magnetic field

[Jan Panteltje]

On a sunny day (29 Jun 2005 15:13:58 -0700) it happened "emma"
<mrandmrsrelativity@yahoo.com> wrote in
<1120083238.071640.207610@g49g2000cwa.googlegroups.com>:

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).
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.

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?
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?

amplifier --- 3 Ohm -- 1 Ohm coil -- ground

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.
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 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 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).

If you want only frequencies in the audio range, use the PC sound card,
and a good sound editor.

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.
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.

What do you call power amps where it is not used for audio only
but also for function gen amplification??
I would call a 10MHz power amp a power amplifier with 10MHz bandwidth....

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??
Correct.

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?
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 think the perfect setup for me is to get a variable frequency
sine wave power inverter.
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?

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.
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?

 

[Vidar Lřkken]

Jan Panteltje wrote:

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.

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.

--
MVH,
Vidar

www.bitsex.net

 

[Vidar Lřkken]

cheian07@yahoo.com wrote:

good day!!!

when i test my newly installed lan, i have a problem setting up its
connections.I thought i can view all the stations or they will
recognize each other. but when i open network neighborhood, what
happened was there were groups of computer that recognized each others.
say S-1, S-2, S-3,S-5. The other groups of computer recognized
themselves. ex. S-4,S-6,S7.They are all connected to a single ethernet
hub but why is it that they seemed to group into 2?
We have 3 hubs here interconnected with the 3rd hub connected to a
router (connected to a DSL line).When i try typing IPCONFIG at the
command prompt, the first group of computers has no default gate way in
them so it did'nt recognize other computers connected to other hubs.
The second group of computers has a DEFAULT GATEWAY IN IT so it
recognizes other computers connected to other hubs including the hub
which is connected to the DSL line.
Any help would be gladly apprciated. thank you

Just to clearify some terms:
IP: The adress of the host. Can be anything, but should be either a
address range you own, or one of the private (192.168.0.0/255.255.0.0 or
10.0.0.0/255.0.0.0)
Netmask: Netmask is included above. Think as the netmask as a filter.
A ip is a 32 bit address, so is the netmask, so a mask of 255.255.255.0
filters out all, but the last bit. If the address your box is trying to
reach for is outside the last part, it will go to the gateway. So the
gateway is ONLY applied whenever you're trying to reach for something
outside the scope of the netmask.
To help, it'd be usefull if you said what IP range you used, and your
netmask.
Normally, you'll only ask for GW whenever you want to pass on to the
internet. Typycailly, the GW is the DSL modem.

Btw, I hope you didnt buy new hubs...

ian

--
MVH,
Vidar

www.bitsex.net

 

[John Fields]

On 29 Jun 2005 07:13:58 -0700, "Manny" <cktmanny@yahoo.com> wrote:

I have a 12 volts 100 watts DC halogen lamp. From P=VI,
I (current) = P/V = 100/12 = 8.33 Ampere

I used a multi-meter to check the resistance of the
lamp. It is 0.7 ohms.

---
That seems awfully high.
---

Now since it's 12 volts. From V=IR, I=V/R=12/0.7= 17A.

How come it's 17A when the calculated one in the first
sentence is only 8.33A?? Which is the right one?

---
They're both right. Tungsten (the metal the filament is made of) has
a positive temperature coefficient of resistance, so the hotter the
filament gets, the higher its resistance. If it has a cold resistance
of 0.7 ohms and you connect it to a 12V source, it will draw 17 amps
at the instant the connection is made, but as time goes by and the
filament gets hotter and hotter, its resistance will increase to the
point that it will be

E 12V
R = --- = ------- = 1.44 ohms
I 8.33A

When it's dissipating 100 watts.
---

When you use a multi-meter to test the resistance
in ohms of the lamp, is the reading accurate?

---
If you do it properly, yes.

For instance, when you measured 0.7 ohms as the cold resistance of the
lamp, did you subtract out the resistance of the test leads?

I have a 50 watt 12V halogen lamp I just checked, and it has a cold
resistance of 0.22 ohms, so I suspect your hundred watt lamp should
have a cold resistance of about half that, or around 0.1 ohms.


--
John Fields
Professional Circuit Designer

 

[John Fields]

On Wed, 29 Jun 2005 19:57:15 -0500, "Dave" <db5151@hotmail.com> wrote:

Not only that, but the filament is a COIL, which has inductive reactance to
AC. Your multimeter uses DC to measure resistance. Not the same. Trust
the calculations. They tell the truth.

---
What calculations???

It's essentially an air-wound coil with an inductance of less than a
microhenry, so the difference between the resistance and the impedance
of the lamp due to the inductive reactance at 60Hz is going to be so
insignificant that it can be ignored.

Even if it was a microhenry, we'd have, for the reactance:


Xl = 2pi f L = 6.28 * 60 * 1E-6 ~ 3.77E-4 ohms


and for a filament with a measured DC resistance of 0.1 ohm, an
impedance of:


Z = sqrt (R˛ + Xl˛) = sqrt (0.1˛ + 3.77E-4˛) = 0.1000007 ohms

--
John Fields
Professional Circuit Designer

 

[CWatters]

See thread "Emma".

 

[John Fields]

On Wed, 29 Jun 2005 13:18:41 -0500, "Old Man" <nomail@nomail.net>
wrote:

"Tm" <smiller615@nospamcast.net> wrote in message
newsI6dnfn9gcjPvV_fRVn-iw@comcast.com...

Low Pass will do it. But you will need more than a 110 volt square wave to
start with if you expect to get a 110 volt sine.

No loss in RMS power.

In principle (R_ series = 0, R__parallel = infinite), an LC
low pass filter is lossless.

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

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.

3. 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

 

[Jan Panteltje]

On a sunny day (30 Jun 2005 07:50:11 -0700) it happened "emma"
<mrandmrsrelativity@yahoo.com> wrote in
<1120143011.829616.107460@g44g2000cwa.googlegroups.com>:

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.
OK

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??
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 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.
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.

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.
You are welcome.

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.
Neither do I, what is a 3D sensor?

Ask him some time, I'd like to know.

 

[CWatters]

"emma" <mrandmrsrelativity@yahoo.com> wrote in message
news:1120143011.829616.107460@g44g2000cwa.googlegroups.com...

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

Na that won't pass the Turing test.

 

[CWatters]

"Manny" <cktmanny@yahoo.com> wrote in message
news:1120120151.483526.278950@g44g2000cwa.googlegroups.com...

Thanks for the help guys.

Another inquiry. I'm confused about something. Say I have 2 watts
resistor. If I plug this directly to the ac outlet with voltage of 110
volts. What would happen.

That depends on the value of the resistance R which you don't mention.

How do you tie it to V=IR?

Well once you have decided on R try inserting it in I=V/R.

Also what must be the resistance?

The value of the resistor R will depend on what you are trying to achieve.

 

[CWatters]

Lets try an example.

Suppose we play with 9V DC as it's safer...

Late say you have a 1W light bulb rated for 6V that you would like to
connect to the 9V supply. What you need is a resister in series with the
lamp to drop the voltage from 9V down to 6V (eg by 3V). Here is how you
calculate the value and power rating for the resistor....

First calculate how much current the lamp will draw at 6V....

Power (W) = Current (I) * Voltage (V) so...

I = W/V

Now W=1 and V=6 so
I=0.1667 Amps.

Now in the new circuit on 9V with a dropping resistor.... We want the
Resistor (R) to drop 3 Volts (V) when the same current (I) is flowing
through it and the lamp...

V=I*R or
R=V/I

V = 3
I = 0.1667 so

R = 18 Ohms approx.

Now what power rating does the resitor need to have...

The power (Wr) burnt in the resistor is given by

Wr = V*I

V = 3
I = 0.1667 so

Wr = 0.5 Watts

Best use a 1W resistor because a 0.5 watt resistor would be operating at
it's limit and that needs good airflow.

 

[CWatters]

"me" <me@here.net> wrote in message
news:Xns968474FCAB639meherenet@38.119.71.210...

"CWatters" <colin.watters@pandoraBOX.be> wrote in
news:Utwwe.132966$si1.6974535@phobos.telenet-ops.be:

If we were talking about 110V DC (which you probably aren't) you could
built a DC-DC converter to turn the 110V into say 275V.


um, that's changing the source...

So add a bridge rectifier and a capacitor up front. You want fries with
that.

 

[CWatters]

See thread "Emma"

 

[ehsjr]

kell wrote:

redbelly wrote:

CWatters wrote:


http://www.vishay.com/document/70596/70596.pdf

Thank you.


view in proportional font (if you are using google, click on "show
original")

this circuit gives very tight current regulation:
load n-channel
(thermistor) mosfet R (see below)
_____/\/\/\_________ _______________/\/\/\____________
| _|___|_ _|_ | |
| _ / \ | |
| |_______/ \____________| |
| | c e |
| / npn transistor |
| \ |
| / 1K |
|_____________________\ |
| ground
__|__
_
_____ battery
_ or power supply
_____
_
|
|
ground

for R start with a value of .6 divided by desired current and adjust
until you get the current you want

This is a sink. If you want a source you could turn everything upside
down and do it with a p-channel mosfet and a pnp transistor, and put
the thermistor grounded.

The base-emitter voltage of the current-sensing transistor varies with
temperature -- the only source of any significant inaccuracy in the
circuit, as long as your power supply has stable voltage. Regulation's
pretty independent of load. You should look up the b/e temperature
coefficient for a bjt, I don't remember offhand.

It's better, not simpler. None of the circuits discussed
in this "simpler" subthread will meet the OP's specs. He
wants .1 mA, and AFAIK the best he can expect with a
gate-source connected FET is about .2 mA. It gets worse.

He's going to have to face some hard realities. At 100 uA,
if it can be achieved, physical construction becomes critical.
Assuming 5V Vcc, a 100 meg resistance draws 50 uA. His spec
was .5% accuracy. 100uA * .005 = .5 uA
So if dirt/humiditity/whatever on his pc board yields a 100
meg or lower path for current to follow, he's blown away by
two orders of magnitude, or more.

Frankly, I'm clueless as to how to meet his specs. You
could use an op amp to get even better current regulation,
but I don't know how to accomodate that on a PC board with
those specs. Thompson could design it inside a chip,
combining the thermistor function with regulation and
whatever else, and yielding an output that is not critical.
I don't think it can be done with discretes.

Ed

 

[John Fields]

On 30 Jun 2005 10:10:46 -0700, curiousjohn4@yahoo.com wrote:

emma wrote:
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 ]

---
No, the equation is:

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

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.

---
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?
---

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.

---
Since your filter has a series capacitor (7000ľF, remember?) won't
that block DC?
---

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.

---
No, it isn't, since both stages are continuously biased into
conduction in order to eliminate crossover distortion. That bias
doesn't exist in class B, so that quiescent power is never dissipated,
making class B more efficient than class AB.
---

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

--
John Fields
Professional Circuit Designer

 

[Thomas Magma]

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

"emma" <mrandmrsrelativity@yahoo.com> wrote in message
news:1120042632.423989.148510@z14g2000cwz.googlegroups.com...

Jan Panteltje wrote:
On a sunny day (28 Jun 2005 18:37:11 -0700) it happened "emma"
mrandmrsrelativity@yahoo.com> wrote in
1120009031.592710.272250@g43g2000cwa.googlegroups.com>:

Hi,

I learnt that square waves for examples produced by power
interver has high frequency components.
Any square wave can be thought of as an infinite amount of sine waves.
In practice you will have some harmonics, f, 2f, 3f, 4f etc.
Look up Fourier analysis.

Some inverters make a near square wave, some a modulated pulse width
that is filtered into an approximate sine wave, some just combine some
smaller square waves to make something that looks a bit like a sine.
Some make really good sine waves.


What's the typical
value of the frequency supposed the source is 110 volts,
60 hertz.
120, 180, 240, 300, etc, with varying amplitude.

Is there no way to remove the high frequency
components? How does it affect the load?
Yes, LC filter, but you will lose some power.

So the high frequency can be removed. I thought they can't.
How complicated is the LC filter design to remove 100% of
the high frequency component in the square wave?


If you for some reason need a pure sine wave, get a converter that does
that.

Many applications / appliances do not need a real good sine wave, as
these
have their own power processing / power supplies - those could handle
square
waves even.
In fact in some cases square wave could be beneficial.

In pure line Alternating Current (sine wave) from power companies.
Is there any high frequency component in the sine wave or is it
pure 100% 60 Hz AC??

emma

 

[John Fields]

On 30 Jun 2005 10:18:41 -0700, "DJ Craig" <spit@djtricities.com>
wrote:

Forgive me if I've posted this message in the wrong group. I'm not
really sure where I should post this.

I'm a mobile DJ. I often DJ outside people's houses, or at hotels,
camp grounds, appartment blocks, churches, etc. These places usually
will let you draw about 2000 watts from one of their circuits before
you blow the circuit breaker. My equipment requires between 4000 and
5000 watts. The lighting effects, fog machine, etc, requires about
2000-2500 watts, and the audio equipment requires about 2000-2500
watts. I have about 1000 watts-worth of equipment that *must* remain
on througout the event, such as CD players, mixer, amp and speakers.
It is a nightmare trying to distribute my equipment across multiple
circuits so that I dont blow fuses, which I often do. It is not too
bad if the lights and fog machine turn off during an event, but it is
important that my audio equipment keeps running thoughout an event.

I thought that it may be possible to build a gadget that would hook
into 2 circuits, where i would plug my audio equipment into the
"reliable" plug, and the lighting into the "unreliable" plug. If
either of the two circuits blow, then the box would switch the audio
equipment over to the remaining circuit, and turn off the lighting.
Does such a gadget already exist, or would it be possible, and
practicle to build?

One issue is that if my CD players are turned off, even for an instant,
the music will stop and I have to re-cue the music. So a box that just
uses a relay to switch the circuits may not work. I'm not sure, but
the CD players may not be able to handle the delay caused by the relay.
Especially if the voltage from the circuit that is blowing fades
before it goes right out, creating even more delay and causing the CD
player's condensers to empty. Could it be done with a triac instead?
Would a triac be able to handle the high wattage?

Can anyone think of another solution to my problem? I always try my
best to hook into as many circuits as possible, and distribute my
equipment evenly, but I still have trouble.

---
I like the relay idea to switch the lights and audio stuff, but I'd
get an uninterruptible power supply for the CD players and wire its
input to the audio side of the relay. That way, even if you lose
power and it takes the relay a few tens of milliseconds to switch the
audio stuff to the good line, the CD player would have power supplied
to it all the time and it would just keep playing.

I'd use a DPDT relay with contacts rated for 30 amps and hook it up
like this:



L2>----+----------------------------+
| |
| |
L1>----|----------------+ |
| | |
O-->\ <--O O-->\ <--+
NO \ NC |NO \ NC
O | O--+----[AUDIO STUFF]
| | | |
[LAMPS] [COIL] +--[UPS]---|---[CD PLAYERS]
| | | |
| | | |
NEUT>--------+----------+-------------+-----+

That way, when you plug the thing into line 1 (L1) the relay coil
(120VAC) will be energized and will cause the common contacts to move
over to the normally open (NO) contacts. That will cause the lamps to
be connected to L2 and the critical stuff to L1.

Now, if L2 drops out the lights will go out, but the critical stuff
will keep going. However, if L1 drops out the coil will no longer be
energized, which will cause the common contacts to fall back on the NC
contacts. That will disconnect the lamps from L2 while connecting the
critical stuff to L2, and the UPS on the CD players will keep them
from getting the slightest hint that anything happened, which is what
you want.

There is one small snag though, and that's that there's no guarantee
that neutral and hot won't be reversed on on of the sockets you plug
into, so uless you keep everything separate, nasty shit could happen.
Fortunately, there's an easy way out and that's to use another relay
to switch the neutrals, like this:

L2>----+----------------------------+
| |
| |
L1>----|----------------+ |
| | |
O-->\K1A <--O O-->\K2A <--+
NO \ NC |NO \ NC
O | O----+----[AUDIO STUFF]
| | | |
| +--+----+ | |
| | | | |
[LAMPS] [COIL1] [COIL2] +--[UPS]---|---[CD PLAYERS]
| | | | |
| +-------+--------+-+-----+
| |
O O
NO / NC NO / NC
O-->/K1B <--O O-->/K2B <--O
| | |
N1>----|------------------------+ |
| |
| |
N2>----+------------------------------------+

A good relay would be an OMRON MGN2C AC120, $29.70 each at Digi-Key.

--
John Fields
Professional Circuit Designer

 

[John Fields]

On 30 Jun 2005 11:04:53 -0700, curiousjohn4@yahoo.com wrote:

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.

---
Since your filter has a series capacitor (7000ľF, remember?) won't
that block DC?
---

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

---
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.

--
John Fields
Professional Circuit Designer

 

[John Fields]

On 30 Jun 2005 12:46:58 -0700, "kell" <kellrobinson@billburg.com>
wrote:

He's going to have to face some hard realities. At 100 uA,
if it can be achieved, physical construction becomes critical.
Assuming 5V Vcc, a 100 meg resistance draws 50 uA.

---
E 5V
I = --- = ------ = 5E-8a = 0.05ľa
R 1E8R
---

His spec
was .5% accuracy. 100uA * .005 = .5 uA
So if dirt/humiditity/whatever on his pc board yields a 100
meg or lower path for current to follow, he's blown away by
two orders of magnitude, or more.

---
No, at 100 megohms he's an order of magnitude better than the spec.
---

Frankly, I'm clueless as to how to meet his specs. You
could use an op amp to get even better current regulation,
but I don't know how to accomodate that on a PC board with
those specs. Thompson could design it inside a chip,
combining the thermistor function with regulation and
whatever else, and yielding an output that is not critical.
I don't think it can be done with discretes.

I can't find the original post. What purpose does the OP have in mind
for this circuit... maybe he'll find a solution off the shelf, like a
temperature sensor chip.

--
John Fields
Professional Circuit Designer

 

[Vidar Lřkken]

emma wrote:

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.


--
MVH,
Vidar

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