Audio clipping detector

Audioguru

Joined Dec 20, 2007
11,248
An amplifier or opamp has extremely high gain. Negative feedback is added to reduce the gain to a reasonable amount which also reduces the distortion.

The volume control feeds a divided signal level to the amplifier and the gain of the amplifier remains the same as when it was built. Then you simply attenuate the output of the amplifier to almost the same level as the input of the amplifier.
You attenuate with a voltage divider made with two series resistors.

The clipping LED lights whenever the attenuated output level is less than the input level.
You probably want to add precision rectifiers and peak detectors ahead of the comparator.
 

Ron H

Joined Apr 14, 2005
7,063
With op amps and negative feedback, you can add signals, or subtract signals, or amplify (or attenuate) a signal or combinations of signals. You can't make one signal affect the gain of another one.
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
An amplifier or opamp has extremely high gain. Negative feedback is added to reduce the gain to a reasonable amount which also reduces the distortion.

The volume control feeds a divided signal level to the amplifier and the gain of the amplifier remains the same as when it was built. Then you simply attenuate the output of the amplifier to almost the same level as the input of the amplifier.
You attenuate with a voltage divider made with two series resistors.

The clipping LED lights whenever the attenuated output level is less than the input level.
You probably want to add precision rectifiers and peak detectors ahead of the comparator.
The volume knob changes the voltage from the pre-amp output which changes the level the amp is driven as far as I know. Also the voltage out on the amp changes so I don't see how a fixed voltage divider will be correct?

With op amps and negative feedback, you can add signals, or subtract signals, or amplify (or attenuate) a signal or combinations of signals. You can't make one signal affect the gain of another one.
I'm having a hard time following you guys. Let me do a simple schematic in a program I use and maybe I can demonestrate my idea. I'm not sure if we are on the same idea or not. I'm going to post a schematic in a little. Anyway I am willing to pay anyone who can really help me out to the end of this because I'm really trying to get it done...
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
I attatched a png below...I had more that I put together before but I cannot figure it out now. This is just to show, the above voltage source is at 55v while the bottom is at 4v, but the outputis adjusted to the 4v..the problem here I guess is that it will be exactly the same reguardless of clipping.

Audio guru, you mentioned that when clippping, the clipped signal will be less then the input? What I can't figureout is the voltage division with fixed resistors where the amp output will attenuate to the input at whatever the input is(it changes too!). Is there a way to connect the input and output together through a voltage divider and have the output attenuate in proportion to the input value?
 

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Ron H

Joined Apr 14, 2005
7,063
I attatched a png below...I had more that I put together before but I cannot figure it out now. This is just to show, the above voltage source is at 55v while the bottom is at 4v, but the outputis adjusted to the 4v..the problem here I guess is that it will be exactly the same reguardless of clipping.

Audio guru, you mentioned that when clippping, the clipped signal will be less then the input? What I can't figureout is the voltage division with fixed resistors where the amp output will attenuate to the input at whatever the input is(it changes too!). Is there a way to connect the input and output together through a voltage divider and have the output attenuate in proportion to the input value?
Your circuit has an input connected to the op amp output. That doesn't work.
Regarding your last question: That is exactly what I told you is impossible to do with a simple linear circuit (op amps, resistors, etc.).
 

Audioguru

Joined Dec 20, 2007
11,248
The amplifier in your example has a voltage gain of 50V/4V= 12.5 times.
Use two series resistors to attenuate the output down to about 4.1V when the input is 4V.
When clipping causes the attenuatred output to be less than the input then a comparator circuit will light an LED.

The volume control feeds the input of the amplifier. When the amplifier is not clipping then the output level is ALWAYS 12.5 times higher than the input level.
If the input is 0.4V then the output is 5V. The attenuated output is 0.41V.
If the input is 0.04V then the output is 0.5V. The attenuated output is 0.041V.
If input is 5V but the output might clip at maybe 52V so the attenuated output is 4.16V which is too low then the comparator circuit will light an LED.

There is an article about comparing the negative feedback input signal of an amplifier to its actual input signal (the negative feedback input signal is already attenuated to the same level as the actual input). The article is here:
http://www.sound.westhost.com/project57.htm#top
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
The amplifier in your example has a voltage gain of 50V/4V= 12.5 times.
Use two series resistors to attenuate the output down to about 4.1V when the input is 4V.
When clipping causes the attenuatred output to be less than the input then a comparator circuit will light an LED.

The volume control feeds the input of the amplifier. When the amplifier is not clipping then the output level is ALWAYS 12.5 times higher than the input level.
If the input is 0.4V then the output is 5V. The attenuated output is 0.41V.
If the input is 0.04V then the output is 0.5V. The attenuated output is 0.041V.
If input is 5V but the output might clip at maybe 52V so the attenuated output is 4.16V which is too low then the comparator circuit will light an LED.

There is an article about comparing the negative feedback input signal of an amplifier to its actual input signal (the negative feedback input signal is already attenuated to the same level as the actual input). The article is here:
http://www.sound.westhost.com/project57.htm#top
Okay I understand now. I may have a problem, the amplifiers that will be used have an adjustable gain knob. This tool is actually mostly used so that when one can set it properly. If the LED lights then the user backs off the gain as long as the output of the amp beeing tested matches the other amp stages properly.

So what can I do about an adjustable gain, won't the proportion of voltage from the input to the output change?
 

Ron H

Joined Apr 14, 2005
7,063
The problems with the differentiator scheme are:
1. You have to have an AGC amplifier in front of the differentiator, so that the amplitude into the differentiator is independent of the amplifier output amplitude. A good AGC amp is a PITA to design.
2. If the clipping is "soft", the differentiator will not work well, if at all.

If you have the capability, I think the best method would be to do spectrum analysis on the output.
If you can drive the input with a pure sine wave (no harmonic distortion), and the amplifier has low THD until it is overdriven, then you would simply crank up the output level until the THD increases dramatically.
The description is simple. The devil is in the details, as you have already discovered.
 

Thread Starter

chunkmartinez

Joined Jan 6, 2007
180
I may have lost all confidence in this project...It seemed simple at first but idk anymore. Let me propose one last idea.

There is a "ripple remover" circuit that uses an op amp to filter out any AC ripple. What it does is it uses a capacitor in the loop of a "common mode" input so that AC is present on both inputs but DC is filtered from one side by the cap, so the differential of the DC is seen and the AC is cancelled out. I figure this circuit could maybe make it easier to make obvious when there is clipping, or "dc". This was my first idea but I doubt it will work because it goes back to the idea of using a capacitor to try and catch or filter too little a rate of voltage change.

Using reactive components to filter clipping dosn't seem to work, but why? Is it because they aren't that good at telling the difference between a short portion of little rate of change and the rest of the larger rate of change? If timing was an issue can't we insert a sort of delay or modify what we are working with?
 
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