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In Reply to: RE: What tubes to use with this Interstage Transformers? posted by megasat16 on September 01, 2016 at 20:58:28
Well, we can look at this from either side.
Let's see what the grid of the output tube sees looking back through the IT to the driver stage.
The 5k output impedance of your driver tube will look like 80k to the grid of the output tube.
AFAIK, 80k will not properly drive the Miller of any DHT I am aware of.
If you had a driver stage with 500 ohms output impedance that would look like 8k ohms to the output tube and that would be fine in most instances.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Follow Ups:
Hi Tre,
Thanks for making me see the coins on the flip side.
So, I do some more reading on the Cmiller effecting the total Cin of the Triode.
I know you gave me an example before and the this pretty much seems the same except this paper has no mentioning the Grid Current Requirement.
http://www.aikenamps.com/index.php/what-is-miller-capacitance
Since I don't have any Output tube in mind, I can't do an actual Cin calculation for this discussion yet. So, let's presume the Cin is 1uF of an unknown DHT which is a unrealistically pretty big for a tube.
So, in the Triode, the 1uF appears at the Grid of the Tube and termed Cin, yeah?
Let's say the 600V AC (from my last example yesterday) appears from the Grid Pin of the IT and present to the 1uF Cin of the Tube.
At this point, what happened to that voltage at the Grid? Will it not go inside the Grid of the tube? Will the Grid of output tube draw a lot of current from the IT so the voltage becomes very small?
Since the Grid is loaded on 1:4IT, will it load the driver tube from the ideal Rl? Ideally, the Rl is higher than Rp of the driver tube not to excessively load the driver tube during Grid Current demand.
But here is what I am thinking. If a 1uF Coupling cap is put in between the Plate of the Driver Tube and the Grid of the Output tube, does it present the same condition as Cin to the Driver tube? Even a very small amount of AC seems to pass through the coupling Cap fine. Then, why it won't it pass through another 1uF Cin inside a glass tube? It'll just be two 1uF caps in series in the circuit.
Thanks,
James
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Miller capacitance is not series capacitance, it's shunt capacitance.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
There is some current through 1:4IT but how much is required?
A simple 1st Order RC filter will filter at 6dB/octave or the 20dB/decade beyond Fc. It doesn't matter how much current you are supplying at Fc because Voltage will just be attenuated due to the Xc. The only thing to reverse is to supply a bigger Vin to negate the roll off or uses a local feedback to reduce the gain (thus changing Ctotal value to shift the Fc).
The drop is 3dB at Fc. So, ideally we want to see Fc at 20KHz or beyond. If Fc is at 20KHz, the FR roll off probably starts at around 18KHz.
The nice thing about the 1:4IT I have is that it'll slightly adds about 0.5dB gain from 14KHz and beyond so it'll somewhat negate the effect of the Triode roll off.
As to the Grid Current Requirement from the driver stage, it would be impossible for the circuit below to work.
But it was published in the MJ magazine so it must work. I really like to see how it measures and where the Triode Roll off begins.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"A simple 1st Order RC filter will filter at 6dB/octave or the 20dB/decade beyond Fc. It doesn't matter how much current you are supplying at Fc because Voltage will just be attenuated due to the Xc. The ONLY THING to reverse is to supply a bigger Vin to negate the roll off or uses a local feedback to reduce the gain (thus changing Ctotal value to shift the Fc)."
That is not true. A lower drive impedance means the Fc is shifted higher, out of your way. That fixes the problem at it's core.
"The drop is 3dB at Fc. So, ideally we want to see Fc at 20KHz or beyond. If Fc is at 20KHz, the FR roll off probably starts at around 18KHz. "
One octave below the Fc is the -1db point. So a Fc of 20kHz means you are still -1db at 10kHz.
The phase is disturbed for a decade.
So, ideally we want to see Fc at 200KHz or beyond.
That is why, for a low pass filter, if you want 20kHz undisturbed you need the -3db point to be 200kHz.
And at the other end, for a high pass filter, if you want 20Hz undisturbed you want the -3db point to be 2Hz.
All of that aside, you should look closer into Miller capacitance and what it takes, current wise, to drive shunt capacitance.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
I am a little rusty on the filters and I stand corrected on the roll off points.
However, I want to standby my own account of increasing Vin proportionately to the attenuating Slope passing Fc point to remain the constant Vout from a low pass RC filter circuit. Since R is fixed value, it can only supply a maximum current to the shunt C when Xc is effectively becoming a short at very high frequency. R is always been fixed and Vin is fixed so how much maximum current (Imax) it can supply through the R is fixed too. The only thing changes is that the load current demand which is based on the Xc. It's just that Xc comes into play based on how much current it gets from series R and how often it gets charged.
Even though Xc is the reactance components, it still is really a cap and the charge across the cap at various frequency.
Before the Fc point, the Xc is high so it does not present a very current demanding load to the R. Voltage drop across R is small or none due to the Xc being high.
Passing the Fc point, the Xc value gets smaller and smaller thus shunting more current to the Ground. This creates a bigger voltage drop across the R and smaller voltage appears at the C.
But When the series R reaches the max current it can supply due to the Fixed Vin divided by it's own value, the Imax is reached regardless of how lower Xc goes.
To overcome this attenuation, you need a larger Vin passing Xc point since the R is fixed to supply more current. You can't get more current through a Fixed value R and a fixed value Vin / Vsource regardless of how much current is made availalbe at the Vin / Vsource or anything in front of the R. The R value itself it the limiting factor here when Xc is effectively becoming a short circuit to Ground.
So, let's get back into the RC filters and shunt cap drive requirement.
In RC series filter, R is usually fixed. And the C is fixed as well since it's inside the tube.
To get more current into the C, you can either increase the voltage (resulting more current flowing through the R) or lower the value of R (as your suggestion).
Lowering R sounds good if it's possible since it's the simplest and easiest. But this defeats the whole point of this thread which is to find the usable tubes with the 1:4IT.
Or we can reduce the C value for a fixed R value? Since the R is fixed in my case, we can find the tubes with low gain since the open loop Gain has the most effect on the total C. It's exactly what I am hoping to find.
.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
Where is "R"?
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
R is the fixed R between the plate of the Driver and the Grid of the Output tube in the Direct Coupling Circuit.
Or if it's eliminated, it's the Rp of the Driver / Source.
And you are saying 160Kohm impedance of the IT as the R in this case.
I don't think Z(IT) is in series with the Shunt C at all in the IT circuits. It is effectively Rp.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
And with this, we are now back at Square 1.
Reducing Rp will have more current available to the C. I don't think I disagree with that. It just meant a different driver tube other than 5K Rp in the case of 801A.
But really, how much current a Grid needs? Elimate the IT and the fixed resistor R in the direct coupling circuit and substitute with the short so the Vsource is the same as the low Rp tube like 500ohms. Fc will shift due to the lower Rp (or R in the RC) value but passing that Fc point, the attenuation will occur the same. Rp values are fixed too. In a way, it is more current avilable till Imax is reached. But beyond Xc point, it'll all be just the same all over. Fc is shifted due to the R being smaller.
So, you can't make more current available without changing Rp (the tube itself) or the IT. I think it's all you have been saying that 1:4IT won't supply enough current due to it's high Z value and the Fc point will occur within AF band.
I also think I have been saying that Triode Roll Off will happen. But when you say it won't work, I started misunderstanding why won't it work.
But I think we are all saying the same thing in a different way.
Essentially, your choice is not to use the 1:4IT which I understand and respect that.
But my choice is to use the IT I have and to work around it and find a suitable tubes for it. I think you have mentioned that it would be very difficult if not next to impossible?
.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"Fc is shifted due to the R being smaller."
Correct but with a low enough Rp (and it not being multiplied by a step up IT) Fc is high enough so it no longer matters.
"Essentially, your choice is not to use the 1:4IT which I understand and respect that.
But my choice is to use the IT I have and to work around it and find a suitable tubes for it. I think you have mentioned that it would be very difficult if not next to impossible? "
Yes, and I will say it again.
If you had a driver stage with (let's say) a 500 ohm output impedance, that 500 ohms would look like 8k ohms to the output tube's grid and Miller C. As long as the output does not have too much Miller C then the 8k ohms would drive it just fine. That is to say the Fc would be high enough as to not matter.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Yes! I agree to that.
That's truly make it difficult to design a circuit for Fc out of 200KHz not to be effected by the RC filter slope.
So, let's say your Fc moves out to the 40KHz instead of 20KHz, the low pass attenuation starts at around 4KHz?
Now, here is the advantage of the 1:4 IT I have. It slightly adds more Voltage gain at the rising frequency so it'll negate that effect of the Low Pass filter loss and puts an overall FR of the final output a little flatter. You normally don't have that advantage with a constant voltage sources.
Now back to the transformer impedance, it was specified by the manufacturer at a certain (unknown to me) frequency as the starting point. But it is the Audio IT, I can safely assume that is the the optimum load it present to the Source during the AF Band.
Since I can't measure and don't know the exact value of it, I must use it as an optimal loading point to the Driver Tube and find an appropriate driver tube. Lower Rp driver tube does not pose a problem but Higher Rp driver more than 10K Rp will be loaded by the Transformer Primary.
It's the same reason you can't put 600Ohm Primary IT on the 5K Rp tube since it'll just load the Voltage in the primary and drop most of it on the primary side and less current (or AC Flux) is available to the core and to the secondary side. It's the opposite of Saturation (current starving core) which looks like a saturated core. I think Dave and I had quite a lengthy discussion of that.
Anyway, I appreciate you taking seriously to make me see through this.
Sometimes, things got lost in translation and my apologies for that. But I think we are essentially on the same page with a different approach to the solution.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
"And you are saying 160Kohm impedance of the IT as the R in this case."I am not saying that. I never said that.
"I don't think Z(IT) is in series with the Shunt C at all in the IT circuits. It is effectively Rp."
Correct, if the IT is 1:1 then Rp (not really Rp but the output impedance of the driver stage) is reflected to the secondary and that is the source impedance and the "R" used to calculate the effect of the Miller Capacitance.
But with a 1:4 winding ratio IT we have a 1:16 impedance ratio IT so "R" is the output impedance of the driver stage times 16. That's is what I have been saying.
BTW Transformers do not have impedances. They only have winding ratios.Transformers are made to work with certain impedances. The inductance and winding capacitance dictate that and that is what the "impedance rating" of a transformer is all about.
But that is not to say that they have impedances of their own. They don't.
Transformers only "reflect" impedance according to their impedance ratio.
The "R" presented to the primary will reflect to the secondary and the "R" presented to the secondary will reflect to the primary.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 09/04/16
Published in MJ dosn't mean it works good.....
Look at the inputstage stu-001. 100k output and driving a 211, also a 100Ohm potentiometer infront of the stu 001. TerribleAnd also 2 output transformers in serie, wow! Even more terrible.
These circuits are for MyFi projects, with weird high colored speakers systems. Not the usual Hifi home stereo.
Edits: 09/04/16
I agree to your on the cover assessment of how that circuit works or sounds.
But the MJ magazine is not a small Internet Magazine. It's not known to most people outside of Japan but very highly respected in Japan. If it's comparable, the Japanese audiophiles holds MJ in the same regard as that some of us do for the Stereophile magazine in the U.S.
http://www.preservationsound.com/?p=1286
Anyway, the circuit you've seen was done by Sakuma. He has possibly made tens or even hundreds of these eyebrow raising circuits. I can only wonder how they sound but he is well known inside Japan for his nontraditional circuits and the self made DHT amps.
I don't think Mr. Susumu Sakuma holds an Engineering Degree but he is no small fry in the DHT circuits.
http://www10.big.or.jp/%7Edh/
I wonder how it really sounds though. May be No Highs, No Lows, just the perfect midband? I don't know but Bose has a pretty good fan base too.
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.Thou shall not stand where I type for I carry a bottle of Certified Audiophile Air and a Pure Silver Whip.
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