Chip with simple program for Toy

[Bob Masta]

On Mon, 14 Apr 2008 22:10:49 -0700 (PDT), Delsol
<Aaronofkent@gmail.com> wrote:

On Apr 14, 3:11 pm, mrdarr...@gmail.com wrote:
So I'm looking at this ($149.99 at Sears)

http://www.sonystyle.com/webapp/wcs/stores/servlet/ProductDisplay?cat...

Manual says 8-16 ohm impedance, and I'm wondering how bad will it be
if I connect my 4-ohm subwoofers to it?

Are we talking Blue Smoke here?

Michael


No blue smoke, but you wont be sending the maximum power to the
speakers (and thus hurting yourself in the long run).

Check out "maximum power transfer", it basically says that impedance
in must equal impedance out for power to be transferred effectively.

http://en.wikipedia.org/wiki/Impedance_matching

The old "impedance matching" thing almost certainly doesn't apply here
because the amp specs are not based on output impedance. (Unless this
amp has an output transformer... which pretty much went out with
vacuum tubes.) Modern solid state amps have output impedances that
are near zero at low frequencies.

The "8-16 ohms" rating is more likely based on power handling
capability of the output stage. If you use a lower impedance load, it
takes more current at lower voltage to get the same power, which
means the output stage has to dissipate more power. Like a pair of
voltage regulators with changing setpoints, the positive and negative
output devices have to pass the output current while dropping the
voltage difference between the supply voltage and the output voltage.

Let's take a simple example: Say the supply is 40 VDC and at some
instant the positive output device of the amp has to deliver 50W into
its load. (No RMS here, just instantaneous power.) So with an 8 ohm
load the output is 20V to give 20^2/8 = 50W, and the device itself has
40-20 = 20V across it, while the current through it is the same as the
current through the load = 20/8 = 2.5A. So the device has to
dissipate 2.5 * 20 = 50W.

Now consider a 4 ohm load. First of all, you need to turn the volume
control down to get the same 50W into the load, since now the output
voltage must be 14.14 since 14.14^2/4 = 50W. So the device now has
40-14.14 = 25.86V across it. But its current is 14.14/4 = 3.535 and
the power it dissipates is 25.86*3.535 = 91.4W.

So if the power rating of the amp is based on the power that the
output devices can handle, you'd need to derate it significantly to
use a 4 ohm load. Some amps list separate power ratings for 4 ohms
(and sometimes 2 ohms). If the 4 ohms rating is less than the 8
ohm, the output devices are the limiting factor. If the amp is more
sturdily built (a pro amp, or better grade home amp) the 4 ohm rating
will be higher, indicating that the limit is more likely just the
supply voltage. (+/-40V gives 28VRMS, which into 8 ohms gives 100W,
or into 4 ohms gives 200W.)

But this amp wasn't rated for 4 ohms, so you may have to derate it.
And you'd be taking a chance that someday somebody didn't crank up
the volume too high.

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[Bob Masta]

On Mon, 14 Apr 2008 22:10:49 -0700 (PDT), Delsol
<Aaronofkent@gmail.com> wrote:

On Apr 14, 3:11 pm, mrdarr...@gmail.com wrote:
So I'm looking at this ($149.99 at Sears)

http://www.sonystyle.com/webapp/wcs/stores/servlet/ProductDisplay?cat...

Manual says 8-16 ohm impedance, and I'm wondering how bad will it be
if I connect my 4-ohm subwoofers to it?

Are we talking Blue Smoke here?

Michael


No blue smoke, but you wont be sending the maximum power to the
speakers (and thus hurting yourself in the long run).

Check out "maximum power transfer", it basically says that impedance
in must equal impedance out for power to be transferred effectively.

http://en.wikipedia.org/wiki/Impedance_matching

The old "impedance matching" thing almost certainly doesn't apply here
because the amp specs are not based on output impedance. (Unless this
amp has an output transformer... which pretty much went out with
vacuum tubes.) Modern solid state amps have output impedances that
are near zero at low frequencies.

The "8-16 ohms" rating is more likely based on power handling
capability of the output stage. If you use a lower impedance load, it
takes more current at lower voltage to get the same power, which
means the output stage has to dissipate more power. Like a pair of
voltage regulators with changing setpoints, the positive and negative
output devices have to pass the output current while dropping the
voltage difference between the supply voltage and the output voltage.

Let's take a simple example: Say the supply is 40 VDC and at some
instant the positive output device of the amp has to deliver 50W into
its load. (No RMS here, just instantaneous power.) So with an 8 ohm
load the output is 20V to give 20^2/8 = 50W, and the device itself has
40-20 = 20V across it, while the current through it is the same as the
current through the load = 20/8 = 2.5A. So the device has to
dissipate 2.5 * 20 = 50W.

Now consider a 4 ohm load. First of all, you need to turn the volume
control down to get the same 50W into the load, since now the output
voltage must be 14.14 since 14.14^2/4 = 50W. So the device now has
40-14.14 = 25.86V across it. But its current is 14.14/4 = 3.535 and
the power it dissipates is 25.86*3.535 = 91.4W.

So if the power rating of the amp is based on the power that the
output devices can handle, you'd need to derate it significantly to
use a 4 ohm load. Some amps list separate power ratings for 4 ohms
(and sometimes 2 ohms). If the 4 ohms rating is less than the 8
ohm, the output devices are the limiting factor. If the amp is more
sturdily built (a pro amp, or better grade home amp) the 4 ohm rating
will be higher, indicating that the limit is more likely just the
supply voltage. (+/-40V gives 28VRMS, which into 8 ohms gives 100W,
or into 4 ohms gives 200W.)

But this amp wasn't rated for 4 ohms, so you may have to derate it.
And you'd be taking a chance that someday somebody didn't crank up
the volume too high.

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[Bob Eld]

"Delsol" <Aaronofkent@gmail.com> wrote in message
news:8a269971-72f6-4072-b8c9-c37a729dc381@2g2000hsn.googlegroups.com...

On Apr 14, 3:11 pm, mrdarr...@gmail.com wrote:
So I'm looking at this ($149.99 at Sears)

http://www.sonystyle.com/webapp/wcs/stores/servlet/ProductDisplay?cat...

Manual says 8-16 ohm impedance, and I'm wondering how bad will it be
if I connect my 4-ohm subwoofers to it?

Are we talking Blue Smoke here?

Michael


No blue smoke, but you wont be sending the maximum power to the
speakers (and thus hurting yourself in the long run).

Check out "maximum power transfer", it basically says that impedance
in must equal impedance out for power to be transferred effectively.

http://en.wikipedia.org/wiki/Impedance_matching

Wrong! Modern amplifiers just like your wall plugs are constant voltage.
Impedances are NEVER matched. The output impedance of a modern amplifier
like your wall plugs is near zero ohms, a few hundredths of an ohm.

Matched impedances do transfer maximum power but they also dissipate as much
power in the source as is delivered to the load. That condition is very
wasteful and unnecessary. As an example consider operating a 100 Watt light
bulb on a matched impedance basis. The impedance would be: Z = V/(sqrtP) =
120/10 = 12 ohms and the wall plug would also have to be 12 ohms if it was
matched. A 100 Watts would go to the light and another 100 Watts would be
wasted in the 12 ohms at the generator. That's absurd! The same thing is
true of amplifiers. You are trying to deliver power to the speakers not
dissipate just as much in the amplifier.

Matched impedances are ONLY used to avoid reflections of energy in RF work
and to properly load vacuum tube plates. Delivering power is best done
constant voltage.

 

[Tim Woodall]

On Tue, 15 Apr 2008 17:25:37 -0500,
John Fields <jfields@austininstruments.com> wrote:

On Tue, 15 Apr 2008 12:16:17 -0700 (PDT), chesemonkyloma@gmail.com
wrote:


OK you guys have helped me but what I still don't get is: Why does the
equation V^2/R produce a different power than I^2*R,

---
It doesn't.

And it's easy to prove from V=IR

V^2/R = V * V / R = IR * IR / R = IR*I = I^2*R

Tim.

--
God said, "div D = rho, div B = 0, curl E = - @B/@t, curl H = J + @D/@t,"
and there was light.

http://tjw.hn.org/ http://www.locofungus.btinternet.co.uk/

 

[Tim Woodall]

On Tue, 15 Apr 2008 16:34:45 -0500,
John Fields <jfields@austininstruments.com> wrote:
<snip>

and to extend this a bit more:

If we want to deliver 10A to a one ohm load, then the circuit looks
like this:

I--->
+----------------------------------------------------+ <--------+
| | |
[SOURCE]10A [wire]r [LOAD]R [E]
| | |
+----------------------------------------------------+ <--------+

Now we've got some very long wires connecting our source to our load
with non-negligible resistance r.

So the source needs to have a voltage of 10*(1+r)

and the power we waste heating the wires is I * Vwire = 10 * 10 * r

We want to minimise this. One option is to make the wire thicker and
reduce r. But this gets expensive buying all that metal for the wires.

So instead we'll do the following with two identical (assumed perfect)
transformers

I--->
+---------+ +-----------------------------+ +------+ <--------+
| P||S S||P | |
[SOURCE]10A R||E [wire]r E||R [LOAD]R [E]
| I||C C||I | |
+---------+ +-----------------------------+ +------+ <--------+


No assume that SEC has 10x as many turns as PRI. So the voltage in the
middle section will be 10x what it is at the source and the load.

Because the voltage is 10x and energy must be conserved, the current in
the middle section must be 1/10 the current from the source.

When we convert back we go back to 1/10 the voltage, so we must have 10x
the current, again to conserve energy.

So now the voltage drop in the wire is 10*1/10*r and the power loss is
10*1/10 * 10*1/10*r = r

So we've reduced the power lost from 100r to r just by adding two
transformers.

The main problem with using higher voltages is that they are harder to
keep in the wires. You're going to have to have better insulators
supporting or surrounding the wire etc. So there comes a point where the
money you save by reducing the current is offset by the extra money you
spend insulating the wires.

I think the grid goes up to 650000V. So (approximately) 3000x the wall
voltage. That means that the current in those wires is about 1/3000 of
what the end user is using and so the power wasted heating the wires is
about 1/9000000 of what it would be if it was 240V all the way from
powerstation to home.



One final point. It's important to realise that transformers conserve
energy. In the ideal case, if the secondary is open circuit and no
current is flowing then no power is being drawn from the transformer so
no energy is going into the primary either.

So here:

+---------+ +----------
| P||S
[SOURCE] R||E
| I||C
+---------+ +----------

There is no power being drawn from the source. (It's not true to say
there is no current flowing in the primary, just that there is no power
being drawn from the source)

In practice there are losses in the transformer which is why a wall wart
still gets warm if it is left plugged in with no load attached. But you
may have noticed that it gets warmer when there is a load attached - of
if you plug one into a power meter, you'll be able to see the power
being drawn from the mains increase as the power drawn from the load
increases.


Tim.

--
God said, "div D = rho, div B = 0, curl E = - @B/@t, curl H = J + @D/@t,"
and there was light.

http://tjw.hn.org/ http://www.locofungus.btinternet.co.uk/

 

[Doug Miller]

In article <48051F90.B637A2DA@hotmail.com>, Eeyore <rabbitsfriendsandrelations@hotmail.com> wrote:

robert.janczak@poczta.fm wrote:

And you just *had* to repost the whole spam, didn't you?

The original spam didn't make it past my filters, but your repost did.

Thanks a lot, dumbass.

--
Regards,
Doug Miller (alphageek-at-milmac-dot-com)

Get a copy of my NEW AND IMPROVED TrollFilter for NewsProxy/Nfilter
by sending email to autoresponder at filterinfo-at-milmac-dot-com
You must use your REAL email address to get a response.

Download Nfilter at http://www.milmac.com/np-120.exe

 

[Bob Masta]

On Tue, 15 Apr 2008 17:59:47 +0100, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:

mrdarrett@gmail.com wrote:

Would sound quality suffer if I put two speakers in series?

Yes. Reduction of damping factor.

Graham

Yes and no.

Technically, the damping factor is the rated load impedance divided by
the actual output impedance. It's just a way of specifying the output
impedance of the amp, which (as has been pointed out already) for
modern amps is very low. So if an amp has a damping factor of (say)
100 with an 8 ohm load, then the output impedance is 0.08 ohm.
(Note that damping factor almost always falls with frequency, but the
OP is talking about subwoofers so there is no need to worry about that
here.)

Putting 2 drivers in series doesn't affect the amp's output impedance.
However, each speaker is now being driven through an additioanl 4 ohms
(nominally), so from the speaker's point of view the effective damping
factor has dropped substantially. Nevertheless, this may or may not
be a problem, depending on the particular speakers. If the OP already
has the amp and speakers, he should try it. Otherwise, why buy this
amp? The stated price doesn't look all that attractive.

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[Bob Masta]

On Tue, 15 Apr 2008 17:59:47 +0100, Eeyore
<rabbitsfriendsandrelations@hotmail.com> wrote:

mrdarrett@gmail.com wrote:

Would sound quality suffer if I put two speakers in series?

Yes. Reduction of damping factor.

Graham

Yes and no.

Technically, the damping factor is the rated load impedance divided by
the actual output impedance. It's just a way of specifying the output
impedance of the amp, which (as has been pointed out already) for
modern amps is very low. So if an amp has a damping factor of (say)
100 with an 8 ohm load, then the output impedance is 0.08 ohm.
(Note that damping factor almost always falls with frequency, but the
OP is talking about subwoofers so there is no need to worry about that
here.)

Putting 2 drivers in series doesn't affect the amp's output impedance.
However, each speaker is now being driven through an additioanl 4 ohms
(nominally), so from the speaker's point of view the effective damping
factor has dropped substantially. Nevertheless, this may or may not
be a problem, depending on the particular speakers. If the OP already
has the amp and speakers, he should try it. Otherwise, why buy this
amp? The stated price doesn't look all that attractive.

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[Paul E. Schoen]

"Hammy" <spamme@hotmail.com> wrote in message
news:vudc04ptl32u3iaeqctm8umu9lfg0k5962@4ax.com...

On Wed, 16 Apr 2008 17:07:12 GMT, Hammy <spamme@hotmail.com> wrote:

Here are the simulated waveforms if it explains more clearly what I'm
doing.

http://i31.tinypic.com/15sao2g.png

Top trace (RED ID3) the switch is closed : 50mA current pulse for 1mS
steady state 20mA.Through the first load D2 and D3.

Second trace ( RED I(U12:4)) is the mosfet drain of U12 conducting
from the battery until Q20 is biased and turns it off. The 7805 is now
supplied by a DC/DC converter represented by V2 pulsed source in the
schematic.

Bottom trace (GREEN I(R12)) shows the second load conducting at 135mS
(switch open).

It would help if you explained what you are trying to do. It seems like a
lot of extra circuitry to switch among various sources and loads, and I
don't understand why you are seemingly trying to turn on a transistor to
set Vgs of a PMOS to zero to turn it off, rather than just have a resistor
from gate to source and pull the gate to GND to turn it on.

If you can post the circuit in LTSpice ASCII, it might help.

Also, I did not see any power supply or battery (other than the pulse
sources), and there is only one capacitor, which seems to be a bypass for a
power source that is not identified. The 7805 should have bypass
capacitors.

LTSpice has a voltage controlled switch component, but it only works for DC
in one direction. And you must set some of the parameters for it to work.

Paul

 

[Paul E. Schoen]

"Hammy" <spamme@hotmail.com> wrote in message
news:vudc04ptl32u3iaeqctm8umu9lfg0k5962@4ax.com...

On Wed, 16 Apr 2008 17:07:12 GMT, Hammy <spamme@hotmail.com> wrote:

Here are the simulated waveforms if it explains more clearly what I'm
doing.

http://i31.tinypic.com/15sao2g.png

Top trace (RED ID3) the switch is closed : 50mA current pulse for 1mS
steady state 20mA.Through the first load D2 and D3.

Second trace ( RED I(U12:4)) is the mosfet drain of U12 conducting
from the battery until Q20 is biased and turns it off. The 7805 is now
supplied by a DC/DC converter represented by V2 pulsed source in the
schematic.

Bottom trace (GREEN I(R12)) shows the second load conducting at 135mS
(switch open).

It would help if you explained what you are trying to do. It seems like a
lot of extra circuitry to switch among various sources and loads, and I
don't understand why you are seemingly trying to turn on a transistor to
set Vgs of a PMOS to zero to turn it off, rather than just have a resistor
from gate to source and pull the gate to GND to turn it on.

If you can post the circuit in LTSpice ASCII, it might help.

Also, I did not see any power supply or battery (other than the pulse
sources), and there is only one capacitor, which seems to be a bypass for a
power source that is not identified. The 7805 should have bypass
capacitors.

LTSpice has a voltage controlled switch component, but it only works for DC
in one direction. And you must set some of the parameters for it to work.

Paul

 

[Bob Eld]

<chesemonkyloma@gmail.com> wrote in message
news:786a05af-57b3-44b8-80f8-e408f04cdeb7@m44g2000hsc.googlegroups.com...

My sister would like to use a light on the end of a string, and I was
wondering how that could be made. I was thinking of putting a bright
LED, and resistor, and some hook-up wire in a clear, plastic ball or
even in some Saran Wrap with a battery to power the circuit. This may
be completely implausible, I don't know.

When I was in College one of the guys in the dorm hung a bare 60 watt
incandescent light bulb, 120Volts, from the concrete ceiling of his room on
a piece of string. The bulb would just hang there lit, fully bright, but
looking at it you couldn't figure out how it was powered. It was a great
conversation piece and kept a lot of people guessing. A bare bulb just
hanging there on a sting without a socket or apparent wires burning away, it
was great.

What he had done is twisted some very thin magnet wire, like 36 Ga. into the
string and soldered the ends to the bulb base where the string was tied. The
wires went up the string then across the concrete ceiling hidden in the
natural cracks and roughness of the painted concrete. The string was taped
to the ceiling. You could not see the wires without climbing up and very
closely inspecting the situation. The wires came down one wall to a wall
plug where other stuff was plugged in concealing the termination.

I wouldn't recommend this as it could be a fire hazard but it sure would be
a hell of a lot more impressive than an LED. It takes a clever artistic
person to weave and conceal the thin wires. If not carefully done it would
be a dead giveaway if wires can be seen. If you try it be sure to never
leave this kind of thing plugged in unattended.

 

[Bob Eld]

"Archimedes" <shelton.dcruz@gmail.com> wrote in message
news:b3b30960-27bc-4d59-924f-4312caffe7c2@x41g2000hsb.googlegroups.com...

Hi all

If I change the tank circuit components (reduce the capacitance and
inductance) and change the transistors to appropriate VHF ones, will I
be able to pick up ATC (Air Traffic Conversations) using this
circuit ?

http://www.electronics-lab.com/projects/rf/006/index.html

Thanks
Shelton.

Not likely. I doubt you'd ever get that circuit to work at VHF frequencies
no matter what you did with the resonant circuit portion. There are many
problems including wrong impedances for the various parasitic capacitances.

Secondly air traffic stuff is FM I believe.

 

[Phil Allison]

"Archimedes"

" Thanks for all the reponses - yes i can calcualte the required
parameters to get this into the ATC band - but im now confused - some
say it will work some say it wont? So will it or will it not ? "


** IMO - very poorly, if at all.

You will need to be very close to an airport to hear anything and then you
will regularly hear several voices together as TRF sets have poor
selectivity.

VHF communications listening really requires the receiver to be a superhet -
preferably a double conversion one.

A small " scanner " receiver is ideal.



....... Phil

 

[Claude]

"Michael Black" <et472@ncf.ca> wrote in message
newsine.LNX.4.64.0804170103550.24809@darkstar.example.org...

On Wed, 16 Apr 2008, Bob Eld wrote:


"Archimedes" <shelton.dcruz@gmail.com> wrote in message
news:b3b30960-27bc-4d59-924f-4312caffe7c2@x41g2000hsb.googlegroups.com...
Hi all

If I change the tank circuit components (reduce the capacitance and
inductance) and change the transistors to appropriate VHF ones, will I
be able to pick up ATC (Air Traffic Conversations) using this
circuit ?

http://www.electronics-lab.com/projects/rf/006/index.html

Thanks
Shelton.

Not likely. I doubt you'd ever get that circuit to work at VHF
frequencies
no matter what you did with the resonant circuit portion. There are many
problems including wrong impedances for the various parasitic
capacitances.

Secondly air traffic stuff is FM I believe.

Military may use FM, I don't know, but airplane related communication is
unique in that it does use actual AM.

A project that saw publication a number of times in the old days took
advantage of that, a "crystal radio" that tuned VHF. It was nothing
more than a tuned circuit and a diode detector feeding an earphone, not
sensitive but useful near airports and since it didn't radiate anything,
even useable (though maybe not legally) on an actual airplane.

The description of the circuit says it's a regen receiver, and those
were never popular at VHF, I'm assuming instability came into play.
You did see superregen receivers there. Either type will radiate, and
that's not a good thing in the aircraft band.

Michael

Here are the facts

Military ground troups and close air support
30 Mhz to 87.975 FM only

Air traffic control
117.975 to 156.000Mhz, AM only

Maritime
156.000 to 173.975 ( with some reserved for Sonobuoy operations) FM only

Military
225 Mhz to 399.975 Mhz AM or FM ( they can actually choose wich one for any
freq in this band)
Not interesting in this band as there is a lot of encryption, freq hopping
and other hush hush stuff.

Claude

 

[Claude]

"Michael Black" <et472@ncf.ca> wrote in message
newsine.LNX.4.64.0804170103550.24809@darkstar.example.org...

On Wed, 16 Apr 2008, Bob Eld wrote:


"Archimedes" <shelton.dcruz@gmail.com> wrote in message
news:b3b30960-27bc-4d59-924f-4312caffe7c2@x41g2000hsb.googlegroups.com...
Hi all

If I change the tank circuit components (reduce the capacitance and
inductance) and change the transistors to appropriate VHF ones, will I
be able to pick up ATC (Air Traffic Conversations) using this
circuit ?

http://www.electronics-lab.com/projects/rf/006/index.html

Thanks
Shelton.

Not likely. I doubt you'd ever get that circuit to work at VHF
frequencies
no matter what you did with the resonant circuit portion. There are many
problems including wrong impedances for the various parasitic
capacitances.

Secondly air traffic stuff is FM I believe.

Military may use FM, I don't know, but airplane related communication is
unique in that it does use actual AM.

A project that saw publication a number of times in the old days took
advantage of that, a "crystal radio" that tuned VHF. It was nothing
more than a tuned circuit and a diode detector feeding an earphone, not
sensitive but useful near airports and since it didn't radiate anything,
even useable (though maybe not legally) on an actual airplane.

The description of the circuit says it's a regen receiver, and those
were never popular at VHF, I'm assuming instability came into play.
You did see superregen receivers there. Either type will radiate, and
that's not a good thing in the aircraft band.

Michael

Here are the facts

Military ground troups and close air support
30 Mhz to 87.975 FM only

Air traffic control
117.975 to 156.000Mhz, AM only

Maritime
156.000 to 173.975 ( with some reserved for Sonobuoy operations) FM only

Military
225 Mhz to 399.975 Mhz AM or FM ( they can actually choose wich one for any
freq in this band)
Not interesting in this band as there is a lot of encryption, freq hopping
and other hush hush stuff.

Claude

 

[Jon Slaughter]

<ghurki@gmail.com> wrote in message
news:4da5e605-d207-494a-a9e6-39ef650416bf@b64g2000hsa.googlegroups.com...

hi there!

few months back i decided to set up a new line of business and with
research and etc. i decided to enter into energy sector.. more
specifically i decided to jump into solar power. and importing solar-
powered products into my country and market them so as to reduce the
load on electricity..

for start up. i decided to work on solar powered street lights. with
some homework i came to know that LED is the latest thing in light and
its making its way very quickly and replacing the conventional
flourescent, incandescent and sodium bulbs.

however my problem is with the understanding to their power and
wattage and lumens. .i am really confused with all that.. top it all
if i dont understand these all calculations i cannot propose the
required solar panel and battery for any system..

can anyone please help me with this in very very simple language...

i have learned the first formula of (volt * amp = watt)
also i have discovered that there are 5mm and 10mm led pieces that are
commonly used in LED lights. however the availability of 1 watt one
piece, 3 watt one piece and 5 watt one piece LEDs is also
interesting..

also i dont know nothing about the circuits and etc.. have studied
physics in scshool which i hated the most.. but with few bits of
knowledge i have gained now i think it is not that bad.. i must be the
way i was taught.. anyways can anyone help me with that?

according to my research and email with the LED suppliers i have
learnt that 5mm LED produce 2-4 lumens (measuring scale of light) of
light and to produce 1 watt of LED light you will have to put 14
pieces of 5mm LED pieces.. Similarly one watt of LED is equvilent to 4
watt of normal light (the concept of energy savers) i.e. using lesser
energy and producing more light... can anyone validate that?

also with my friend electrician.. i have made some research which
concluded as: 5mm of LED with some power of resistnace consumes
12mAmp. now i want to know that how much amp would it require to lit
up 856 LED in either series or parallel.. i have just used the words
series and paralled.. just know a little bit about them..

similarly how many piece of 10mm LEDs are required to produce one watt
of LED light.. what is their curernt consumption?

what should i understand if one supplier of solar street light quotes
me that the lamp head has a 50 watt LED light.. how much light does
this LED produces when compared to normal light? how can i calculate
the curernt consumption of this light. and also how much power of
battery is required to light up this lamp head for 10 hours..?

my queries may sound silly to most,, but i am a new entrant into this
all..
help will be greatly appreciated?

haha... how bout get out of the market? Surely you should have some basic
understanding of what you are trying to do before you decide to jump in?
Maybe for the next 6 months learn some basic electronics(these are pretty
basic questions)? That would be the bare minimum.

 

[Ecnerwal]

In article
<261d06fa-a259-4162-90d1-778e8b79ea3a@p25g2000hsf.googlegroups.com>,
Brotherwarren <twarren@kesgrave.suffolk.sch.uk> wrote:

I have to pass a VGA display cable through a wall. To get the
connector through I'd have to drill a hole 1inch in diameter.

Big whoop - get a hole saw/core drill and have at it. Better yet, drill
2 inches and cover the next set of connections you'll think of later all
in one hole.

Is this advisable/possible?

Depends/Yes.

Am I able to do this?

Well, that depends a lot upon you. If you feel the need to ask a
newsgroup about it, I'd guess probably not. A degree of competence at
soldering is required, and if you already possessed it, I doubt you'd be
asking...

--
Cats, coffee, chocolate...vices to live by

 

[Ecnerwal]

In article
<261d06fa-a259-4162-90d1-778e8b79ea3a@p25g2000hsf.googlegroups.com>,
Brotherwarren <twarren@kesgrave.suffolk.sch.uk> wrote:

I have to pass a VGA display cable through a wall. To get the
connector through I'd have to drill a hole 1inch in diameter.

Big whoop - get a hole saw/core drill and have at it. Better yet, drill
2 inches and cover the next set of connections you'll think of later all
in one hole.

Is this advisable/possible?

Depends/Yes.

Am I able to do this?

Well, that depends a lot upon you. If you feel the need to ask a
newsgroup about it, I'd guess probably not. A degree of competence at
soldering is required, and if you already possessed it, I doubt you'd be
asking...

--
Cats, coffee, chocolate...vices to live by

 

[Ken Fowler]

On 17-Apr-2008, Brotherwarren <twarren@kesgrave.suffolk.sch.uk> wrote:

Hi folks:

Is it possible to solder 15-pin D-Suub connector onto a cable?

I have to pass a VGA display cable through a wall. To get the
connector through I'd have to drill a hole 1inch in diameter.

Instead what I'd like to do is pass a narrow cable through, then
solder the connector on once it reaches the other side.

Is this advisable/possible?

I thinbk I've found the cable and connector I need to use here:-

http://www.maplin.co.uk/Module.aspx?ModuleNo=1310&Criteria=15%20D%20Sub&C=GKW&U=Auto_2003513&T=15%20D%20Sub&gclid=CI_ExM2_4pICFQSU1Aodm1xb-Q

and here:-
http://www.maplin.co.uk/module.aspx?ModuleNo=230&doy=17m4


Am I able to do this?

thanks for any and all suggestions!

Tony

You could solder new connectors to the wire but it might be easier to cut a
cable in the center, push one end through the hole, strip back the wires
and splice the corresponding colors together. Use insulated sleeving over
the soldered splices and a layer of tape over the wires.

Ken Fowler, KO6NO

 

[Lord Garth]

"Brotherwarren" <twarren@kesgrave.suffolk.sch.uk> wrote in message
news:261d06fa-a259-4162-90d1-778e8b79ea3a@p25g2000hsf.googlegroups.com...

Hi folks:

Is it possible to solder 15-pin D-Suub connector onto a cable?

I have to pass a VGA display cable through a wall. To get the
connector through I'd have to drill a hole 1inch in diameter.

Instead what I'd like to do is pass a narrow cable through, then
solder the connector on once it reaches the other side.

Is this advisable/possible?

I thinbk I've found the cable and connector I need to use here:-

http://www.maplin.co.uk/Module.aspx?ModuleNo=1310&Criteria=15%20D%20Sub&C=GKW&U=Auto_2003513&T=15%20D%20Sub&gclid=CI_ExM2_4pICFQSU1Aodm1xb-Q

and here:-
http://www.maplin.co.uk/module.aspx?ModuleNo=230&doy=17m4


Am I able to do this?

thanks for any and all suggestions!

Tony

You can but it is a pain. The RGB signal lines within the plug are coaxial
cables.
Soldering can be done but it will require patience. You might opt for a 15
pin
D sub that accepts crimp connectors.

There are devices available that can send the VGA as well as mouse, keyboard
and
audio over one or more CAT5 cables. Sometimes these devices use their own
specialty cable to interconnect the equipment.
http://www.svideo.com/vgacat5.html one such item.