What's the easiest way to power 9v devices from your 12v car battery ?

Can I just use a resistor in series ?

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You'll want a regulated 9 volt output. Just putting a resistor in series will drop some of the voltage, but how much voltage gets dropped will depend on the load you are connecting.

If you are looking for a DIY approach, simple 3 terminal regulators like the LM7809 type are very easy to use. The three leads of this type of regulator are unregulated input, ground, and regulated output. Its also a good idea to to put a diode in series with the unregulated input, to protect the regulator and your 9V device in case you mis wire the thing.

Some of these regulators require certain amounts of capacitance on the output and/or input. The datasheet for the actual part you use will ususally contain suggestions for the capacitance if it is required.

Otherwise, there are off the shelf units to produce several different regulated output voltages from a car battery. They usually have a cirgarette lighter type plug and an assortment of different types of output power jacks.

"Its also a good idea to to put a diode in series with the unregulated input..."

Hi! What kind of diode? Knowing the input is 12V/3A and I need 9V/1A as output. Thanks.

andy707075 years ago
Please don't use a voltage divider, with the sorts of currents you will need to draw, you will need some very large resistors, probably 10w or so. I used one mistakingly on a car battery and it drained it within a day. Diodes aren't really a good way to drop voltage either, especially as someone quite rightly said, the battery may hold as much as 14v when fully charged and as low as 10v when empty, so not very good for sensitive electronics. As someone else said, the 7809 voltage regulator is the way to go. 11-35v input, regulated 9v output, and you only need 3 external components, a 1000µF electrolytic, a 220nF and 100nF. The 7809 does 1A, but you can buy a 2A, 3A, and 5A version if you need more power.
lemonie5 years ago

What/which 9V devices?

Absolutely. Depending on the device, 12v might be okay.
gmoon5 years ago
Try a trick that some unregulated wallwarts use: several diodes in series.

Each silicon diode (1N4001, 1N4007, etc.) will drop approx 0.6V. Five diodes drop about 3V total.

Keep in mind that a 12V car battery might be has high as 14V when fully charged.

(If precision is needed, then a regulator is best...)
Actually, you could use 2 resistors if you're prepared to put up with poor efficiency; google "potential divider" for how. The trick is to make sure that the resistors in the potential divider are much higher impedance (at least 10x that of the load), and rated for sufficient current.
Nah...a divider is really only practical as a voltage reference, and only if the source is constant, which isn't the case with a car electrical system.

Even zeners aren't really sufficient as anything more than a reference. You need to use a regulator for 99% of real world loads...
I think you mean the divider impedance is 1/10 of the load.
That is where the wasting efficiency comes into play.
Otherwise the load would massively unbalance the voltage divider.

Not just the efficiency, the circuit regulation goes to hell in a handbasket too.
seandogue5 years ago
No, a resistor is completely insufficient.

radio shack and other outlets sell devices that will plug into your cigarette lighter or "accessories" outlets. Many are "programmable output", ie, they have switches that one can set to a variety of common output voltages.

If you need some special load capability, you can build one for your self using a voltage regulator, a few capacitors, some wire, and a small enclosure in which the house the circuit.