Clipping LEDs

[Electric_Death]

Finally, I have some either UV or IR leds I bought a year or two ago for remote-control building.  Unfortunately, I failed to label them and they sit in a plastic bag, unusuable because I don't know if they give off light or respond to light.

Actually, I'm pretty sure that all LEDs do BOTH. They give off light and they also respond to light shone upon them.

Bingo, IR/UV LED's work as both sensors and transmitters.

[brefmint]

I had a non-working Alesis 3630 Compressor that I attacked with a solder gun last night. I retrieved about 50 LED's. Most are red, some green and a few yellow. Could I use these as clipping diodes? They seem to be a standard brightness, not ultra. Here's a pic of the front. Thanks!

[JDoyle]

This is an ideal silicon diode. If you were to put a pair of these ideal diodes in a Tube Screamer, the output wave form would be squared off; the output would abruptly and absolutely stop at, and continue to be, 0.6 Volts when the output is sufficiently large (0.6V or larger) to turn on the diodes. I don't think this would be sonically pleasing as the sharp corner would create a lot of high harmonic content.

Actually in a Tube Screamer circuit it would follow the input once the diode began conducting. It is basically a short after the .6 volts which would make the op amp a voltage follower.

Don't mind me, I'm just picking nits....

--john

John, this is great, I never feel like I really understand what is going on in this situation. I hope you have the patience to follow me as most of this is me thinking out loud...

So you are saying that when the diode conducts and it's effective resistance becomes 0 (or close to it), the gain of the op amp drops to one as the diode becomes a short, and THEN:

1 - If the input signal is larger than the diode drop without the op amp's normal gain, the diode continues to conduct and the signal at the op amp output is a buffered version of the input.

2 - If the input signal is NOT larger than the diode drop, the diode shuts off, raising it's effective resistance to infinity, taking it out of the circuit, and the op amp's gain kicks back up to the ratio of the resistors in the feedback loop.

But in reality, in case number 2, as long as the input times the gain of the op amp is sufficient to turn on the diodes, the op amp will be forced to stay in the sweet spot.

Because the input signal multiplied by the op amp's gain was sufficient to cause the diodes to conduct in the first place, we know that the diodes will always be on. But we also know that they CAN'T be on because that drops the gain of the op amp to one, lowering the output signal sufficiently to take the diodes out of conduction (and therefore the circuit) which brings us around to the beginning, where we know that the gain of the op amp times the input signal is sufficient to bring the diodes into conduction.

So when the diodes are on, their effective resistance drops to zero, dropping the gain to one and turning the diodes off. This drops the signal level at the op amp's output, bringing the diodes out of conduction, raising their effective resistance, and therefore raising the gain back up, which brings the op amp output signal back up to a level that turns on the diodes, which lowers the diodes effective resistance, lowering the gain, bringing the diodes out of conduction, which raises the effective resistance of the diodes, which raises the gain, which causes the diodes to conduct, which lowers the gain, which brings the diodes out of conduction, which raises the gain, which brings the diodes back into conduction... Etc.

If we put our theoretical ideal diode in there, I think this proves my point that it would sound pretty harsh. An ideal diode would have two states and would either be conducting (the op amp is a buffer) or not conducting (the op amp's gain is ratio of the resistors in it's feedback path), and because the math won't work in either state for the op amp, it will bounce back and forth as fast as it can between the two states trying to find something in between that works (and creating massive amounts of high harmonics as the output continuously and abruptly transitions from no gain to full gain). It continues to do this until the input signal is either large enough to keep the diode conducting on its own when the op amp is a buffer, or decays to the point where its amplitude multiplied by the gain of the op amp isn't enough to turn the diodes on anymore. And because the compensation cap can only reduce a signal at the output to a minimum of 1, the speed and magnitude of the bouncing back and forth would only be limited by the slew rate of the op amp and any capacitance to ground and would also be completely unrelated in frequency to Vin.

I'm going to go out on a limb here and say that I don't think that would sound very good.

But luckily real diodes are not ideal. Real diodes have a portion of their transfer curve near the threshold voltage where they pass some, but not all, current. In this zone the diode acts as a nonlinear resistor, (the ideal diode's absence of this zone makes it effectively a switch) which allows the op amp to find a point on that curve where the effective resistance of the diode, in parallel with the resistor from the output to the negative input (the gain pot), is a combined resistance that DOES allow it to output a current that can make it's two inputs be the same voltage (this is the 'math' the op amp is always trying to make work; to the op amp, the absolutely and only thing that matters is keeping the voltage on its two inputs equal and it will do anything it can at it's output to ensure that, even if it means dying while attempting to do so). With the ideal diode that resistance is never available, the ideal diode's effective resistance is either zero when on, or infinity when off. A real diode has a small area in between those extremes that the op amp can utilize.

SO if the input is smaller than the headroom of the diodes, independent of the circuit, and the gain of the stage is sufficient to bring the output above 0.6Vp-p, then the diodes are never fully on or fully off; the op amp is constantly trying to find the spot in the diode's conduction curve which turns the diode on just enough to have the correct effective resistance, that in parallel with the gain setting resistors, will allow it to output a current that makes the math of it's feedback network correct. If the input signal is sufficiently large to conduct the diodes itself without additional gain from the op amp, once the diode is conducting fully, the op amp goes to the buffer state and stays there until the input signal's amplitude starts to drop below the diode's threshold, at which point the gain of the op amp starts to rise and the op amp's balancing act begins again. If the input signal level falls to the point that even with the gain from the op amp it isn't sufficient to turn on the diodes, then the diodes are effectively out of the circuit and it is just an op amp gain stage.

Now, maybe I'm confused, but in case number one above, while the diode is conducting and the input voltage is sufficiently large to conduct the diodes without any gain, the op amp may output a current that drops the correct voltage across the resistor to ground (or Vbias) on the negative input so that it appears TO THE OP AMP that it is buffering the input signal and it's gain is one. But it is passing this current through a diode, not a resistor, so instead of the voltage across the diode varying proportionally with the output current like a regular resistor (or the actual resistor on the negative input where that IS occurring), the voltage across the diode is essentially held constant at the threshold voltage of the diode regardless of the current through it.

So while the output of the OP AMP may be acting like a buffer and outputting a current that drops the input voltage across the resistor on it's negative input, the output waveform of the whole CIRCUIT is actually fixed at 0.6V (or Vth) while the diode is on because all of the current at the output of the op amp is going through the diode so it can't drop a voltage across anything else until it gets to through the diode (where it drops the input voltage across the resistor on the negative input of the op amp).

And because the diode is effectively a short when it is fully on, it would be near impossible to even sample a little of it to develop an output voltage because a short circuit in parallel with any resistance is still a short circuit.

Does this make sense?

And thanks to anyone who made it through that mess...

Jay Doyle

[johngreene]

Because the input signal multiplied by the op amp's gain was sufficient to cause the diodes to conduct in the first place, we know that the diodes will always be on. But we also know that they CAN'T be on because that drops the gain of the op amp to one, lowering the output signal sufficiently to take the diodes out of conduction (and therefore the circuit) which brings us around to the beginning, where we know that the gain of the op amp times the input signal is sufficient to bring the diodes into conduction.

So when the diodes are on, their effective resistance drops to zero, dropping the gain to one and turning the diodes off. This drops the signal level at the op amp's output, bringing the diodes out of conduction, raising their effective resistance, and therefore raising the gain back up, which brings the op amp output signal back up to a level that turns on the diodes, which lowers the diodes effective resistance, lowering the gain, bringing the diodes out of conduction, which raises the effective resistance of the diodes, which raises the gain, which causes the diodes to conduct, which lowers the gain, which brings the diodes out of conduction, which raises the gain, which brings the diodes back into conduction... Etc.

Whew.
Close but not quite right. Once the diodes start conducting the gain is effectively one for all signals where Vin * gain > .6V. Lets say you have an input signal of .01 volts and a gain of 100. With no diodes the output would be 1V. With the diode in the feedback loop any input voltage less than .6/100 = .006V experiences a gain of 100. Any input voltage above .006V experiences a gain of 1. So with a diode in the feedback loop and .01 volts on the input the output will be: (.01 - .006) + .6 = .604V. Even with an ideal diode.

Maybe it is the unity thing that is throwing you off. Try thinking about several diode networks in the feedback loop. The diode pair with the lowest Vf has the largest resistor in series with it. Now you have an amplifier with a stepped gain response and several break points (one for each diode pair). You can create any degree of 'soft knee' you want this way by adjusting the series gain setting resistors and choosing diodes with appropriately spaced Vf voltages. Or use the Vbe multiplier that was mentioned earlier and set the Vf's to where ever you want.

--john