High Frequency Component in Square Wave questions

I think you're still missing the point. A square wave has "high-frequency components" simply by virtue of being a square wave. Take those away, and it's not a square wave any more. Take ALL of them away, except for the base, or what's called the "fundamental" frequency, and you have a pure sine wave.

Basically, you get additional components in ANY signal in which something is changing more rapidly than it would in the case of a sine wave at that frequency. (That's a terrible oversimplification, but I'm hoping you'll see the point from it.) In short, if you see any sort of "wave" with "sharp edges," it has high-frequency components.

A sine wave is a "pure" signal in the sense that it exists, or has components, only at one specific frequency. A 60 Hz sine wave is a "pure 60 Hz tone," in more musical terms.

Bob M.

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No. A _square_ wave is composed of an infinite series of _odd_
harmonics.

Well. Let's just say that I need to know all the magnetic field configurations of all kinds of signals... square waves with high freq components, triangular waves, sine waves because I want to build a radio or other circuit powered by induction (without contact). For example, a radio put near a computer monitor that can power itself (by induction).

Do you think all oscilloscopes can show even the high frequency components in the square waves. Do all have the same sensitivity. What particular feature must I look for in oscilloscopes? Single or dual trace, etc?

Yes I know this configuration.

Won't a 1 Watt 2 ohm resistor explode if I use it in conjunction with a 2 ohm coil in series. I can't predict the voltage produced by the power amp since the manufacturers are accurate on the power rating. Suppose a buy a 100 watt power amp... connect the 4 ohm resistor and coil at the output.. put the function gen at the input... initiate a 200 hz sine or square wave signal... what if the voltage produced is say 50 volts and the current is more than the load can handle... unless the 4 ohm load would only draw the current it needs irregardless of the voltage. Is this what you mean. But I need high current to cause high magnetic field in the coil so I can measure it easily and representative of computer monitor magnetic field strength (at the sides).

I don't want to design my own. There are so many power amps there. I'm bad in soldering and often produced cold soldering points causing endless hours of debugging.

Hmm... yes.. a good idea.. Pc sound card... it can also produce current that can cause magnetic field in the coil, right?? What's the typical amperage of the pc sound card, I need very high amperage so I can easily measure the magnetic field and typical of monitor magnetic field strength.

Oh no... if power amp has bandwidth of 10-20 khz. Then I can't use higher signals in the function gen above 20 khz??

I tried building a power inverter with variable frequency. It's designed for 60 hz but I replace some parts so I can use frequency as low as 1 hz to as high as 10 khz. But after some use, my transistors always gets fried and have to replace them. Know the reason why?

I think the perfect setup for me is to get a variable frequency sine wave power inverter. Know any commercially available ones where you can adjust the frequency?? I don't want to construct one from kits as so many parts need to be soldered and I always get cold solders.

emma

emma wrote: [snip]

[snip]

"Best efforts will not substitute for knowledge," W. Edwards Deming.

--
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

Can I recommend you try placing an ordinary radio next to a computer first. Try picking up a selection of different stations.

On a sunny day (29 Jun 2005 15:13:58 -0700) it happened "emma" wrote in :

Interesting. But remember that the EM field, or rather magnetic field you use, decreases at the third power of the distance! So, even as the deflection current in the horizontal scan coil of a CRT monitor maybe 10A pp, the magnetic field lines on the OUTSIDE of that coil are extremely weak (on purpose by design), and at say

10cm from the coil (flat against the screen) VERY weak. A tape recorder playback head wil pick these up, AFTER AMPLIFICATION you can hear the 100Hz scan (or whatever is used). In case of a LCD monitor there is hardly any field, maybe from switch mode supply or back light HV generator perhaps.

This is vague, you need to mention a frequency, say the wave is 32kHz sawtooth (as in CRT monitor), say we want to be able to see 10th harmonic (would give sufficient true waveform display), then 320kHz is enough. Simplest scopes are analog 10MHz, then 20 MHz, 50MHz digital sampling storage to 1G samples / second and 100 MHz, 1 GHz, 1 GHz analog.... it all exists. Unless you want to power from your microwave oven, anything over 10MHz should be enough in your case. If you want it for the future to play electronics, go for a digital one and 1G samples. As for the sensitivity, what do you expect? You want to power a radio. Say the radio uses 100mW (very low power speaker), so you need to get that from the external magnetic field. You need a LOT of magnetic field! You can use a normal scope with 10mV per division. BUT it also depends on what you use as sensor, I asked this before, what do you use as sensor?

OK some basics: A resistor is measured in Ohms, the resistance. When current flows through that resistor, it gets hot. Above a certain temperature it will melt or burn. The resistor manufacturer will specify how many Watts you can 'dissipate' in the resistor so it just does not go kaput. From this you can -calculate- the current. P (watt) = U (volt) x U (volt) / R (Ohm) or P (watt) = I (Ampere) x I (ampere) x R (Ohm) And U (volt) = I (Ampere) x R (Ohms)

So let us take some examples how to use this: Say the amplifier can output 20 V sine. And say you have a 3 Ohm resistor and a 1 Ohm coil as shown above. The power in the total output load will be

20 x 20 / (3 + 1) = 400 / 4 = 100 W The resistor gets 3/4 of this, so 75 Watt, the coil 1/4 so 25 W So the resistor you need to buy is 3 Ohm 75 W, or 3 1 Ohm 25 W resistors in series.

If you only have a bunch of 1 W 1 Ohm resistors, than we can reverse the calculation, total load allowed now is 4W, so U x U / 4 = 4 or U is 1V sine. Not a lot of voltage! Anyways with Ohm's law you can calculate what components you need. One remark: a 20 V sine wave is really a voltage swing from 20 x 2 x sqrt(2) =

56V, such an amplifier will have a total supply voltage of 60V or so, to make the output possible.

I hope I have explained that, BEFORE you attach any load to the amp, check the correct calculated output level with the scope (or an AC voltmeter).

Well, it depends, my creative sound cards only give line level output, you still need an audio amplifier. Some sound cards have 1 W or more.

Correct.

If it has a transformer in it, and is designed to run at 60Hz, lowering the frequency will reduce the impedance of the transformer to its resistance in Ohms, and your output transistors will die of too much current. Do not mess with designs like that, it will only work in the specified range.

My personal opinion on this idea is that powering via magnetic induction will not work (unless you are sitting under a power line perhaps). The simple 1 W or up amplifier with PC sound card will do for the audio range, else use the one I provided the link to, it will do 40 kHz. Else you get into very expensive stuff, although 1MHz wide 1 W power amp is easy to make yourself. Have you ever considered using a ring core (ferrite) transformer in the output of an amp to step up the current?

Get some RF signal generator from ebay perhaps, but limit yourself to the audio range or just above it. What makes you think there are strong 10MHz signals (harmonics) around? EVERY piece of equipment (except cell phones haha) these days is certified NOT to emit any of these fields.

And what sensor do you use for detecting the magnetic fields?

ES1371 soundcards have a TDA1517m which provides 2x6W. That is the most powerfull soundcard I'm aware exicsts. Typically it gives 4-5W.

--
MVH,
Vidar

www.bitsex.net

I didn't say I'm gonna power a convensional radio. But something nanotech like molecular circuitry (which I'm still exploring). For now I just want to master the different magnetic field variations produced by different current and voltage waveforms.

I can't find any 3 ohm 75 Watts at any electronics stores in my place. The most they have is 1 watt. If I'm gonna use a load such as bulb or heater. Are you aware of anything that is only 3 ohms??

I fried 4 pcs of MJ15015 Power Transistors and 9013 ordinary transistors already. My transformer is 12-0-12 primary and

110 volts secondary. I can't understand what you mean the transformer will reduce the impedance to its resistance and the output transistors will die of too much current. Can you just mentioned what is the principle called (for example, Lenz Law). I'll just research about it so you don't have to type and explain a lot.

I thank you so very much for the replies and all the information, Jan. I've gained the necessary information I needed in my project.

My friend has a very sensitive 3D sensor where he can image the entire magnetic field intensity and harmonics. I don't fully understand it well yet.

emma

See thread "Emma".

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1. There is no such thing as "RMS power"

1. Since, for a square wave, RMS and peak voltage are the same and since for a sine wave they're not, a lowpass filtered 120V 60Hz square wave will yield a 120V _peak_ 60Hz sine wave. That's about an 85VRMS sine wave.

2. Of course there's a loss in power. Where do you think all the energy in the harmonics went, into the fundamental?

-- John Fields Professional Circuit Designer

On a sunny day (30 Jun 2005 07:50:11 -0700) it happened "emma" wrote in :

OK

Using light bulbs as load works, but light bulbs have this peculiar thing that if cold, the resistance is about 1/10 of when hot. Say you have a 35 W car headlight (available form any garage), 12 V So (12 x 12) / R = 35, so R = 144 / 35 = 4.11 Ohm. However when cold it is more like 0.4 Ohm. It will perfectly protect your power amp though, light bulbs (with a filament) act as a constant current source. Most transistor amps have output power limiting. You can increase resistance by 2 by using the dim and main light connections only (series), divide by 2 by using both in parallel.

No it is very simple. Suppose the transistors deliver a sine wave to the transformer. The impedance of the primary coil is R + jwL The trick is in 'w', if you use 1/10 of the frequency, then the impedance is also 1/10, and the current that flows is 10x.

In reality it will likely just switch the primary across the supply. For any inductor goes: I (Amperes) = U (volts) x t (seconds) / L (Henry) This is a very very important formula. It means that if you connect a coil of 1 Henry across a 1 V battery, the current in the coil wil *linearly* rise to 1 Ampere after 1 second, (and 10 Amp after 10 seconds etc), only to be limited by the resistance in Ohms of the coil (normally quite low). So if the switcher switches at say 50 Hz, after 1 / 50 = 20 mS a current I will flow, if you go to 5 Hz, 1 / 5 = 200 mS, the current will be 10x higher! Here is where your transistors die. The number of turns in the primary of the transformer determine the inductance (and the core material has effect too). For a lower frequency you need a LOT more turns, more iron in the core perhaps, so you can only use a 60 Hz transformer for 60 Hz, not also for

6 Hz, or lower.

You are welcome.

Neither do I, what is a 3D sensor? Ask him some time, I'd like to know.

Na that won't pass the Turing test.

The high frequency comes from the sudden change in di/dt in the each square wave. You will notice that each square wave starts and ends with a sudden pulse. That pulse is the sudden change in di/dt. A pure sine wave is caused by a smooth changing di/dt where the peak di/dt is at zero amplitude and the zero di/dt occurs at peak amplitude.

There are countless options. It depends on the of amplifier *class* you are using. Here's a cheap and simple idea. In series with the output of your amplifier you could place a 7000 uF capacitor in series with a 1 mH inductor. Again, that's a dirty cheap method though. I only mention this so perhaps you can understand what's happening. An L & C in series form a resonant circuit. The equation is f = 1 / [ 2pi * (LC)^2 ] Also, such a dirty method would not waist that much energy because the LC circuit resists all current outside 60Hz, but it's reactive resistance. Pure reactive resistance will not consume energy.

For better options you'll need to do some study on different class amplifiers. Try this for starters ->

I would stay away from class A amplifiers since they generate dc current. Therefore it would be difficult to filter out the unwanted frequencies, including dc, without corrupting the amplifier. Class B would work but are generally not used for high quality because of crossover, which generates some unwanted high frequency. Class AB is more efficient. Class D is extremely efficient. Class E is even more efficient. Class G & H are less efficient than E. Here's a brief explanation of some different class amplifiers ->

See thread "Emma"

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No, the equation is:

               1
     f = --------------
          2pi sqrt(LC)

I don't think you should have included EM in that statement because the magnetic and electric field from EM falls as 1/r. A closed loop dc current causes magnetic field that falls as 1/r^3-- of the equation is not linear at close distances and in such a case there are more complex equations. Magnetic field from a current segment falls of at 1/r^2

Thanks! That was just a typo on my part, ^2 should have been ^1/2 You can see that I used the correct equation in my numbers; i.e., 7000uF &

1mH ~=3D 60Hz resonance frequency -- 1/[2pi * (7000uF * 1mH)^1/2] =3D 1/60 Hz.

There was no energy because it was just a voltage penitential generated by the amplifier. As mentioned, it is a dirty method mostly to demonstrate what is happening since you can't perfectly filter out all unwanted frequencies.

Yes, which is why I clearly said no Class A amplifiers :-)

Power companies do not produce pure sine waves. Along with the sinusoidal distortions there are variations in amplitude, phase and frequency. But because of the low pass nature of power lines and other components, the spectrum of the AC is somewhat cleaned up.

In fact, it is debatable whether pure sine waves even exist in our physical world.

Thomas

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You missed the point, which was that since the capacitor will block DC
it doesn't matter whether the filter's being driven by a class A
amplifier or not, no DC will get into the load since it's being
"filtered" out by the cap.