Diode substitution question

[thelonious]

i asked a friend at work to give me some 3mm red leds but the ones he gave me have a clear lens so i asummed they where clear ones. well im shocked when i tried it with the meter in diode test mode it lights up red.
how does that happen?

Didn't see anyone else answer your colo(u)r question yet. When a LED is biased on, electrons from one side jump to fill "holes" on the other side, and as they do, they release energy in the form of light. The color of that light depends on what semiconductor material is used inside the LED. A red LED actually produces red light regardless of whether the case is clear or red... usually the case is dyed to diffuse the light and make the LED still look its pretty color when it is biased off.

[thelonious]

i asked a friend at work to give me some 3mm red leds but the ones he gave me have a clear lens so i asummed they where clear ones. well im shocked when i tried it with the meter in diode test mode it lights up red.
how does that happen?

Didn't see anyone else answer your colo(u)r question yet. When a LED is biased on, electrons from one side jump to fill "holes" on the other side, and as they do, they release energy in the form of light. The color of that light depends on what semiconductor material is used inside the LED. A red LED actually produces red light regardless of whether the case is clear or red... usually the case is dyed to diffuse the light and make the LED still look its pretty color when it is biased off.

[Mike Burgundy]

@Midwayfair: my question still holds - the way I understand the circuit, Fv determines the threshold. Lower Fv, lower threshold - so compression kicks in earlier, at a lower level input. Higher Fv needs more input voltage for the diodes to turn on, charge C10 and start compressing. So that's a actually less compression, a larger part of the envelope is unaffected. That is why I asked for some more explanation on highFv = high comp, cause I'm thinking in exactly the opposite direction.

[midwayfair]

@Midwayfair: my question still holds - the way I understand the circuit, Fv determines the threshold. Lower Fv, lower threshold - so compression kicks in earlier, at a lower level input. Higher Fv needs more input voltage for the diodes to turn on, charge C10 and start compressing. So that's a actually less compression, a larger part of the envelope is unaffected. That is why I asked for some more explanation on highFv = high comp, cause I'm thinking in exactly the opposite direction.

I couldn't tell you the electronic reason for it. Does it make a difference that it's creating a negative voltage swing, not positive?

What I do know is that I tried many different diodes, and the higher the Fv, the more extreme the compression. The voltage swing with a Schottky maxed out at about -2v with single coils; with 1n4001 it was several volts. I can't remember which thread it was (or even which forum), but the question of substituting diodes in this circuit came up recently elsewhere and someone else mentioned right off the bat looking at the circuit that they thought 1N4001s would get crazy. That backed up what I'd already observed.

I do my best with theory, but most of the time it's brute force experimentation. I'm sure someone like RO Tyree, PRR, or RG Keen could tell you exactly why. Or they might tell me that what I observed was not what I think I observed. :shrug:

[Kipper4]

I've bought some 1N60
just 6 so i have some in stock for future projects

[ashcat_lt]

More voltage into the gate of the transistor means less gain from the opamp means more extreme compression.  With higher forward voltage in those diodes and more gain into them to make them turn on you get more compression.  I'm having trouble figuring out how else to put it.  The thing is built specifically to compress more when the voltage through that loop is higher.  You are talking about purposely increasing that voltage.  What outcome are you expecting?

Like I said before, you could increase the gain, use si diodes, and then divide the voltage back down before the transistor ti get it back to "normal" operation.  Or you could just use a shottky.

Edit - yeah, this voltage is all negative, so when I said "more" or "higher" I guess it should be "less" or "lower" or "more negative", but the point is the same.

[Mike Burgundy]

Makes sense - but: those diodes do not conduct until there is enough voltage potential there. Then the start charging the cap, on the peaks of said voltage, and only on the peaks. All energy below Vforward is blocked. The *charge on the cap* is what drives the gain control, doesn't matter if you use the negative or positive side. So if you *increase* the forward voltage of the diodes (move to regular Si), they turn on "later" and for less of the waveform. Doesn't the cap get charged *less* that way, reducing the amount of, well, reduction?  
I think I can see the ratio changing if you up the Vforward and compensate with more gain, on the same principle. Even though you force the diodes into conduction at the same point of the waveform, there is more power going into the cap (more area if you graph the voltage) so differences will be enhanced > ratio goes up.