Voltage regulation for car

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wojeepster

New Member
Voltage in my car runs about 14v is there a recommeded way to set the voltage for a pic? 14 to 5 seems like a big drop to me is there an advantage to cascading regulators 14 to 12 12to 9 then 9to 5? or is that just a waste of time? Can I go straight from 14 to 5? If I do that does that increase the heat the regulator has to dissapate? Can anyone point me in the direction of necessary circuit or do you just need regulator and capacitors? Thanks for your help in advance. I am thinking of using pic for theft deterent circuit, and another for rear window defrost timer.
 
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boriz

Senior Member
Where possible, always separate the ‘power’ hungry components from the logic. IE: Use independent supplies.

Dropping 14v to 5v means losing 9v. Think of a regulator as the top leg of a potential divider and your load as the bottom leg. As the load current varies (equivalent to varying the resistance of the lower leg), the top ‘resistor’ needs to also change it’s value to maintain the required voltage at the junction (in this case 5v). Or another way to think of it is as a heater. It converts unwanted power to heat. It makes no difference if you lose this 9v in one step or in 3. The same amount of power will be dissipated. The only possible reason to use more than one stage is to limit the heat dissipation in one regulator to within it’s specified maximum by sharing the load.

EG: Your 5v load will require a maximum of 200mA say. The same current flows through the regulator and the 9v dropped across it. That’s 9v*0.2A = 1.8 Watts. So the regulator will be radiating 1.8 Watts as heat.

@50mA, you get 9v*.05A= 0.45W radiated as heat from the regulator.

Etc.

The power dissipated as heat needs to go somewhere. The rule of thumb is: If it gets too hot to hold your finger on it, then it needs more heatsinking.

If you need to radiate (say) 5 Watts, and your regulator is limited to (say) 3 Watts, then a common solution, assuming you have enough voltage overhead, is to use another regulator to provide an intermediate voltage. In this case you’d use a second regulator to drop from 14v to something between 14 and 5. To spread the load equally (drop the same voltage across each regulator), the second regulator would need to regulate at 4.5v above your final 5v voltage. (half of 9v). So it would need to be a 9.5v regulator. But I’m sure a 9v one would suffice. Two regulators would rarely be needed. Picaxe and associated parts should use only a small amount of current.

In short. Read the datasheet. Find out what is the max power dissipation of your 5v regulator. Find out what is the max current your 5v circuit will EVER require. Calculate the dissipation like in the example above. (Watts dissipated = Voltage dropped * Max required current). Use a second regulator only if it’s really required.
 
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InvaderZim

Senior Member
I'd buy a universal DC adapter with a car socket; it takes a lot of worry out of it all, and they're cheap and small. Set it for 9VDC output, and use a 5V linear regulator near the picaxe. If you're keeping in the 1-200mA range, heat issues should be minimal. Re-think if you need more than that.

Might be worth adding some basic protection as well, like a zener diode on the 5V rail. Just in case your car spits out some nasty voltage spikes.
 
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moxhamj

New Member
You are going to lose 9V whether you do it from 14 to 5V in one step or in many steps.

The energy you lose (heat, in Watts), is equal to Volts times Amps. V=9, so if, say, you were running something at 1 amp, you would lose 9 Watts. That is a waste of power, and would need a heatsink, and would flatten the battery after a while.

But a picaxe only uses 0.003A. So that is only 0.027W. That is hardly worth worrying about, especially when you bear in mind that the human body, when resting, is using 100W.

So if you just want to run a picaxe (and maybe flash a led intermittently or other low power things), use a low power 5V regulator eg LP2950.

If you are switching something high power, maybe use a BC547 transistor to drive a relay, and make the relay a 12V coil one. Drive the gate of the transistor with a 2k7 resistor - that uses hardly any power. Try to make all the high power things 12V rather than 5V - which is pretty easy in the automotive world anyway. Motors, lightbulbs etc are all 12V.

If you are really stuck and you have a 5V high power device and you cannot do it with 12V, then look at a switching regulator. But even then, I'd buy a pre built module rather than trying to build one from components.

If you can afford a few dollars, maybe look at mosfets. I've used BUK555s in the past and they can switch up to 60 amps and the gate can be driven from 5V directly from a picaxe and they use no power. The picaxe manual contains some fet examples and more modern (cheaper) mosfets. A mosfet would be the simplest way to get a picaxe to turn on (say) a car headlamp or defroster, because it is just one component.

As for power supply spikes, yes they are there. But I've had a circuit controlling my car radiator fan and it has been going 2 years now. It is a LP2950, picaxe, BC547 driving a relay which turns on the fan. I put 470uF 25V across the input to the LP2950, 22uF 6V across the output, and 0.1uF across the picaxe power pins. And it sits in the engine bay just near the battery, and it is in a plastic box. With a retrospectoscope I'd probably put it in a metal box. If you are really keen you could put in some filtering with inductors,resistors and capactitors.
 
hippy

hippy

Ex-Staff (retired)
There have been a number of discussions on automotive power supplies in the past so a Forum Search should turn up additional useful information. To search on short terms, append a '*'; eg, "PSU*".
 
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