Matching Transitors for Ross Compressor Clone?

[koa-dogg]

I am planning to build a Ross Compressor Clone and wanted to know wether or not there is a need to match the Transistors (2N3904). If so, How does one go about matching them?
Thanks,
Scott

[Jay Doyle]

No need to match them really. 2N3904s should be fine.

If you want to match them, use a multimeter with a hFE test on it and find 5 with near the same gain (over 100).

If you want to match the ones that appear to need matching the most, match the ones in the rectifier that share the 150k collector resistor, but that isn't all that important because they are running at such a high gain.

If you want to spend a lot of time trying to hear slight differences in tone and/or noise, audition a number of transistors for the input buffer transistor and the transistor that goes to the level pot. But again, this may not be worth your time; changing resistors to metal film in the right places will get better results in less time.

There is a lot of hype and are some "mojo" myths that appear in builders write ups about the Ross Comp. Don't believe everything you read, especially if someone is trying to sell something.

Search the forum for my recent posts, one linked to an old write up I did on getting the best performance and lowest noise out of the Ross circuit.

Jay Doyle

[Mark Hammer]

All compressors use an envelope follower circuit of some kind so that changes in input signal level can translate into appropriate changes in attenuation or gain in the compressor circuit to maintain a relatively constant, or at least less dynamic, output signal level.

Envelope followers can attempt to "track" the signal level by shaving off one half cycle and, in effect, measuring the amplitude from the midpoint of the waveform to *either* the negative or positive peak as it progresses.  This is referred to as "half-wave rectification".

Unfortunately, there is the matter of what to tell the gain cell or control element during those brief, but long enough, periods between each little peak/blip on the oscilloscope.  At low notes, especially, the up and down transition between the tops of half-cycles and zero, is long enough to be heard as a very very fast tremolo or burr as the note decays.  

Fortunately, if you use both negative and positive halves of the input waveform to measure amplitude, the little momentary dips between blips only last half as long (i.e., doubling the frequency), which can often be enough to make them much less audible or objectionable.  This "full-wave rectification" is a more complex circuit, so less often used, but it pays dividends in terms of less envelope ripple effect.  

There are a number of ways to do it, and the Dynacomp/Ross/Pearl/CS-2 circuit is one of them.  Looking at transistor Q2 in the Tonepad "Compreross" layout, it is sort of hard to tell, but it is used as a phase-splitter.  When a bipolar transistor is implemented with equal value resistor on the emittor and collector, what you end up with are equal and opposite phase versions of the same signal at the collector and emitter outputs.  You can see in the Compreross that the emitter and collectors of Q2 each go through a .01uf cap to the base of another transistor.  Although there is a diode to ground oriented in the identical fashion in each of these transistor subcircuits, remember that the signal each one carries is 180 degrees out of phase with the other, so in reality they are each carrying the opposite half-cycle of the input signal.  The outputs of the Q3/Q4 collectors are summed together so that now we have an envelope signal which is comprised of the input signal "folded over" - a full-wave rectified signal.  The briefer and tinier dips between blips are smoothed out a bit by the 10uf cap to ground so that what Q5 sends to the 3080 to control its gain is relatively ripple free.

Incidentally, full-wave rectification is often what is used to produce octave-doubling.  Look at the Foxx Tone Machine, Univox Superfuzz, Dan Armstrong Green Ringer and Fender Blender, and you'll see something analogous to Q2, with equal value resistors on each end of a bipolar transistor and their diode-processed outputs combined to yield twice the frequency.

Why am I telling you this?

Simple.  If the Q3 and Q4 subcircuits are not perfectly matched they will *still* provide a full-wave rectified envelope signal, albeit with potentially a little more ripple than a perfect FWR circuit (because one half-cycle is not the same amplitude as the other) but still much less than a half-wave rectifier circuit.  To be "perfect" of course, you'd probably want to assure not only that the transistors matched, but that the 1M resistors and diodes to ground matched, and that there were trimmers on Q2 to assure that collector and emitter outputs were truly identical in amplitude.

So ultimately, while it could *potentially* make a difference in perceptible ripple (which many people "hear" as distortion), practically speaking it is lost amidst so many other sources of a less than perfect envelope signal.

Bottom line: Jay is absolutely correct.  No pressing need to match them.

[R.G.]

Bottom line: Jay is absolutely correct. No pressing need to match them.

You guys just don't understand - or have not succumbed to MBA disease. The pressing ened for matching them is the desperate need for *advertising copy* that moves the products. 8-)

Necessity is not the mother of invention. Desperation is the mother of invention.

[Mark Hammer]

Jimmy Carl Black was one of the mothers of invention too (and the "Indian" of the group).  Wonder what he's doing now?   :lol:

But okay, okay, I will yield to destiny.  Ladies and gentlemen, I command you to buy.....

stuff

....especially stuff that is the latest stuff or most frequently advertised stuff or stuff touted as really good stuff by people who have a lot of stuff.   Mmmmmm, stuffff (insert Homer Simpson trance-like drool here).  Yeah, stuff is good, especially when you buy it, although I hear tell stuff is even better when you're selling the stuff and other people give you stuff (like money) for your stuff.

Stuff, hard to beat.

There, are you happy?  :wink: and am I the only one to recognize its the Ides of March today?

[Russ]

I used unmatched 2N5088 trannies (per RG Keen at geofex.com) in my Ross comp clone, and it sounds great.
Russ

[Dan N]

This forum is so cool!

[DaveTV]

Jimmy Carl Black was one of the mothers of invention too (and the "Indian" of the group).  Wonder what he's doing now?

And Ian Underwood was the straight guy in the group. He's doing lots of stuff now.

[Jay Doyle]

Mark Said:

If the Q3 and Q4 subcircuits are not perfectly matched they will *still* provide a full-wave rectified envelope signal, albeit with potentially a little more ripple than a perfect FWR circuit (because one half-cycle is not the same amplitude as the other) but still much less than a half-wave rectifier circuit. To be "perfect" of course, you'd probably want to assure not only that the transistors matched, but that the 1M resistors and diodes to ground matched, and that there were trimmers on Q2 to assure that collector and emitter outputs were truly identical in amplitude.

Your write up was spot on but I think that it misses one main aspect of the Ross/Dyna circuit and that is the integration of the rectified signal by the 10uF cap after the FWR. With this in mind, it is easy to see why the components do not need to be matched.

What that cap does is to take the "folded over" signal from the FWR and completely remove any ripple by quickly discharging the cap at the onset of a signal and then slowly recharging it when no signal, or a signal of less strength than the original, is present.

Taking a closer look, lets consider what is happening at rest. First off, the gain of the 3080 OTA, is highest when the current into its control port (pin 5), or Iabc (which stands for amplifier bias current) is at its highest. Since the current is coming through a resistor, the highest current available to us in when the signal on the other end is at V+ (yes, there is a transistor driving this resistor, but the situation is the same, we can ignore that transistor for the purpose of this explanation), and the gain is at it's lowest when the signal at the other end is at ground. In between, the gain fluctuates depending on the amount of current into pin 5. Because current is voltage through a resistor, and we always have the 27k resistor attached to pin 5, we can talk about voltage at the other end of pin 5.

Now, in the Full Wave Rectifier (FWR) section, as Mark said, we have two transistors being driven by equal, but opposite phase signals. The diodes on the bases of each transistor do two things. First, again like Mark said, they shunt half of the signal to ground; but as the signals to each transistor are out of phase, one is shunting while the other is amplifying and vice versa, and again because they are out of phase, one is amplifying the positive swing, the other the negative swing, of the signal into the effect. Thus, we have a full wave rectifier.

The other reason the diodes are there is to keep the transistors on when no signal is present. Think about the situation when no signal is present. Leakage current through the base of the transistor, keeps the bias at the base at .6V due to the voltage drop of the diode. This keeps the base .6V above the emitter, keeping the transistor on without a signal present. But it is just barely on, the only direction the transistor can go is to turn on harder, allowing more current through the collector and thus lowering the voltage at the collector.

Because the transistor is just barely on, the signal on its collector is sitting at close to 9V when no signal is present. This makes complete sense as without a signal, a compressor is running at its highest gain. But more importantly, the cap is charged through the 150k resistor to be at 9V.

Now let's apply a signal. We apply a large pulse, which drives the transistor into saturation, turning it on completely. The collector of the transistor is now at ground (probably a little above but close enough). What happens to the cap? Because it was at 9V, it "wants" to discharge its current (which kept it at 9V) to match the voltage that it is now presented (which is the signal at the collector). It cannot discharge through the 150k resistor, it cannot discharge through the transistor attached to the SUSTAIN pot (current doesn't flow that direction); the only path it can discharge is through the collector-emitter paths of the transistors in the FWR. The time this takes is very short, the resistance of the c-e path is very low, and because two are parallel even smaller still, which means a lot of current can pass through this channel and drain the cap very quickly. What happened is the gain went from its highest when the cap was at 9V to its lowest with the onset of the signal very quickly. This is the compressor's "attack".

Once that large pulse is removed, the base goes back to .6V and the collector returns to 9V (or close enough). Now the cap needs to charge to the voltage at the collector. The path for current to charge the cap is through the 150k resistor on the collector. It takes about 2 seconds to charge the cap up to 9V. Here the gain of the compressor went from it's lowest when the cap was at ground, and then the gain slowly climbs back up to it's highest as the cap charges back up to 9V. This is the compressor's "decay".

With the above in mind, consider a "real world" situation by considering an input of a sine wave, instead of an on/off pulse. If we consider the sine wave as starting at zero, it then goes upward. As the sine is going upward, the voltage at the collector of whichever transistor is amplifying the positive phase moves toward ground at the rate of the sine times the gain of the stage. The cap will charge or discharge to whatever voltage it is presented with at the collector. So the larger the sine wave, the more the collector of the transistor moves toward ground, the more the cap discharges, lowering the gain of the compressor. This all happens nearly instantaneously as the attack is very quick.

Once the sine wave reaches its apex, it starts going back down towards its starting point. Without the presence of the cap, the gain would start going up again just as fast as it went down, essentially following the signal. But with the cap, as the signal goes down the cap can only try to follow the signal, and the cap can not do a very good job because it charges very slowly through the 150k resistor. With a two second decay, the cap barely starts to charge again before the sine wave dips "below" zero.

Once the sine dips below zero, the first transistor shuts off and the other transistor takes over. This is the "folding over" of the input signal Mark talked about. One transistor handles the positive half of the input signal swing, the other the negative half, with both working the same in unison with the cap. So, the other transistor starts at zero and moves upward to 9V attempting to reduce the gain by charging the cap. But because the resistor on this transistor's collector is the same 150k resistor as before, nothing changes. The cap is still slowly charging as the sine wave goes even lower, which, because the sine is folded over, causes the base to go higher, which in turn causes the collector of the transistor to go down toward ground again. Somewhere along the travel of the collector, it passes the voltage that the cap has charged to and discharges the cap again, raising the gain which did not have much of a chance to fall anyway.

As the signal fades, the cap continues its seesaw up and down, but the down (decreasing the gain) does not go as far; the signal strength is not enough to drive the collector all the way to ground.

The point I am getting here is that while the attack is fast, the decay is very slow, so slow that any little differences in the two transistors is smoothed out by the cap. If one can't pull all the way to ground because of less gain than the other, it doesn't matter because when the other one pulls to ground the cap wouldn't have had enough time to charge again to V+, in fact it probably wouldn't have charged much at all. The other transistor would take over, discharge the cap completely and the cap would start its charge/discharge process all over again. This would cause ripple but because all guitar signals happen a lot faster than 2-second intervals, you would not notice it.

The short of it: the cap not only smoothes out the ripple in the control signal, it smoothes out any differences between the transistors as well.

This is hard to explain without using graphics, if there is something that you do not get please ask and I will try to answer the best I can.

Hope this helps,

Jay Doyle

[Mark Hammer]

Hey....Hey!.....HEY!!!  That's MY soapbox, Doyle!  Move yer damn lemonade stand somewhere else!  Yer stealing my customers!!

:lol:

Seriously, nice work.  This is now officially stamped as a really good thread for anyone who wants to understand the inner workings of this category of compressor.  It may be drawn differently, but you will see this same rectifier circuit crop up again and again and again.  Learn it, understand it, master it, control it.  You'll be glad you did.

Props to JD for a nice explanation.

[Jay Doyle]

Hey....Hey!.....HEY!!!  That's MY soapbox, Doyle!  Move yer damn lemonade stand somewhere else!  Yer stealing my customers!!

Very sorry Mr. Hammer, I am moving to another corner now...  

But before I get off of the soapbox, I find it amazing (and enlightening) that those who sell clones of the Ross don't seem to understand the inner workings of the effect they are selling! At least that is how it appears to me by reading all of the marketing hype...

Jumping down and giving Mr. Hammer his spot back.

Jay

PS - Between you and me Mark we could have three long winded posts each and end up with a thread that is word for word longer than any of the moral/cloning/ethics threads.

[Peter Snowberg]

PS - Between you and me Mark we could have three long winded posts each and end up with a thread that is word for word longer than any of the moral/cloning/ethics threads.

....And MUCH more fun to read!

Kudos to you guys!



Take care,
-Peter