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High Frequency Component in Square Wave questions

V

Vidar Løkken

Jan 1, 1970
0
Jan said:
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.

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

emma

Jan 1, 1970
0
Jan said:
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.

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.

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.

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??
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.

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

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
 
C

CWatters

Jan 1, 1970
0
emma said:
I didn't say I'm gonna power a convensional radio. But something
nanotech like molecular circuitry (which I'm still exploring).
I can't find any 3 ohm 75 Watts at any electronics stores in my
place.

Na that won't pass the Turing test.
 
emma said:
Now how does the high frequency component got generated??
Also when it is in the maximum amplitude, how
does it generate the high frequency component in the square
waves?

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.

What would it take to build a square wave power inverter that
totally eliminate the high frequencies riding in the square wave??

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 ->

http://www.bcae1.com/ampclass.htm

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 ->

http://www.absoluteastronomy.com/encyclopedia/e/el/electronic_amplifier.htm
 
Jan said:
But remember that the EM field, or rather magnetic field you use,
decreases at the third power of the distance!

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
 
John said:
The equation is f = 1 / [ 2pi *

---
No, the equation is:

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

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

---
If you started with a 60Hz square wave and wound up with a 60Hz sine
wave, what happened to all the energy in the harmonics if it's not
wasted?
---

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 :)
 
T

Thomas Magma

Jan 1, 1970
0
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
 
E

emma

Jan 1, 1970
0
Jan said:
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.

I've got other idea. I'll get meters of high gauge wires and measure
3 ohms and use it as resistor. How do I calculate how much will it
heat up? I've got Resnick and Holliday 800 page book on
Electromagnetisms and still slowly going thru it but need the
information asap. Thanks.

Say. Is there a difference in performance if I have say a 20 meter
thin cooper wire measuring 3 ohms versus a thick 5 meter cooper
wire measuring 3 ohms also?
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.

Thanks. It explains why I fried my power transistors 2 times
already. Well if I never go below 60 Hz but go higher to 10Khz.
Would it also fry the circuit?? I notice a high pitch sound in
the circuit when the frequency is increased linearly from 60 hz
to 10 Khz.
You are welcome.

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

Some kind of non-linear detection system that is based on the
quantum potentials of the vectors. I can't explain it. I'll ask.

emma
 
V

Vidar Løkken

Jan 1, 1970
0
emma said:
I've got other idea. I'll get meters of high gauge wires and measure
3 ohms and use it as resistor. How do I calculate how much will it
heat up? I've got Resnick and Holliday 800 page book on
Electromagnetisms and still slowly going thru it but need the
information asap. Thanks.

Say. Is there a difference in performance if I have say a 20 meter
thin cooper wire measuring 3 ohms versus a thick 5 meter cooper
wire measuring 3 ohms also?

Go read that book. Thicker wire == less resistance pr m cable, since
there's far more copper to conduct the current.
Think of it as water. Bigger pipe equals lower resistance equals more water.
Higher pressure equals more water trough the same pipe.
 
S

St. John Smythe

Jan 1, 1970
0
Jan said:
That depends on a lot of factors, for example when you wind it like a coil
it will heat up a lot more then when you leave it laying about.

You fellows are being trolled big-time. Can't you see it?
 
John said:
You missed the point.

Perhaps.

Do you know much about low temperature electronics? I am looking at an
ADC chip, which has a temperature range of -30 to 80 C. Why wouldn't
the chip work with say -50 C? I would like to work with low fields and
eliminate as much thermal noise as possible. Any ideas?
 
E

emma

Jan 1, 1970
0
Vidar said:
Go read that book. Thicker wire == less resistance pr m cable, since
there's far more copper to conduct the current.
Think of it as water. Bigger pipe equals lower resistance equals more water.
Higher pressure equals more water trough the same pipe.


--
MVH,
Vidar

www.bitsex.net

Thanks for this crucial information. I know thicker tungsten means
lesser resistance because there are more spaces in the lattice where
the electrons can move or vibrate. So it is also valid in pure
wires. You save me the trouble of getting very thick wires and
more length required because of lower resistance. So I'll just
get thinner wire.


emma
 
E

emma

Jan 1, 1970
0
Jan said:
I've got other idea. I'll get meters of high gauge wires and measure
3 ohms and use it as resistor. How do I calculate how much will it
heat up?
That depends on a lot of factors, for example when you wind it like a coil
it will heat up a lot more then when you leave it laying about.
I've got Resnick and Holliday 800 page book on
Electromagnetisms and still slowly going thru it but need the
information asap. Thanks.
Try it!
There is special resistance wire, that you can use to make your own
resistance. Google:
http://www.surplussales.com/Wire-Cable/Resistance.html
Then you can use shorter length.
Say. Is there a difference in performance if I have say a 20 meter
thin cooper wire measuring 3 ohms versus a thick 5 meter cooper
wire measuring 3 ohms also?
Well, since you want to measure magnetic fields, I would use a light bulb,
because that wire (especially when the 20 meter is wound) will be a coil
of its own, with its own magnetic field, interfering with your setup!
I presume you meant 20 meter thick versus 5 meter thin.
Thanks. It explains why I fried my power transistors 2 times
already. Well if I never go below 60 Hz but go higher to 10Khz.
Would it also fry the circuit?? I notice a high pitch sound in
the circuit when the frequency is increased linearly from 60 Hz,
to 10 kHz.
At higher [switching] frequencies the switch time of the transistors
becomes important (they will dissipate heat while not 100% on or off),
and also the losses in the core of the transform will increase.
Then there is 'skin effect' in the wire (electricity only flows in outside).
For this reason above say 3 kHz you will often see ferroxcube (ferrite)
cores.
And possibly litze wire (wire made up of many strands).


Many thanks for this important information. I'll attempt to
build my fried circuit for the third time and just use frequency
of 60 hz and above (to 20Khz). I hope it won't fry anymore.

So many thanks for all the information. I have enough to carry
on and read the Resnick book with perspective.

emma
 
E

emma

Jan 1, 1970
0
St. John Smythe said:
You fellows are being trolled big-time. Can't you see it?

You thought that way because I asked too many questions. Well.
I have gotten the necessary and important information already
so I'll carry on myself with the help of Resnick thick book on
electromagnetism and not ask too many questions (that is starting
to irritate some folks).

Sorry and many thanks.


emma
 
S

Strange Indeed

Jan 1, 1970
0
Clearly matches one model but has been amateurish in execution when
compared to the other trolls who are regulars here.

You thought that way because I asked too many questions. Well.
I have gotten the necessary and important information already
so I'll carry on myself with the help of Resnick thick book on
electromagnetism and not ask too many questions (that is starting
to irritate some folks).

Resnick is too advanced for you.
Sorry and many thanks.


emma

Nice, this, but futile. It is clear you didn't actually understand
any of the answers.
 
O

Old Man

Jan 1, 1970
0
Semantics. Write the equation.
2. 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.

No. Simple addition doesn't cut it. In a series LC circuit,
voltages are added in quadrature. The voltages across L
and C (+ R_load) aren't in phase.
3. Of course there's a loss in power.

No. R_ series = 0, R__parallel = infinite. Give the equation.
All of the power goes into the resistive load. None elsewhere.
Where do you think all the
energy in the harmonics went, into the fundamental?

Exactly. Energy and power are conserved.
John Fields
Professional Circuit Designer

Professional ? Fields displays delusions of competence.

[[Old Man]
 
O

Old Man

Jan 1, 1970
0
Only for a resistor wherein current and voltage are in phase.
The voltage across C / L leads / lags the current by pi / 2.
Here's the equation that Fields should have written

Power = V * I * cos( phase)

for L or C phase = + pi / 2 or - pi / 2. In both cases,
Power = 0. Write the equation. Go figure.
No. Simple addition doesn't cut it. In a series LC circuit,
voltages are added in quadrature. The voltages across L
and C (+ R_load) aren't in phase.

---
You still don't get it, huh? Who said anything about a series LC
circuit? What we're talking about is using a lowpass filter to
eliminate/attenuate harmonics in a square wave in order to extract
the fundamenal. (That is, how to turn a square wave into a sinusoid
with the same period as the square wave) That means the filter is
going to look like this: (View in a non-proportional font.)

SQIN>--+--[L]--->>--+
| |
[C] [RL]
| |
GND>---+-------->>--+

No way. Cockeyed low pass filter. The capacitor is in parallel
with the load, not in parallel with the source. It is then a series
LC circuit with the load in parallel with the capacitor.
Since the reactance of the capacitor will decrease, and the reactance
of the inductor will increase as frequency increases, the high
frequency components of the input square wave will be attenuated by
the inductor and shunted to ground by the capacitor, allowing the
desired low frequency(ies) to pass through the inductor and into the
load.

The capacitor in fields cockeyed circuit dissipates no power.
Neither does the inductor.

Resonant circuit is Fields straw man.
---


---
Of course they're "conserved", but that doesn't mean that the energy
in the harmonics will be magically converted into the fundamental.
What will happen is that it will be shunted to ground through the
capacitor and converted into heat. Wasted, in other words.

Bullshit. Write the equation. "shunted to ground" is magic.
"ground" isn't a physical entity. The current flows back to
the source, pi / 2 out of phase with the source voltage. There
is no net power consumption. Go figure.

[Old Man]\
John Fields
Professional Circuit Designer

Professional ? Fields displays delusions of competence.

[[Old Man]
 
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