Tom Scholz's Rockman Compressor with two release time constants

All about modern commercial stompbox circuits from Electro Harmonix over MXR, Boss and Ibanez into the nineties.
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VivMeLol
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Post by VivMeLol »

In patent 4627094 of December 2, 1986, Tom Scholz described a guitar compressor.

He also highlighted an issue with the release times of the compressor.

Tom described the ADSR curve of a guitar signal as follows:

"On a guitar, the first sound or pulse that comes out can be a huge peak which is almost always much stronger than the signal that follows after a few milliseconds."

Then he described the problem with his earlier versions of the compressor:

"The operation has not been found to be optimised....When a loud note is followed by a soft note, the transition does not follow (correct compression) curve exactly due to the time constants. The (compressor) tends to act as a fixed gain amplifier when a hard note is quickly followed by a weaker note; there is a tendency for the weak note to start quietly and ride up to the previous note volume"


If the compressor has one slow release time
Lots of hard notes will raise the voltage on the integrator cap which will reduce the gain,
Then if the next note is soft, but the integrator cap is fully charged and discharging very slowly, it will keep the gain of the compressor low even though the current note is low volume.
As the integrator cap slowly discharges, the compressor gain starts to increase and the weak note starts to get louder.

Basically the issue is that a fast release is needed initially, followed by a slow release. Ie a compressor with two release time constants. This would track the amplitude changes in guitar signals in a better way.

Which obviously means two discharge paths for the integrator cap.

Tom solved that problem very elegantly in the Rockman Sustainor and Ultimatum by adding various diodes in parallel to the integration capacitor.

Image

Here the output signal of an Opamp goes to R1 (which sets the attack time)

D3 rectifies the signal and charges integration Capacitor C1.

C1 normally discharges via R3

Tom placed diodes D4-D7 as the additional discharge path. He took advantage of the nonlinear VI curve of diodes to basically make a variable resistor, the value of which depends on the imposed voltage.

When C1 has a higher voltage, the diodes D4-D7 appear as lower resistors and discharge C1 at a fast pace.

When C1's voltage drops, the diodes appear to be higher value resistors, and C1 continues to discharge via R3 as usual.

This leads to 2 release time constants, a fast one initially and a slow one later on.

Image

The green curve is a sine wave input signal that starts at t=0.5s and stops at t=1.5s

The red curve shows the exponential discharge of C1 in case there were no D4-D7. It is basically a discharge via R3 with time constant R3.C1

The blue curve shows the discharge of C1 with D4-D7. We can see around t=1.5 till 3.5s, the blue curve shows faster discharge than the red curve. That is because initially C1 is discharging via D4-D7 since they appear to be low value resistors.

However after about t=3.5s, the time constants of the blue and red curves appear to be similar. That is the zone when the diodes D4-D7 appear to be high value resistors.

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Post by mr coffee »

VivMeLol said:
The blue curve shows the discharge of C1 with D4-D7. We can see around t=1.5 till 3.5s, the blue curve shows faster discharge than the red curve. That is because initially C1 is discharging via D4-D7 since they appear to be low value resistors.

However after about t=3.5s, the time constants of the blue and red curves appear to be similar. That is the zone when the diodes D4-D7 appear to be high value resistors
Your analysis is very good, with the exception of one minor point. Actually, the resistance of D4-D7 is very low when conducting, and quite high when not conducting, forming a knee in this high resistance RC circuit. The resistance of the diodes doesn't change significantly in this circuit because the resistance of the whole release RC is so very high. The shorter time constant of the initial release is due to the fact that D4-D7 are in series with a 1 megohm resistor, which is 1/10th the value of the 11 megohm resistor which sets the slower release TC when D4-D7 are not conducting.

Now whether the capacitance of the LED (D4) has any effect on charging C1 from the very brief, but high level initial transient, well, that is beyond my skill level to guesstimate. LEDs are not like the IN914 signal diodes with very low capacitance and abrupt knees. Depending on the LED, the junction capacitance can be fairly high and possibly significant above 1.8 volts where the signal diodes turn on and put the LED capacitance in the RC circuit. :scratch:

And determining that won't be done with a simulator, ... and it would take some pretty sophisticated testing. But who cares? If you like the way it sounds, just do it Tom's way!

Nice post! :applause:

mr coffee

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Post by VivMeLol »

Thanks for your comments.

There are indeed two ways to look at this

A. The diodes are switches and they change abruptly from high resistance mode to low resistance mode, so they abruptly switch in or out the 1M resistor

Or

B. The diodes act as voltage controlled resistors and gradually shift from high resistance to low resistance modes. The 1M to there to control the minimum resistance of that leg when the diodes are conducting



It is easy to get the simulator to plot out the effective resistance of the diodes by plotting the R = V/I curve over time

Alternatively, we can work out effective time constant for first half second of discharge, and from that guess the effective R in the RC equation.

Let me get back to my PC to make these plots.

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Post by VivMeLol »

Image

Above is effective resistance of the diode string D4-D7.

The input audio signal stops at 1.5 seconds. At that moment, when integration cap is charged high, the diode string D4-D7 acts like a 5Megs resistor

The effective resistance of the diodes slowly goes up as the voltage across them drops. It is not an instantaneous switch.

At time 2s ie 0.5sec after the audio signal stops, the diodes D4-D7 have effective resistance of 17.5Megs whereas at 5s, the effective resistance is around 175Megs.

The diodes D4-D7 never have very low resistance since they are never driven very hard on, due to clamping effect of D1 and D2.

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Post by mr coffee »

The input audio signal stops at 1.5 seconds. At that moment, when integration cap is charged high, the diode string D4-D7 acts like a 5Megs resistor
Can you add some more details about how you are getting this "diode string D4-D7 acts like a 5Megs resistor"?

Also I can't read the grey numbers on your graph. Can you do high contrast black and white?

When you say, "integration cap is charged high", what voltage are you referring to?

Thanks for the interesting discussion. I have never spent much time thinking about the above-the-knee equivalent resistance of a signal diode because the knee is such a small voltage change (not instantaneous or hyper-abrupt as you say, but a very small voltage change). I have used the knee resistance change in quite low impedance circuits where the diode dynamic resistance is changing in the 50 to 500 ohm range. It can make quite a gradual change in resistance over a small voltage change relative to the nominal 0.6v threshold of silicon.

Have you actually measured this high megohm varying resistance of the diode string, or is this based on a simulator result? I haven't looked at 1N914 spice models, but I wasn't aware they actually model the device in the knee in such detail.

Thanks for sharing.

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Post by jhergonz »

thank you for sharing this. but can you share the images again? I think they're broken.

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Post by bmxguitarsbmx »

jhergonz wrote: 05 Oct 2022, 15:39 thank you for sharing this. but can you share the images again? I think they're broken.
I can see them. Try your pop up settings?

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Post by jhergonz »

bmxguitarsbmx wrote: 05 Oct 2022, 19:30
jhergonz wrote: 05 Oct 2022, 15:39 thank you for sharing this. but can you share the images again? I think they're broken.
I can see them. Try your pop up settings?
can see it now. Thank you.

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Post by Carvindc125 »

mr coffee wrote: 10 Aug 2022, 05:27


And determining that won't be done with a simulator, ... and it would take some pretty sophisticated testing. But who cares? If you like the way it sounds, just do it Tom's way!

Nice post! :applause:

mr coffee
The OP is not describing the way Tom did it. The OP uses SPICE to try and find the bias of the compressor by plugging in the values and outputs we found from X100s that Myself and a few others provided to him.
Look up JFET resistor divider biasing. That’s the first peace of the puzzle and the part the OP left out.
Once that’s understood then add the components in the op to the feedback loop of an opamp then you will have a potential compressor/limiter. But it won’t work like OP states yet. Because every Fet is different we need to manually tune the resistor divider network I spoke of earlier. To do it the way Tom did you use an oscilloscope and a tone generator. This process is documented on YouTube believe it or not in a random video.
To get a starting point you also need to know info on The JFET. Mainly the VGSOFF.
The process is not as simple as it seems and we found that SPICE allowed us to find a starting point but more times than not the circuit needs many component values tweaked before it has the characteristics of the Rockman Compressor.

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Post by Carvindc125 »

mr coffee wrote: 13 Aug 2022, 04:13

When you say, "integration cap is charged high", what voltage are you referring to?
Here is what he means.
I made this video back on 1-23-21
The OP did not believe yet that a compressor could have two time release constants.
In the video you can see the two release time constants. Hitting the guitar charges the cap to around -1.9 then it drops to -.9 in like 1ms that’s the first release the second release is much slower from -.9 down to 0 which is exactly what Tom explains is needed for the guitar. A compressor with two release time constants

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