??? I assume your talking about the cable that conects the power

supply to

the machine itself??

No...sorry. I didn't mean to make you type out that long
explanation; I hope it helps someone else. The actual *input* for
the AC cable is messed up. Like on the cheap older keyboards - I
have to position the jack that goes into it just so to get it to
work. Is there a way to fix this?

?
So it's been a long week and I find that I have time on my hands. Firstly here are the solutions to the Simple Ohms law problems I set at the end of the last Email.

??? These questions are fairly straight forward, so I'm only going to show the working out on question one, and give the answers to the rest. If you got them wrong, and can't figure out where, let me know and I'll see if I can help you. Any way the solution for question one is as follows:

??? The remaining answers are as follows: Q.2 = 0.2A, Q.3 = 3.6mA, Q.4 = 50mA

??? It has occurred to me that If we have some true beginners on the list a brief explanation of the resistor color code might be useful. I'll start off with a quick picture of the color code, explanation to follow.

??? On most Resistors you see in electronics kits you will see four bands of colors. These colors allow us to know the value of the resistor. The above picture may or may not be totally confusing to you, but isn't really that hard to understand. The important part of the color band system, at least for most people actually building something, is the first three bands. I'll deal with the forth band and what it means later.

?????? The first two bands, are the ACTUAL "value" of the resistor. say we have a resistor with the following colors:????

??? So, we have?? Red - Red. If you look at the chart above, we will see that Red corresponds to the value "2", so Red - Red equals 2 - 2. The third band is Green. This is what is called the "Multiplier". In this case Green equals 5, so we need to add 5 zeros, or Multiply by 10^5. We have now gathered enough information about the resistor to tell us the value. Using the first two colors we know that the "Value" is 22, Using the third color we know the multiplier is 5. Lets put this together. 2200000. That is 22 with 5 zeros added to the end. That makes the calculated value 2,200,000 Ohms. or more commonly, 2M2 Ohms. (That is 2.2 Mega Ohms).

???? Here are some more examples, and the solutions:????

On the first resistor we have Red - Black - Black????? That means it is 2 - 0 and no multiplier.....???? or 20 Ohms
On the second resistor????????? Brown - Black - Red???? That means it is 1 - 0 and 10^2 multiplier????? or 1000 Ohms
On the third resistor???????????? Yellow Violet Green???????????????????????? or? 4 - 7? and 10^5?????????????????? or 4700000 Ohms
on the forth resistor????????????? Green - Red - Brown?????????????????????? or 5 - 2? and 10^1??????????????????? or 520 Ohms

??? One further thing, the forth band. This is the "Tolerance" band. Remembering that these devices are manufactured in bulk. Like any mass produced object there are unavoidable? inaccuracies in their manufacture. If you take you normal run of the mill Resistor and measure it with an Ohm meter, you will of course find there is some difference between the value written on the resistor and the ACTUAL measured value. Manufacturers use the forth band on the resistor to indicate how far outside this value the ACTUAL value might be. On a four band resistor, typically this band will be either gold or silver. To give an example say we have a resistor with Brown - Black - Red - Silver. The value is 1 - 0 - and 10^2 multiplier. Or 1000 Ohms. How does the silver band affect this?? A silver band means the component has a 10% tolerance. 10% of 1000 is 100, so the possible manufactured error is 1000 Ohms + or - 100 ohms. This means that the actual measured resistance will be some where between 900 Ohms and 1100 Ohms. On the average every day flashing light or siren kit from tandy, this is "close enough". Sometime however it is necessary to have the resistance a bit closer to the desired tolerance, so you might use a resistor with 5% tolerance (gold band), or even 1% these resistors have five bands but are read basically the same way.

??? Another thing to worry about is interpreting the various ways you will find the value for a set resistance written down. it is obviously inconvenient to continually write 1,000,000 Ohms all the time. or even 1,500. So we use the common Kilo and Mega prefixes to designate thousands and millions. So 1k Ohm equals 1000 Ohms and 1 Mega Ohm equals 1,000,000 Ohms. Things get a bit more complicated here how ever. In the old times, (when I was a lad) it was common to talk about Ohms or decimal parts there of with a decimal point. So 1.5k was 1500 Ohms. 2.2k was 2200 Ohms. I imagine there is a good reason, but nowadays they have changed this, and now the convention is: 1k5 equals 1500 Ohms, and 2k2 would be 2200 Ohms. Like wise with Mega Ohms. 4M7 would be 4,700,000 and 5M6 would be 5,600,000. I imagine The reason could be for clarity. I imagine that it might be really easy to miss a small decimal point and use the wrong resistor. Even looking at what i've just typed here on my 17" monitor it is awful hard to see the decimal points and so easy to confuse 1.5 with 15. However it is a little more harder to miss a "k" or an "M".

??? Something else that you may see. It is now becoming very common to see Surface mount devices in even hobbyist electronics. Surface mount resistors come in several forms. There are small legless barrel looking types, imagine a small resister without legs,? these packages are called Melf, or the smaller version, MimiMelf. These are from what i've seen usually labeled with color bands, and can be read in much the same way as normal resistors. Another version you might run across is the 1206, or 0805 "Chip" resistor. These are small rectangular flat looking things, with silver ends. Usually, they have the value written on the top surface, in the case of the 1206 package (at least as I know it to be called) it is fairly easy to read the writing. However the 0805 package is allot smaller and you might wish to use some optical assistance to read it. When you look at these resistors, you will see four numbers on the top. The first three represent the "Value" and the forth the multiplier. So 4703 would be 470 multiplied by 10^3 or 470,000 that is 470k.Like wise 1002 would be 10,000 Ohms or 10k Ohms. Things get a little different when you go bellow 1k Ohm. This would be 1001, or 1,000 Ohms. When you get below 1000 Ohms,? they change the designation slightly. 100 Ohms is designated 100R likewise 47 Ohms would be written 47R0. This is fairly easy to understand if you can actually read the bloody thing. An 0805 package for instance is around 2mm long and 1mm wide. This makes the writing VERY small.

??? I'd love to show you some pictures, but fate is conspiring against me EVER buying a scanner to suit my aging computer it seems. Hopefully soon.......

??? O.K. so I'm not a teacher, but I hope that you have been able to understand at least SOMETHING of what I have written here without being totally confused out of your brains. Next time, as promised i'll have a quick look at series/parallel resistors and how to have all sorts of fun solving these problems..........

--
)O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(

???? The Sinister Dragon
?

Hey, Just because I'm EVIL,
???? Doesn't make me a bad person......??? >:-)

Visit My sites at Geocities:
? Blood Moon -
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akirarpg@... wrote:

Is there a way to fix a messed up AC input???? It's the 9V input for my
Roland TR-606 analog drum machine.??? It still works, but I have to
wrap the cable around the 606 and get it at a certain angle (at which
it doesn't like to stay.)??? Any ideas?

????????? I assume your talking about the cable that conects the power supply to
the machine itself?? If thats the case, it sounds like the wire is "Broken".
This is a comman problem with plug packs, the cable gets bent over and over
again, and the wires running up the middle fatigue, eventually breaking. You
will no doubt find that the problem is caused when you bend the cable right
up close to the little plug that goes into the drum machine. If thats the
case, it's just a matter of cutting about 6 inchs or so off the cable and
putting a new plug on it. I don't know what sort of plug the plug pack has on
the end, but there is a fair range available at most electronics hobiest
stores such as R.S., Tandy, or Dicksmith. You mentoined that the supply is
A.C., but make sure of this. If the supply is D.C. you will need to be sure
to conect the wires around the right way. This is no problem if it is A.C.
but make sure or you may damage your equipment. If you have next to zero
experiance with this sort of thing, do you have a freind who knows a bit more
or has acsses to a soldering iron????? This is really a fairly simple operation
as long as you have SOME electronics knowledge.

)O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(

???????????? The Sinister Dragon
???

Hey, Just because I'm EVIL,
???????????? Doesn't make me a bad person......????????? >:-)

Visit My sites at Geocities:
??? Blood Moon -
??? Little Herbal -
??? Zenit Photographic -

Join my Egroup:
??? Inverted Pentagram -
??? Zenit-EM -
???

Oh and it's DC, not AC....sorry.....

By the way, my electronics knowledge is next to nil...

Is there a way to fix a messed up AC input? It's the 9V input for my
Roland TR-606 analog drum machine. It still works, but I have to
wrap the cable around the 606 and get it at a certain angle (at which
it doesn't like to stay.) Any ideas?

--- Gururaj V <gururaj@...> wrote:

Could anyone pls help me about the following
doubts......

1. What is meant by dual rail format?

This refers to the power supply. Many analog circuits (op amps as an
example) are designed requiring +12 or 15 VDC and -12 or 15 VDC. the
power supply points are sometimes called rails. Dual rails refers to
two supply voltages.

2. What is the difference between
Complementary and Differential of
two signals?

If I understand this correctly: Differential circuits present an
ouput based on the difference of the two input signals.
Complementary circuits produce an output equal but opposite to the
input signal.

Larry Hendry


__________________________________________________
Do You Yahoo!?
Make international calls for as low as $.04/minute with Yahoo! Messenger

--- In Electronics_101@y..., ramiroi@e... wrote:

I want to send a video by telephone line, but i dont know how to
start im lookin for a circuit that modulate the telephone line with

a

viedo sinal and at the other end it demodulate to see this imagen

in

a tv. Please if any one can give me an idea or help, send me taht

If you're not sending it over the public telephone system, maybe you
could consider using CAT5 cable instead of straight phone wire. It's
not that much more expensive and has a lot more bandwidth. Keith
Doxey at sells some modules for sending video and
audio over CAT5 cable. He sends it as an analog balanced line
signal. You could build one yourself using ICs from
if you live outside the UK (he only ships to
the UK).

If you are sending it over the public phone system, you have to
consider that it's going to be messed around with a great deal
between the sender and receiver (converted to digital for one thing).

开云体育

???? The Sinister Dragon
?

Hey, Just because I'm EVIL,
???? Doesn't make me a bad person......??? >:-)

Visit My sites at Geocities:
? Blood Moon -
? Little Herbal -
? Zenit Photographic -

Join my Egroup:
? Inverted Pentagram -
Zenit-EM -
?

To unsubscribe from this group, send an email to:
Electronics_101-unsubscribe@...



Your use of Yahoo! Groups is subject to the .

This is great! Too basic for me but I trust you'll get to parts I'm unsure about.
Keep it coming.
ralph

?
??????? Just for our beginners, Here is some basic information on ohms law, and resistance calculations. I hope they are helpful. I'm starting here as I don't know what you all know in the way of electronics, and I hope to be able to advance to some more complicated information, such as amplifiers and digital stuff. However I want to establish a base line first. So if this is a bit basic for you, wait, I hope I'll eventually get to something that will interest you a bit more.......
?

??? Above, is about the most fundamental electronic circuit you will ever encounter. When it comes down to it nearly all electronic circuits can be expressed in this form or similar. Basically, you have a power source, and a load. In this case the power source is a twelve volt battery, and the load is a 1200 ohm resister, or a 1.2 kilo ohm resistor expressed as 1k2. Though the above circuit taken in it's literal form is not very useful, other than to flatten the battery, is serves as a basic platform for discussion into Ohms law. Lets assume that the above battery for instance can deliver a 1 amp current for 1 hour before it flattens. So how long will it take the battery to flatten in the above circuit?? This can be very easily calculated using a bit of common sense and Ohms law. Basically, Ohms law states:

??????????????????????? Current flowing in a resistor (or Conductor) is proportional to the supplied voltage, and inversely proportional to the resistance.

??? Just what the hell does this mean exactly?? It means, that what ever you do to the voltage, will happen to the current. so for the same resistance, if you double the voltage, the current will double. That covers the first part of the law, now the next bit. Assuming you have a fixed voltage and start playing with the resistance, what happens to the current??? What happens when you increase the resistance, is that the current is divided by the same amount as the resistance is increased. For example, if you double the resistance, you divide the current by two, or halve it. If you triple resistance, you divide the current by three, and so on. Very soon you will be able to demonstrate this with ohms law yourself, but for now just trust me. In any case, we come to how this law is applied in the real world.

??? In the above circuit, we know the applied voltage is 12V. We know that the circuit resistance is 1200 ohms. Now using the formula below, we can calculate the current:

??? Lets do the current calculation for our little problem.......

??? So we now know how many amps are running through our little circuit, 0.01 Amps, or to express it another way ten thousands of an Amp, or more commonly 10 milliamps.

??? Now, just a small side step from the ohms law issue, we know our battery can deliver 1 Amp for 1 hr. In theory that means it can deliver 0.1 Amps for ten hours. By extension I'm sure you can see that it will hopefully be able to, in theory at least be able to deliver 0.01A (10 milliamps) for 100 hr. Things in real life are actually just a bit more complicated than this. As the power in a battery is used, it generates internal resistance, which affects it's ability to supply a fixed voltage and current. A fair amount of sophisticated chemistry and internal mechanical design goes into preventing this happening by battery manufacturers. However it serves as a general rule of thumb for calculating how long a given battery will supply the necessary power to a circuit.

??? With our calculated current now placed in the circuit, we now have the below information:

??? Just a bit of further information:

??? Just what is current?? Current as we all know is the flow of electrons through a circuit. To be precise, One Ampere is the rate of flow of electric charge (electric Current) when one coulomb of charge passes a set point in one second. One Coulomb equals the charge in 6.24X10^18 electrons. So for those of you not familiar with this notation, that is 6,240,000,000,000,000,000 Electrons flowing past a fixed point in one second equals one Ampere.

??? When we calculate the amount of current flowing in a circuit, we talk about electrons as if they have are moving in a specific direction. Now just what direction do electrons travel around the circuit. Back in the days when Electronics was in it's beginnings, it was believed that electrons carried a POSITIVE charge, and for various then observable reasons it was decided that electron flow from the positive terminal, through the circuit, to the negative terminal. This is termed CONVENTIONAL current flow, and is used in nearly all electronic calculation, unless specifically stated otherwise. As is now known, however, Electrons are negative and travel from the negative terminal through the circuit and to the Positive terminal. This is called ELECTRON flow. UNLESS SPECIFICALLY STATED, we will ALWAYS be referring to Conventional flow when we do our calculations.

??? For those of you who are still awake and haven't just deleted this little lesson, here are some simple examples for you to practice with. I'll post the solutions in the next couple of days:

Remember that 2k5 = 2500 ohms, and 100R = 100 ohms. It a good idea to try and get familiar with this notation, as it is used fairly regularly through out modern electronics.

??? I'll post the solution to these problems in the next week or so, and then we will have a look at what happens when we have more than one resistor in our little circuit. check out the two circuits below:

??? How do we figure out the current flow?? We'll deal with it in the next week or so.........
?
?

)O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(? )O(

???? The Sinister Dragon
?

Hey, Just because I'm EVIL,
???? Doesn't make me a bad person......??? >:-)

Visit My sites at Geocities:
? Blood Moon -
? Little Herbal -
? Zenit Photographic -

Join my Egroup:
? Inverted Pentagram -
Zenit-EM -
?

Hi,

Could anyone pls help me about the following doubts......



1. What is meant by dual rail format?

2. What is the difference between Complementary and Differential of two
signals?




Thanks and Regards,
Gururaj.

O.K., For those of you who are beginers, you may find some interesting info at:

?????????????????????

There is some basic kit building info for those who are starting out.

For those of you who are into something a bit more deeper, Ray seems to be into ameture
radio, and has quite a bit on his site about this, plus a bit of other stuff you might find
interesting. Check it out at:

?????????????????????

If anyone else has any non comercial Electronics sites they wish to tell us all about feel
free to send the URL to the group. TAKE NOTE though, NON-Comercial. If you have any questions
about wether a site qualifies as non-comercial, send me (the moderator) the URL off list, and
I will asses it.
???

)O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(

???????????? The Sinister Dragon
???

Hey, Just because I'm EVIL,
???????????? Doesn't make me a bad person......????????? >:-)

Visit My sites at Geocities:
??? Blood Moon -
??? Little Herbal -
??? Zenit Photographic -

Join my Egroup:
??? Inverted Pentagram -
??? Zenit-EM -
???

Hello,
Well, I'll give a stab at it. I havn't actually done anything
like this but it sounds pretty simple. Off the top of my head I
would think you could take your 12V battery run it through a 5V
converter and tie the 5V and ground to the same terminal connections
as where your solar pannel connects up. I think I would also add a
diode to each line going to the solar panels, not sure on that but
seems like a good idea just in case the battery tries to overpower
the solar cells. Like I said, not sure on this idea but is sounds
good in my head.

Good luck
Michael

--- In Electronics_101@y..., markallen <markallen@m...> wrote:

Hello everybody.

I recently put together a small solar powered project. It draws

about

300ma at a regulated 5v. It runs great here in LA, but now I want

to

bring it to London. Obviously I'm going to be getting a lot less
power out of my panel. What I want to do now is attach a 12v lead
acid battery that can power the project when the sun is behind the
clouds (hopefully not all the time!). Is it complicated to power a
circuit and charge the battery at the same time? Should I try and
find an off the shelf solution to this problem, or is this

something

I can put together myself (with some help from the list)?

Any advice greatly appreciated.

Mark Allen

Can someone explain the basics of a hall effec sensor to me? I
believe if I understand correctly that they sense changes in magnetic
fields. Is that correct? If so are they sensitive enough to sense
the magnetic field that is produced by a wire that has been
energized. If not what sensor should I use for this type of
application. I am wanting to sense when a spark plug wire on a small
engine is firing and then send that back to a F-V for a read-out.

Thanks

Hi all

I am a newbie to the art of electronics so please be gentle with me :-)

I am trying to run 16 LED's from a single 12v power source. The LED specs are as follows:

Parameter Value Units
DC Forward Current[1] 30 mA
Peak Forward Current 100 mA
Average Forward Current 30 mA
Power Dissipation 120 mW
Reverse Voltage (IR = 100 mA) 5 V
LED Junction Temperature 100 ?C
Operating Temperature Range ?40 to +80 ?C
Storage Temperature Range ?40 to +100 ?C

What I would like to know is what would be the best way to hook these things up and what value resistors I would need to do it.

Cheers
Pedro



_________________________________________________________________
Get your FREE download of MSN Explorer at

Hello everybody.

I recently put together a small solar powered project. It draws about 300ma at a regulated 5v. It runs great here in LA, but now I want to bring it to London. Obviously I'm going to be getting a lot less power out of my panel. What I want to do now is attach a 12v lead acid battery that can power the project when the sun is behind the clouds (hopefully not all the time!). Is it complicated to power a circuit and charge the battery at the same time? Should I try and find an off the shelf solution to this problem, or is this something I can put together myself (with some help from the list)?

Any advice greatly appreciated.

Mark Allen

A quick rant from your freindly moderator,

????????? Just a question asked on another list that I am on, and a quick reply.
I'm posting this for the "Beginers" on the list so those of us that know
everything can of course instantly delete this post and go on with our
perfect lives.
???

???>> What is the sponge used for, I have seen people wipe the tip before
???>> and after they use it on a wet sponge?

???

????????? The sponge is used to clean the tip between soldering. When the Iron has
been standing for any period of time wilst on, a layer of oxides will form on
the
tip. Also, any flux remaining on the tip will deteriate. It is important to
remove these before the next use as it will contaminate the joint. Just to
put this into
percpective. This will perhaps have little affect on a "Tandy" flashing light
kit, that you can resolder if the joint goes bad, but might break things if
the aircraft's
computer dies. In any case I tend to have the habit of wiping my tip clean
before use whenever I'm soldering, as it is not a bad one, and though the
telecomunications equipment I work on isn't going to kill anyone if a "bad"
joint has problems, it does affect the reliabillty.

????????? Just a few words about using sponges to clean:

????????????????????? Firstly, it is a VERY comman error for people to have WAY to much
water in the sponge. If you can wring water out of the sponge you are using
then
there is WAY to much in it. When you hear of people talking about damp
sponges they mean DAMP. If you have to much water in the sponge, when the tip

touchs the sponge, it imediatly cools to below the melting point of the
solder, thus effectivly making it imposible to remove the oxidized solder,
and potentially
making the next joint soldered cold. This especially happens when the tip is
long and narrow. Sure, you have to re wet the sponge after a couple of hours,
but
this is really isn't going to hurt you now is it??

????????????????????? Second, I have seen some REALLY discustingly filthy sponges in my
time. Your sponge should be cleaned, thoughly rinsed under fresh water before

each use. (Be carefull not??? to let scrap solder go down the drain, it
contains lead don't forget) Otherwise when you wipe the tip. you end up
puting more dirt
on it than you are removing. It takes about a whole minute to do this not a
lot of time in the grand scale of things, and no time at all compared to the
time it
can take to find a dodgy joint in that prized project you've been working on
for weeks that now doesn't work.

????????????????????? Likewise I have seen sponges used WAY past their sensible used by
date. A sponge costs what, a dollar?? So why use some ratty, dog chewed peice

of crap that you wouldn't normally think twice about throwing away?? If it
starts wearing out replace it for the above reasons.

????????????????????? One final piece of advice, is only to use sponges designed for the
purpous. I dont know how true or false the info is, but sponges for this use
are suposed to
have been designed to drag a hot iron across. They can stand the heat, and
suposedly have no funny additives. I do know there are quite a few varities
of
synthetic sponge around. Alot of these have additives to help with cleaning
kitchen surfaces, and they may have low melting points. These sponges will
once
again contaminate the tip. In any case once again the proper thing just ain't
that expensive, so why take the risk.

????????? O.K. my rant is now over, I can't believe I wrote so much about tip
cleaning sponges. Maybe I need a life or something. In any case, I hope the
info was
helpfull to our beginers.......

)O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(??? )O(

???????????? The Sinister Dragon
???

Hey, Just because I'm EVIL,
???????????? Doesn't make me a bad person......????????? >:-)
???
???

but cant we charge them in parallel i.e connect all the batteries in
parallel
and then apply 1.5 volts.

is configuration will require only 1.5 volts no matter how many
battries
are charged at a time.

Well, it would work, but you'd need a current source that could handle it. With batteries charging in series, you'll always deliver the same current (well...not exactly...not all batteries have the same internal resistance). One battery at 1.5V versus two batteries at 3V will give you the same current because you've doubled your resistance along with the voltage. So your power supply will only need to deliver that much current no matter how many batteries you're charging.

I'm not saying it's impossible, though, try it.

No matter how your batteries are configured, series or parallel, it's still going to be the same wattage dissipated in the power supply.

-Tavys

hi

i just got a idea yesterday and here it is.

normally we charge batteries in series i.e if we want to charge 2
batteries
of 1.5 v then we connect them is series and apply 3 v and so
on..changing
the voltages...as the no. of batteries increase.

but cant we charge them in parallel i.e connect all the batteries in
parallel
and then apply 1.5 volts.

is configuration will require only 1.5 volts no matter how many
battries
are charged at a time.

can this idea work.
please reply .
thanks.

aditya n