I'm asking this question because of audio amplifiers and diode clipping.
I've been thinking about this situation:

Does a pair of diodes cause all the frequencies to clip, even if some frequencies are too low in voltage to activate the diode? I imagine two pure sine waves at 100 hz and 1000 hz, which have 0.5 volt and 1 volt respectively. When the 1000 hz hits 0.7+v to activate a silicon diode, the frequency experiences clipping. In this region where the 1000 hz is activating the diode, is the 100 hz frequency also being clipped during this region, even if the voltage of the 100 hz isn't high enough to activate it on it's own?

Is it more like a toggle switch, where only the 1000 hz clips and the 100 hz is unaffected? I don't quite understand how or if different frequencies affect each other, or if they are all treated as separate objects within a circuit, for processing purposes.

 

There is only one unique voltage being input to the circuit at any one instant, which we call the instantaneous voltage.
But this voltage varies with time ie it is a function of time.

The input waveform you describe is composed of two components, which are additive or in series and there is also a phase angle between them.
Since you have chosen harmonically related frequencies the waveform will be repetitive at the lower frequency, modified by the higher tenth harmonic.

You should sketch out this waveform given by the formula

The instantaneous voltage Vi = 0.5*sin( 100t*2pi) + 1.0*sin( 1000t*2pi) and t is time in seconds.



Clipping will occur wherever this waveform voltage is greater than about 0.7 volts in your circuit, as you note. (don't forget this applies to the positive half cycle only for one diode).

 

That is a very good question.

In terms of audio frequencies, what is the result of clipping?
A clipped sine wave contains odd-integer harmonics of the fundamental frequency which gives it harsh (fuzz) tonal qualities.

If you add a lower frequency of lower amplitude, when the sum of the two exceeds the clipping threshold they will both be clipped. It is difficult to anticipate the resulting audio. Try it and find out.

 

There is only one unique voltage being input to the circuit at any one instant, which we call the instantaneous voltage.
But this voltage varies with time ie it is a function of time.

The input waveform you describe is composed of two components, which are additive or in series and there is also a phase angle between them.
Since you have chosen harmonically related frequencies the waveform will be repetitive at the lower frequency, modified by the higher tenth harmonic.

You should sketch out this waveform given by the formula

The instantaneous voltage Vi = 0.5*sin( 100t/2pi) + 1.0*sin( 1000t/2pi) and t is time in seconds.



Clipping will occur wherever this waveform voltage is greater than about 0.7 volts in your circuit, as you note. (don't forget this applies to the positive half cycle only for one diode).

Ah thank you. I had considered that but it seemed like if they added together, then presumably the negative peak of one could completely nullify the half positive of the other leading to a gap, as I imagined it, where there is no sound, due to summing to zero volt difference, and it seemed that this would create lots of holes in audio (which didn't make sense, but more proper thinking about it says this should be true of audio anyway), but the way you describe it seems to say that my first instinct was correct; they add together at any given time and if that means they add up to more than diode drop, both get clipped, if the add together to less than diode drop, neither get clipped.

It seems like the only tonal difference between trying to keep lower frequencies out of clipping range prior to a diode clip stage is the effect of a low pass filter that makes low frequencies quieter; the harmonics added by the distortion of clipping diode will still appear. I think the next step is to test different levels of low frequency voltage with strict 1v high frequency to see how the wave form(s?) change, how often harmonics show up in a period, and all that.

Time to play with a breadboad

 

Edit please note I have put the formula right (I have corrected the omega = 2pi f)

 

If you want to take it a step farther, look at diode mixing circuits. (Google will show you a bunch.)