I'm planning to build a van Waarde 6as7 headphone amp, maybe substituting two 3a5's for the 6922 preamp stage. 3a5 has a directly heated cathode, and lately I've been reading about how the cathode is in the signal path -- the signal wants to go to ground, and it can go thru the cathode, either to ground or get soaked up by the cathode heater's filter caps.
The TentLab folk point out that what you really want in the heater circuit is low DC resistance (to heat the cathode) and high AC impedance, which makes the signal see a "brick wall." Plz tell me if the following idea has any merit -- I know very little theory, so I'm just thinking out loud here.
Consider a plate resistor. If you removed it so the plate was hooked directly to the p/s filter caps, they would still provide B+ but would also soak up the signal appearing on the plate. Adding the plate resistor still passes the DC, but also provides AC impedance to prevent the signal from going that way, que no?
Therefore: how about another tap at the p/t's B+ secondary, rectified to, let's say -55vdc - 0 - +55vdc, filtered, then knocked down to cathode heater voltages (I think +1.4vdc -- -1.4vdc for 3a5) with some high-value power resistors. Doesn't this block the loss of the signal the same way that a plate resistor does?
What's wrong with this picture? 😕
Could there be much too much current available? (3a5's need 100 mA each).
The TentLab folk point out that what you really want in the heater circuit is low DC resistance (to heat the cathode) and high AC impedance, which makes the signal see a "brick wall." Plz tell me if the following idea has any merit -- I know very little theory, so I'm just thinking out loud here.
Consider a plate resistor. If you removed it so the plate was hooked directly to the p/s filter caps, they would still provide B+ but would also soak up the signal appearing on the plate. Adding the plate resistor still passes the DC, but also provides AC impedance to prevent the signal from going that way, que no?
Therefore: how about another tap at the p/t's B+ secondary, rectified to, let's say -55vdc - 0 - +55vdc, filtered, then knocked down to cathode heater voltages (I think +1.4vdc -- -1.4vdc for 3a5) with some high-value power resistors. Doesn't this block the loss of the signal the same way that a plate resistor does?
What's wrong with this picture? 😕
Could there be much too much current available? (3a5's need 100 mA each).
Ummm... no, the plate resistor does not "block" anything.
Rather it sees the current drawn by the tube (although technically everything actually flows from the cathode but let's keep it simple...). This means that there is a voltage drop across the resistor - that is what makes the signal appear as a voltage. Even without the plate resistor, if the tube is modulated properly, there is a current change, it's just hard for you to see it on the plate side. If there is no plate resistor, and just a cathode resistor we call that a "follower". In which case the signal is taken from between the cathode and cathode resistor.
One could derive a voltage from the B+ to run the heaters, but that's not a good idea since the heaters can draw a fair amount of current, and the Vdrop required will waste a lot of power.
Hope this starts to explain...
_-_-bear
Rather it sees the current drawn by the tube (although technically everything actually flows from the cathode but let's keep it simple...). This means that there is a voltage drop across the resistor - that is what makes the signal appear as a voltage. Even without the plate resistor, if the tube is modulated properly, there is a current change, it's just hard for you to see it on the plate side. If there is no plate resistor, and just a cathode resistor we call that a "follower". In which case the signal is taken from between the cathode and cathode resistor.
One could derive a voltage from the B+ to run the heaters, but that's not a good idea since the heaters can draw a fair amount of current, and the Vdrop required will waste a lot of power.
Hope this starts to explain...
_-_-bear
With current source heating (the best sounding way) the impedance is high even at dc, and this avoids problems with low frequency audio.
The high impedance is beneficial to keep cathode signals from leaking into the heater circuit (where dielectric materials are usually very poor). The signal currents involved are really tiny, so to make an audible difference, the impedance must aim to be 500K .. 10M - as high as possible, and certainly a lot higher than a plate resistor. Using an IC or (better) a transistor CCS for DHT heating makes a wonder of a difference, and takes little time to build.
The high impedance is beneficial to keep cathode signals from leaking into the heater circuit (where dielectric materials are usually very poor). The signal currents involved are really tiny, so to make an audible difference, the impedance must aim to be 500K .. 10M - as high as possible, and certainly a lot higher than a plate resistor. Using an IC or (better) a transistor CCS for DHT heating makes a wonder of a difference, and takes little time to build.
If impedance is high (500K - as high as possible) even at DC, how do we get much current to flow with a reasonable supply efficiency?
Hey ET,
The 3A5s consume 0,22A as you should parallell the halves and also the heaters The one below is designed as an interface for a Monica/TDA1545 DAC but shows the principle for the heaters. So just look below the tube.
By letting the CCS heater-current go through the cathode-resistor it gets so small that no decoupling is needed. It is easely calculated. Go for ca 8-9mA/90V=-2V. 2/0,22 is ca 9ohm. 1,4+2V=3,4V needed.
Consider the parallelled 3A5 halves as a tiny, tiny, tiny dualplate 2A3.
When parallelled Ri gets halved, Gm doubled and mu stays the same.
Maybe I should add that the CCS can be more sofisticated. You could also use a resistor from B+ but this would ask for a fat 20-30W resistor.
The guru of DHT pre-amps is Andy Evans.
The 3A5s consume 0,22A as you should parallell the halves and also the heaters The one below is designed as an interface for a Monica/TDA1545 DAC but shows the principle for the heaters. So just look below the tube.
By letting the CCS heater-current go through the cathode-resistor it gets so small that no decoupling is needed. It is easely calculated. Go for ca 8-9mA/90V=-2V. 2/0,22 is ca 9ohm. 1,4+2V=3,4V needed.
Consider the parallelled 3A5 halves as a tiny, tiny, tiny dualplate 2A3.
When parallelled Ri gets halved, Gm doubled and mu stays the same.
Maybe I should add that the CCS can be more sofisticated. You could also use a resistor from B+ but this would ask for a fat 20-30W resistor.
The guru of DHT pre-amps is Andy Evans.
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The impedance in this case is the small-signal impedance: that's to say it applies to variations..
Think of walking up to a 1A transistor current source (a proper one, with the current delivered by a BJT collector or a FET drain). The CCS is supplying a 5V filament, or a resistor etc. Apply a voltage source, like a car battery across the filament, and the filament voltage increases to 12V. How does the CCS respond? Providing it has enough supply voltage, it carries on pumping 1A into the load network just as before, or maybe changes by 10uA.
Then, its small-signal dc impedance is:
delta-V/delta I
7V/10uA
= 700K
OK, so that was a large signal change really, and the impedance may be worse for 7V cathode signals. But for normal cathode signals, the impedance is very high. I wanted to mention that it was high at dc, to illustrate that there is no low-frequency roll-off mechanism. At high frequencies, the CCS impedance is degraded by parasitic capacitance in the collector (or drain), so for this reason be sure to choose a transistor with a low value of Cob. BJTs do better than power-FETs in this regard.
The other point to illustrate is that the high impedance (small-signal) of a collector is intrinsic. This is not the case if you make a CCS sourcing current from an NPN (ie current drawn from the emitter). This kind of CCS derives its high impedance from feedback, and degrades according to the performance of the feedback circuit: e.g. goes low impedance at high frequency.
The mechanics and impedances of transistor current source is best dealt with in:
Horowitz & Hill: "The Art of Electronics" 2nd Ed Cambridge University Press 1989
The transistor chapter, pp 74..77
Think of walking up to a 1A transistor current source (a proper one, with the current delivered by a BJT collector or a FET drain). The CCS is supplying a 5V filament, or a resistor etc. Apply a voltage source, like a car battery across the filament, and the filament voltage increases to 12V. How does the CCS respond? Providing it has enough supply voltage, it carries on pumping 1A into the load network just as before, or maybe changes by 10uA.
Then, its small-signal dc impedance is:
delta-V/delta I
7V/10uA
= 700K
OK, so that was a large signal change really, and the impedance may be worse for 7V cathode signals. But for normal cathode signals, the impedance is very high. I wanted to mention that it was high at dc, to illustrate that there is no low-frequency roll-off mechanism. At high frequencies, the CCS impedance is degraded by parasitic capacitance in the collector (or drain), so for this reason be sure to choose a transistor with a low value of Cob. BJTs do better than power-FETs in this regard.
The other point to illustrate is that the high impedance (small-signal) of a collector is intrinsic. This is not the case if you make a CCS sourcing current from an NPN (ie current drawn from the emitter). This kind of CCS derives its high impedance from feedback, and degrades according to the performance of the feedback circuit: e.g. goes low impedance at high frequency.
The mechanics and impedances of transistor current source is best dealt with in:
Horowitz & Hill: "The Art of Electronics" 2nd Ed Cambridge University Press 1989
The transistor chapter, pp 74..77
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