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oldspark 
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Posted: March 06, 2010 at 5:28 AM / IP Logged  
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 detailed circuit - Page 2 -- posted image. they are really smoke sinks!) detailed circuit - Page 2 -- posted image.
seasonalskier 
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Posted: March 06, 2010 at 4:48 PM / IP Logged  
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.
detailed circuit - Page 2 -- posted image.
Novice tinkerer, master of mistakes :)
seasonalskier 
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Posted: March 06, 2010 at 5:04 PM / IP Logged  
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).
Novice tinkerer, master of mistakes :)
seasonalskier 
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Posted: March 06, 2010 at 8:05 PM / IP Logged  
And since I apparently can't modify messages my car is a 2006 Ford Fusion.
Novice tinkerer, master of mistakes :)
oldspark 
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Posted: March 06, 2010 at 8:15 PM / IP Logged  
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.
KPierson 
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Posted: March 07, 2010 at 12:02 AM / IP Logged  

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).

Kevin Pierson
seasonalskier 
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Posted: March 07, 2010 at 6:29 PM / IP Logged  
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.
Novice tinkerer, master of mistakes :)
oldspark 
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Posted: March 07, 2010 at 6:49 PM / IP Logged  
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.
seasonalskier 
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Location: Michigan, United States
Posted: March 07, 2010 at 7:18 PM / IP Logged  
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.
Novice tinkerer, master of mistakes :)
KPierson 
Platinum - Posts: 3,527
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Joined: April 14, 2005
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Posted: March 07, 2010 at 7:29 PM / IP Logged  

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. 

Kevin Pierson
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