help me understand how this sustainer (supa sustain) works?

[Derringer]

here's the shem I'm looking at
http://www.generalguitargadgets.com/diagrams/cssussc.gif?phpMyAdmin=78482479fd7e7fc3768044a841b3e85a



the sensitivity pot sets how much signal the 1st gain stage sees
Q1 is the heart of the first gain stage with a power-supply-filtering? arrangement I haven't seen before.

Why is that voltage divider there (82k + 220k) with the 10uf cap to gnd? Is it just filtering the supply a little more?
The first stage offers a lot of gain with a 4k7 emitter R and about 300K on the collector.

The base of Q2 is DC coupled to the collector of Q1 and creates another gain stage, employing a NP bypass cap on the emitter.
Does this bypass cap need to be bi-polar?

There is some negative feedback coming from Q2's emitter to the base of Q1 ala a fuzz face design. Is this just to give some distortion or does it have another use in this circuit?

And then we have what I think is positive feedback coming from the collector of Q2, passing through another NP capacitor (does it need to be bi-polar here too?) passing through the LDR and a limiting resistor 100R,  back to the emitter of Q1.
I am guessing that this is where the 'sustaining' comes from?

And then I get really confused. The output of Q2's collector is AC coupled to the base of a darlington pair. This 0.047uf cap with the 220K to GND also creates a LP filter. The signal goes through the HP 0.047 filter through another resistor and then out through the volume pot.

But back to the darlington pair that powers the LED.
How does this power the LED?

I'm guessing that the amplitude of AC signal is on the base of the darlington pair controls the brightness of the LED.

But I'm also imagining that the lower the AC signal, the brighter the LED needs to be so that more positive feedback is produced?



Thanks for any info you can give me

-Bill

[earthtonesaudio]

You've got it mostly right.

Initially (no signal), the darlington is biased off (0V on base), the LED is off, and the LDR is at max resistance.  When the output of Q2 goes high, then positive (AC) current enters the darlington and turns the LED on, which reduces the resistance of the LDR and lowers the gain.

Conversely when the output of Q2 goes low, nothing happens.  This is a half-wave system.


The first stage just has some power filtering, yes.  The bypass cap on the second transistor can be polarized or not, doesn't matter.


The .047/220k is not much of a filter.  For simplicity I'd just call it AC coupled for all (guitar) frequencies.

Finally, I would call the LED/LDR feedback "negative," not positive.

[Derringer]

how is the LED/LDR feedback negative though?

The phase of the signal at the collector of Q2 is equal to the phase of the signal at the base and emitter of Q1.

But I don't really know what happens when a signal is injected into the emitter of Q1 there.


thanks

[earthtonesaudio]

how is the LED/LDR feedback negative though?

The phase of the signal at the collector of Q2 is equal to the phase of the signal at the base and emitter of Q1.

But I don't really know what happens when a signal is injected into the emitter of Q1 there.


thanks

The emitter is a non-inverting input as well as an output.  Signal in Q1's emitter comes out the collector, where it goes into Q2's base, is inverted, and out Q2's collector, where the process repeats.  Due to the (single) inversion that takes place in Q2, the effect of this DC feedback loop is to decrease the gain (compared to open-loop).

Also, when the output of Q2 turns on the Darlington/LED section, it reduces the resistance of the LDR, which reduces the gain.  So a BIG output signal makes the gain small, and a small output signal allows the gain to remain BIG.

[Derringer]

awesome!

thank you !

[neldom]

Hey Fellas,
Just wondering if anyone would have any ideas what I could check here.
I just got one of these sustainers for free, because of course it's not working.
When you engage the pedal there is a very noticeable volume drop and of course only the output pot has any effect.
You really have to hammer on the strings even with humbuckers to get the led to light, and get the voltage up over 0.5v.
I have checked Q1 and Q2 and the voltages seem okay with the base about 0.5V higher than the emitters.
Just wondering if you know what a fellow might check next?
Thanks.

[Earthscum]

Neldom, close the case, lol... I'm kinda kidding. Try hitting the strings and putting your thumb over the LDR to block the light. With the light blocked, you should have a smaller amount of feedback (higher resistance), and thusly a larger signal. This is what I would check next. After that, if still no, maybe check the LDR. Pull it out and try running a signal... you should have a ton of gain. That means the LDR died. If not, then signal probe each transistor with the LDR out to see where the signal chokes off at.

[deadastronaut]

@derringer...ive seen this around for years...i almost bought a battered up old one about 10years ago... but never did.....does it really give good sustain...is it like a souped up compressor or something?..

in short, what's it like?..

[neldom]

Thanks I will try that. I did remove the LDR and the resistance was changing from anywhere between about 40 and 400 ohms depending on the proximity to light (I never tried in absolute darkness) So it does seem to be working.
I will see what I can probe on the transistors.