Reducing 12v to 5v with resistors

[Riccarr]

Hello,

I have a home made circuit (arduino) that I want to use in my car, powered from a 12v source. I already have a lm7805 voltage regulator in my circuit, and it works when hooked up to the 12v car power, however even with a heat shield on it, it still gets too hot to touch after a few minutes. The circuit carries very little load, it basically just powers a 2 line VFD display.

I was think to reduce the 12v down to something manageable like about 9/8 volts before feeding that to my circuit, using a resistor based voltage divider.

Does this seem reasonable to reduce the 12v power down before feeding to my circuit by using a resistor based divider? I already have my circuit all soldered together and the LM7805 is soldered in place as well, so I'm trying to avoid tearing it apart.

Here's wiki pic of a divider (I know you all know it, but for reference sake) ...

https://upload.wikimedia.org/wikipedia/commons/d/db/Resistive_divider.png

This is silly, but I'm not sure how to hook this up. The Vin would be my car's 12v. The Vout would be my circuits power + wire. And my circuits negative - wire would go ...? Do I just connect my circuit negative together with the dividers ground, and connect both those to the ground wire on my car?

Thanks for any feedback.
Cheers.

 

[alec_t]

You don't need a voltage divider; a single series resistor will drop a few volts, but to calculate its value you need to know the current draw of your circuit. Be aware that the same total power will be wasted; some in the resistor and some in the 7805.

 

[Mosaic]

With a 7805 linear reg with a 12.5 V supply a voltage dropping resistor works like this:
with a 100mA draw we have the 7805 dropping 7.5V. Power lost as heat= 7.5 V*0.1 A= .75Watt.
select a resistor to handle about 5V of that drop, this leaves a bit of Voltage headroom for the 7805 to work with.

The resistor value should be R=V/I = 5/.1 = 50 ohms.

The resistor then dissipates a 1/2 Watt by the same P=VI calc. leaving about 1/4 Watt for the 7805 or just 33% of the original heating.
Your resistor should be a 1Watt rated unit or a pair of half watt units in parallel. You should find 51 or 47 ohm units easily. But a pair of 100 ohms, 1/2 watt units, in parallel is also easy.

It's worthwhile to remember that when driving the alternator charge circuit runs at up to 14.4 V, so you should design with that in mind. But these example values should still give a reasonable result. Mount the resistors a bit 'off' the board for cooling.

It might be easier to slap on a TO220 heatsink or a piece of 1/8" aluminum to heat sink the 7805.

FYI a buck regulator will also do the job and dissipate minimal heat for a bit more cost.

 

[heydonms]

You can get buck regulators (not just the IC, a complete module with caps, etc.) off ebay for about $1. It will run much cooler and be far less messing about. Or get a cigarette socket phone charger for a few dollars, it will do the same thing in a neat package.

IMHO, unless you need low ripple or are pulling very low current, there are few situations where a linear reg would be better than a switcher. Cost used to be a factor but that is no longer the case.