12DC Sockets

[4pyros]

So now I'm wondering if I should get one of these & some componants to play with & make these circiuts with leds/resistors & transistors so I can feel in my bones & help me remember
then as we progress I can play with stuff as we go along

Get somthing like this;
https://www.amazon.com/Vintage-Sports-Cards-MX-906-Electronic/dp/B00005K2SY

 

[cowboybob]

Yes, by all means get a breadboard. Playing with REAL parts is a great learning experience.

A virtual simulator is nice, I will say. They sort of "prove the point" and help you not waste time. But nothing is more satisfying that having REAL parts flash, or make noise, or occasionally get very hot and smoke.

And it was my pleasure to help.

73 ("Good luck and farewell" in Ham-speak)

CBB

 

[Muttley600]

Yes, by all means get a breadboard. Playing with REAL parts is a great learning experience.

A virtual simulator is nice, I will say. They sort of "prove the point" and help you not waste time. But nothing is more satisfying that having REAL parts flash, or make noise, or occasionally get very hot and smoke.

And it was my pleasure to help.

73 ("Good luck and farewell" in Ham-speak)

CBB

Wow, your 73, amazing & thanks for your help so far

I would hope that you will stick around as last night turned my world backwards. Eek!

I knew when my dad was explaining atoms to me that I should have paid attention but when your 21yrs old & have cars & girls to daydream about (seeming more important at the time.....sigh)
My attention returned when he got to amps/watts etc. so I knew about calculating resistors, how to fuse safely but never until last night understood my theory was all back to front.

So I went back right to the beginning, a student again but now 42yrs old, logic tells me that's a backwards step but now I understand more so have moved forwards, hows that for confusing. Lol

So all this electrickery stuff has sent my world into turmoil.
Basically Ben & his kite got + & - the wrong way around meaning I have been living under an illusion that + was live when in fact - is the current (supply)

So if your happy to still help this fool named muttley, can I still ask q's
I'm thinking that once I grasp this notion truly I can then go back to projects, yes, my q's will change to that of componants without q your diagrams all the time.lol

So we have an atom that splits into 3, we fire a proton along the + lead that pushes an electron back along the - lead at the speed of light which supply's the current to devices.
So basically we are sending electrons to battery which would eventually fill it & leave us with no protons left to fire to destabilise atoms at ground & send current back.

So we have to keep replacing protons in battery by the alternator, but the alternator produces current which would equal more electrons????

Where have I misunderstood the atomprocess?

This is fun isn't it

Going for walk today but will be eager to understand tonight
Graham

 

[cowboybob]

I always found it easiest to understand the basic concept of plain old electricity, i.e., power, wires, generators and loads, as a water plumbing system. Its not a perfect simile, but it worked for me. Somebody out there will find fault with this, but I don't really care.

The wires are pipes. The electron flow or current in the wires (it is entirely electrons doing the work. More on that later) is the water in the pipes. The generator is the water pump (alternator, battery, whatever) that creates the pressure (potential, or voltage) that causes the electrons to move. Resistors ( or capacitors, transistors, CPU, etc) are pressure regulators. and, let's say, the load is a waterwheel at the end (something doing some work).

Electrons (the "-" symbol", generally ground, or cathode) tend to go where there are fewer electrons (the "+" symbol, or cathode).

In a battery system, to replace the electrons, the direction of the current flow is reversed. This forces electrons back into the battery, thereby charging it. Electrons will travel to wherever there are fewer electrons, however that potential difference may be created.

All atoms (except for Helium) are composed of Protons (a "+" potential), Neutrons (no potential) and Electrons (a "-" potential). The protons and neutrons make up the nucleus, or center and are bound together very strongly. The nucleus does not contribute to the current flow. The electrons orbit around the nucleus in various patterns. Think of the Sun's planetary system. The electrons are "free" to move around from one atom to the next, so long as they are instantly replaced by a neighboring electron from another atom. Thus they can move relatively easily, especially in a conductor such as copper. And electrons are ALWAYS attracted to a positive potential (and are equally repulsed by a negative potential).

Thus the juice flows.

Interesting you bring up old Ben and his kite. I'm a Great, Great, Great, Great Nephew of his. Wowsers, you say, right??!!

 

[Muttley600]

Interesting you bring up old Ben and his kite. I'm a Great, Great, Great, Great Nephew of his. Wowsers, you say, right??!!

Cool so if you can't help me I'm doomed.lol

First of all, thanks for the explanation of above, so just to make sure.lol
The electrons are in both leads +&-
The voltage pushes them about between poles & the generator keeps the voltage up, that's it In a nutshell, hence it was unimportant which side of componants things are as long as they are doing what you want In the circuit you want? So electronics is a matter of manipulating energy into whatever you want, hence you can change voltages, stabilze it etc

Not a bad outcome for me in 24hrs, Paul, I think you deserve to be put on my Christmas card list.
Thanks a million mate

An open circuit is a dead circuit & a closed circuit is a producing/live circuit

Have I finally got basics right?

Had good walk, bobby (dog) now fast asleep on my lap

 

[cowboybob]

Yes. You've got it.

Just keep in mind that the current flow is always from - to +. This is why many components (such as transistors, certain capacitors, LEDS, etc.) are "polarized". That means that they have parts or leads that are marked either - or + and must be attached to that particular polarity in order to work correctly. And some that are not, notably resistors.

Looks to me like you're on your way. Congratulations.

 

[Muttley600]

Yes. You've got it.

**broken link removed**

Just keep in mind that the current flow is always from - to +. This is why many components (such as transistors, certain capacitors, LEDS, etc.) are "polarized". That means that they have parts or leads that are marked either - or + and must be attached to that particular polarity in order to work correctly. And some that are not, notably resistors.

ok, thanks for the warning on polarization **broken link removed**

Looks to me like you're on your way. Congratulations.

That was great, I came here with intention of learning & to be honest really enjoyed the last 24hrs **broken link removed**

It was nice having you there to check my theory & explain where I had misunderstood it **broken link removed**

I really cannot thank you enough **broken link removed**

I was going to ask a long question about resistors but I'm determined to crack this, so now I have basic electric theory, I'm going back to ohms law to see if I can solve the short circuit dilemma I'm having in my head

Hopefully I'll be able to come back & explain what I wasn't getting about how you can short circuit a + & - with power on with just R1 as power is on in box

If not I may need asimple answer but would be nice to have tried my best to understand first **broken link removed**

 

[KeepItSimpleStupid]

Yes. You've got it.

Just keep in mind that the current flow is always from - to +. This is why many components (such as transistors, certain capacitors, LEDS, etc.) are "polarized". That means that they have parts or leads that are marked either - or + and must be attached to that particular polarity in order to work correctly. And some that are not, notably resistors.

Looks to me like you're on your way. Congratulations.

CBB: OOps CONVENTIONAL CURRENT goes from + to -; electrons go from - to +. Usually you don't have to know which way the electrons are moving.

 

[cowboybob]

KISS,

I was trying to explain electricity. NOT electronics.

Yeah, I know that conventional wisdom says that "holes" are involved where current is involved, etc. But I'm old-school. I started out with tubes (my fav being the 6146A, RCA only). Current flow was cathode to plate. Period. Back then there weren't no "holes" in a tube.

And I keep it straight now (in my mind, anyway) with solid state (does anybody even use that term anymore?) by visualizing the current going against the arrow (in a diode, or an NPN transistor, for instance).

As a prime example, I give you the battery. Convince me that the electrons (what I still call current) flow, at ANY time, from the anode to the cathode. I suppose that you could, but I won't believe it. Just my way of thinking.

And a another thing. When I took a class on OP AMP theory, It drove me crazy (CRAZY) that they had to introduce an "imaginary" number (the square root of -1, [which doesn't exist, hence it's not real] twice) into the formula in order to change the sign of the answer. Yeah, I guess it made the formula look more elegant (none of those pesky parentheses around the whole thing), but an "imaginary" number that is then multiplied against another "imaginary" number to give you a -1? Which then changes the sign of the answer? Huh?? What ever happened to putting a -1 in front of the whole formula to do that?

AC current goes both ways. I think we can agree on that.

It's like New Math. Hate it. Give me that ol' time religion, er, sorry, electricity. It still works for me.

CBB

PS. Tnx for the speed test info. VERY helpful.

 

[KeepItSimpleStupid]

I use the other convention, that conventional current (positive to negative) flows in the direction of the arrow. My course in Solid State Physics and Chemistry makes you think the other way. So does vacuum tubes. Then there are holes which we won't get in to.

But, when you hookup an ammeter you get a positive deflection with conventional current flow.

At one time, this concept confused me. So did orbits of electrons. Orbits are not orbits, they are probability functions.

 

[cowboybob]

Speaking of confusing (that is the topic of this thread, isn't it??) I remember a microwave "phenomena" back in the late 60's that had to do with the propagation speeds of E and H fields in a waveguide (which we called plumbing: there's that simile again).

We observed (or thought we observed) that one of them, but not both (don't remember which one) appeared to traverse the waveguide faster than the speed of light.

Huh?? you might ask. We certainly did. But we were told to ignore that little weirdness and get on with the rest of the experiment. We were not told we'd done something wrong. My best guess at the time was that the math we were using was a tad flawed. That or the instrumentation.

Those guys at the CERN facility, no doubt, got told the same thing.

 

[Muttley600]

Speaking of confusing (that is the topic of this thread, isn't it??) .

It is, breadboard purchased along with FUSES as running it off my bench power pack

 

[cowboybob]

Graham,

I'm sorry. I was responding to a post by KISS (just before my post above).

Anyway, good going. Have fun with the breadboard.

Let us know how it goes!

CBB

 

[Muttley600]

Graham,

Anyway, good going. Have fun with the breadboard.

Let us know how it goes!

CBB

I'm still struggling with this short circuit in my head, going to do some tests tomorrow,
I've read ohms law again but surely if you have voltage & circuit, you have a current?
Why would a wire melt if you put it across poles? Isnt that the same as short circuit?

Can you use your simulator to show battery to gnd with just 470r as that would be the same thing wouldn't it?

As you can see I'm still Ingrossed in theory.lol

More ideas tomorrow & I'll let you know results of my testing

Eek!

 

[KeepItSimpleStupid]

Breadboard:

When I was a kid, I had one of those from radio Shack. The experiments are rather limited. It was definately worth something when I was 12 Years old.

You sort of have to decide what might be the kinds of experiments you would want to do.

At work, I had a powered breadboard, not like these, but somewhat close: https://www.globalspecialties.com/electronic-trainers/powered-breadboards/item/89-pb-203.html

This is something close too: **broken link removed**

Mine had debounced switches, leds, a +-15 V power supply and a 5 V supply and a meter. It was useful for getting your feet wet.

I also had at my disposal a two 0-20 V tracking/independent supplies and a 5V power supply made by heathkit which was nice. This was the one I had: **broken link removed** The supply does not have an independent current limit.

There are wall/wart powered 3.3/5V power supplies that fit on the end of a breadboard

What I can say is that way back when linear stuff was either+-12 of +-15 and digital was +5. Now it's not. The digital stuff is 0.8/3.3/5V and 12V and the analog stuff is low voltage too. Usually around +-2.5V. 12 and 24 V could be common as well.

I can't say that the experimentor's kit that I had taught you anything. It was just a collection of circuits, but now how they worked.

I guess what I'm saying is that a power supply with a current meter is really convienient. Current limit is even better. Displaying current and voltage at the same time is better yet.

At home I have a few supplies:
1. +-12, +5 V unmetered supply that I picked up for $8.00, 30 years ago. It doesn't like being backfed. 723 regulator based. 0.5 to 1 Amp or so. Some trimmer adjustments.
2. A 0-40 V, 0-2A analog metered (V or I) supply that I got free
3. A +- something that I got off of ebay (one meter) for about $60; Bel Merit no documentation; needs a binding post or two.
4. Now I have a couple of multi-input (wall wart, USB and LiPo to +5 or 3.3 V supply.

It just depends on what your focus might be. I needed the high power 12 V supply for car radio stuff.
4. A 0-30 V, 10 Amp Selectable fixed voltage power supply that I picked up for $15 broken. I invested $15 and all is fine agin in the 70's. No schematic. Harrison Labs.

 

[KeepItSimpleStupid]

Everything has resistance, including wires. Wires are goverened by the formula R=pL/A where p is Rho and is a material property in example units of ohm-cm. Aluminum, Copper, Silver etc would have different values of p.

L is the length. If you were calculating the resistance of say a 100' wire going to a light bulb 100' away, d would be 200 feet (the total path length). To be obsessionally correct with p being in ohm-cm, the length would be in cm.

A is cross-sectional area. If the wire was square (which is possible) or the wire could be circular. There is a unit called Circular Mills just to make the numbers easier.

You may find in a catalog 0-ohm resistors. What? they are just jumpers in a resistor body so that insertion equipment can insert them.

Measuring extremely low resistances (500 milliohms) and high resistances like 100 M-ohms requires special techniques.

 

[cowboybob]

Graham,

Below are sims of the LED circuits and the related current and voltage values for the un-shorted LED and shorted LED conditions.

View attachment 60458 View attachment 60459

Note that R1 is 4.7 k (4700) ohms. Also note that with LED1 shorted, there is 0 (zero) volts across it AND the switch. And don't forget that the LED has resistance also.

A voltage is generated when current passes through a resistance. Simple as that. No resistance, no voltage generated.

This involves the general assumption that a short , essentially, has zero resistance. Understand that it does, in fact, have some resistance, but is is so small as to be inconsequential in a circuit of this sort.

What does happen with a short across a 12VDC battery, is that the current goes to infinity (I = E/R, or in this case, I=12 divided by 0). In reality, of course, the maximum current the battery can produce runs through the wire, and if it cannot carry that load, it, or the fuse protecting it, blows.

As to the 470 ohm resistor, again, I = 12 divided by 470, which equals 0.026 mA. See sim below to confirm:

View attachment 60461

Clear as mud, right??

CBB

 

[KeepItSimpleStupid]

In simulations, the voltage sources are ideal. e.g have 0 resistance. Batteries have resistances too. Connections also have resistances.

There will be a big difference if the short is a wrench or a screwdriver or a piece of 14 AWG wire. In the former, the battery could explode and in the latter, the wire would simply melt.

 

[cowboybob]

Way back when, I was an ET in the Navy. Calibrated and repaired precision electronic test equipment (nuke submarines).

Dead gear that came to us with perfectly good fuse(s) were at first sorted into the group we called the "pile with components that had sacrificed themselves to save the fuses."

Anyway, interestingly, the TINA sim I use allows you include an internal resistance in the DC power source(s), as well as introducing a DC component (or noise, if you want it) into the AC source(s). I usually, however, leave the internal resistance(s) at the default, i.e., 0 Ohms. Just easier to deal with.

KISS is correct to point out that a little hidden resistance (or a not so hidden adjustable wrench) in a circuit can cause big headaches, especially in low voltage/high current systems.

 

[Muttley600]

Good morning Paul

ohms. Also note that with LED1 shorted, there is 0 (zero) volts across it AND the switch. And don't forget that the LED has resistance also.

I have tried to interpret circuit as you have drawn it, is this right?
If so, I have 12V at the switch which is where I'm getting confused, 12v when jack plugged in to pins 1 & 3 only, but then 12v on pin 2 when jack taken out

I think I need to go & make circuit & measure to really understand, I said earlier I could know the theory, but it makes it so much easier if you see it.............hopefully **broken link removed**


**broken link removed**

Clear as mud, right?? CBB

Now you know where I'm coming from **broken link removed**

I'm sure it will all be fine **broken link removed**

other questions would relate to theory too......sigh
If there is 1000 ma in 1 amp, how is that put on paper?

Is it 1.000 = 1 amp

so when you are on about ma shorts 0.026 ma X 4 for all sockets? does that equal 0.104 ma, so does that equal a 0.1 amp short for all four sockets against the earlier transistor short of 1.2 ma X 4 = 4.8 ma = so is that 0.48 or 0.048 amps?

so in theory the transistor looks to have less of a short or are these numbers theorectical because they are so small & in real terms would not make any difference?

Sorry about more questions

* Edit: In theory, I can see the last diagram you did makes sense with meter showing tiny short, guess it is just my head blocking this as my old way of wiring would never to be to short anything out, hence the learning starts here.....that's a nice feeling

Hopefully I'm turning out to be a good student, yes, maybe a little slow as well.lol *

Best Regards as always for sticking with me
Graham