Clipping Diodes 101

[swinginguitar]

Could someone just walk me thru the premise of how diodes work in a clipping context? (in a feedback loop and grounded hard clipping)

I got this from GEOFEX, but dont quite get it:

As a diode turns on, the equivalent resistance of the diode goes down as the diode turns on harder. Effectively, there is a small range from about 0.4V to 0.7V (the exact voltages depending on the type of diode, the package, the doping, etc, etc.) where the diode resistance goes from an open circuit to a very low value, perhaps a few ohms for signal diodes. So as the diode turns on, the gain of the opamp stage changes, going down to just over 1 if the diode can be considered a short circuit compared to the 4.7 K resistor on the (-) input. Even if the Drive control is set for a gain of 100, the diodes conducting cause the gain to drop to 1 for only those portions of the signal where the product of the input signal and the gain exceeds the diode threshold. This means that the signal is "clipped" at the forward voltage of the diode; because there are two opposing diodes, this happens for both signal polarities.

How is a gain of one going to clip a signal?

[R.G.]

It has to do with how opamps work. Opamps work by balancing current at the inverting (-) input. An input signal tries to shove current into/remove current from the inverting input. The input is high impedance and can't "eat" or source any current on its own. But the output can. And it's tied to the inverting input by a feedback element. The inverting input's voltage moves in the direction it's pushed by the input current. The output moves in the other direction until the current sucked/sourced by the output exactly balances what the input resistors are trying to do to the inverting input. When that happens, it all balances. I realize that's a confusing look at opamp gain, but it's the way it really works, and it's a useful thing to know when the input and feedback elements are not simply Rs, Ls and Cs.

Let's say the circuit is set up for a gain of 100 if we ignore the diodes.
We put in a 1mV (peak) signal, and the output obligingly goes to 100mV peak.
We put in a 4mV signal and the output goes to 400mV.
We put in a 7mV signal and ... the output tries to go to 700mV, but there are those diodes. At about 700mV (roughly, depends on the diodes, etc.) one of the diodes conducts. Instead of the output having to move by a voltage of V = I*Rf to balance a little bit more current (I) at the input, suddenly the diode is letting a huge amount more current through with only a little more voltage increase. This satisfies the current needs of the input, so the output voltage quits increasing (or decreasing, as the case may be).
So if we put in 50mV, we don't get 5V out; the diode conducts more current without needing the output to change voltage very much, so the output may only go to 750mV, not 5V.

From the outside, the gain below the threshold where the diodes conduct looks like 100. Above the level where the diodes conduct, the voltage is clamped by the diode conduction.

It's not the basic gain that has changed. It's the incremental gain; for signals over the level where the diodes start to conduct, the gain for each additional millivolt of signal decreases for that incremental millivolt.

I've probably made it more confusing, not less. The effective gain of the circuit is whatever the resistors say it is until the diodes conduct; after that, you get less and less gain as the diodes conduct more and more.

[swinginguitar]

confusing, yes...

Another question in the same vein - do more diodes in series mean less clipping?

[R.G.]

confusing, yes...

I'll try again.

Gain is not equal to clipping or distortion. If I have a pair of clipping diodes that conduct at 0.7V each connected to ground, and a resistor in series with the pair above ground. I put a signal into the resistor and look at the resulting signal at the junction of the diodes and resistor. As long as the signal is less than the 0.7V threshold of conduction for the diodes, they may as well not be there, and the signal is not clipped at all. So a 0.3V peak signal passes through unclipped. If I use a bigger signal, like 1V peak, then the diodes conduct whenever the instantaneous signal gets above 0.7V, and the diodes "eat" all of the available signal above 0.7V. So the bigger signal is clipped at 0.7V. Notice that there is no gain here at all, only bigger or smaller signals.

Does that small part make sense?

Another question in the same vein - do more diodes in series mean less clipping?

In a roundabaout way, yes, kind of.

What more diodes in series do is to make the threshold where the diodes conduct be bigger. If one diode clips at 0.7V, two in series don't conduct until 1.4V.

So if you have a fixed amount of signal voltage, four diodes, two diodes in series each way, will clip off less of the tops and bottoms a smaller percentage of the total signal level. The signal will be bigger in absolute volts, but less percent clipped off. It will sound less clipped and distorted, but will be louder because less of it got cut off.

I know I need to go draw pictures, and you'd see it instantly. Forgive me, I'm not set up to draw pictures at the moment.

Also notice that I'm dramatically simplifying diode conduction, using a nominal 0.7V for silicon; germanium diodes conduct at 0.2-0.3V, Schottky at 0.4-0.5V, and LEDs from 1.2V to as high as 4V. The transition into conduction is not instantaneous. Just insert a mental footnote that there are multiple kinds of diodes that can be used to clip, and they have varying thresholds and abruptnesses of starting conduction.

[Loudthud]

R.G. makes it pretty clear but you may still not understand how opamps work with feedback loops on a gut level. I didn't understand feedback until I studied regulated power supplys. No, not the kind made with a chip that does it all, ones that used descrete transistors or an opamp. Something about DC is just easier for me to understand.

[swinginguitar]

Does that small part make sense?

Yes that helps for sure! Glad you used the diodes to ground example - I was wondering about that too (hard clipping a la Rat). In that instance, I understand that the diodes turn on at .7v and whisk (some of) the signal to ground abruptly. But in a feedback loop it's not quite that simple, eh?

Thanks....I really appreciate how this forum takes time out for noobs.

Sounds like I need to study opamp feedback a little more. I understand that the gain is set by dividing the feedback, but when diodes, caps, etc are added, it gets a little more difficult to visualize...

So gain is amplification factor, not clipping, but would an opamp eventually clip itself if the gain were high enough?

[R.G.]

So gain is amplification factor, not clipping, but would an opamp eventually clip itself if the gain were high enough?

Gain is by definition a simple multiplication. Gain of 2, twice as big. Gain of 100, 100 times as big.

That's the theory.

In the real world, there are always limits. For amplifiers, it's (at least) the power supply voltage. An amplifier cannot produce an output voltage as big as its power supply.

So an opamp output will follow its input (mostly) faithfully right up until it can't any more. Then it stops because it is banging up against its power supply. Yes. All amplifiers clip at some point. It clips if the OUTPUT is big enough, not if the gain is high enough. High gain only participates in that it makes big output signals.