detailed circuit

Ok so I am an expert in low voltage circuits and I have created a circuit I want to run in my car, but the catch is my cars wiring that I need to hook it up to is 13.8V, 1.5mA and the circuit that I am working on I have set an operational rating at 5V and .5mA so how do I make it so I can change the single input down to those specs and then take the 3 outputs and raise them back up to the 13.8V and 1.5mA. Thanks so much for all help on this!

~Kenny

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Novice tinkerer, master of mistakes :)

use a 5v or an adjustable voltage regulator to drop the input voltage.
the output of your circuit can be used to activate 12v relays via transistors.
from you description, i believe you will need nPn transistors.

Ditto!

There are 7805 (5V 1A) or LM317 (variable, 1.5A) voltage regulators...
Might be worth reading ANs (Application Notes) etc for extra protection if you add capacitors etc (ie, reverse polarity diodes (IN4004 etc) across output, from input to output, and as with most car circuits, across input).
And a fuse (say 1A).

Outputs as per 91stt - commonly called "Open Collector" outputs.
When "on", the output is shorted/connect to ground.
This might be through transistors, FETs, relays....
Because it is switching to ground (0V) else "Open" circuit, the "next system voltages" don't matter. (Assuming the open-collector output can handle the "off" voltage of the next "input".)

Computers & logic (CPUs and interfacing) commonly use open collector outputs.
Some car systems too - like horns, ignition coils (ie, both points and ignitors), wiper motors. Some use it for lighting - like old VW Beetles and Datsuns - and they often suffer lighting problems as a result.... but that's a different story!

Sorry if too much detail but not enough....
I just wanted to endorse, and provide some useful hints.

Ok so I see voltage regulators sometimes need heat sinks when pulling stuff down from that high of voltage, will I need a small one for this? Also what do you mean by 91stt? Basically I need to take the power from a turn signal on my car, knock it down to run through my special circuit to create patterns on the 3 bulb output that will be run at the same power as originally sent to the one bulb. And would relays be able to stand up to the high voltage and current that I would run through it on a long term basis?

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Novice tinkerer, master of mistakes :)

"91stt" - the first Replier (2nd Poster) in this thread.

The power dissipated by the regulator is its voltage drop x the current through it.
EG - 12V to 5V, but assume max auto voltage of 16V, hence 16-5 = 11 Volts dropped - let's say a nice round 10V (ie - allows for 15V max input).

10V @ 1.5mA: P = VI = 10V x 1.5mA = 15mW
10V @ 150mA = 1,500mW = 1.5W.
etc.


BUT - maybe you should explain your aim (preferably) and then/else your circuit....
If you are using a flashing signal to power the regulator.... NO!!

If the regulator is powered of +12V but the 5V circuit gets it signal from the flashing signal (voltage limited to 5V of course!), then that's okay.
The +12V could be from the Ignition, or the flasher switch before the flasher can. (That is NOT common here except where the indicators are shared with other bulbs - in our case reverse bulbs; in USA - it's stop bulbs.)

oooooo lol I never even noticed his name lmao.


Naw they are side markers, I am using power from the turn signal switch on the stalk to activate the circuit, this will then in turn power 3 flashing bulbs that will be the side marker. But I need to take the electrical charge from the signal switch and down it enough for the circuit. Then it will be boosted so that it can turn on the 3 bulbs on the side marker.
I know the power drops involved, I was just saying I need to down the entire power of the circuit instead of the voltage or just the current, I was just saying that I know I could down the voltage and up the current but for this circuit it would not work because it would blow it out with high currents or high voltages.

But overall your suggestion is to take a voltage regulator to drop the initial charge then use relays to bump it back up to the original charge?

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Novice tinkerer, master of mistakes :)

can you post up a diagram of your schematic?
we could be more specific with the circuit details if we knew how it is all going to come together.

Sure I can definitely get you a small schematic of it tomorrow, I have work all morning then I have some stuff to do but I will do my best to get one up. Thanks so much for both of your help! Like I said I am good on the same power but I get lost when it comes to adjusting volts AND amps.

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Novice tinkerer, master of mistakes :)

It they are 12V sidemarkers, I don't understand why you need a 5V circuit - there are several ways of using 12V - some simply using ground connections or diodes.
If it's some 5V logic chip use a CMOS equivalent (3-15V).
Otherwise what the heck is it?

But post the circuit.

And find out if your indicator-switch switches 12V to either side's flasher circuit, or if it switches the flasher can (output) to either side.
If it's the former, I doubt you'd need anything other than (a) relay(s) - not voltage converters.

what year and model is you car?

oldspark, skier is saying that the circuit he built will only accept 5v that is why he needs to lower the voltage. probably because he used cmos components.

Yeah - his turn signal switch energises a 12V relay whose contacts supply or connect his circuit's 5V power supply.

His circuit might use your/my suggested ground switching to ground the 3 side-bulb GNDs (assuming the above relay also supplies their +12V if not normally +12V, and also that any normal bulb grounding is broken and isolated).
Or it could control three 5V or 12V relays that connect +12V to each bulb.
In both cases, diodes may be needed to isolate each bulb's supply or GND.

I regard the 12V regulator as part of the 5V circuit - especially if the circuit is electronics rather than relays etc. Hence the VReg is married to existing caps and filtering etc.
But if it's a TTL or similar circuit, why not convert to CMOS if available?
But if 5V etc PICs & CPUs, their outputs may be open collector and powerful enough to ground the 3 bulbs (LEDs?).


Whether the +12V turn signal switch supplies the circuit direct via a relay is academic, though it may need a relay to break any existing bulb grounds if bulb-GND control is used....


But if the turn signal switch supplies a "chopped" flashing signal - that's a different issue.
In that case I'd assume an RC timed relay if the flasher can can handle the intial cap charging; if not, an RC transistor or FET to drive a relay.

I guess the schematic will tell - that should narrow the implementation.


BTW Shredder, I mean, Skier - usually you only supply the volts - the Amps take care of themselves (you just have to supply enough, hence the ). sizing of the components, or capacitors, and as you wisely pointed out, heatsinks. (Don't believe the term "heatsinks" - trust me they are really smoke sinks!)

The part marked KCIC is a CMOS circuit that will only handle up to 5.5V MAX but I want to keep it at a steady 5 so I have some leeway. I am not too sure at what all the CMOS chips are rated amp wise but I believe that it is 5mA so I can't just drop the volts and increase the amps with a transformer. But here is the circuit diagram, trust me you do NOT want to see the CMOS circuit labled KCIC, its quite big and would take a very long time to draw out.



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Novice tinkerer, master of mistakes :)

Oh and I forgot, the bulbs are 5 LED bulbs per that run on 12V feeds, I got them from autolumination.com (if anyone was interested).

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Novice tinkerer, master of mistakes :)

And since I apparently can't modify messages my car is a 2006 Ford Fusion.

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Novice tinkerer, master of mistakes :)

I wonder what it is in the CMOS circuit that limits it to 5.5V. Their normal range is ~3V to ~15V or higher.

So where did the KCIC circuit come from - surely there is a site or info somewhere.... Maybe even some chip numbers.

For 12V @ 1.5mA you could just use a resistor under 3.9k-Ohms feeding a 4.9 or 5.5V zenor diode.
But that doesn't include the output drive current.
It looks like my second control method is being used - so either sink some 12V output relays or source some 5V output relays that supply +12V to the 12V bulbs. Or transistors or FETS instead of the relays.
Unless the bulbs/LEDs are within the switching capability of the KCIC outputs - but you need 12V, and you circuit indicates you will not be ground switching them. (If LEDs, you might only need its onboard 5V - ie for 3.3 or series 1.7V LEDs.)

And your input control is non-flashing power so it it probably a domestic USA car, not an import.
Hence, as befire, it just supplies 12V to whatever 12V to 5V converter you want.
For simplicity and versatility, I'd recommend a 7805 on the CMOS board with appropriate protection.

What is the point of this circuit?

I would power all logic circuits with a 100mA 7805 as mentioned above.

If you need a + input and + outputs you'll need to drive NPN transistors in to PNP transistors.  The NPN will take the high voltage and turn it in to a ground.  The ground out will drive the PNP and cause it to output a 5vdc signal (referenced to the voltage regulator).

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Kevin Pierson

Ok so the point of the IC is to have the LED's blink in a special manor, its hard to describe but this is not a IC that has been manufactured, it is an IC that I am working on that I will most likely be getting a patent on to put it into production. But the circuit involves several IC's:

74LS193D
74LS42D
4049BP
And a few others, I don't have the numbers on me right now but they are a few different various logic gates, and, nand, and nors.

But a question about the 7805, how would I have to heat sync that so that I can completely waterproof the circuit. I was planning on dipping the entire circuit in silicone to seal it but it sounds like that wont work now that I have to have the 7805 on there. Lastly, what do you guys mean when you say "proper protection"? Are you saying put a compactor across the output and ground? I am confused at what you are saying.

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Novice tinkerer, master of mistakes :)

All those chips have CMOS equivalents.

And I think you'd need a Registered Design rather than a Patent - the Design being whatever pattern the LEDs undergo.
I don't see the point of a Patent except to protect that particular hardwired design, and to circumvent, you'd just change a wire or chip etc.
Besides, the lot could be done easily with a PIC etc - even if your were using non-digital chips.
Besides, will you have enough money to defend the Patent (eg, if allegedly I say I'm ripping off your design and I have $1m to defend it)?

If you need a heastink, you may have to galvanicaly (electricaly) isolate it from whatever chip. But there is no reason something like that should require a heatsink - remember, max heat can't exceed its P = VI.

For the circuit & PSU protection, read the ANs for the 78xx series or the LM317 series - I mentioned some of the key issues like protecting for power off with a charged (output) capacitor, also for negative output and input transients, and input spikes.
If you intend to sell this, you will want to have similar protection for your outputs anyhow.
Not to mention however oyu were intending to supply unit's power.

If their are 12V chips I would much rather use those as it would take care of so many problems. But I have no idea where to find them or how I would start to look. I am all up for swapping this to a 12V circuit. And thanks for the Registered Design idea, I will have to look into that. I have yet to go into the information about the patents since the entire circuit is not completed and I do not have a working model installed on the car, etc. But if you could help me on finding 12V equivalent IC's I would greatly appreciate it.

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Novice tinkerer, master of mistakes :)

You won't need a heat sink, but if they make a potting compound that is thermally conductive.  However, I wouldn't recomend using it unless you are absolutely confident in your design (and warranty).  Once sealed, the circuit is no longer accessible!

Also, it seems like an awful lot of work just to flash some lights.  Your overall product will be smaller, cheaper, and more reliable if you center it around one uC chip. 

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Kevin Pierson

Based on my experience, you won't Patent it.

And I would question what it is about your pattern that makes it so different and appealing that you will secure a market...
And what makes it so special that it cannot be imitated.

It is all too simple to put detectors to each LED/bulb, record what they are doing, and program a chip to do it.

If yo ca do what you are doing with a programable chip, then progam it. There is no way a discrete IC design is going to compete.

And you best protection is the "locking" of that chip's software, and selling enough before the imitator's jump on board.

Understand that a Patent costs $thousands, makes the design "public", and there is no "Super-Patent-Man" or Woman that comes to your rescue when somebody rips it off. And in 16 years (usually), everyone is free to rip it off.

Sometimes a simple "Copyright" may be all that is required (as in music & written stuff), else Registered Designs etc.
But you "design" will not be your chips & hardware - it will be what they do.


Your 4049 is a CMOS chip.
Most of the 74xx series chips have their equivalent 40yy CMOS variants.
You get CMOS chips from the same place as the 74's.

Hunt for datasheets by function - eg Hex inverter - CMOS, LS, TTL. Try freescale etc. (Or google!)

Unfortunately my system will be offline for a few days (maybe) so I can't give you specific info...

Ok, well thank you for the information I will search for the CMOS chips. I am going to try to get protection on this circuit because its not out there right now. If I can get a small protection on it and then sell it to a company on a per kit basis I should be able to make some supplementary money to help me pay for my education. I realize that this is not going to be a circuit that will make millions, thousands or probably even hundreds, that is not what I am out for. I want to make this circuit and then get it out for the public so I can see it on the road and say hey I invented that circuit! But of course I want some return on my work. I have been screwed by a company in the past that is why I am going to protect this one so they can't just steal it. If yall want I can post some completed designs and videos when I finally finish it.

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Novice tinkerer, master of mistakes :)

You don't need a patent for that - just a Confidentiality Agreement.

Have you discussed the marketability with anyone?
I find that many think they have some new innovative idea or breakthrough product, unfortunately they don't.   

As to ideas, a general rule is the prototype & idea is 1% of the final production/marketing cost (and the idea/design 10% of the prototype) - but that's too vague for this....

And again, unless you can present the final package with marketing, distribution etc (ie, a Business Plan), you may only have a circuit. And if that isn't with an appropriate PIC etc, then you are merely selling a lighting sequence idea.
And it may not even be legal. (I had to remove my mirror indicators and things that these days are accepted & common...)

I don't mean to kill your idea, but don't be too unrealistic.
As I said, it must be quite novel - not just a new flavor for an old product....

Maybe too the first step is explaining what it does - not how.
But if what it does gives it away, then there is no point patenting or protecting the circuit itself....

I would describe to you what it does, but I am currently in talks with a company that would like to buy the design off of me and they specifically told me to not discuss the product, but I took an exception for this website because I did not know how to successfully complete the project within the time range that I told them I would have a working prototype. I realize you are trying to keep me down to earth about the circuits possibilities and I am a very down to earth person and wouldn't take it this far without reason. :)

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Novice tinkerer, master of mistakes :)

Did they complete a non disclosure agreement before discussing the project?

Why would they want YOU to not talk about it?  You should be talking to as many people / companies that you can get to sign the non disclosure.  It will only make the project more valuable (ie sell it to the highest bidder).

If they are telling you to not talk to other people they obviously don't have your best interest in mind (and you wouldn't expect them to, they are trying to make the most money they can).  I wouldn't be surprised if they took your idea, possibly even with a non disclosure, outsourced it to China and had the same thing made for 25% of what you are going to have in it while you are sitting by not talking to other people about it! 

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Kevin Pierson

seasonalskier wrote:

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...that would like to buy the design off of me and they specifically told me to not discuss the product

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So you must have some sort of agreement or non-disclosure agreement in writing.

Correct there is an agreement as of right now. But nothing is in writing yet therefore I am not going to give them the final product until there is. If they refuse to put it in writing then I will go about getting some form of design protection so that I can market it to other companies.

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Novice tinkerer, master of mistakes :)