Diode clipping mod, Ge and Si in parallel?

Keith Hilton · Post by **Keith Hilton** » 15 Jul 2018 2:18 pm

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 â€“ for both positive and negative cycle"

Keith Hilton · Post by **Keith Hilton** » 15 Jul 2018 5:40 pm

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?

Keith Hilton · Post by **Keith Hilton** » 16 Jul 2018 8:42 am

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?

Keith Hilton · Post by **Keith Hilton** » 16 Jul 2018 9:24 am

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.

Keith Hilton · Post by **Keith Hilton** » 16 Jul 2018 9:36 am

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?

Keith Hilton · Post by **Keith Hilton** » 16 Jul 2018 2:54 pm

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.

ajm · Post by **ajm** » 16 Jul 2018 3:00 pm

Georg: Have you actually built this?
Do you have any wave forms?
How does it sound?

Keith Hilton · Post by **Keith Hilton** » 16 Jul 2018 5:14 pm

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.

Keith Hilton · Post by **Keith Hilton** » 16 Jul 2018 7:35 pm

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.

ajm · Post by **ajm** » 17 Jul 2018 8:04 am

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.

Keith Hilton · Post by **Keith Hilton** » 17 Jul 2018 11:59 am

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?

Steve Sycamore · Post by **Steve Sycamore** » 17 Jul 2018 8:46 pm

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.

ajm · Post by **ajm** » 18 Jul 2018 7:39 am

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.)

Keith Hilton · Post by **Keith Hilton** » 18 Jul 2018 7:38 pm

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.

Keith Hilton · Post by **Keith Hilton** » 19 Jul 2018 9:26 am

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.