MOSFET switch time and harmonics - Page 1
Forum section: electronics / beginners

author: fonograph
author_rank: Frequent Contributor
timestamp: November 14, 2017, 03:30:02 am
content: If I have a FET with 100 nano second switch time,what will be the highest frequency harmonic generated in that switching transient region?100 nano second is equivalent to period of 10 MHz sine wave.Will that switching transient create only harmonics up to 10 MHz and nothing beyond it,or will there be harmonics above it? If yes,how many and how sharply will they fall off ( db per octave ).

author: ejeffrey
author_rank: Super Contributor
timestamp: November 14, 2017, 03:56:26 am
content: There is no hard cutoff -- all the harmonic components will exist, but the amplitude of each will fall off with frequency. They will begin falling quickly above approximately 1/(2 * rise time), or 5 MHz (for a perfect ramp edge).The exact rolloff depends on the shape of the edge. A triangle wave has harmonic amplitudes that fall off as 1/f^2. But the high frequencies are attributable to the cusps at the top and bottom. So if your signal has a bit of rounding at the beginning and end, the high frequencies will fall off much faster -- a higher power law, or even exponentially.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 14, 2017, 09:27:13 am
content: ^To run through that in detail:Look at the Fourier series of a few typical waveforms.Square wave: harmonics go as 1/N.Triangle wave: go as 1/N^2.Trapezoid wave (a square wave with triangle rise/fall segments): goes as 1/N until a cutoff frequency of about (1 / 2 * rise time), beyond which they go as 1/N^2.Parabola wave (a triangle wave, but with parabolic sections): go as 1/N^3.Rounded trapezoid wave (trapezoid with the corners rounded off using parabolic segments -- piecewise continuous first derivative): goes as 1/N until first cutoff, then 1/N^2, then 1/N^3 after second cutoff corresponding to 1/(2*length of parabolic segments).Suppose you continue this rounding-off process infinitely. The corners begin to approximate a Gaussian curve, and the spectrum therefore takes on the shape of a Gaussian as well (because the Fourier transform of a Gaussian is a Gaussian). That means, if you use an exponential function (like this) to round off the corners, you get an exponential frequency cutoff, too.Which is fantastic news, because real signals don't go on to infinity! Nor do they cut off suddenly (as Gibbs phenomenon demonstrates so well). The real answer lies somewhere inbetween, and it's an excellent lesson in the subtlety of the real world, to understand that this is the case.In short: a real square wave (or any squarish waveform, if not a perfect, 50%, constant frequency, square wave) has harmonics that go more or less as 1/N, until the rise time, where they drop... faster. How much faster, depends of course, but it's at least 1/N^2 until a higher cutoff, and so on until there's no signal left at extreme frequencies.And this is why you don't see light pulses emitted from just any square wave oscillator.Tim

author: fonograph
author_rank: Frequent Contributor
timestamp: November 14, 2017, 10:05:22 am
content: Hey all of you,what does that "1/2" and "1/N^2" and "1/N^3" mean? I am effectively illiterate when it comes to any mathematics higher than divide and multiply.I know that millionth harmonic of square wave have one millionth amplitude of fundamental,and millionth harmonic of triangle wave have one quadrillionth amplitude of fundamental,so if we had a wave with harmonics that decrease by "1/N^3",then millionth harmonic would be one nonillionth ( 30 zeros ) amplitude of fundamental?

author: rstofer
author_rank: Super Contributor
timestamp: November 14, 2017, 03:19:30 pm
content: N is the harmonic number and the ratio is the harmonic amplitude relative to the fundamental. So, if a signal contains only odd harmonics (square wave) and the amplitude falls off as 1/N then the 3rd harmonic will have 1/3 the amplitude, the 5th will have 1/5th the amplitude and so on.https://en.wikipedia.org/wiki/Square_waveUnder Fourier Analysis, you can see the series expansion and the fractions out from of each term (1, 1/3, 1/5) are the relative amplitudes. You can think of 2*pi*f as the fundamental frequency and 6*pi*f as the 3rd harmonic and so on.There is also an excellent animation showing how sine waves at the proper frequencies add up to a square wave.If harmonics fall off as N^2 then odd harmonics fall off as 1, 1/9, 1/25, 1/49 (for the 1st, 3rd, 5th and 7th)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 14, 2017, 03:31:28 pm
content: Read "^" as "raised to the (number) power", so, squared, cubed and such.So, 10th harmonic at 1/10^2 = 1/100th amplitude, and so on.For round 10^N (one followed by zeroes) numbers, just multiply the number of zeroes. That is: (10^N)^3 = 10^(3*N). So, for N = 6 (million), the cube has 18 zeroes (1, uh, pentillion?).Note that that's US millions (in steps of thousands: thousand, million, billion, trillion, etc.), not millards (steps of 10^6) sometimes seen in Europe.Tim