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#21 | |
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diyAudio Moderator
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Quote:
Now, one can swamp any current changes in the LED by using my resistor-from-B+-to-cathode trick, but for the sake of discussion, let's not. The AC impedance of a red LED will be on the order of 7R at a "normal" plate current for 12AT7. So the voltage swing across it will be about 1.1mV, which is about 0.25% of the input voltage. If the LED nonlinearity at that tiny current swing were 10% (it's not), we're talking about 0.025% added distortion, which is considerably smaller than that contributed by the tube. The actual number will be even smaller. If we do my resistor trick so that the LED current is 10mA, the impedance is now 4R. So we've dropped the added distortion by nearly half. And at smaller signal swings, the effect of the LED nonlinearity gets even smaller, as is true with higher plate loads. Thus the crazy-low distortion of my phono stage.
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#22 |
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diyAudio Member
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OK,
So how does that compare with a resistor in place of the LED? ![]() Regards M. Gregg
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#23 |
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diyAudio Moderator
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Lower distortion, higher gain, lower output impedance. Compared to bypassed resistor, faster recovery from overload, flat response to DC.
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#24 | |
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diyAudio Member
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Quote:
![]() This has also confirmed my findings with the better drive. Regards M. Gregg
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#25 | |
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Clueless
diyAudio Member
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Quote:
Thanks Charlie
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#26 |
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diyAudio Moderator
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I've posted it a few times before (and I notice that it has found a place in the latest edition of Valve Amplifiers); it's pretty self-explanatory, and I don't claim originality.
Just run a resistor from B+ to the cathode to get a higher LED current. For example, let's say you have a red LED (1.7V) biasing a 12AX7 stage running 1mA, with a 200V supply. To get the LED to a flatter part of the Vf versus If curve, you might want the LED to have 10mA running through it. We supply the extra 9mA from a resistor, which can be calculated by Ohm's Law- R = V/I = (200-1.7)/0.010 ~ 200/0.01 = 20k.As a practical matter, I haven't seen much difference when the plate loads are reasonably high (i.e., running close to constant current).
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#27 | |
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diyAudio Member
Join Date: Mar 2012
Location: Melbourne
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Quote:
So - a resistor (unbypassed) in the cathode provides local feedback and reduces gain. Bypassing resistor with a cathode will retain same quiescent bias point, but will tend to reject signal voltage changes, hence providing a psuedo voltage source - so no local feedback, higher gain but also higher distortion, due to reduced feedback and capacitor non-linearities. A LED mimics the voltage source (assuming small dynamic impedance), retaining the gain of the bypassed resistor, but ditching the yucky electro. So for tubeheads who say that they don't want semi-conductors in their circuits, the options are a battery or an electro bypass cap. A simple unbypassed resistor is NOT equivalent. |
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#28 |
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Clueless
diyAudio Member
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Thanks, I look forward to VA4 coming out again so I can read up on it.
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#29 |
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diyAudio Moderator
Join Date: May 2003
Location: Palatiw, Pasig City
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i blame it all on SY, now i use only led's on my small tube amp cathodes....
![]() ![]() ![]() i still do use resistor/capacitor combo on my guitar amps......
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#30 |
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diyAudio Member
Join Date: Aug 2009
Location: Johnson City, TN
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Likewise, I've run 6N1P, 6N2P, 6N3P, and 6N23P with LED bias and current source as anode load and the distortion figures were consistently lower than the equivalent biased resistor load, resistor or resistor and cap in cathode circuit.
I used both HP spectrum analyzer and PC with sound-card for distortion measurements and to look at harmonic distribution. |
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