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Friend,
This is pretty difficult to explain without maths and a diagram, but I will give it a try.
If you have seen a sinusoidal wave, which is like the "electrical component" of an electromagnetic wave. (Well, just imagine a wave), from zero it goes up to a certain level (called a crest) and to zero level and goes to a negative level (called trough).
In some applications, like voltage regulators, you do not need the wave to the highest level. So what you do is to chop off the topmost part of the wave so that it meets your needs. The level of the clipping (chopping of the topmost part) is determined by the circuit.
If you chop off the negative extreme (trough), it is called negative clipping.
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What is the relationship between the clipping level and the dc voltage?
eat me >>>Corleone<<<
What is the clipping words of Tamil and English?
tamillanda
What is the relationship between RMS and peak voltage for a square waveform?
RMS and peak voltage for a square waveform are the same. There is a small caveat, and that is that you'd have to have a "perfect" square wave with a rise time of zero. Let's have a look. If we have a perfect square wave, it has a positive peak and a negative peak (naturally). And if the transition from one peak to the other can be made in zero time, then the voltage of the waveform will always be at the positive or the negative peak. That means it will always be at its maximum, and the effective value (which is what RMS or root mean square is - it's the DC equivalent or the "area under the curve of the waveform") will be exactly what the peak value is. It's a slam dunk. If we have a (perfect) square wave of 100 volts peak, it will always be at positive or negative 100 volts. As RMS is the DC equivalent, or is the "heating value for a purely resistive load" on the voltage source, the voltage will always be 100 volts (either + or -), and the resistive load will always be driven by 100 volts. Piece of cake.
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mitre peak.............i thinksdyu
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