Lower clippig treshold with LEDs?

[cordovaplata]

I built a MXR Dist+ with GE diodes. The voltage drop is quite low compared to LEDs. I want both (I set a DPDT to select both sets) have unity gain between those 2 sets. Diodes (LEDs) in parallel will low the voltage drop? I ask you this because if you put them in series, the voltage drop adds up. If I put them in parallel will the Voltage drop lower?

[Mark Hammer]

Although the temptation is there to think "Well if putting them in series is additive, then they must behave like resistors, right?", as far as I am aware, diodes do not behave in that manner...or else we would have seen pedals with quartets of LEDs in parallel by now.

The current essentially "looks" for the easiest path to ground from those available.  If you were to place a pair of Ge and a pair of Si in parallel, it would clip according to the forward voltage of the Ge diodes.

[cordovaplata]

Thanks for the reply. So, maybe using a 3PDT to add a low value resistor in the output rail to gnd will equalize the gain between these 2 sets of diodes? (shunting the output signal to ground)

[Perrow]

Use series ge diodes to raise the clipping threshold of those to about the same as the leds.

[Mark Hammer]

Thanks for the reply. So, maybe using a 3PDT to add a low value resistor in the output rail to gnd will equalize the gain between these 2 sets of diodes? (shunting the output signal to ground)

Two ways to approach it.  One is to "conserve" the signal under low forward-voltage conditions by inserting a fixed resistor between ground and the ground lug of the output level pot.  The switch would shunt that resistor with a straight-wire connection when switching over to the higher FV diodes.  Presumably, the added resistor would jack the volume level up when the clipping threshold was lower, and shunting it would be like turning the volume down when switching to higher FW diodes.

The complement is to stick a fixed resistor in series with the input of the volume pot, so that the pot acts like one of a higher value that has been turned down.  E.G. a 100k pot with a 22k resistor in front behaves like a 122k pot that can never be turned up past about 5/6.  When switching from Si (or any other type/combo with a higher FV) to Ge diodes, you again bridge the fixed resistor so that the maximum volume is higher.  They both work the same way and have the same effect.

[cordovaplata]

I took your ideas and suddenly I came with this:

Basically what this does is while Ge diodes are activated, the resistor (trimmer, actually) in series with the output is "jumpered", bypassing the resistance. When the LED pair are activated, the resistor override is deactivated, forcing current pass trough the trimmer (250k trimmer). I was thinking using a trimmer so it could be easier to achieve (by ear) unity. What you think? will this work? (In my circuit simulation it works)

[Mark Hammer]

Yup.  That'll work.  If it were me, I'd use a 2+2 set of Si rather than LEDs, simply because the Dist+ does not really produce all that much gain (stock is 213x).  You might end up with a bit of bite at pick attack, but nothing in the distortion spectrum at anything less than max gain.  Indeed, I imagine most of the audible clipping would not be from the diodes at all but from the chip itself.  Many have said on these very pages "I don't get it.  I don't have any diodes in there and it still distorts."

You might have more success with a lower forward voltage than what LEDs produce.  So, consider a pair of 1N914s in series with a pair of 1N34s.  Together, they yield a forward voltage of around 750-800mv, around half what LEDs do, but 3x what a set of GE diodes do.  Enough contrast to make a difference.

[amptramp]

The lowest LED voltages are in infrared LED's which are readily available in optocouplers.  You can get as low as 0.8 volts with some of them.  You could make some interesting effects by using the LED portion as a clipper and using the phototransistor elsewhere in the circuit where it would be turned on above the clipping level.  Dual optos would enable you to clip both sides of the signal in one device.

[Earthscum]

^^^ +1 on optocouplers. I fried the only one I had that was below 1V. I just measured an actual IR LED and got 1.3V. Also measured a PC123, same 1.3V. The red LED's I have all creep awfully close to 2. Everything else hits 2.25-3.5V, lol.

I've got some interesting results playing around with a H11AA1. It is basically a pair of LED clippers and a transistor. They will detect both sides of the waveform and use that to control the base of a transistor. The one I have in front of me measures a 1.3V forward voltage, both ways (both 1.331, precisely). One trick that seemed promising is setting up the transistor side (it has a pin for the base) as a gain stage and taking the resulting crossover distortion (what it basically ends up as) and mixing it at the end.

I know the last part is a bit removed from the Dist+, but if you run across any kind of optocoupler, don't hesitate to give it a shot. And then while you're at it, may as well see if (like Ron suggested) using the other half for some other purpose.

One last thing I just thought of... JFet clippers have a softer turn-on like LED's, if you have a spare JFet or 2 laying around, try them in place of the LED's. Gate is Anode (+), tie Source and Drain together for the Cathode (-). They only have a V_F of about .6V (like Si diode), so it won't be as much of a strain to actually clip them, still get some "wide-open-ness" over the Ge's, and the volume difference won't be quite so drastic. Also, Jfets are kind of like Ge diodes in the way that a J201 may have a slightly different flavor than a 2n5457 or an MPF102.

[cordovaplata]

I have access to iR leds. do you think they will give a good sound? what about transistors? (using the base-colector juntion)
thanks for your replies

[Earthscum]

The B-E junction is drawn similar to a diode because essentially that's what it is. A Si transistor B-E junction is a Si Diode, and a Ge B-E junction is a Ge Diode. That's why you can always rely on that diode drop. So, figure Si is going to sound like Si, Ge like Ge. JFets use a different process, and also create a diode junction, but with a softer clipping knee (enough to be audible in most situations).

As a side, if you look at the Green Ringer (R.G. has schems at Geofex), the original actually used a transistor as 2 diodes. The B-C junction, when reversed biased from the typical application, also forms a diode junction. So, when the base goes (.6V) higher than the Collector, it will start to forward conduct. NPN is like 2 diodes with common Anodes, and a PNP is like a pair with common Cathodes.

[brett]

Hi
the forward conduction "knee" with voltage is almost identical for ALL diodes. (maybe because it's *doping* that turns insulators like Ge, Si into semiconductors?)
In any case, the only thing that differs is the Vf. Relative to the size of the knee, the Vf of a Ge diode is low and the VF of a diode is high. Kirchoff's law says that two Ge diodes in a row will double the Vf and give somewhere around 2 "knees", preserving that characteristic shape the signal with "soft" Ge clipping.

I'm no expert, but I'm not sure about the trimmer idea.  The idea of equalising the gain with a trimmer might seem attractive initially, but I think the results will depend on what the device is plugged into, because you are making the output impedance very high. For example, to waste half the signal going into a tube amp (high input resistance), you'll need a trimmer of about 500k, whereas plugged into a typical stompbox you'd only need 100k (or less). You'll also get loss of highs if the next stage (or your cables) has significant capacitance (fairly common).

cheers

[Perrow]

I'm no expert, but I'm not sure about the trimmer idea.  The idea of equalising the gain with a trimmer might seem attractive initially, but I think the results will depend on what the device is plugged into, because you are making the output impedance very high. For example, to waste half the signal going into a tube amp (high input resistance), you'll need a trimmer of about 500k, whereas plugged into a typical stompbox you'd only need 100k (or less). You'll also get loss of highs if the next stage (or your cables) has significant capacitance (fairly common).

cheers

You can swap diodes with a spdt, with a dpdt you can parallell (or not) the feedback resistor of the opamp.

Edit: but that'll only get you even clipping, but you could swap in a resistor (trimmer) before the volume pot, 3pdt to do both?

[Earthscum]

Hi
the forward conduction "knee" with voltage is almost identical for ALL diodes. (maybe because it's *doping* that turns insulators like Ge, Si into semiconductors?)
In any case, the only thing that differs is the Vf. Relative to the size of the knee, the Vf of a Ge diode is low and the VF of a diode is high. Kirchoff's law says that two Ge diodes in a row will double the Vf and give somewhere around 2 "knees", preserving that characteristic shape the signal with "soft" Ge clipping.

So all diodes clip the same at all impedances, and in the same fashion? Just curious... I'm not an EE. That just seems quite different than info and opinions floating around everywhere else.

[cordovaplata]

That's what initially tought about increasing resistance=more output impedance but I think it's not a big issue since I always connect my pedal to a mixing board or my pc input (with selectable input impedance) through an amp simulator (guitar rig 4) then to the output (which I think is low impedance). If someday I plan to connect it to a real amp, I could easily connect a buffer before the amp. Remember that it (the high Z issue) only occurs in Si mode, not in Ge. I like more the germanium diodes anyway so silicon is in "special cases" where harsher distortion is needed.