I suppose it depends what you mean by 'correctly biased'. Most valves produce less distortion the hotter you bias them (until grid current clipping takes over, obviously) but more LED current means at least slightly cooler bias.
The voltage across an LED changes by such a small amount with changes in current that I wouldn't expect someone to 'tune' the bias voltage in this way, anyway.
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How do you determine what current is actually flowing through the diode without knowing its R value when in operation?
Either by calibrating the particular LED itself externally (which isn't too hard), or by using the proxy of measuring voltage-drop across the anode resistor, supposing you've elected to use one of those.
GoatGuy
The knee voltage of a LED is set by physics, the bandgap fixes the optical wavelength
Correctly biasing your valve is a compromise - running hard for low distortion reduces lifetime
Correctly biasing your valve is a compromise - running hard for low distortion reduces lifetime
I'd just open the circuit and clip-lead my meter in there to check it for accuracy under working conditions. But there seems to be no discussions on finding the accurate current flows or the impact of bypassing or not and whether or not an LED provides the degenerative feedback an Rk alone provides or whether the LED will affect any GNF applied by changing the K circuit R.
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In valve circuits it is rare to need to know exact current flows. 20% error is fine, except in balanced stages.
LED bias does not need bypassing, as the dynamic impedance of the LED will be much lower than 1/gm in almost all cases. Hence it does not provide any significant degeneration.
An LED alone should never be used as the NFB return point as its nonlinearity will become part of the feedback network and so force nonlinearity on the circuit as a whole. Instead, a series cathode resistor can be used to swamp the LED resistance - but in most cases you might as well then just use the resistor to do the bias and ditch the LED.
LED bias does not need bypassing, as the dynamic impedance of the LED will be much lower than 1/gm in almost all cases. Hence it does not provide any significant degeneration.
An LED alone should never be used as the NFB return point as its nonlinearity will become part of the feedback network and so force nonlinearity on the circuit as a whole. Instead, a series cathode resistor can be used to swamp the LED resistance - but in most cases you might as well then just use the resistor to do the bias and ditch the LED.
Actually, you're making a very good observation, about the equivalence of the LED in-circuit to a self-adjusting resistance, not terribly dissimilar to 'standard' cap-bypassed cathode-resistors.
Remember in the context of DIY audio, there is a sizeable fraction (majority?) of persons who ardently believe that all capacitors in the signal path are the downfall and ruination of an otherwise esquisite sound stage (or some such bull). So, anything-at-all that can be done to eliminate those nastyboys (caps) is godly.
OK. Out of the clear blue, some enterprising person had the light bulb (ouch!) go off, “An LED might reasonably act as a nearly-fixed voltage drop device”, thus simulating a 1.0 to 3.0 volt bypassed cathode resistor bias supply. Sure enough it worked. And it works mostly without needing a bypass cap. Because the V*I curve of solid-state diodes is so exponential, the V hardly changes at all with modest changes in I.
However, if we think of the LED as a voltage-controlled resistance, but a highly nonlinear one, then its nonlinearity can be erased by using exactly the same principle as the cap-bypassed cathode resistor idea. Use a cap across it. Large value (like 2200 μF). It can be a 5 volt part too, since the diode is never going to let more than its forward voltage develop.
The result would be that the I*V curve would be A/C tamed… and the LED would behave much more like the perfect constant-voltage source.
Your insight is excellent. Now, go forth and conquer. And remember, that you can always put diodes in series to get other voltages. A 1.7 volt LED plus a 0.4 volt Schottky gives 2.1 volts forward voltage drop. Etc.
GoatGuy
What you'll find in reality is that it makes not even a little bit of difference. So skip the cap.
Oddly, we disagreed about the bypassing recommendation. I've been bypassing diode-bias for years; if you install a little switch to bring the cap into and out of circuit (for A-B comparison), its pretty obvious that there is something that changes. Now whether you like it or not… you and I have always agreed about that. Up to the ear of the listener.
GoatGuy
I've done the experiment, looked at the noise and distortion. Nope, not even a scintilla of difference.
OK. I'll trust you as much as you trust me (which is to say not at all). I've done the experiment, and have heard a difference. Could be a lot of things, but I suspect how steeply the I(V) curve rises for a particular LED process more than anything else. LEDs with steep I(V) curves will not be 'improved' by bypass caps, because their equivalent impedance is so low to start with. LEDs with more relaxed, broader I(V) curves will have their intrinsic quiescent impedance changed. I suspect my use of 1970s era LEDs may well have been the root of the actual observation.
GoatGuy
I think SY means he measured the differences (or lack of). You can trust measurements. You can't trust your ears any further than you can throw them.
That is exactly what I meant. 😀
To be fair, I only have measured systems with red, green, and IR LEDs, all having relatively low impedance. Things could well be different with higher impedance devices, but it doesn't make much sense to use them in this circuit position.
… another amusing hour with numerical analysis … and I understand your position much more now, SY.
I found some I·V data for a common red LED, and decided to model the small signal equivalent impedance based on [E = IR] → [R = E/I] → [Zeq = dE/dI]. Chose 5 ma, 1.52 Vd, and dE = 0.001V. The LED had a Zeq of 6 to 8 Ω. Nice and low.
At 40 Hz, a 2200 μF has a Zeq of [1/(2 π FC)] → 1.8 Ω. The Zeq decreases from there as F rises through the audio band. Clearly, this lower Zeq will in parallel with the Zeq of the LED, will drop the equivalent impedance. Will it make a material difference to the amplification of the stage? I rather think it won't. (In other words, agreeing with you).
Then I invested a bit more time with one of the LEDs that I still have hundreds of, bought in the early 1970s. (Remember Mike Quinn's? Bag of 1,000 for $17. I'll never be rid of these things!). Hooked it up to a 9V battery, a 5 K pot, and the led. Measured voltage drop with a 4½ digit DVM. The Zeq is about 20 ohms, at I = 5 ma.
Maybe we're both right! My crummy old LEDs just had higher Zeq. Hence the largish C-bypass had a real world effect. Modern LEDs seem tighter, sharper.
GoatGuy
I found some I·V data for a common red LED, and decided to model the small signal equivalent impedance based on [E = IR] → [R = E/I] → [Zeq = dE/dI]. Chose 5 ma, 1.52 Vd, and dE = 0.001V. The LED had a Zeq of 6 to 8 Ω. Nice and low.
At 40 Hz, a 2200 μF has a Zeq of [1/(2 π FC)] → 1.8 Ω. The Zeq decreases from there as F rises through the audio band. Clearly, this lower Zeq will in parallel with the Zeq of the LED, will drop the equivalent impedance. Will it make a material difference to the amplification of the stage? I rather think it won't. (In other words, agreeing with you).
Then I invested a bit more time with one of the LEDs that I still have hundreds of, bought in the early 1970s. (Remember Mike Quinn's? Bag of 1,000 for $17. I'll never be rid of these things!). Hooked it up to a 9V battery, a 5 K pot, and the led. Measured voltage drop with a 4½ digit DVM. The Zeq is about 20 ohms, at I = 5 ma.
Maybe we're both right! My crummy old LEDs just had higher Zeq. Hence the largish C-bypass had a real world effect. Modern LEDs seem tighter, sharper.
GoatGuy
That's exactly it- it is the opposite of a CCS, so you don't use them in the same circuit position. Quieter and lower impedance than a Zener, though; the HLMP6000 that I used in my phono stages have less than a 3 ohm impedance and a noise voltage density less that 0.4nV/rt Hz.
The combo of a CCS as a plate load and LED biasing the cathode yields a tube with its distortion pretty close to its absolute minimum.
One point to be potentially aware of is that many "white" LEDs are actually blue LEDs with the addition of a yellow phosphor. There isn't a specific operating voltage for white LEDs since it depends on the actual LED chip type(s) used. The graph is a useful guide though, especially in terms of red vs green vs blue, but should not be regarded as an absolute in terms of LED colour vs LED voltage drop.
You don't need to feed extra current to the LED. It only increases the bias voltage a bit, which increases rather than reduces distortion.
Its interesting,
to measure the voltage drop under power before you use them..
things aren't always as they seem
The other thing is do they sound good or do they sound different (current dependant)..
Regards
M. Gregg
to measure the voltage drop under power before you use them..
things aren't always as they seem
The other thing is do they sound good or do they sound different (current dependant)..
Regards
M. Gregg
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If you read this very thread I think you will see that the consensus is that modern LEDs used for bias don't sound of anything.
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