Diode-based compressor / expander

[object88]

Within the last two weeks (maybe even just this last week), someone posted a link to a circuit on another website... and try though I have, I've managed to loose it.  The circuit involved one op-amp, one pot, and four diodes.  The pot swept between two sets of two diodes in a clipping arrangement in the feedback loop of the op-amp (like in a tube screamer).  I *think* that one set of diodes went to +V, and the other to ground.  The ultimate effect was that you could dial in a compression or expansion.  I can almost picture it in my head, but I would like to see it again for verification; can someone point me back there again?

Thanks!

[object88]

Found the page I was referring to:

http://www.audiodesignguide.com/doc/ac_logarithmic_amplifier.jpg

(I somehow left it up on my work machine...)  The circuit is an "AC Logarithmic amplifier".

[polaris26]

Neat circuit - but this will modify the waveform itself - the graph of output vs. input is instantaneous, not averaged over some time.

Dave

[Hides-His-Eyes]

It's a cool idea though; It would certaintly work as a kind of limiter.

I think you'd get a unique fuzz tone out of it if it went to clippers.

[cpm]

The bad news:
- Signal must be kept below the diodes threshold voltage for the exponential part to work. Also, the linearity of the diodes is important, i find Ge diodes to be the softer transition, other Si diodes may have a steeper curve.
- Diode junction can add thermal noise (hiss), which would aggravate with the low signal levels required.
- Diode conduction is not temperature stable, expect a noticeable drift in amplitude due to temperature changes.

[object88]

Thanks for everyone's feedback!

[Jazznoise]

A threshold of 0.7 volts is fine - especially considering guitar signals are so low anyway!

What if instead of the pot you used a VCR? But you'd need a way to feed it an RMS voltage of the signal plus a way to vary the length of the average voltage.

Could be done, just not by me. There's probably alot nicer ways to build a compressor, but that circuit has interesting applications! It could be a sort of super soft clipper.

[Steben]

super super sof clipper is not always nice.
Can sound muffled / LoFi. Reminds me of Germanium tube screamers... way to soft.
Tube-like tone is not about super super soft. Most Tube amplifier stages are in fact quite lineair when driven softly (below clipping). Of course the clipping is rather soft, but not really from 0V on. Dynamic compression comes from the sag.

[Hides-His-Eyes]

If the signal's under 0.7V how does it get through at all?

[Taylor]

I simulated it, seems to do the exponential thing well with small signals. Not sure about the soft clipping or whatever we want to call it, though. Seems to be subtle on that side of the pot's control.

I'll build it tonight and see if it's fun as a waveshaper - simple enough to give it a go. At least it would be a simple way to make exponential LFOs, which is something I've sometimes wanted.

[cpm]

posted in this thread:
http://www.diystompboxes.com/smfforum/index.php?topic=91213.msg777513#msg777513

thats one half of a sine through a log-amp, (diode in front)

[earthtonesaudio]

If the signal's under 0.7V how does it get through at all?

ideal: http://upload.wikimedia.org/wikipedia/en/thumb/f/fe/Diode_Modelling_Image11.png/300px-Diode_Modelling_Image11.png
real: http://openwetware.org/images/7/77/Realidiode.jpg

[Johan]

If the signal's under 0.7V how does it get through at all?

ideal: http://upload.wikimedia.org/wikipedia/en/thumb/f/fe/Diode_Modelling_Image11.png/300px-Diode_Modelling_Image11.png
real: http://openwetware.org/images/7/77/Realidiode.jpg

also, the circuit above uses the "-" input, wich can also be thought of as a current sensing input. so the opamp only does what it allways do, it's the behaviour of the diodes when you run a tiny current through them that decides the behavior of this circuit
J

[Steben]

If the signal's under 0.7V how does it get through at all?

The given "standard" treshold voltage in datasheets is always a compromise. There is not such a thing as "on" or "off". Even at tiny bits of current there will be a tiny voltage drop.
There is a zone where large variations in current result in only small variations in voltage drop. This happens mostly around 0.6-0.7V with a silicon diode. That's why they speak of a "treshold". But keep in mind that some diodes designed for very large currents will result in a drop 1.0V or more when given those large currents.
. A diode will always have a certain amount of impedance. It is common to see this as zero impedance as long as other impedances determing the circuit are much larger. But in some cases it is relevant indeed.

Because of this non-ideal patern, diodes can be used as musical sounding clippers. Since there is always an amount of voltage change with a current change, even small, the result will never be flat cut signal, resulting in less harsh harmonics.

[teemuk]

The easy way to make an expander with diodes is connecting two antiparallel diodes in series with the input resistor of an inverting amp. With input signal threshold below the diodes' Vf the gain is determined by the inverting amp's input and feedback resistors (as usual in inverting amp), when Vf threshold is exceeded the decreasing impedance of the diodes in the input will begin to increase gain. Crude and simple.

I believe this is what the example circuit was trying to do but I don't see how it could work properly if signal below diode Vf can't pass through - this appplies both to input diodes and to diodes in the feedback loop. They both require parallel resistor for gain definition at signals belov Vf.

The "logarithmic" compressing amp is nothing but an ordinary diode clipper.

[Steben]

if signal below diode Vf can't pass through - this appplies both to input diodes and to diodes in the feedback loop.

Again there is always some current passing through. The diode acts as a very very very very large resistor.
Since the feedback resistance is equally very very very large, gain is likely to be around unity. But ... with a lot of noise.

[sault]

Again there is always some current passing through. The diode acts as a very very very very large resistor.
Since the feedback resistance is equally very very very large, gain is likely to be around unity. But ... with a lot of noise.

So to reduce noise and increase signal pass-through you'd want resistors parallel to the diodes.

Adding parallel resistors would decrease the effectiveness of your pot, to a certain point, would it not?
For instance, adding 1k paralleled across each set of diodes would mean  Ri 1k -> 48k pot -> Rf 1k

So

Rf 36k / Ri 12k = 3x    ... becomes ...  Rf 37k / Ri 13k = 2.8x
Rf 12k / Ri 36k = 0.3x   ...  becomes ...  Rf 13k / Ri 37k = 0.36x

After a bit of calculator work, the difference becomes more pronounced at higher values of resistance, ie at
10k resistors

Rf 36k / Ri 12k = 3x   ...  becomes ...  Rf 46k / Ri 22k = 2.1x
Rf 12k / Ri 36k = 0.3x   ...  becomes ...  Rf 22k / Ri 46k = 0.48x


If I'm understanding this correctly, this means that the signal below Vf would be partially linearized...
.... wouldn't this also mean a "softer" clipping?


Saul t

[cpm]

look AN174 (NE571 application note) for a controlled compressor/expander) figure 18

[Steben]

all this can results in softer clipping yes, but there is no reason not to search for slightly more complex combinations of resistors and diodes only in the feedback loop to achieve the same with less calculations.
The response/clipping will be easier to design.