Ampeg Scrambler : trying to understand how it works !

[ricothetroll]

Hi !

I just love that fx, but I never fully understood how it's internally working !
http://www.freeinfosociety.com/electronics/schematics/audio/pictures/ampegscrambler.gif

Q1 is a buffer.

Q2 looks like some kind of emitter follower with a feedback or retroaction from the diode bridge.

The diode bridge (with the texture pot at its center) reminds me of some diode ring modulator (http://upload.wikimedia.org/wikipedia/commons/c/cd/Ring_Modulator.PNG), that would explain the ringy sound of the unit when the texture pot is turned clockwise.

Q3 and Q4 seem to form a differencial amplifier.

How does that all work together ? Any clues are welcome !

Thanx in advance !

Best regards.

Eric

[PRR]

The www.freeinfosociety.com direct link does not work. You have to sneak-up on it, drilling through menus, and I can't find the path.

I found this image and cleaned it up a bit:



> Q2 looks like some kind of emitter follower

But.... no signal is taken at emitter!?

It's a simple hi-gain common-emitter stage. With a fancy feedback path.

> The diode bridge ...reminds me of some diode ring modulator

Recall that two of the diodes shown in the factory plan are not really there.

> the ringy sound of the unit

Ring Modulators don't ring. ("Bell-like" tones are possible, but not this easy.)

D1 D2 are a rectifier. Half the wave goes through D1, the other half through D2.

D1 feeds a 0-10K variable resistor, D2 feeds a 8.2K resistor.

Now the paths come together, and couple back to the base of Q2.

Q2's gain is different one half of the wave or the other half. The D2 side gain is roughly 8.2K/2.2K. The D1 side gain is variable from 10K/2.2K to 0/2.2K. Near-equal one way, half-wave chop the other way.

> Q3 and Q4 seem to form a differencial amplifier.

But look what is going in. Positive half-waves go in the non-inverting input. Negative half-waves go in the inverting input. This is a full-wave rectifier.

The net result is a spray of even-order harmonics with adjustable flavor.

It helps if you abstract the building blocks and re-draw the topology:



The analysis shows the pot at half-turn, 5K versus the 8.2K on D2. That causes the high/low on alternate half-cycles. If pot is set to zero, half the wave goes away. Also the original fundamental goes away.

[Joe Kramer]

Now the paths come together, and couple back to the base of Q2.

At first I thought this was some sort of feedback as well, but after re-drawing the circuit several times, I have come to see otherwise.  I would now say, rather than a feedback network, the output of Q1 feeds not only the base of Q2, but also the bases of Q3/Q4 through the 8.2k/10 pot network.  In other words, the inputs of the diff amp see an inverted signal output from Q2, rectified through D1/D2, as well a non-inverted signal from Q1.  To follow all the signal flipping is a headache, but it's essentially a sort of cancellation/reinforcement scheme.  You could (as I have done) simply use a phase-splitter to arrive at the same results as long as you provide enough gain to the diff amp.  It's a confusingly drawn schemo, but take another look and see if you find the same thing I did. . . .

   

[PRR]

> also the bases of Q3/Q4 through the 8.2k/10 pot network

Not "through" the 8.2K/10K network.

You have caught me in a slip. Q3 Q4 are in parallel (not series) with the 8.2K/10K. Does that matter?

What is the base impedance of Q3 or Q4? We have 4V over 4.7K, call it 1mA, split 2 ways, 0.5mA per transistor. Hie is about 60 ohms. What is the hFE? Is that really a 2N5306?? It is another handle for MPSA14, or favoriter Darlington. At 2mA the hFe is minimum 7,000, typically far higher, and remarkably flat with current. Say 10,000. Then 60*10,000 is 600K base impedance. "No" shunting effect on 8K/10K. And even if the signal to the bases flowed "through" 8K/10K, no effect (2% drop).

I cling to my interpretation. Q2 has open-loop gain near 50. Closed-loop gain is limited by the 2.2K in front and the 8K/10K-0 parts to be 4 or less.

Note also that this connection gives "perfect diodes". The 0.5V diode drop is inside the loop and ideally vanishes. Because the loop-margin is maybe only 10, the residual drop is like 0.05V at the Q3 Q4 pair, near 0.01V referenced to the input.

[Joe Kramer]

Hey PRR, not trying to "catch" you or trip you up in any way, and certainly can't argue with your explanation of the circuit.  Thanks for that.

I came to my conclusion via breadboard tinkering.  I found in practice that, if you use a matched xstr pair, matched diodes, and adjust the 10k pot to exactly 8.2K (or just use two 1% 10k resistors) you arrive at a "null point."  At that point, the FWR effect is perfectly symmetrical and there is a very sensitive gating effect as well; anywhere outside this null point, the symmetry and gating effects begin to diminish.  After pulling things apart a bit, I found I could get the same results without "feedback," but using a phase-splitter to the diff amp.  Then again, maybe I should face the breadboard anew and take better notes this time.      

But if you wouldn't mind saying, what then is the purpose of the feedback?  The signal is rectified by D1/D2, summed via 8.2K/10k pot, AC coupled, and returned to the base of Q2 in order to do what?  Correct crossover distortion?

 

[ricothetroll]

Thanks a lot for those great answers ! Things are a lot clearer now for me.

It's an amazing luck for us (the "not so learned") that you guys take time to answer our questions...

So, the goal of the feedback is to get rid of the diode drop right ?

Best regards.

Eric

PS : sorry for the late answer but I've been absent a few days.

[Joe Kramer]

Okay, after a new round of breadboarding, I learned this: shut up and listen to PRR.   

Turns out my phase-splitter approach was only another form of feedback, since I was returning the rectified signal to the emitter.  This does work, but not in the way I assumed, nor as well as in the stock circuit.  So yes, I concur/concede, as PRR said, it's a feedback loop that offsets the diode drop.

BTW, another observation: feedback or not, the low-level FWR of this circuit is not good (i.e., symmetrical/accurate).  But this doesn't matter because the diff amp overloads so easily that practically any signal goes into full squarewave, obliterating non-symmetry.  Similar to the FWR in the MXR/Ross compressor, good for generating a DC control voltage, obviously the original intention of the circuit.  Using it for audio probably started as sort of a goof. . . .

       

[PRR]

> Okay, after a new round of breadboarding

Trying-it is the only "truth".

However sometimes the path to "truth" is not simple nor direct.

> the low-level FWR of this circuit is not good

No rectifier is perfect to zero signal. Low-level errors are the bane of "precision" rectifiers.

> this doesn't matter because the diff amp overloads so easily

Right. Didn't I estimate that? "near 0.01V referenced to the input".

And "to the ear" it is very reasonable for the harmonics to fade-out as level drops.

> good for generating a DC control voltage, obviously the original intention of the circuit.

I've never seen it used that way; but it IS interesting.

[Joe Kramer]

> good for generating a DC control voltage, obviously the original intention of the circuit.

I've never seen it used that way; but it IS interesting.

Well, maybe my terminology isn't proper, but that's what I'd say it's doing here (except for the feedback) with the diff amp FWR in the Ross/MXR compressor:

http://fuzzcentral.ssguitar.com/ross.php

And here, in a variation sans diodes, used to drive an opto device:

http://www.godiksennet.com/images/sch/CS1.jpg

At any rate, here the end result of the FWR is a DC voltage which is being used to control something, in these cases gain.  I've been thinking the same circuit block might make a good envelope follower CV for a filter, but I haven't actually tried that yet. . . .