12DC Sockets

[Muttley600]

Don't ask me how but it works.....yahaaa

Time for cuppa & smoke before getting meter on it so I can get my head round this,
I have just had the lift I needed to carry on
Excited. Me......cause I am. This makes me want to understand it even more now

Thanks Paul

 

[cowboybob]

Graham,

I loaded up the wrong sim for the first (not-shorted LED) example which has the switch shorted and the wrong value current limiting resistor is in the circuit.

Here is the correct one:

View attachment 60466

Sorry for the mistake.

(One of these days I'm going to submit a reply that I don't then have to come back in and re-edit...)

 

[Muttley600]

ok, have readings & know they work, I've just got to understand now

If I'm working out ma right, I would have a 0.12 Amp draw if all sockets were not used but likely to be using 3 all the time & fourth in winter so just would be 0.003 Amps in summer which seems like nothing.

Those of you who are eagled eyed will spot the mistake at the bottom **broken link removed**

**broken link removed**

so does that mean you can short as long as leads & componants can take the short, thats what you've been trying to tell me isn't it **broken link removed** so that means if I'd have shorted wire direct it would have melted hence me not getting my head around it, so it isn't the volts that kill wires, it's the current which the resister was subduing

So if I were to:

I = V / R

13.9 / 470 = 0.29

It works **broken link removed** see, told you it did **broken link removed**

ok, so now I'm starting to understand this, it's good fun isn't it **broken link removed**

Please tell me I have understood it right now, feels better now I have got to play

Paul, you are a star **broken link removed**

I'm still not sure about my conversions between ma to A ?

 

[Muttley600]

so, when a man came along & fitted a diode on my motorbike alarm when fitting it, so current was only one way, could I fit one where I tested the current to stop any short or is that not how they work?

 

[cowboybob]

Graham,

Not necessarily in the order that you asked them (I'm still figuring out the "insert quote" function), here are some answers:

1. Yes. It's OK to short a component, depending on its affect on the circuit. And yes, it's current that kills.

2. One amp (1 A) is equal to 1000 mA (lowercase "m" for "milli" (or one thousandth, 1/1000), uppercase A for amps).
Thus, 1 mA is 0.001 A. 10 mA is 0.01 A and 100 mA is 0.1 A.

3. Using the LED/voltage/current sim schematics provided:

Shunt switch CLOSED (no plug in jack, LED off), 2.6 mA (or 0.0026 A)
Shunt switch OPEN (plug in jack, LED on), 2.3 mA (or 0.0023 A)

The current is higher with a closed shunt switch because the total resistance in the circuit is R1 ONLY.
With the LED in the current path (shunt switch open), the resistance is slightly higher, thereby reducing the current.

It should be noted that total current usage (with four LED circuits) is only about 10 to 15 mA, which pretty insignificant but, always there.

And finally:

4. In the original sim (only one LED circuit is being used for all four jacks), I arbitrarily connected ALL the shunt switches in parallel. That configuration would cause the LED to light up ONLY if ALL four jacks had plugs in them.

A series shunt switch wiring configuration would cause the LED to light if ANY ONE jack had a plug in it. Inserting more plugs would have no additional affect.

If you want each jack to have its own LED, then, of course, each jack must have its own, separate LED circuit.

The diode question: Diodes allow current to flow through them in only one direction. This is useful, as regards your example, in preventing damage to the alarm system were you to accidentally hook it up to a power source "backwards", i.e., if the battery terminals were reversed.

5. The first schematic you drew of the LED circuit is spot on. Very nice.

Hope I covered most of it.

CBB

 

[Muttley600]

Not necessarily in the order that you asked them (I'm still figuring out the "insert quote" function), here are some answers:

This makes a nice change **broken link removed** now it's my turn to help you, I bet if I asked five different people about electrics I would get five different answers to their interpration, but heres mine for multi quoting

I was asked the same thing on the bike forum I use, see where I have highlighted in bold

**broken link removed**

Copy & paste bold bits as many times as you like to break into sections so you can answer each item

**broken link removed**

It is easier to put a space betweenI first two **broken link removed**

1. Yes. It's OK to short a component, depending on its affect on the circuit. And yes, it's current that kills.

I'm getting there, feels like I'm starting to understand now

2. One amp (1 A) is equal to 1000 mA (lowercase "m" for "milli" (or one thousandth, 1/1000), uppercase A for amps).
Thus, 1 mA is 0.001 A. 10 mA is 0.01 A and 100 mA is 0.1 A.

Thanks **broken link removed**

3. Using the LED/voltage/current sim schematics provided:

Shunt switch CLOSED (no plug in jack, LED off), 2.6 mA (or 0.0026 A)
Shunt switch OPEN (plug in jack, LED on), 2.3 mA (or 0.0023 A)

The current is higher with a closed shunt switch because the total resistance in the circuit is R1 ONLY.
With the LED in the current path (shunt switch open), the resistance is slightly higher, thereby reducing the current.

I even understood that

It should be noted that total current usage (with four LED circuits) is only about 10 to 15 mA, which pretty insignificant but, always there.

That's good, it will only draw power when ignition on as power is off switched relay so shouldn't affect anything **broken link removed**

And finally:
If you want each jack to have its own LED, then, of course, each jack must have its own, separate LED circuit.

That is my plan

The diode question: Diodes allow current to flow through them in only one direction. This is useful, as regards your example, in preventing damage to the alarm system were you to accidentally hook it up to a power source "backwards", i.e., if the battery terminals were reversed.

Could I use one on this then **broken link removed** could one be placed to stop short?

5. The first schematic you drew of the LED circuit is spot on. Very nice.

I'm getting better slowly **broken link removed**

Hope I covered most of it.
CBB

As always **broken link removed**

 

[cowboybob]

Copy & paste bold bits as many times as you like to break into sections so you can answer each item

Thank you, thank you, thank you, Graham.

CBB

 

[Muttley600]

Ok, spurred on by this working, I can go onto a harder project question wise as I think there is going to be a lot of questions & experiments that could well take some time.
So I can carry on & finish this box after playing with transistor to see the difference.

So I have decided that after checking bike supply previously, that I should limit both boxes to 10A as not to push bike circuits & leave 25% free play.

So my rear box needs to have 10A limit as it will supply front tank at box as well
I have a lot of questions regarding this as I would like to make it capable of:
Protect from short circuit
Protect from overheating/current
Have a stable regulated 12V supply
Relay
Fuse to each socket
Using same sockets, make some of them switchable between 5V & 12V
& anything else that could make it harder to achieve, again the idea is to learn not the outcome.

It doesn't matter how long it takes, how many mistakes I make as long as I can understand reasoning & componants for doing so.

So how about a pcb *big grin*

 

[Muttley600]

Thank you, thank you, thank you, Graham.

CBB

Least I could do really

So how do you find out what is max amp on a 3.6mm jap connector ? For leads

 

[4pyros]

So how do you find out what is max amp on a 3.6mm jap connector ? For leads

You Google it.

 

[Muttley600]

You Google it.

I tried, can't find info, never mind, think it is just my OCD.lol

Paul, when you say short off readings is 2.6mA or 10-15 over all four sockets, how can that be when I had 29.3 mA off one or 0.003 A * confused *

 

[cowboybob]

Those connectors you have should handle anything you might hang off of them. And since they'll be fused (right?) I wouldn't worry overly much about it.

There still is the issue of the 5 VDC output. The voltage regulator will get hot. It'll need to have a heat sink.

The 29 mA figure you're getting is with a 470 ohm resistor. The (corrected) sim had a 4700 ohm (4.7K ohm (in Europe they use 4K7 I think), resistor. Your math is fine, just the resistor's value was in error.

The multiplier conventions are: M=times 1,000,000, K= times 1000, m= times 0.0001 and u= times .0000001. Capital "K" and lower case"k" are often interchangeable.

And on that note, R1 is the resistor you'll want to play with (vary its value) to get the brightness you want out of the LED you chose for the project.

 

[Muttley600]

Can't link off phone, 5v reg was black plastic block in picture, I'll hook it up to power pack & see how hot it gets, think they said 45degrees on spec I posted

Yes all sockets will be indivually fused

 

[cowboybob]

Make sure you put a load on the regulator that represents the most current you think you will ever use. That's what heats it up. And they can get very hot.

 

[Muttley600]

Make sure you put a load on the regulator that represents the most current you think you will ever use. That's what heats it up. And they can get very hot.

There's an interesting project in itself, how about 2.5 A max which is over the top really. how do you make a dummy load to test circuits?

Have been contemplating how I got resistor value wrong
When back on computer I'll show you how I came to 470 but what your saying makes perfect sense
I know your right again

Still really enjoying learning
Thanks Paul

 

[cowboybob]

There's an interesting project in itself, how about 2.5 A max which is over the top really. how do you make a dummy load to test circuits?

What is the current capacity of the 5 VDC regulator that you have now? If it's greater than 2.5 A then you're golden.

Easiest dummy load would be to use one (or more, if necessary, to achieve the desired load) of the devices you intend to hook up to the regulator. They should have, clearly marked on them (or in the documentation), what they draw. If it's listed as watts, divide that by the voltage (5 VDC in this case) and that will give you the amperage value.

 

[Muttley600]

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

**broken link removed**

Spec of 5V regulator, I can't put phone on as I need to cut cigarette socket off first or am I being to cautious? It needs to be taken off at some point.lol
Oh, guess now's a good time

Camera is 0.57A while recording or 0.25A while charging battery
Have not tested phone yet but can go play

 

[Muttley600]

The 29 mA figure you're getting is with a 470 ohm resistor. The (corrected) sim had a 4700 ohm (4.7K ohm (in Europe they use 4K7 I think), resistor. Your math is fine, just the resistor's value was in error.

The multiplier conventions are: M=times 1,000,000, K= times 1000, m= times 0.0001 and u= times .0000001. Capital "K" and lower case"k" are often interchangeable.

And on that note, R1 is the resistor you'll want to play with (vary its value) to get the brightness you want out of the LED you chose for the project.

ok, now for this resistor thing again **broken link removed**

**broken link removed**& if I look at specs, how do you know the difference?????? edited as picked wrong ones......sigh

**broken link removed** So I did get the wrong ones **broken link removed** at least I'm learning, so now I just need to go get some of these **broken link removed**

**broken link removed**

I calculated them by taking information from led

then doing sums as follows: (although I have since changed leds to low current ones)

Calculation example I followed:
If = 20 ma
Vf = 2.5v
12v dc supply

R=(12 – 2.5)9.5 / 0.02 = 475 r nearest is 470
Other consideration working out what wattage resistor to get:
12v x 20ma = 240ma = 0.24 w

Should now be:
If: 30mA
Vf: 1.65v
DC: 13.9
R=(13.9 - 1.65) 12.25 / 0.03 = 408.33 but again nearest one is 470 where am I getting mixed up now you can see what I'm doing?
13.9v x 30 mA = 417 mA converted to w = 0.417 w which is just under 0.5 w resistor & I have 0.6w ones

now that I'm looking at it again I notice the forward voltage is meant to be 1.65V now but I still 13.9V across circuit *confused again* unless the new resistor will put that right


The other thing I noticed a mistake on my schematic I have put the resistor inline with socket, which was ok for testing if this worked but assume it will need to be on led wire/off current to socket for normal purposes **broken link removed** ignore me, after making mistake I think I am over thinking things, resistor is on led wire so schematic is right

 

[cowboybob]

Should now be:
If: 30mA
Vf: 1.65v
DC: 13.9
R=(13.9 - 1.65) 12.25 / 0.03 = 408.33 but again nearest one is 470 where am I getting mixed up now you can see what I'm doing?
13.9v x 30 mA = 417 mA converted to w = 0.417 w which is just under 0.5 w resistor & I have 0.6w ones

All your math looks good. On the last line above (13.9 x 30 mA = 417 mA) etc, the underlined mA should be mW. Minor point.

now that I'm looking at it again I notice the forward voltage is meant to be 1.65V now but I still 13.9V across circuit *confused again* unless the new resistor will put that right

Keep in mind that the 13.9VDC source voltage is divided across the resistor and the LED, i.e., the sum of their voltage "drops" of 1.65V (for the LED) and and 12.25V (for the 470 Ohm resistor) add up to 13.9 VDC.

The other thing I noticed a mistake on my schematic I have put the resistor inline with socket, which was ok for testing if this worked but assume it will need to be on led wire/off current to socket for normal purposes ignore me, after making mistake I think I am over thinking things, resistor is on led wire so schematic is right

Since we are grounding out the LED, its negative lead must must be connected to ground, with the shunt switch grounding the LED positive lead when closed (no plug in the jack). The resistor is going from LED + to the source positive lead. That's the only way this circuit can work because of the construction of the jack.

 

[Muttley600]

Hi Paul
So why aren't I understanding the difference if 470 is right in sum
What I'm really saying is I can see the 47k will make short less but then sum is wrong.
Only asking so I understand the theory, hopefully learning how not to make mistakes in future?????

Thanks for picking up on mA - mW issue as well

Thanks for helping
Graham