wonder what Tone Burst testing would look like . . . probably quite pretty.
Another major advantage of Tri - Amping is much lower Intermodulation Distortion in all 3 amplifiers.
And the drivers are controlled by the amplifier design and amplifier output impedance, etc.
No crossover elements that change the impedance to each driver, and no resonance of one crossover to another crossover (like low frequency crossover LCR to mid frequency crossover LCR, etc.
Clean as a whistle.
Makes me wonder what Tone Burst testing would look like . . . probably quite pretty.
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6AU6 gives you 1000,000 Ohms Rp and 3.5 Watt Pdiss. gm 5200
Putting 4 in parallel would give you 250K Ohms Rp and 14 Watts Pdiss. gm 20400
5 in parallel would give 200K Ohms Rp and 17.5 Watts.
Usually an abundantly available tube. 12AU6 around too.
Then there is that earlier cross coupled R compensation scheme for a P-P design. Should be able to null the residual Rp out.
Putting 4 in parallel would give you 250K Ohms Rp and 14 Watts Pdiss. gm 20400
5 in parallel would give 200K Ohms Rp and 17.5 Watts.
Usually an abundantly available tube. 12AU6 around too.
Then there is that earlier cross coupled R compensation scheme for a P-P design. Should be able to null the residual Rp out.
There is a fashion in Russia now, single tube 1W amps using 6J43P tubes, with paralleled anodes.
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