| Author |
Topic: Diode clipping mod, Ge and Si in parallel? |
Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 15 Jul 2018 12:15 pm |
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I have drawn a basic four-stage softclipper circuit with common/GND offsets for the actual clippers. It is for +/-12-18Volt supply, but can be used as base for circuits using any voltage. Haven't added any component values, as they depend on supply voltage, etc.
Think that v4 = +.5V, v3 = +.25V, v2 = +.12V and v1 = +.05V (v4' to v1' are same values but negative).
Capacitors are there to keep these values stable, and conduct signal to common/GND via diodes and resistors.
The stronger the signal out of the OP AMP, the more diodes conduct. Given that 1N4148 diodes break down at about .65V (+/-.05V), clipping will take place at about .15V, .4V, .53V and .6V – for both positive and negative cycle, with more diodes with resistors adding load on the resistor on the OP AMP's output.
As the clipping is dampened by the resistors in series with the diodes, and the various diode/resistor stages start clipping at different levels, the effect with well-balanced components is that of a softer clipping than with regular one-stage diode clippers. As more diode/resistor stages can be added, more "fine-grained" clipping levels can be introduced for even softer-sounding clipping. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 15 Jul 2018 2:18 pm
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Great diagram Georg. I understand now what you are talking about. It is a totally different arrangement of the resistors and diodes than I was thinking about. What I was describing was not hooking the diodes and resistors to the power rails, but to the common ground.
From your diagram, it appears the resistors and diodes are hooked to the the plus power rail, and the minus power rail. I see the capacitors go to signal ground, not the power ground. Is that correct?
I don't understand what you said below?
"Think that v4 = +.5V, v3 = +.25V, v2 = +.12V and v1 = +.05V (v4' to v1' are same values but negative)".
Also, could you explain the math that you figured the below out with?
"clipping will take place at about .15V, .4V, .53V and .6V – for both positive and negative cycle" |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 15 Jul 2018 3:34 pm
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Keith,
Yes, capacitors go to signal ground, to short clipped signal and stabilise the offset-voltages. Playing with small values for those capacitors will affect clipping vs frequency, and offset-stability, which may result in interesting distortion sounds/effects.
As you can see/read from the basic diagram and description, in my example the positive offsets affect negative signal-swing, and negative offsets affect positive signal-swing.
The resulting "math" is that +.5V ("v4" offset) -.65V (diode's approx voltage drop) = -.15V (negative signal-swing needed to make diode in that stage conduct). Same for "v4'" offset, but then +.15Volt positive signal-swing is needed. And so on for the other clipping stages, that gradually need slightly greater signal-swing to make their diodes conduct.
The result should be a slightly more sinus-shaped clipped signal, rather than the more square-shaped seen and heard from most clippers. Adding more clipping stages into the "cascade", allows for improved sinus shaping and "softer" distortion. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 15 Jul 2018 5:40 pm
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Georg, I have been looking at the circuit diagram, and have some questions. Over on the right hand side, there is a line coming from the plus power rail, and one coming from the negative power rail. Where these two lines come together, there looks to be a gap where the plus and minus rail does not touch the signal output line. Does the positive rail touch the negative rail at this point? If they don't touch, then why did you draw the two lines like they might be coming together?
If they do come together this would be a direct short and result in smoke.
In looking at the direction of the diodes, the positive rail will conduct to the signal output line of the op amp.
Likewise the negative rail will conduct to the signal output line of the op amp. George, this is positive DC electricity meeting negative DC electricity at the point where they are connected to the output of the op amp.
Georg, with all due respect to your knowledge and ability, connecting things this way would cause a direct short and smoke. It would be like connecting the positive of a DC battery to the negative of the DC battery. Maybe there is something I am not understanding about the diagram. I have looked at the diagram over and over, and it looks to me like it would burn up. With that said, I respect you electronic skills, so maybe I am not aware of something. Or maybe there is something missing in the diagram? |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 15 Jul 2018 8:35 pm
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| Keith Hilton wrote: |
| Georg, I have been looking at the circuit diagram, and have some questions. Over on the right hand side, there is a line coming from the plus power rail, and one coming from the negative power rail. Where these two lines come together, there looks to be a gap where the plus and minus rail does not touch the signal output line. Does the positive rail touch the negative rail at this point? If they don't touch, then why did you draw the two lines like they might be coming together? |
Keith, there's no chance of smoking circuits here, and I'm sorry if my personal circuit-drawing style causes any confusion.
That's how I always draw one connection crossing, but not in contact with, another connection. Some use a curved line to indicate crossing lines with no connection, and (maybe) a dot for every (otherwise not perfectly clear) connection. I quit doing that back in the -70s, except when I add to and/or modify existing circuit drawings made by others where such conventions are followed.
The upper row of offset voltage resistors is connected to the lower row of offset voltage resistors, to form one continuous and self-balancing voltage divider. The output is not connected to it, the horizontal output line just passes the vertical offset voltage divider line in the drawing.
I could also have left out one of the resistors – at right side of "v1" or "v1'" – and then simply give the remaining resistor twice the value. Makes no difference, as once incorporated in a complete circuit the values of all those resistors have to be calculated for that circuit anyway. Thus, I just indicated their presence in my example drawing.
As for my knowledge and ability in electronics: I learn something new every day, just as I did when I picked it up as a hobby 57 years ago. Decades of working for others as serviceman, electrician, and professional troubleshooter on just about anything with or without electronics from 25KV down, and running my own mechatronic development and design business for years (until a medical condition caught up with me and reduced my work-capacity), have taught me that there's always another, and maybe better, way to put together a stand-alone circuit and/or total solution. And, although I am in early retirement now (at 65), I won't quit learning … life would be too boring  |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 16 Jul 2018 8:42 am
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Georg, thank you so much for taking the time to explain the circuit. I understand the plus and minus power rails, with the resistors, create a voltage divider. I understand now that the power rails are not connected at the right of the diagram. If they "were" connected, there would "not" be smoke. The voltage at that "point" would be divided in half, if the resistors on the plus and minus rails were equal. This is the way a artificial ground is created.
The power rails connect through resistors and diodes to the output of the op-amp. The output of the op amp sees the connection as the artificial ground that has been created.
Is this the correct way of thinking Georg?
I wonder if putting a small capacitor in series with the resistors and diodes would help or hurt? The AC signal would still pass, and the capacitors would block DC. Do you think that would help the circuit? |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 16 Jul 2018 9:24 am
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One other thing Georg. Could you give me the suggested resistor values on the plus and minus rails? Plus the resistor values used with the 4148 signal diodes? Plus the value of the resistor on the output of the op-amp?
If the Plus rail was 4.5 volts and the minus rail was 4.5 volts. Considering the op-amp would have a gain of 1-to-60.
Could you also give me the values of the components if the plus rail was 12 volts and the minus rail was 12 volts. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 16 Jul 2018 9:36 am
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Georg, I should of added that the resistors would be 1/4 watt resistors.
One other question: One resistor on each power rail will create the needed voltage divider. Is there a reason why you put resistors between the connection of each resistor/diode connection? |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 16 Jul 2018 10:46 am
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Keith, you're close enough for comfort, but…
| Keith Hilton wrote: |
| I wonder if putting a small capacitor in series with the resistors and diodes would help or hurt? The AC signal would still pass, and the capacitors would block DC. Do you think that would help the circuit? |
No, as the circuits rely on stable/well-controlled DC-levels. Capacitors value-tolerance and DC-leakage are impossible to predict, and since capacitors in series with the diodes will create a form for "capacitor-pumps" these DC-levels can/will drift all over the place. Can create nice-sounding effects, but no two identically built circuits will ever sound the same.
One way to make the circuit more "universal", is to add a non-polarised capacitor in series with the resistor on the OP AMP output (see Cx below), and couple off the output after this to the center of the offset voltage divider (se Rx below). This makes the DC level in the signal-way "auto-center" to the clippers offset voltage divider, making it pretty easy to incorporate the softclipper into any circuit by reducing the number of components that must be "tuned in"…
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 16 Jul 2018 10:52 am
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| Keith Hilton wrote: |
| One other question: One resistor on each power rail will create the needed voltage divider. Is there a reason why you put resistors between the connection of each resistor/diode connection? |
Sorry, but are you joking?
Without the stepped offset voltages created by those resistors in the divider, there will be no soft-clipping.
Any number of diode stages – one or a thousand – referencing the same offset voltages, will end up as just another regular hard-clipper variant. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 16 Jul 2018 2:54 pm
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George, I wondered about a capacitor on the output of the op-amp. Seeing capacitor Cx in your diagram is good.
The Rx resistor in the diagram tied to the output of the op-amp, with a equal value resistor on the positive rail seems like a good idea. That establishes a center point.
I think I can figure the resistor values. I will breadboard the circuit and let you know how it sounds. Georg, I respect you electronic skills. |
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ajm
From:
Los Angeles
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Posted 16 Jul 2018 3:00 pm
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Georg: Have you actually built this?
Do you have any wave forms?
How does it sound? |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 16 Jul 2018 4:14 pm
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| ajm wrote: |
| Georg: Have you actually built this? |
Yes. I picked up the idea from a Scandinavian techno-magazine – Danish I think – in the late -70s, and put together a few variants of it. Used on 6-string guitar they sounded like distorted sine-waves run through a limiter – less harsh than simple diode-clippers.
Experimented with and modified them from time to time for a couple of years, but never took them outside my home-studio. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 16 Jul 2018 5:14 pm
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Georg, the key to his thing is the math concerning resistor values. Resistor values depend on the voltage of the power supply. The other factor not mentioned is the output of the Op-Amp. The output of the Op-Amp is what drives everything.
Since you did not give resistor values, may have to resort to using a volt meter at different points.
Georg, I have looked at hundreds of distortion circuits, and what you presented is different. Different is good and you have a sharp mind. |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 16 Jul 2018 6:10 pm
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Keith, I would normally use a dual OP AMP set up as phase-inverters with variable gain, and rig the soft-clipper between the two amp stages.
As for supply voltage (and current): just choose what is right for what it is supposed to be used for and work with – battery fed stomp-box or fixed supply unit, and calculate/measure resistor values from there. I used to eyeball component values when breadboarding, and tune values once I had signals running through the circuits. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 16 Jul 2018 7:35 pm
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Strange you mentioned using a dual op amp, because that is what I was thinking also. First op-amp to create gain, 2nd op-amp to buffer the soft clipper. Soft clipper in between the two op amps. I don't agree with using phase inverters. I would arrange both op amps to be non inverting. I never did like working with phase inverters. Maybe I am just used to working with non inverting op amps.
Math will get you the approximate values. Then as you say,
tune values once signal is running through.  |
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ajm
From:
Los Angeles
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Posted 17 Jul 2018 8:04 am
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Keith: You could use two non-inverting op amp stages.
Or.......
You could use two inverting stages, the end affect being a non-inverted signal. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 17 Jul 2018 11:59 am
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ajm---yes I know I could use two non inverting, and end up with a non-inverting. I prefer using the non-inverting method for several reasons.
If you will notice, in Georg's explanation, the resistance values drop the voltage, with V4 being the highest voltage, and V1 being the lowest voltage. The signal from the op amp must raise each of the individual 1N4148 diodes above where it conducts--somewhere around 6.5 volts each. So the signal at V4 will conduct at a lower signal coming from the op amp. Likewise it will take a stronger signal coming from the op amp to make V1 conduct.
Here is something Georg failed to mention; Let me explain:
Using one 1N4148 diode will require a resistor, to limit current. Max current of this diode is 150mA. Signal diodes can be put in series to lower voltage. The forward voltage drop across a silicon diode is almost constant. Individual voltage drops across each diode in series are subtracted from the supply voltage. Each diode has a junction resistance relating to the small signal current flowing through it. Meaning the number of diodes in series will have that many times the junction resistance of each diode. The more diodes in series the greater the voltage reduction will occur. You can also connect diodes in parallel with a load resistor, to act as a voltage regulator. Getting the voltage drop you want may be easier than trying to figure the voltage drop across each resistance. Then again, I suppose a volt meter measured at each point of resistance, with power connected, would also determine the voltage present.
Georg, had you thought about using diodes in series to get close to the needed voltage at each stage? |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 17 Jul 2018 2:07 pm
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| Keith Hilton wrote: |
| Georg, had you thought about using diodes in series to get close to the needed voltage at each stage? |
Nope, I find that totally unnecessary. Back when I played with, and through, this type of clipper, I found it much easier to modify the resistor values in the voltage divider, to make the diodes connect at desired levels.
Note that the actual voltage at "v+" and "v-" can be any voltage – "v+" can be fed a lower voltage than "v-" if you want. This lets you set the clipping-levels wherever you want them to be over a wide range, from clipping at zero signal (which isn't very useful) to clipping at signal peaks near the supply rail voltages (which will result in OP AMP stages being overdriven before clipping takes place).
It is all about balancing the offset voltage divider to where the soft-clipper sounds right, which in my experience can only be done by playing through it. However, the voltage at "v4" should not be higher than about +.55Volt, and "v4'" not lower than about -.55Volt, or else those first diodes will conduct and permanently clip the signal. |
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Steve Sycamore
From:
Sweden
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Posted 17 Jul 2018 8:46 pm
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| If I remember, the major difference between the inverting and non-inverting opamp configurations is that much, much more gain is possible in the inverting configuration. Possibly distortion and noise are significantly lower too. But it's been a long time since I was designing opamp circuits and was fresh on the details. |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 18 Jul 2018 2:36 am
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| Steve, on a basic level that's pretty much it |
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Georg Sørtun
From:
Mandal, Agder, Norway
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Posted 18 Jul 2018 3:46 am
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To expand a little on how the soft-clipper in my examples work, here follows a simplified diagram that shows how voltage at the offset voltage divider affects clipping levels…
Shown only for the "v4" to the "v1" levels – in black, with actual clipping levels about .65V (including diode-drop) on the negative side of the signal shown in red.
Mirror those red levels to see clipping for "v4'" to "v1'" for positive signal peaks.
As can be seen: setting the "v4" (and "v4'") to common/GND/identical voltage, will make all clipper-stages break down simultaneously for about 1.3Volt peak-to-peak signal – +/- .65Volt. That is pretty much how a hard-clipper works.
Setting "v4" voltage lower (and "v4'" higher) than common/GND/identical voltage, will again make it work as a soft-clipper, but now it will clip for much higher peak-to-peak signal levels. (That the diode-stages seem to work kind of in reverse, only makes a difference if the resistors between the "v4" to "v1" stages are not identical.)
Just varying the "v4" voltage over this +.5V to -.5V range, and mirror it for "v4'", will therefore make the clipper circuit emulate a whole range of clipper-arrangements as found on the market today, and in addition create some unique clipping-sounds of it own.
Add to that that "v4" and "v4'" do not have to mirror each other, and one can see that this basic circuit can be set up for symetrical or asymetrical soft and/or harder clipping with minimal adjustments and/or modifications.
One can of course also play with the capacitor values connected to "v4" to "v1" (and "v4'" to "v1'"), to affect frequency response and DC-drift for each stage.
One can even modulate the voltage at "v4" to "v1" (and "v4'" to "v1'") over the most usable range – +/- .5Volt, to create a form for unique tremolo effects.
There are no "wrong ways" or "right ways" to create distorted sounds, and it is pretty much up to the imagination, and taste for distorted sounds, of anyone with the most basic knowledge in electronics, to make a circuit like the one I have described work in context.
Good luck |
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ajm
From:
Los Angeles
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Posted 18 Jul 2018 7:39 am
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Keith Hilton wrote: "I prefer using the non-inverting method for several reasons."
Could you give a couple of examples?
I could see where it would come into play if you were using split paths for stereo processing. But that usually doesn't happen with distortion/OD circuits.
Steve Sycamore wrote: "If I remember, the major difference between the inverting and non-inverting opamp configurations is that much, much more gain is possible in the inverting configuration. Possibly distortion and noise are significantly lower too."
How is more gain possible with one versus the other? And if it's possible, is it a practical consideration? (Not that it wouldn't be good to know, mind you.)
Where are the noise and distortion specs for the inverting versus non indicated?
And this is good to know, except: My assumption is that we are not using the op amp here for distortion or huge amounts of gain.
If true, then is this a factor in this case?
Also, with regards to noise: Wouldn't the choice of op amp used have a bigger influence on this? (The answer would be yes if based upon my limited first hand experience.) |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 18 Jul 2018 7:38 pm
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ajm I will try to answer your questions. If you use a inverting op amp, the output is inverted. If you don't want a inverted signal you have to invert the signal again with a 2nd op amp.
I don't agree with Steve Sycamore about gain or noise. You can get huge amounts of gain out of either hookup method. You can get noise out of either method if the hookup design is flawed.
Each op amp is built for a intended purpose. Make sure the op amp you chose will work for what you intend to build.
Some intended uses of op amps seem to work better with one of the two mentioned methods. For example, most people use a inverting op amp when building a summing amplifier.
If you are building audio, use a dual polarity op amp. You will find using a single supply amp more involved, and more noisy.
I prefer using the non inverting hookup method in most cases. That does not mean it is the best method. It simply means that is what I prefer. If the design calls for it, I have no problem using the non inverting method. |
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Keith Hilton
From:
248 Laurel Road Ozark, Missouri 65721
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Posted 19 Jul 2018 9:26 am
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Georg's circuit is not as simple as it appears. Let me explain; When trying to select resistances to arrive at the voltages for V4 and V4', which are approximately +.55v and -.55v, what do you reference for ground? The voltage after the resistances on both the plus and minus power rails are below the .65v required to make the diodes conduct. When there is a sine wave signal from the op amp, the signal will conduct through one of the 1N4148 diodes. Conduction from which diode depends on at which point the sine wave is positive or negative.
Therefore when figuring resistance values the sine wave ground for the plus rail, and minus rail, is 1/2 of the plus and minus power rails. Meaning the signal ground of the dual polarity power supply must be used to calculate resistances. This means to figure the resistances required, you must use the signal plus and minus rails referenced to the signal ground.
Sounds simple enough--right? Not so fast! Remember, all DC power taken off of the plus and minus power rails, when using a dual polarity power supply must have something that limits too much current. Meaning you can burn up one side, or both sides, of a power rail splitter--the device that creates the signal ground.
Therefore, to figure resistances, calculate from each power rail to signal ground---making sure each rail of the rail splitter is protected from too much current.
Now, if we look at V1 and V2' the resistance voltage is .05v, much lower voltage than V4 and V4'. Still not enough voltage to make .65v required to turn the diodes on.
So what is happening? When the sine wave conducts, the signal sees more voltage reverenced to ground at V4 and V4' than it does at V1 and V1' One of these will clip the signal more because one has a higher voltage reverence to signal grtound.
Georg, if I am telling people anything wrong, please correct me. |
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