12DC Sockets

[Muttley600]

Go with Diver 300. You don't need useless lights, you will know when it is working but need to know when it fails.

That is another way of looking at things I suppose

As for fusing, remember, in general you are protecting the wire, the device will normally clear to protect the fuse. And unlike the idiots on that TV bike show, please use grommets or armored cable.

When you say 'the device will clear to protect the fuse' what does that mean **broken link removed**
I don't watch TV so am at a loss as to what show you are on about but I'll put some photos up at the end of this post so you can see **broken link removed**

Ok sounds like fun.
As always the skys the limit, knowing when to stop will come from experience.

Fun, I'm loving this, was out of bed at 7.30 this morning with coffee & smoke looking at site on phone waiting for better half to go to work, I don't start until 12.30. I'm loving this **broken link removed**
Like you say, once I understand things, I can make decision based on knowledge

As for the fuses;
The main fuse should be at the power source, IE battery to protect the wire run to the load.

ok, first photo, showing fuses on bike lead that come to front of bike

**broken link removed**

The 5 volt regulator should have a fuse on the input to protect the regulator.

So how do you decide what to fuse **broken link removed**

The outputs have to be fused.

I agree that each toy needs it's own fuse **broken link removed**

If you put a LED with a dropping resister across a fuse it will light if the fuse blows.

How does that work then **broken link removed**

If you put two diodes in series with each connector power supply. They will light a LED if a load is present, IE something plugged in.
this can all be done with diodes and maybe better suited for a beginner.
Let me know if you would like more info on the diode approach.
Andy

Go for it Andy, I don't care if I make MKI, MKII & MKIII of boxes, this is about me learning what componants do, if you are happy to explain principles of diodes to me, I'm happy to learn, this isn't about the final result, more so the learning **broken link removed**

ok, I seem to have been replying for ages, so I shall post this then carry on, time for cuppa & smoke

 

[Diver300]

If you put a LED with a dropping resister across a fuse it will light if the fuse blows.

How does that work then?

The LED needs a dropping resistor or it will blow with 12V, but that is just detail. Just regard the LED with dropping resistor as a small 12V powered light.

A fuse is a short length of wire. There can never be a large voltage across a working fuse, so there is no voltage and the LED does not light.

When the fuse blows, the wire isn't there, so there is no connection. The 12V from the battery is on one side of the fuse, and there is around 0V on the other side, where the load is still trying to take current.

The 12 V difference in voltage across the blown fuse will light the LED.

 

[Muttley600]

Refreshed & raring to go
**broken link removed**

Once you done the hand drawing (schematic) of one jack wired and configured it the way you like it, all you've got to do is "clone" it and it's components/wires (with the appropriate power) as many times as you like. Give yourself lots of room to draw the thing. Hopefully, you can see some of the standard symbols for various components normally used. All part of the learning curve.

ok mate, I got up this morning, bingo, 90% understood diagram so without drawing I feel I can explain what I've understood so far **broken link removed**

Looking at the diagram with a fresh pair of eyes F1 is my fuse on bike

The little dot after would be where my lead comes into tank bag then box

F2 would be my 4 way fuse block, I need to understand why 5v reg needs fusing?

T1 is a transistor, this is basically an electro mech switch that supplys power to led but when there is power to pin 2, a small current holds switch open & stops led lighting hence the 0.1ma drain to hold switch open. Then when power is dropped off pin 2 it lets the switch go & the current lights the led

Please correct me where I have misunderstood, so I get the principle of this transistor, but this itself begs many Q's **broken link removed**

How do you know which transistor to choose
Are the 3 legs marked/identified on unit
why are there metal & plastic versions

Then moving onto resistors, these limit current to said objects
I take it a 470r is half of a 1K one so I can understand power markings on them, this is not important at this stage until power putput is shown so I won't Q reasons for choosen ones yet **broken link removed**

on the diagram I have following Q's

Why is the resistor after the led on R4, if they are used to limit current to stop led blowing, I don't understand that bit ?

R1 is totally baffling me as it seems like there is a direct link from + to - ???

R2, why is this going from gnd to gnd with resistor

I can clearly see where wires are going now, it is starting to click, hope that isn't too many Q's in one go **broken link removed**

the voltage regulation (and the associated current requirements) issue is yet to be determined.

ok, the voltage regulator is rated at 3A lets show you my plans

**broken link removed**

I will be supplying power as follows:
1: Sat nav (nicknamed 'Dora' (the explorer) consumes 0.83A off 12v socket

2: Heated gear will comsume roughly 3A off 12v socket but want both rated at 5A in case I ever change gadgets at any point in future

3: to iphone (haven't measured this yet but lead at the moment has 1A fuse

4: Bike camera consumes 0.57 amps while recording or 0.25 while just charging battery

That is it, nice simple project before moving onto rear box which will run this one

And yeah, I like simulator I have. It's pretty versatile (TINA Basic, ver. 9).

Great peiece of kit, have you got a link to that **broken link removed**

I applaud your goals. As you will see, not all that difficult once it starts to click.

I'm even starting to understand what each symbol means **broken link removed**

This can be an incredible fun.

Ace **broken link removed** I'm loving every moment

I'd like you to look here: **broken link removed**

Those Procar lighter plugs are the BEST I have ever seen.

They look like my bike power for tank bag

**broken link removed**

I have one of these into tank bag

so box will plug in here

**broken link removed**

For the fuse blown indicator, I would consider the LED/Resistor combination across the fuse although I would probably make this only work at "Lamp Test" part of the ignition circuit. This part of the "LAMP TEST" would indicate a failure instead. You would also have a separate push button to check for a fuse failure as well.

Just a suggestion.

if your happy to explain another way of doing things & it helps my understanding of electronics, go for it, consider me to be a bank canvas of knowledge & hopefully I will learn lots

So with all these different ideas banding around, is it worth me getting a test board or the like to play with componants before final installation, that way I can try loads & understand different concepts & why each one works

 

[Muttley600]

Componants I'm playing with before I had this brainwave of learning a new dark art **broken link removed**

**broken link removed**

**broken link removed**

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**broken link removed**

 

[Muttley600]

The LED needs a dropping resistor or it will blow with 12V, but that is just detail. Just regard the LED with dropping resistor as a small 12V powered light.

A fuse is a short length of wire. There can never be a large voltage across a working fuse, so there is no voltage and the LED does not light.

When the fuse blows, the wire isn't there, so there is no connection. The 12V from the battery is on one side of the fuse, and there is around 0V on the other side, where the load is still trying to take current.

The 12 V difference in voltage across the blown fuse will light the LED.

so if theres power in box, won't the led light or is this way about choosing a different resistor somehow??

sorry for being slow to take it in

I know a 470r will light led without blowing it, how do you make it so it does not come on normally?

& would that tell me if something was plugged in the jack??

God I'm thick **broken link removed**

I was thinking well if I can't see my toys how would I know they are working as camera is in seperate compatment of bag, but what your actually saying is that instead of having lights on all the time to show there is power to toys, the light only comes on if there is no power alerting to me to the fact that fuse has blown **broken link removed**

can you draw a basic diagram so I can understand how it works & I can't see how it wouldn't light up normally?

Thanks for your patience

 

[cowboybob]

OK. By my estimation, you need to leave shortly (if your haven't already), so here we go.

Yes, F1 is the fuse already in the bike that I'm assuming is dedicated to accessories.

In the schematic, the dots are physical (soldered or otherwise) connections. Whenever a wire crosses another wire, a dot indicates that the two (or three or a million) are connected together. NO dot means they may cross one another, but they are NOT connected.

I fused the 5 VDC regulator circuit to protect it from an overload. It's easier to replace the fuse than to replace the regulator. Some components have built in short protection, in which case, fuses are not necessary. All a matter of component selection and one's willingness to assume that Murphy Law is all a big lie.

The top lead of T1 is the "collector" (if your familiar with the electron tube , or valves as you guys call'em, a collector is similar to a "plate"), The bottom lead (with the arrow) is the "emitter" (or cathode in tube talk [TT]) and the lead at a 90 degree angle jutting to the right is the "base", or "grid" in TT. Basically, the base controls the flow of current through the transistor from emitter to collector (or "holes" from collector to emitter for you solid state theory purists).

T1, with its BASE held at ground by the shunt switch in the jack, (zero potential), will not conduct. Once a plug is in the jack, thereby opening the shunt switch, current can flow through R2 thereby generating a positive voltage (bias) on T1's base, causing T1 to conduct, which in turn causes the LED to light. The small drain (0.1 mA) in the circuit, with R2 shorted, is the current flowing through R1 to ground (with a tiny tiny bit flowing through T1, though not enough to light the LED.

Transistors come in a huge variety. I chose T1 (2N2222) because it is widely available (in fact all the electronic components are available at Radio Shack). It's cheap and reasonably rugged. We can go deeper here if you like.

R4 is, indeed, a current limiting device. It also acts, in conjunction with the inherent resistance of the LED, as part of a voltage dividing network. So, it serves as both a voltage "reducer" and a current "reducer". Marvelous little devices.

I strongly suggest, at this point, that you take a look at an Ohm's Law tutorial. A huge (HUGE!!) element of practically ALL (and I mean ALL) electronic circuits requires a firm understanding of Ohm's Law.

See www.electronics-tutorials.ws/dccircuits/dcp_2.html , for example.

R1, with R2, represent a "voltage divider". What that means is together they are taking the 12 VDC and dividing it into two 6 VDC "pieces". One of 6 VDC "pieces" is applied, as a bias, to the base of T1.

How do they do that, you say? They are both 10K ohms (20k ohms in series). In essence, the 12 VDC is multiplied by the ratio of the value of R1 divided by the SUM of R1 plus R2 (or 1/2) , thereby dividing the 12 VDC into 6 + 6 VDCs AND reducing the current by 1/2.

The juncture of R1 and R2 is only going to ground when there is NO plug in any of the jacks, thus causing the "shunt" switch to be closed, which shorts out R2. This means that the base (conduction controlling element of the transistor) of T1 is being held at zero (or a minus voltage potential, if you like), thereby keeping T1 from conducting (and LED1 from lighting).

Hope I explained all this satisfactorily. AND accurately.

I should add that my wife (of 32 years) often complains that when she asks me for the time, I explain, in great detail, how a watch works...

 

[Muttley600]

OK. By my estimation, you need to leave shortly (if your haven't already), so here we go.

I had gone to work but saw your post on phone, nearly fainted **broken link removed**

still early where you are I see **broken link removed**

Yes, F1 is the fuse already in the bike that I'm assuming is dedicated to accessories.

Yes from the relay, the lead upfront has it's own fuse

In the schematic, the dots are physical (soldered or otherwise) connections. Whenever a wire crosses another wire, a dot indicates that the two (or three or a million) are connected together. NO dot means they may cross one another, but they are NOT connected.

Got that, I was just using it as a reference point but understand what dots mean

I fused the 5 VDC regulator circuit to protect it from an overload. It's easier to replace the fuse than to replace the regulator. Some components have built in short protection, in which case, fuses are not necessary. All a matter of component selection and one's willingness to assume that Murphy Law is all a big lie.

ok, I took the easy route on that one **broken link removed**

**broken link removed**

The top lead of T1 is the "collector" (if your familiar with the electron tube , or valves as you guys call'em, a collector is similar to a "plate")

No sorry, me knows nowt.......but I get the T1 bit **broken link removed**

The bottom lead (with the arrow) is the "emitter" (or cathode in tube talk [TT]) and the lead at a 90 degree angle jutting to the right is the "base", or "grid" in TT. Basically, the base controls the flow of current through the transistor from emitter to collector (or "holes" from collector to emitter for you solid state theory purists).

with you so far

 

[Muttley600]

T1, with its BASE held at ground by the shunt switch in the jack, (zero potential), will not conduct. Once a plug is in the jack, thereby opening the shunt switch, current can flow through R2 thereby generating a positive voltage (bias) on T1's base, causing T1 to conduct, which in turn causes the LED to light.

so it is effectivly moving the ground point, is that correct **broken link removed** please let it be

The small drain (0.1 mA) in the circuit, with R2 shorted, is the current flowing through R1 to ground (with a tiny tiny bit flowing through T1, though not enough to light the LED.

so why have it there, I've missed something important here haven't I **broken link removed**

Transistors come in a huge variety. I chose T1 (2N2222) because it is widely available (in fact all the electronic components are available at Radio Shack). It's cheap and reasonably rugged. We can go deeper here if you like.

ok, so it can only run a small current, why do they make different ones?

I'm going to regret asking that aren't I **broken link removed**
wait til I've understood the other bit first **broken link removed**

R4 is, indeed, a current limiting device. It also acts, in conjunction with the inherent resistance of the LED, as part of a voltage dividing network. So, it serves as both a voltage "reducer" and a current "reducer". Marvelous little devices.

but how can it do that on wrong side of led **broken link removed**

I strongly suggest, at this point, that you take a look at an Ohm's Law tutorial. A huge (HUGE!!) element of practically ALL (and I mean ALL) electronic circuits requires a firm understanding of Ohm's Law.

See www.electronics-tutorials.ws/dccircuits/dcp_2.html , for example.

I was so dreading opening that link, I was so relieved when I did, got that **broken link removed**

 

[Muttley600]

R1, with R2, represent a "voltage divider". What that means is together they are taking the 12 VDC and dividing it into two 6 VDC "pieces". One of 6 VDC "pieces" is applied, as a bias, to the base of T1.

need a smoke.......ok, power makes a circuit when it grounds, it will take the easiest route to ground, switch or led

so your providing a 6v charge to work switch on TI at the base which is grounding through R2, why are you calling it a bias?
why doesnt it ground through jack if that's the easier route**broken link removed**

I did say I would be asking lots of numpty Q's **broken link removed**

How do they do that, you say?

**broken link removed** How did you know my next Q
That'll teach me to read ahead

They are both 10K ohms (20k ohms in series). In essence, the 12 VDC is multiplied by the ratio of the value of R1 divided by the SUM of R1 plus R2 (or 1/2) , thereby dividing the 12 VDC into 6 + 6 VDCs AND reducing the current by 1/2.

getting it, or that bit anyway

The juncture of R1 and R2 is only going to ground when there is NO plug in any of the jacks, thus causing the "shunt" switch to be closed, which shorts out R2. This means that the base (conduction controlling element of the transistor) of T1 is being held at zero (or a minus voltage potential, if you like), thereby keeping T1 from conducting (and LED1 from lighting).

So is it grounding R1 which makes it minus voltage because it's being grounded out or have I misunderstood that bit?

Hope I explained all this satisfactorily. AND accurately.

erm......I'm getting there slowly **broken link removed**
look forward to see how I did....with a bit of nerves, maybe it needs time to sink in

I should add that my wife (of 32 years) often complains that when she asks me for the time, I explain, in great detail, how a watch works...

**broken link removed**

 

[Muttley600]

Had to post before I walk dog, i'vejust got it
Thanks a million
You are providing switch with 6v which then gnds through R2 but if jack unplugged R1 gnds changing switch

That's right isn't it.......yippee

Just thought while walking, that is a 6v gnd so maybe not right :-(

Only R4 to understand now

Thanks CBB

 

[cowboybob]

CBB

R4 is in series with LED2 to limit both the current AND, as a result, the voltage across LED2.

I guess that I haven't explained that, generally speaking, your average LED starts to emit light when 1.3 to about 1.6 VDC is applied across its leads, drawing as little as 30Ma (very low level output) to 150 Ma (brightest).

If you like, I can include a simulation showing how the various values of R4 affect the LED, but as an example, were you to connect the LED directly to the 12 VDC avaiable on the bike, it would be destroyed. R4 "resists" the current thereby reducing it.

 

[Muttley600]

I get led & resistors, just confusingme it is on gnd side not +

How did I do with understanding other bit?
6v gnd doesn't sound right?

 

[KeepItSimpleStupid]

Please correct me where I have misunderstood, so I get the principle of this transistor, but this itself begs many Q's

1. How do you know which transistor to choose
2. Are the 3 legs marked/identified on unit
3. why are there metal & plastic versions

1. Bipolar transistors come in two varieties, NPN and PNP. They have different parameters such as case style, frequency of operation, polarity, How much current can be safely passed, an SOA Safe operating area, Vce(Sat), Maximum voltages, thermal resistance and gain to name a few. Search for datasheet 2n3904, for instance.

2. Only the datasheet will tell you

3. Cost, power dissipation, ability to add a heatsink

The transistor leads are labeled E, B and C. Current Gain is a parameter that varies all over the place. There are diodes associated with transistors. One, of particular interest is Vbe or the Voltage from the base to the Emitter. This is a temperature and current dependent parameter, but 0.6 V is used for a lot of designs for a transistor made of Silicon. There are a few equations that govern the behavior of a transistor. One is Ie = Ib+Ic or the Collector Current is the sum of the base current and the emitter current. Ic = (beta)Ib where (beta) is the Current Gain or Hfe of the transistor.

Many times, it's easy to switch the low side of a load with a transistor. That requires the load to be connected to the + supply and then the Collector of an (NPN) transistor. The Emitter would be connected to ground. There would be a base resistor connected to the base.

To use the transistor as a switch, you have to look at the Vce(sat), Hfe(minimum) and Vbe. let's assume that V(drive) is the amount of voltage applied to the base resistor.

You have to assure that the Load current (Ic) is met and the base current is in the range in which you need. Ib is a small fraction of the emitter current. Rb < (V(drive)-0.6)/Ib and Ib > Ic/(Hfe(min).

That's the general basics on using an NPN transistor as a switch. If the load is a relay a diode is usually placed reverse biased, to protect the transistor when the relay turns off.

 

[cowboybob]

Tonight I'll put together a few schematics on just the concept of LEDs, resistors, positive voltages and grounds. It'll probably be easier to see it that to explain it.

Ohm's Law reigns supreme in the realm of electronic circuits. It gives you the math to understand why you see the values and signals you get from the parts you put together.

Incidentally, the concepts will also apply to transistors.

 

[cowboybob]

CBB,

I owe you an apology.

I have over-engineered things, once again. There is a much simpler method.

Here's your solution for the "LED ON when the plug is in the jack":

View attachment 60357

With the shunt switch closed (no plug in jack), the LED is shorted out: all current goes through R1 ONLY.

With the shunt switch open (plug is in the jack), the LED is no longer shorted and , therefore, illuminates.

R1's value of 4.7 K ohms should work for just about any LED you might buy.

I don't know how I got so side tracked, but I did.

The transistor switch has its uses, but it's major over-kill for the application.

BUT, as promised, I'll be sending along the basic resistive networks schematics shortly.

Again, sorry for the excessively complicated answer to your original question.

Paul

 

[cowboybob]

OK. Here are the resistive network schematics. SW1 is the "Shunt" switch in your connectors. Note how each meter is labeled (voltmeter or ammeter).

These demonstrate Ohm's Law, which, again, I suggest you apply to the examples to confirm the meter readings shown.

This one shows the starting point (no power applied, both R1 and R2 = 1K ohms and both SW1 and SW2 OPEN):

View attachment 60368

OK, now with power applied (I've removed SW2 for simplicity), SW1 still OPEN:

View attachment 60369

Now with SW1 CLOSED:

View attachment 60370

Now with SW1 OPEN and R1 = 2k Ohms:

View attachment 60371

Now with SW1 OPEN and R1 = 5k Ohms:
View attachment 60372

In each case, using Ohm's Law ( E = IR where E is voltage, I is current and R is resistance) you can multiply the reading on the ammeter by the value of the resistor and derive the voltage across the resistor. For example:

2 mA (0.002 amps) times 1K (1,000 Ohms) equals 2 Volts

OR, as in the last example, you could use I = E/R: 2 Volts divided 1,000 Ohms = 0.002 amps.

All you need to know are two of the three values in a resistive network to derive the third.

Try with both resistors!

Hope this helps your understanding of how resistors work.

Paul

(It's a good thing you can edit this stuff, once it's been posted!!!!)

 

[Muttley600]

Thanks both
This is all good, busy for next few hours but then I still have
Q's

Don't worry about over engineering things ;-) I'm sure once I have grasped basic understanding that I will be trying to make more difficult things & it will get harder but makes learning more fun
Quite happy to understand what transistors do.

I shortened Cowboybob to CBB cause of using phone not keyboard

Thanks Graham

 

[JMW]

Clearing the fuse, means blowing or opening the fuse. Fuses have a delay, in addition, depending on the waveform and load, they may be overrated by as much as 100%. This means the device may fail before the fuse has a chance to blow.
You have done some awesome work and you speak "American" like a native.

 

[Muttley600]

R4 is in series with LED2 to limit both the current AND, as a result, the voltage across LED2.

I get led & resistors, just confusingme it is on gnd side not +

ok, realising that I was missing something major in my understanding of how things were working, I spent a hour & half confusing my 72 yr old dad today who was totally baffled as to why I was not getting it, I knew how to convert watts to amps etc, I think we were both frustrated, I told him that I wasn't giving up until I could see what I was missing.
Now please don't kick me too hard or hurt yourselves falling off your chairs laughing at me **broken link removed**

I found the problem **broken link removed** which of course you could already see I missing as you pointed me back in that direction......so based my trust that you were steering with good reason, I went back to ohms law, was all going well, understood it.........then I got to the bit where it says

'Power within an electrical circuit is only present when BOTH voltage and current are present'

doh, how could I forget that, it was driving me crazy as to why you were grounding a live wire out **broken link removed**

yes, it all suddenly clicked into place, everything made sense...........what can I say, I did warn you I was a numpty **broken link removed**

but seriously guys, thanks for bearing with me while I trod old ground, it meant a lot you didn't give up on me

& at least my poor old dad couldn't get out the chair to slap me.lol

I owe you an apology.

I have over-engineered things, once again. There is a much simpler method.

Again, sorry for the excessively complicated answer to your original question.

Paul

Paul if anyone should be saying sorry here buddy it's me, the time you have put into this to get me to understand, I am forever grateful for sticking with me **broken link removed**

& besides the fact, you taught me what transistors do, which is why I'm here really to learn, the end goal of getting a box working is just a bonus to be honest **broken link removed**

So we have covered transistors & resistors in lesson one **broken link removed**

& all the diagrams you spent time doing to get me to understand, made perfect sense, thanks

So now I'm wondering**broken link removed** & 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

Clearing the fuse, means blowing or opening the fuse. Fuses have a delay, in addition, depending on the waveform and load, they may be overrated by as much as 100%. This means the device may fail before the fuse has a chance to blow.

EEK! ok, I can learn fuses, earlier on you said that a device can clear to save fuse, but now it sounds the oppisite way around **broken link removed**

look, I can't not understand this soon, I've only just caught up with you all **broken link removed**

so how do you decide what fuse your putting in & don't say look on the cigarette socket your about to cut off **broken link removed**
is there a way to work it out **broken link removed**

You have done some awesome work and you speak "American" like a native.

was that to me or cowboybob **broken link removed**
It can't be me because all I've done so far is embarrass myself **broken link removed**

 

[Muttley600]

So if you lose your way, go back to basics

I might be some time **broken link removed**