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I've been looking around for an easy way to convert car battery 12v to 5v. I have seen some people saying that a simple resistor is all that is needed.

What I've tried so far is the DC to DC regulator. It works, but it's pretty expensive for such a cheap project. And not to mention it is twice the size of the project.

Following the ohm's law, it should mean that a single resistor with appropriate wattage rating should be fine to convert the car battery into a 5v supply. But the project used a max of 350mA, will the resistor or the project burn?

EDIT: The project will run even with ~100mA supply, but at some points it could go way up to ~350mA. Haven't tested it thoroughly but atleast 500mA would be sufficient. And yes, since it's a Microcontroller, a stable 5VDC would be preferred. And about destroying the OBD2 ports... Really I never had any thoughts about it.

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    \$\begingroup\$ A resistor solution will require constant load. Go for a simple 7805 instead. \$\endgroup\$ – winny Sep 16 '16 at 10:12
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    \$\begingroup\$ At 350mA, make that a 7805 and a suitable heatsink. \$\endgroup\$ – Brian Drummond Sep 16 '16 at 10:14
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    \$\begingroup\$ In that case, get yourself a 12 V -> 5 V COTS converter module. \$\endgroup\$ – winny Sep 16 '16 at 10:28
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    \$\begingroup\$ If you've got a car battery, why are you worried about the size of a voltage regulator? Without a heatsink, at 7V*350mA, a 7805 is dead, a resistor is equally large, a buck converter is the smallest choice because it wastes least heat. \$\endgroup\$ – Brian Drummond Sep 16 '16 at 10:44
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    \$\begingroup\$ You can get DC-DC regulators for $1-5! You are literally asking for a USB phone charger. \$\endgroup\$ – Bort Sep 16 '16 at 14:36
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A single resistor is not appropriate. The voltage a resistor drops is proportional to the current thru it. Even then the resulting voltage will vary with the input voltage.

At 350 mA out, a resistor or linear regulator will dissipate a lot of heat. When the car is running, figure the input voltage could be as high as 14 V (13.6 V is a common value). That means the linear pass element will drop 9 V. At 350 mA thru it, it will dissipate 3.2 watts. That's going to require some space and expense one way or another. That's too much, for example, for a TO-220 in free air. The forced air cooling or extra dissipation surface will be big and expensive.

The best answer is a buck regulator. These are much more efficient, and are therefore smaller and cheaper since they don't have to deal with getting rid of all that heat. There are many commonly available chips from a number of manufacturers (Microchip, ST, TI, Linear, etc) that come with the controller and switch integrated. You add the inductor, input/output caps, and a few extra external parts. A properly designed buck solution will be smaller than anything that can safely dissipate 3 watts.

Consider that car power can have a few 10s of volts spikes on it occasionally. You need to get a buck switcher with a sufficiently high maximum input voltage, or put some kind of clamp in front of it.

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  • \$\begingroup\$ Agreed. So far I've got my hands on a 12v-24v Buck Regulator to 5v. With the car battery rated at 12v, with a spike of 10v it should be somewhere around 22v, nearing its max input. Would it withstand it? It's big, but I don't see any other options as you said the best answer is a buck regulator. It's a KIS 3R33S on a breakout board. \$\endgroup\$ – TenzoNakami Sep 17 '16 at 15:17
  • \$\begingroup\$ @Tenzo: Car voltage spikes can go higher than 24 V. If you're going to use a converter rated for only 24 V in, then you have to put a clamp in front of it. \$\endgroup\$ – Olin Lathrop Sep 17 '16 at 18:22
  • \$\begingroup\$ Do you have any reference as for the car voltage spikes? Like the peak voltage? And sorry, what clamp? If I were to use a converter rated for 37v would it suffice? \$\endgroup\$ – TenzoNakami Sep 19 '16 at 12:05
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Convenience stores and discount retailers sell these things for as little as $1, they are called USB Car Charger Adapters. They are made to plug into the cigarette lighter port and provide output to a USB jack, but you could adapt it for your purpose.

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Firstly I want to make it clear that I agree with Olin that a DC-DC converter is the way to go and that DC-DC converter should be a Buck regulator.

But in the interests on completeness assuming you want to go with a linear solution you will still need to dissipate 3.2W maximum but this does not all need to be in your TO-220 regulator.

If you put a 12 ohm 2W resistor between the battery and the regulator you can reduce the dissipation in the regulator but this will still require a heatsink but you can get away with a smaller one.

Calculations for this:

enter image description here

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  • \$\begingroup\$ First I will try to increase the case size so that the bulk regulator would fit. But if it doesn't work then I'll have to try yours. And yes I've always agreed and would use a buck regulator, but for the size, it just doesn't fit. \$\endgroup\$ – TenzoNakami Sep 17 '16 at 15:21
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try a switching regulator with a small footprint. something like MURATA's OKI-78SR-5/1.5-W36H-C

It costs only 4$ and has a very wide input (7V to 37V) which may allow it work when cranking or when a load dump happens

enter image description here

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  • \$\begingroup\$ I'll try to find that board here. It's pretty hard to find stuff like this >.> \$\endgroup\$ – TenzoNakami Sep 19 '16 at 12:05

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